1 Introduction

A Ricci shrinker is a triple \((M^n, g, f)\) of smooth manifold \(M^n\), Riemannian metric g and a smooth function f satisfying

$$\begin{aligned} Rc+\text {Hess}\,f=\frac{1}{2}g. \end{aligned}$$
(1)

By a normalization of f, we can assume that

$$\begin{aligned} R+|\nabla f|^2&=f, \end{aligned}$$
(2)
$$\begin{aligned} \int e^{-f} (4\pi )^{-\frac{n}{2}} dV&=e^{\varvec{\mu }}, \end{aligned}$$
(3)

where \(\varvec{\mu }\) is the functional of Perelman. As usual, we define

$$\begin{aligned} \mathcal {M}_{n}(A) {:}{=}\left\{ (M^n, g,f) \left| \, \varvec{\mu } \ge -A \right. \right\} . \end{aligned}$$
(4)

Lying on the intersection of critical metrics and geometric flows, the study of Ricci shrinkers has already become a very important topic in geometric analysis. Up to dimension 3, all Ricci shrinkers are classified. In dimension 2, the only Ricci shrinkers are \(\mathbb {R}^2, \,S^2\) and \(\mathbb {R}P^2\) with standard metrics, due to the classification of Hamilton [25]. In dimension 3, we know that \(\mathbb {R}^3,\,S^2 \times \mathbb {R},\,S^3\) and their quotients are all possible Ricci shrinkers, based on the work of Perelman [46], Petersen–Wylie [47], Naber [43], Ni–Wallach [45] and Cao–Chen–Zhu [8]. If we assume the curvature operator to be nonnegative, then the Ricci shrinkers are also classified, see Munteanu–Wang [42]. However, an important motivation for the study of the Ricci shrinkers is that the Ricci shrinkers are models for short time singularities of the Ricci flows. In dimension 3, by the Hamilton–Ivey pinch [25, 26, 30], one may naturally assume that the Ricci shrinker has nonnegative curvature operator. If the dimension is strictly greater than 3, the loss of pinch estimate makes the nonnegativity of curvature operator an unsatisfactory condition and should be dropped. Also, it is well known (cf. Haslhofer–Müller [27]) that most interesting singularity models are non-compact. Therefore, to prepare for the singularity analysis of high dimensional Ricci flow, we shall focus only on the study of non-compact Ricci shrinkers without any curvature assumption. Since M is non-compact, the inequality

$$\begin{aligned} \sup _{M} |Rm|< \infty \end{aligned}$$
(5)

may fail. The failure of Riemannian curvature bound causes serious consequences. Many fundamental analysis tools, e.g., maximum principle and integration by parts, cannot be applied directly without estimates of the manifold at infinity.

In this paper, we shall provide a solid foundation for many fundamental analysis tools in the Ricci shrinkers. We shall mostly take the point of view that Ricci shrinkers are time slices of self-similar Ricci flow solutions. After a delicate choice of cutoff functions and calculations, we show that most of the fundamental tools, including maximum principle, existence of (conjugate) heat solutions, uniqueness and stochastic completeness, integration by parts, etc., work well on the Ricci shrinker spacetime. Then we use these fundamental tools to study the geometric properties of the Ricci flows induced by the Ricci shrinkers. Therefore, we are able to check that most known important properties of the compact Ricci flows, including monotonicity of Perelman’s functional, no-local-collapsing and pseudo-locality theorem of Perelman, curvature tensor strong maximum principle of Hamilton, do apply on noncompact Ricci shrinkers. Furthermore, since the Ricci flows induced by the Ricci shrinkers are self-similar, we obtain many special properties of the Ricci shrinkers. The first property is the estimate of sharp Logarithmic Sobolev constant, which can be regarded as an improvement of the fact that Perelman’s functional is monotone along each Ricci flow.

Theorem 1

(Optimal Logarithmic Sobolev constant) Let \((M^n,p,g,f)\) be a Ricci shrinker. Then \(\varvec{\mu }(g,\tau )\) is a continuous function for \(\tau >0\) such that \(\varvec{\mu }(g,\tau )\) is decreasing for \(\tau \le 1\) and increasing for \(\tau \ge 1\). In particular, we have

$$\begin{aligned} \varvec{\nu }(g) {:}{=}\inf _{\tau >0}\varvec{\mu }(g,\tau )=\varvec{\mu }(g). \end{aligned}$$
(6)

Consequently, the following properties hold.

  • Logarithmic Sobolev inequality. In other words, for each compactly supported locally Lipschitz function u and each \(\tau >0\), we have

    $$\begin{aligned}&\int u^2 \log u^2 dV - \left( \int u^2 dV \right) \log \left( \int u^2 dV \right) + \left( \varvec{\mu } +n +\frac{n}{2} \log (4\pi \tau ) \right) \int u^2 dV \nonumber \\&\quad \le \tau \int \left\{ 4|\nabla u|^2 +Ru^2 \right\} dV. \end{aligned}$$
    (7)
  • Sobolev inequality. Namely, for each compactly supported locally Lipschitz function u, we have

    $$\begin{aligned} \left( \int u^{\frac{2n}{n-2}}\,dV\right) ^{\frac{n-2}{n}} \le C e^{-\frac{2 \varvec{\mu }}{n}} \int \left\{ 4|\nabla u|^2+Ru^2 \right\} dV \end{aligned}$$
    (8)

    for some dimensional constant \(C=C(n)\).

In geometric analysis, it is a fundamental problem to estimate uniform Sobolev constant. When the underlying manifold is noncompact, the uniform Sobolev constant in general does not exist. However, (8) says that there is a uniform (Scalar-)Sobolev constant, depending only on n and \(\varvec{\mu }\). In particular, if the scalar curvature is bounded, i.e., \(\sup _{M} R<\infty \), then there exists a classical Sobolev constant. Namely, for each \(u \in C_{c}^{\infty }(M)\), we have

$$\begin{aligned} \left( \int u^{\frac{2n}{n-2}}\,dV\right) ^{\frac{n-2}{n}} \le C e^{-\frac{2 \varvec{\mu }}{n}} \int \left\{ |\nabla u|^2+u^2 \right\} dV \end{aligned}$$

for some \(C=C(n, \sup _{M} R)\). Note that the term \(e^{-\frac{2 \varvec{\mu }}{n}}\) is almost \(|B(p, 1)|^{-\frac{2}{n}}\) by Lemma 2.5 of [34].

The proof of Theorem 1 follows a similar route as done in Proposition 9.5 of [34], by using the monotonicity of Perelman’s functional along Ricci flow and the invariance of Perelman’s functional under diffeomorphism actions.

Secondly, we can improve the no-local-collapsing theorem of Perelman on the Ricci shrinker Ricci flow. By the fundamental work of Perelman [46], the Ricci flow spacetime can be regarded as a “Ricci-flat” spacetime in terms of reduced volume and reduced distance. Now we can regard Ricci shrinker as a special time slice of the induced Ricci flow. On a Ricci flat manifold, an elementary comparison argument shows that \(\frac{|B(x, r)|}{|B(x,1)|}\) grows at most Euclideanly and at least linearly (cf. [59, 64], and Theorem 2.5 of [35]). This comparison geometry picture has a spacetime version which is used to illustrate the no-local-collapsing (cf.  [46, 53]). Although the comparison argument (even the space-time version) does not apply directly in the Ricci shrinker case, we can still show that similar phenomena hold for Ricci shrinkers.

Theorem 2

(Improved no-local-collapsing theorem) Suppose \((M^n,p,g,f)\) is a Ricci shrinker, \(r>1\). Then

figure a

Here q is any point on \(\partial B(p, r)\), and C is a dimensional constant.

Although the volume estimate (9a) behaves like the Ricci-flat case, its proof is totally different and much more involved. The proof builds on the the Sobolev inequality (8) and an improvement (cf. Remark 8) of the induction argument due to Munteanu and Wang [41]. The non-collapsing estimate (9b) in general does not hold for Ricci-flat manifold. This indicates that Ricci shrinkers are more rigid than Ricci-flat manifold. See Figure 1 for intuition.

Fig. 1
figure 1

Propagation of non-collapsing on Ricci-shrinkers

The proof of (9b) relies on (6) and an effective volume estimate in [53]. The scale \(\rho \in (0, r^{-1})\) is chosen such that \(R\rho ^2 \le C(n)\) inside B(qr). If we further assume scalar curvature is uniformly bounded on M, then we shall obtain that every unit ball on the Ricci shrinker M is uniformly non-collapsed. Theorem 2 can be regard as a special case of Theorem 23 and Theorem 23, which are more general versions of the no-local-collapsing. In particular, it indicates that any Ricci shrinker must be \(\kappa \)-noncollpased for some constant \(\kappa >0\), see Remark 7. The proof of Theorems 222 and 23 can be found in Sect. 9. Note that Theorem 2 indicates that the Ricci shrinkers are similar to the Ricci-flat manifolds. Actually, there exist many other similarities between the Ricci-flat manifolds and the Ricci Shrinkers. For example, in [29, 34], it is proved that each sequence of non-collapsed Ricci shrinkers sub-converges to a limit Riemannian conifold Ricci shrinker. Such results are analogue of the weak compactness theorem of non-collapsed Ricci-flat manifolds, by the deep work of Cheeger, Colding and Naber (cf.  [12, 14, 20]).

Thirdly, the pseudo-locality theorem of Perelman has an elegant version on the Ricci shrinker Ricci flow. The pseudo-locality theorem of Perelman [46] is a fundamental tool in the study of Ricci flow. It claims that the Ricci flow cannot turn an almost Euclidean domain to a very curved region in a short time period. In the literature, it is known that the pseudo-locality theorem hold for Ricci flow with bounded Riemannian curvature, which condition is clearly not available in the current setting. However, using the existence of special cutoff function, we can show maximum principle and stochastic completeness for conjugate heat kernel. By carefully checking the integration by parts, we obtain that the traditional pseudo-locality theorem holds on the Ricci flow spacetime induced by the Ricci shrinker. Furthermore, the pseudo-locality has the following special version for Ricci shrinkers.

Theorem 3

(Improved pseudo-locality theorem) Suppose that \((M^n,p,g,f)\) is a non-flat Ricci shrinker. Then we have

$$\begin{aligned} \varvec{\mu } < -\delta _0 \end{aligned}$$
(10)
Fig. 2
figure 2

\(\varvec{\mu }(g,\tau )\) of a Ricci shrinker with bounded geometry

for some small positive constant \(\delta _0=\delta _0(n)\). Furthermore, the following properties are equivalent.

  1. (a)

    M has bounded geometry. Namely, the norm of Riemannian curvature tensor is bounded from above and the injectivity radius is bounded from below.

  2. (b)

    The infinitesimal functional satisfies

    $$\begin{aligned} \lim _{\tau \rightarrow 0^{+}} \varvec{\mu }(g,\tau )=0. \end{aligned}$$
    (11)
  3. (c)

    The infinitesimal functional satisfies the gap

    $$\begin{aligned} \lim _{\tau \rightarrow 0^{+}} \varvec{\mu }(g,\tau )>-\delta _0. \end{aligned}$$
    (12)

If one of the above conditions hold, we can define

$$\begin{aligned} \tau _0 {:}{=}\sup \left\{ \tau | \,\varvec{\mu }(g,s) \ge -\delta _0, \quad \forall \; s \in (0, \tau )\right\} . \end{aligned}$$
(13)

Then for some positive constant \(C=C(n)\), we have the following explicit estimates

figure b

We remark that the gap inequality (10) is not new. It was first proved by Yokota in [57, 58]. However, our proof of (10) is completely different and is the base for the proof of (11), (12) and (14). Theorem 3 also indicates that the bounded geometry for Ricci shrinkers is equivalent to the gap inequality (12). This criterion has divided all Ricci shrinkers into two categories characterized by their graphs of entropies, which are illustrated by Figure 2 and Figure 3. Note that Figure 2 represents the functional behavior of a typical Ricci shrinker, for example, the cylinder \(S^{k} \times \mathbb {R}^{n-k}\) for \(k \ge 2\). Figure 3 represents the functional behavior of a Ricci shrinker with unbounded geometry. However, it is not clear whether such Ricci shrinker exists. For Ricci shrinkers with bounded geometry, it follows from (13) and (14) that the number \(\sqrt{\tau _0}\) can be understood as the regularity scale. Actually, under the scale \(\sqrt{\tau _0}\), all the higher curvature derivatives norm \(|\nabla ^k Rm|\) are bounded by \(C(n,k) \tau _0^{-1-\frac{k}{2}}\), in light of the estimates of Shi [49].

Fig. 3
figure 3

\(\varvec{\mu }(g,\tau )\) of a Ricci shrinker with unbounded geometry

There exist several other special versions and consequences of the pseudo-locality theorems. The proof of all of them, including the proof of Theorem 3, can be found in Sect. 10.

Fourthly, the curvature tensor strong maximum principle, developed by R. Hamilton, works on Ricci shrinker Ricci flows and also has an improved version. Using the curvature tensor maximum principle, Hamilton shows that the nonnegativity of curvature operator is preserved under the Ricci flow and the kernel space is parallel. Therefore, the manifold splits as product when kernel space is nontrivial. Since different time slices of a Ricci shrinker Ricci flow are the same up to scaling and diffeomorphism, the preservation of curvature conditions is automatic. The interesting problem on Ricci shrinker is to show the strong maximum principle, i.e., the splitting of the manifold when eigenvalues of curvature operator satisfy some nonnegativity condition. On this perspective, we can improve the traditional strong maximum principle of curvature operator to the following format.

Theorem 4

(Improved strong maximum principle of curvature tensor) Suppose \((M^n,g,f)\) is a Ricci shrinker and \(\lambda _1 \le \lambda _2 \le \cdots \) are the eigenvalue functions of the curvature operator Rm. Then the following properties hold.

  • If \(\lambda _2 \ge 0\) as a function, then there is a \(k \in \{0,1, 2, \ldots , n\}\) and a closed symmetric space \(N^k\) such that \((M^n, g)\) is isometric to a quotient of \(N^k \times \mathbb {R}^{n-k}\).

  • If \(\lambda _2 \ge 0\) as a function and \(\lambda _2>0\) at one point, then \((M^n,g)\) is isometric to a quotient of round sphere \(S^n\).

The statement in Theorem 4 should be well known to experts in Ricci flow if we replace \(\lambda _2\) by \(\lambda _1\). In fact, by the work of Munteanu–Wang [42] and Petersen–Wylie [47], we know that the same geometry conclusion hold if we replace \(\lambda _2\) in Theorem 4 by \(\lambda _1+\lambda _2\). Their proof builds on the celebrated work of Böhm–Wilking [5] on the closed Ricci flow satisfying \(\lambda _1+\lambda _2>0\) and also relies on a weighted Riemannian curvature integral estimate \(\int |Rm|^2 e^{-f} dV<\infty \). If \(\lambda _1+\lambda _2 \ge 0\), the Riemannian curvature integral estimate can be deduced from the Ricci curvature integral bound \(\int |Rc|^2 e^{-f} dV<\infty \), which follows from a clever integration-by-parts. In Theorem 4, with only condition \(\lambda _2 \ge 0\), Riemannian curvature integral estimate \(\int |Rm|^2 e^{-f} dV<\infty \) becomes nontrivial. As done in [34], we apply local conformal transformations and the classical Cheeger–Colding theory to study the local structure of Ricci shrinkers. Combining the \(L^2\)-curvature estimate of Jiang–Naber [31] with the improved no-local-collapsing Theorem 2, we are able to show that \(\int |Rm|^2 e^{-f} dV<\infty \) always holds true (i.e., Theorem 26). Consequently, the work of Petersen–Wylie [47] applies and the curvature tensor strong maximum principle holds for Ricci shrinkers. Then we are able to obtain \(\lambda _1 \ge 0\) from the condition \(\lambda _2 \ge 0\). Clearly, the condition \(\lambda _2 \ge 0\) is weaker than \(\lambda _1 + \lambda _2 \ge 0\) and Theorem 4 is an improvement of the results of Munteanu–Wang [42] and Petersen–Wylie [47]. Note that \(\lambda _2 \ge 0\) is a novel condition in the Ricci flow literature. It is not clear whether \(\lambda _2 \ge 0\) is preserved by the Ricci flow on a closed manifold. Actually, in Theorem 4, the same conclusion holds if one replace the condition \(\lambda _2 \ge 0\) by an even weak condition

$$\begin{aligned} \lambda _2 \ge -\epsilon \dfrac{\lambda _1^2}{|R-2\lambda _1|} \end{aligned}$$

for some \(\epsilon =\epsilon (n)\). The details can be found in Theorem 27. The proof of Theorems 4 and 27 appear in Sect. 11.

The proof of the previous four theorems requires some elementary, but delicate, geometric and analytic facts on Ricci shrinkers.

  • The level sets of f are comparable with geodesic balls.

  • A special cutoff function.

  • Special heat solution and conjugate heat solution on the Ricci shrinker Ricci flow.

  • The existence of heat kernel and stochastic completeness of the backward heat solution.

  • The existence and uniqueness of bounded (conjugate) heat solutions.

After the above estimates are developed, we check that the entropy of Perelman is monotone along the Ricci flow induced by the Ricci shrinker, whose proof needs more delicate integration by parts. Then the proof of Theorem 1 follows a similar route as the one in Proposition 9.5 of [34], with more involved technique. From Theorem 1, we can obtain Theorem 2 by repeatedly choosing proper test function u. When integration by parts are assumed, one can formally follows the routine of Perelman to obtain the differential Harnack inequality (i.e., Theorem 21), and then the traditional pseudo-locality theorem. Combining with a standard localization technique, one can deduce Theorem 3. However, as the functional derivatives contain quadratic Ricci curvature term, many terms concerning high order derivatives need to be carefully handled to verify the integration by parts. This causes many technical difficulties. One key difficulty is the delicate heat kernel estimate to derive the differential Harnack inequality. Therefore, the following heat kernel estimate is in the central position for developing fundamental analytic estimates on Ricci shrinker.

Theorem 5

(Heat Kernel estimate) Let \((M^n,g,f)\) be a Ricci shrinker in \(\mathcal M_n(A)\). Then the following properties hold.

  1. (i)

    (Heat kernel upper bound)

    $$\begin{aligned} H(x,t,y,s) \le \frac{e^{-\varvec{\mu }}}{(4\pi (t-s))^{\frac{n}{2}}}. \end{aligned}$$
  2. (ii)

    (Heat kernel lower bound) For any \(0<\delta <1\), \(D>1\) and \(0<\epsilon <4\), there exists a constant \(C=C(n,\delta ,D)>0\) such that

    $$\begin{aligned} H(x,t,y,s) \ge \frac{C^{\frac{4}{\epsilon }}e^{\varvec{\mu }(\frac{4}{\epsilon }-1)}}{(4\pi (t-s))^{n/2}} \exp {\left( -\frac{d_t^2(x,y)}{(4-\epsilon )(t-s)}\right) } \end{aligned}$$

    for any \(t \in [-\delta ^{-1},1-\delta ]\) and \(d_t(p,y)+\sqrt{t-s}\le D\).

  3. (iii)

    (Heat kernel integral bound) For any \(0<\delta <1\), \(D>1\) and \(\epsilon >0\), there exists a constant \(C=C(n,A,\delta ,D,\epsilon )>1\) such that

    $$\begin{aligned} \int _{M \backslash B_s(x,r\sqrt{t-s})} H(x,t,y,s)\,dV_s(y) \le C\exp {\left( -\frac{(r-1)^2}{4(1+\epsilon )}\right) } \end{aligned}$$

    for any \(t \in [-\delta ^{-1},1-\delta ]\), \(d_t(p,x)+\sqrt{t-s}\le D\) and \(r \ge 1\).

We briefly discuss the proof of Theorem 5. Notice that the Logarithmic Sobolev inequality for all scales implies the ultracontractivity of the heat kernel by Davies’ methods (see Chapter 2 of [21]). We prove that the same result (i) holds for Ricci shrinkers. The lower bound of the heat kernel can be estimated by considering the reduced distance (i.e., Theorem 16). We first obtain an on-diagonal lower bound of the heat kernel, in which case the estimate of the reduced distance is straightforward. Then we derive the general off-diagonal lower bound by exploiting a Harnack property (i.e., (200)). To prove the integral upper bound, we consider the probability measure \(v_s(y){:}{=}H(x,t,y,s)\,dV_s(y)\). Following the work of Hein–Naber [28], we show that \(v_s\) satisfies a type of Logarithmic Sobolev inequality (i.e., Theorem 13). The equivalence of the Logarithmic Sobolev inequality and the Gaussian concentration (i.e., Theorem 14) then shows that we can estimate the integral upper bound of the heat kernel by its pointwise lower bound.

Organization of the paper In Sect. 2, we review the definition of the Ricci flows induced by the Ricci shrinkers. We also present the estimates of the potential function and volume upper bound. In Sect. 3, we introduce a family of cutoff functions and prove a maximum principle (i.e., Theorem 6) on Ricci shrinker spacetime. Moreover, we prove the existence and other basic properties of the heat kernel on spacetime. In Sect. 4, we prove the monotonicity of Perelman’s entropy (i.e., Theorem 10). In Sect. 5, we prove Theorem 1. In Sect. 6, we prove the logarithmic Sobolev inequality (i.e., Theorem 13) and the Gaussian concentration (i.e., Theorem 14) of the probability measure induced by the heat kernel. In Sect. 7, Theorem 5 is proved. In Sect. 8, we prove the differential Harnack inequality (i.e., Theorem 21) by using the heat kernel estimates. In Sect. 9, we provide the proof of Theorem 2. In Sect. 10, we prove the pseudo-locality theorem (i.e., Theorems 24) and 3. In the last section, we obtain an \(L^2\)-integral bound of the Riemannian curvature (i.e., Theorem 26). As a consequence, we prove Theorem 4.

2 Preliminaries

For any Ricci shrinker \((M^n,g,f)\), let \({\psi ^t}: M \rightarrow M\) be a 1-parameter family of diffeomorphisms generated by \(X(t)=\dfrac{1}{1-t}\nabla _gf\). That is

$$\begin{aligned} \frac{\partial }{\partial t} {\psi ^t}(x)=\frac{1}{1-t}\nabla _g f\left( {\psi ^t}(x)\right) . \end{aligned}$$
(15)

By a direct calculation, see [18, Chapter 4], the rescaled pull-back metric \(g(t){:}{=}(1-t) (\psi ^t)^*g\) and the pull-back function \(f(t){:}{=}(\psi ^t)^*f\) satisfy the equation

$$\begin{aligned} Rc(g(t))+\text {Hess}_{g(t)}f(t)=\frac{1}{2(1-t)}g(t), \end{aligned}$$
(16)

where \(\left\{ (M,g(t)), -\infty<t<1 \right\} \) is a Ricci flow solution with \(g(0)=g\), that is,

$$\begin{aligned} \partial _t g=-2Rc(g(t)). \end{aligned}$$
(17)

For notational simplicity, we will omit the subscript g(t) if there is no confusion. From (16) and (17), it is easy to show that

$$\begin{aligned}&\partial _tf=|\nabla f|^2 , \end{aligned}$$
(18)
$$\begin{aligned}&R+\varDelta f=\frac{n}{2(1-t)}, \end{aligned}$$
(19)
$$\begin{aligned}&R+|\nabla f|^2=\frac{f}{1-t}. \end{aligned}$$
(20)

Now we define

$$\begin{aligned} \bar{\tau }=1-t ,\quad F(x,t)=\bar{\tau }f(x,t) \quad \text {and} \quad \bar{v}(x,t)=(4\pi \bar{\tau })^{-n/2}e^{-f(x,t)}. \end{aligned}$$
(21)

It follows from (18), (19) and (20) that

$$\begin{aligned}&\partial _tF=\bar{\tau }|\nabla f|^2-f=-\bar{\tau }R, \end{aligned}$$
(22)
$$\begin{aligned}&\bar{\tau }R+\varDelta F=\frac{n}{2}, \end{aligned}$$
(23)
$$\begin{aligned}&\bar{\tau }^2R+|\nabla F|^2=F. \end{aligned}$$
(24)

Now we define

$$\begin{aligned}&\square {:}{=}\partial _t-\varDelta _t, \end{aligned}$$
(25)
$$\begin{aligned}&\square ^*{:}{=}-\partial _t-\varDelta _t+R. \end{aligned}$$
(26)

We have special heat solution and conjugate heat solution:

$$\begin{aligned}&\square \left( F + \frac{n}{2} t\right) =0, \end{aligned}$$
(27)
$$\begin{aligned}&\square ^* \bar{v}=0. \end{aligned}$$
(28)

Note that (27) is equivalent to

$$\begin{aligned} \square F=-\frac{n}{2}. \end{aligned}$$
(29)

Now we have the following estimate of F by using the same method as [9, 27].

Lemma 1

There exists a point \(p \in M\) where F attains its infimum and F satisfies the quadratic growth estimate

$$\begin{aligned} \frac{1}{4}\left( d_t(x,p)-5n\bar{\tau }-4 \right) ^2_+ \le F(x,t) \le \frac{1}{4} \left( d_t(x,p)+\sqrt{2n\bar{\tau }} \right) ^2 \end{aligned}$$
(30)

for all \(x\in M\) and \(t < 1\), where \(a_+ :=\max \{0,a\}\).

Proof

This originates from the work of Cao–Zhou [9, Theorem 1.1]. We follow the argument of Haslhofer–Müller [27]. It follows from [15] that for any Ricci shrinker \(R \ge 0\) since its corresponding Ricci flow solution is ancient. So from (24), we have

$$\begin{aligned} |\nabla F|^2 \le F. \end{aligned}$$
(31)

It implies that \(\sqrt{F}\) is \(\frac{1}{2}\)-Lipschitz, since

$$\begin{aligned} |\nabla \sqrt{F}|=\frac{1}{2}\left| \frac{\nabla F}{\sqrt{F}}\right| \le \frac{1}{2}. \end{aligned}$$

On the other hand, for any \(x,y \in M\), we choose a minimizing geodesic \(\gamma (s), 0\le s \le d=d_t(x,y)\) joining \(x=\gamma (0)\) and \(y=\gamma (d)\). Assume that \(d >2\), we construct a function

$$\begin{aligned} \phi (s)= {\left\{ \begin{array}{ll} s, &{} s\le 1 \\ 1, &{} 1\le x\le d-1 \\ d-s. &{} d-1 \le x\le d \end{array}\right. } \end{aligned}$$

The second variation formula for shortest geodesic implies that

$$\begin{aligned} \int _0^d \phi ^2 Rc(\gamma ',\gamma ') \, ds \le (n-1)\int _0^d \phi '^2 \, ds=2(n-1). \end{aligned}$$
(32)

Note that from the Eq. (16),

$$\begin{aligned} \bar{\tau }\text {Rc}(\gamma ',\gamma ')=\frac{1}{2}-\text {Hess}F(\gamma ',\gamma '). \end{aligned}$$
(33)

Therefore from (24) we have

$$\begin{aligned} \frac{d}{2}-\frac{2}{3}-2\bar{\tau }(n-1)&\le \int _0^d \phi ^2 \text {Hess}F(\gamma ',\gamma ') \, ds \nonumber \\&\le -2\int _0^1 \phi \nabla _{\gamma '}F \, ds+2\int _{d-1}^d \phi \nabla _{\gamma '}F \, ds \nonumber \\&\le \sup _{s\in [0,1]}|\nabla _{\gamma '}F|+\sup _{s\in [d,d-1]}|\nabla _{\gamma '}F| \nonumber \\&\le \sqrt{F(x)}+\sqrt{F(y)}+1, \end{aligned}$$
(34)

where we used (31) in the last inequality. It is now immediate from (34) that F has a minimum point p. It is clear that \(|\nabla F|=0\) and \(\varDelta F \ge 0\) at the point p by the minimum principle. Hence from (23) and (24) we have

$$\begin{aligned} F(p)=\bar{\tau }^2 R \le \bar{\tau }(\bar{\tau }R+\varDelta F) =\frac{\bar{\tau }n}{2}. \end{aligned}$$

For any \(q \in M\) such that \(d_t(p,q)=d\), it is straightforward from (31) and (34) that

$$\begin{aligned} \frac{1}{4}\left( d-5n\bar{\tau }-4 \right) ^2_+ \le \frac{1}{4}\left( d-\frac{10}{3}-4\bar{\tau }(n-1)-\sqrt{2n\bar{\tau }} \right) ^2_+ \le F(q) \le \frac{1}{4} \left( d+\sqrt{2n\bar{\tau }} \right) ^2. \end{aligned}$$

\(\square \)

Note that \(F(\cdot ,t)\) is a pull-back function of \(f(\cdot ,0)\) up to the scale \(\bar{\tau }\), we can choose a base point \(p \in M\) such that p is a minimum point for all \(F(\cdot ,t)\). Now from Lemma 1, \(F(\cdot ,t)\) can be regarded as an approximation of \(\frac{d_t^2}{4}\).

With Lemma 1, we have the following volume estimate whose proof follows from [9, Theorem 1.2].

Lemma 2

There exists a constant \(C=C(n)>0\) such that for any Ricci shrinker \((M^n,g,f)\) with \(p \in M\) a minimum point of f,

$$\begin{aligned} |B_t(p,r)|_t \le {\left\{ \begin{array}{ll}\,Ce^{\varvec{\mu }}r^n \quad &{}\text {if} \quad r \ge 2\sqrt{\bar{\tau }n}; \\ \,Cr^n \quad &{}\text {if} \quad r < 2\sqrt{\bar{\tau }n}. \end{array}\right. } \end{aligned}$$

Proof

We set \(\rho =2\sqrt{F}\) and \(D(r)=\{x\in M \mid \rho \le r\}\). Moreover we define \(V(r)=\int _{D(r)}\,dV_t\) and \(\chi (r)=\int _{D(r)}R(t)\,dV_t\). It follows from a similar computation as [9, (3.5)], by using (23) and (24), that

$$\begin{aligned} nV-rV'=2\bar{\tau }\chi -\frac{4{\bar{\tau }}^2}{r}\chi '. \end{aligned}$$
(35)

If we set \(r_0=\sqrt{2\bar{\tau }(n+2)}\), by integrating (35) we obtain, see [9, (3.6)] for details, that

$$\begin{aligned} V(r) \le 2r^nr_0^{-n}V(r_0) \end{aligned}$$

for any \(r \ge 2\sqrt{\bar{\tau }n}\). Then it follows from Lemma 1 that for any \(r \ge 2\sqrt{\bar{\tau }n}\),

$$\begin{aligned} |B_t(p,r)|_t \le V\left( r+\sqrt{2n\bar{\tau }} \right) \le V(2r) \le 2^{n+1}r^nr_0^{-n}V(r_0). \end{aligned}$$

By definition, we have

$$\begin{aligned} D(r_0)&=\left\{ x\in M \left| F \le \frac{\bar{\tau }(n+2)}{2} \right. \right\} \\&=\left\{ x\in M \left| f(x,t) \le \frac{n+2}{2} \right. \right\} =\left\{ x\in M \left| f(\psi ^t(x)) \le \frac{n+2}{2} \right. \right\} . \end{aligned}$$

Moreover, since \(g(t)=\bar{\tau }(\psi ^t)^*g\),

$$\begin{aligned} V(r_0) \le \bar{\tau }^{\frac{n}{2}} \int _{f(x) \le \frac{n+2}{2}} \,dV \le \bar{\tau }^{\frac{n}{2}}| \{x \mid f(x) \le (n+2)/2\}|. \end{aligned}$$

For any x such that \(f(x) \le (n+2)/2\). it follows from Lemma 1 that \(d(p,x) \le c_0(n)\). Therefore for any \(r \ge 2\sqrt{\bar{\tau }n}\),

$$\begin{aligned} |B_t(p,r)|_t \le C(n)|B(p,c_0)| r^n \le C(n)e^{\varvec{\mu }}r^n, \end{aligned}$$

where the last inequality follows from [34, Lemma 2.3].

Finally, the case \(r \le 2\sqrt{\bar{\tau }n}\) follows from the comparison theorem [55, Theorem 1.2] by using (16). Indeed, for any x with \(d_t(p,x) \le 2\sqrt{\bar{\tau }n}\), it follows from Lemma 1 that \(f(x,t)=\bar{\tau }^{-1}F(x,t) \le C\). Therefore, from (20) we obtain \(|\nabla f|(x,t) \le C{\bar{\tau }}^{-1/2}\). Now it follows from [55, (1.5) of Theorem 1.2] that for any \(s \le r\),

$$\begin{aligned} \int _{B_t(p,r)} e^{-f(x,t)} \,dV_t \le e^{Cr{\bar{\tau }}^{-1/2}}\frac{r^n}{s^n}\int _{B_t(p,s)} e^{-f(x,t)} \,dV_t \le C\frac{r^n}{s^n}\int _{B_t(p,s)} e^{-f(x,t)} \,dV_t. \end{aligned}$$

Then the conclusion follows if we let \(s \rightarrow 0\). \(\square \)

3 Cutoff functions, maximum principle and heat kernel

Now we construct a family of cutoff functions which is important when we perform integration by parts throughout the paper.

Fix a function \(\eta \in C^{\infty }([0,\infty ))\) such that \(0\le \eta \le 1\), \(\eta =1\) on [0, 1] and \(\eta =0\) on \([2,\infty )\). Furthermore, \(-C \le \eta '/\eta ^{\frac{1}{2}}\le 0\) and \(|\eta ''|+|\eta '''| \le C\) for a universal constant \(C>0\). For each \(r \ge 1\), we define

$$\begin{aligned} \phi ^r {:}{=}\eta \left( \frac{F}{r} \right) . \end{aligned}$$
(36)

Then \(\phi ^r\) is a smooth function on \(M \times (-\infty ,1)\). The following estimates of \(\phi ^{r}\) will be repeatedly used in this paper.

Lemma 3

There exists a constant \(C=C(n)\) such that

$$\begin{aligned} (\phi ^r)^{-1} |\nabla \phi ^r|^2&\le Cr^{-1}, \end{aligned}$$
(37)
$$\begin{aligned} |\phi ^r_t|&\le C{\bar{\tau }}^{-1}, \end{aligned}$$
(38)
$$\begin{aligned} |\varDelta \phi ^r|&\le C(\bar{\tau }^{-1}+r^{-1}), \end{aligned}$$
(39)
$$\begin{aligned} |\square \phi ^r|&\le Cr^{-1}, \end{aligned}$$
(40)
$$\begin{aligned} |\square ^{*} \phi ^{r}|&\le C\left( r^{-1} +\bar{\tau }^{-1} + \bar{\tau }^{-2} r \right) . \end{aligned}$$
(41)

Proof

Note that \(F \le 2r\) on the support of \(\phi ^r\), it follows from the assumption of \(\eta \) and (31) that

$$\begin{aligned} \frac{|\nabla \phi ^r|^2}{\phi ^r}&=r^{-2}\eta '^2\eta ^{-1} |\nabla F|^2\le Cr^{-2}F \le Cr^{-1}. \end{aligned}$$

This finishes the proof of (37). Similarly, by using (22), (24), (29) and (31), we can prove

$$\begin{aligned} |\phi ^r_t|&=r^{-1}|\eta 'F_t| \le Cr^{-1}\bar{\tau }R\le Cr^{-1}\bar{\tau }^{-1}F \le C{\bar{\tau }}^{-1}, \\ |\square \phi ^r|&=|(\partial _t-\varDelta )\phi ^r| =|r^{-1}\eta ' \square F-r^{-2}\eta ''|\nabla F|^2|=|-nr^{-1}\eta '/2-r^{-2}\eta ''|\nabla F|^2| \le Cr^{-1}. \end{aligned}$$

So (38) and (40) are proved. Then we have

$$\begin{aligned} |\varDelta \phi ^r|&=|-\square \phi ^r +\partial _t \phi ^{r}| \le |\square \phi ^{r}| + |\phi ^{r}_{t}| \le C(\bar{\tau }^{-1}+r^{-1}). \end{aligned}$$

Hence we obtain (39). Finally, using (24) again, we have

$$\begin{aligned} |\square ^{*} \phi ^{r}|&=|\left( -\partial _t-\varDelta + R \right) \phi ^{r}|=|\left( \square -2\partial _t +R \right) \phi ^{r}|\\&\le |\square \phi ^{r}| + 2|\phi _{t}^{r}| + R \phi ^{r} \le |\square \phi ^{r}| + 2|\phi _{t}^{r}| + \bar{\tau }^{-2} F \phi ^{r}\\&\le C\left( r^{-1} +\bar{\tau }^{-1} + \bar{\tau }^{-2} r \right) , \end{aligned}$$

which proves (41). \(\square \)

Now we move on to show the maximum principle on general Ricci shrinkers. On a closed manifold, maximum principle holds automatically. If the underlying manifold is noncompact, then some additional assumptions are needed in order the maximum principle to hold. For example, in [35, Theorem 15.2], a condition

$$\begin{aligned} \int _a^b \int u_+^2(x,t)e^{-cd^2(x)}\,dV\,dt < \infty \end{aligned}$$
(42)

is needed for the maximum principle of the static heat equation subsolution u. In our current setting of Ricci shrinker spacetime, the metrics are evolving under Ricci curvature. Then the distance distortion of different time slices is not easy to estimate directly without Ricci curvature bound. Fortunately, we can replace \(d^2\) by f and obtain a maximum principle under a condition similar to (42).

Theorem 6

(Maximum principle on Ricci shrinkers) Let \((M^n,g,f)\) be a Ricci shrinker. Given any closed interval \([a,b] \subset (-\infty ,1)\) and a function u which satisfies \(\square u \le 0\) on \(M \times [a,b]\), suppose that

$$\begin{aligned} \int _a^b \int u^2_{+}(x,t)e^{-2f(x,t)}\,dV_t(x)\,dt < \infty . \end{aligned}$$
(43)

If \(u(\cdot ,a) \le c\), then \(u(\cdot , b) \le c\).

Proof

From Lemmas 1 and 2, it is easy to see

$$\begin{aligned} \int _a^b \int e^{-2f(x,t)}\,dV_t(x)\,dt < \infty . \end{aligned}$$

Therefore, we only need to prove the special case when \(c=0\), by considering \(u-c\).

Multiplying both sides of \(\square u \le 0\) by \(u_+(\phi ^r)^2e^{-2f}\) and integrating on the spacetime \(M \times [a,b]\), then we obtain

$$\begin{aligned} \int _a^b \int \left( \frac{u_+^2}{2}\right) _t(\phi ^r)^2e^{-2f}\,dV_t\,dt&\le \int _a^b \int \varDelta u u_+ (\phi ^r)^2e^{-2f}\,dV_t\,dt. \end{aligned}$$
(44)

For the left side of (44), we have

$$\begin{aligned}&\int _a^b \int \left( \frac{u_+^2}{2}\right) _t(\phi ^r)^2e^{-2f}\,dV_t\,dt \nonumber \\&\quad =\int \frac{u_+^2}{2} (\phi ^r)^2e^{-2f} \,dV_b -\int _a^b \int u_+^2\phi ^r\phi ^r_te^{-2f}\,dV_t\,dt \nonumber \\&\qquad +\int _a^b \int u_+^2(\phi ^r)^2f_te^{-2f}\,dV_t\,dt +\int _a^b \int \frac{u_+^2}{2}(\phi ^r)^2Re^{-2f}\,dV_t\,dt \nonumber \\&\quad \ge \int \frac{u_+^2}{2} (\phi ^r)^2e^{-2f} \,dV_b -\int _a^b \int u_+^2\phi ^r\phi ^r_te^{-2f}\,dV_t\,dt \nonumber \\&\qquad +\int _a^b \int u_+^2(\phi ^r)^2|\nabla f|^2e^{-2f}\,dV_t\,dt, \end{aligned}$$
(45)

where we have used \(R \ge 0\), \(f_t=|\nabla f|^2\) and \(u_+(\cdot ,a)=0\). For the right side of (44), we have

$$\begin{aligned}&\int _a^b \int \varDelta u u_+ (\phi ^r)^2e^{-2f}\,dV_t\,dt \nonumber \\&\quad =\int _a^b \int -|\nabla (u_+\phi ^r)|^2e^{-2f}\,dV_t\,dt+\int _a^b\int |\nabla \phi ^r|^2u_+^2e^{-2f}\,dV_t\,dt \nonumber \\&\qquad +\int _a^b \int 2\langle \nabla u_+,\nabla f \rangle u_+(\phi ^r)^2e^{-2f}\,dV_t\,dt \nonumber \\&\quad =\int _a^b \int -|\nabla (u_+\phi ^r)|^2e^{-2f}\,dV_t\,dt+\int _a^b\int |\nabla \phi ^r|^2u_+^2e^{-2f}\,dV_t\,dt \nonumber \\&\qquad +\int _a^b \int 2\langle \nabla (u_+\phi ^r),\nabla f \rangle u_+\phi ^re^{-2f}\,dV_t\,dt \nonumber \\&\qquad -\int _a^b \int 2\langle \nabla \phi ^r,\nabla f \rangle u_+^2\phi ^re^{-2f}\,dV_t\,dt. \end{aligned}$$
(46)

Combining (45) and (46), we obtain

$$\begin{aligned} \int \frac{u_+^2}{2} (\phi ^r)^2e^{-2f} \,dV_b \le I+II, \end{aligned}$$
(47)

where

$$\begin{aligned} I&= -\int _a^b \int u_+^2(\phi ^r)^2|\nabla f|^2e^{-2f}\,dV_t\,dt-\int _a^b \int |\nabla (u_+\phi ^r)|^2e^{-2f}\,dV_t\,dt \nonumber \\&\quad +\int _a^b \int 2\langle \nabla (u_+\phi ^r),\nabla f \rangle u_+\phi ^re^{-2f}\,dV_t\,dt \le 0, \end{aligned}$$
(48)

and

$$\begin{aligned} II&=\int _a^b \int u_+^2\phi ^r\phi ^r_te^{-2f}\,dV_t\,dt+\int _a^b\int |\nabla \phi ^r|^2u_+^2e^{-2f}\,dV_t\,dt \nonumber \\&\quad -\int _a^b \int 2\langle \nabla \phi ^r,\nabla f \rangle u_+^2\phi ^re^{-2f}\,dV_t\,dt. \end{aligned}$$
(49)

From our construction of \(\phi ^r\), it is easy to see that all functions involved in last three integrals are supported in the spacetime set

$$\begin{aligned} K_{r} {:}{=}\{r\le F(x,t)\le 2r, \, a\le t\le b\}. \end{aligned}$$
(50)

Moreover, all the cutoff function terms can be estimated by (37) and (38). For example, we have

$$\begin{aligned} |\langle \nabla \phi ^r,\nabla f \rangle | \le \bar{\tau }^{-1}|\nabla \phi ^r||\nabla F| \le C\bar{\tau }r^{-1/2}\sqrt{F} \le C(1-b)^{-1} \quad \text {on}\; K_{r}. \end{aligned}$$

Plugging (37), (38) and the above inequality into (49), we arrive at

$$\begin{aligned} II\le C\left( (1-b)^{-1}+r^{-1}\right) \iint _{K_r} u_+^2e^{-2f}\,dV_t\,dt. \end{aligned}$$
(51)

It follows from (47), (48) and (51) that

$$\begin{aligned} \int \frac{u_+^2}{2} (\phi ^r)^2e^{-2f} \,dV_b \le C\left( (1-b)^{-1}+r^{-1}\right) \iint _{K_r} u_+^2e^{-2f}\,dV_t\,dt. \end{aligned}$$

Note that the left hand side of the above inequality is independent of r. Letting \(r \rightarrow +\infty \), the finite integral assumption (43) implies that

$$\begin{aligned} \int \frac{u_+^2}{2} e^{-2f} \,dV_b \le 0. \end{aligned}$$

Therefore, \(u(\cdot ,b) \le 0\) by the continuity of u and positivity of \(e^{-2f(\cdot , b)}\). \(\square \)

The condition (43) is satisfied in many cases. For example, if u is a bounded heat solution. The technique used in the proof of Theorem 6 will be repeatedly used in this paper.

Now we control the spacetime integral of \(|\text {Hess}\,F|^2\).

Lemma 4

For any \(\lambda >0\), \(a<b<1\), there exists a constant \(C=C(a,b,\lambda )\) such that

$$\begin{aligned} \int _a^b\int |\text {Hess}\,F|^2 e^{-\lambda F} \,dV_t\,dt \le C. \end{aligned}$$

Proof

From (29) and direct computations,

$$\begin{aligned} \square |\nabla F|^2=-2|\text {Hess}\, F|^2. \end{aligned}$$

Multiplying both sides of the above equation by \(\phi ^re^{-\lambda F}\) and integrating on the spacetime \(M \times [a,b]\), we obtain

$$\begin{aligned}&2\int _a^b\int |\text {Hess}\,F|^2 \phi ^re^{-\lambda F} \,dV_t\,dt \\&\quad =- \left. \left( \int |\nabla F|^2\phi ^re^{-\lambda F} \,dV_t \right) \right| _a^b +\int _a^b\int \square ^*\phi ^r|\nabla F|^2 e^{-\lambda F} \,dV_t\,dt \\&\qquad +\lambda \int _a^b\int \left( \phi ^r(\lambda |\nabla F|^2-F_t-\varDelta F)-2\langle \nabla \phi ^r,\nabla F\rangle \right) |\nabla F|^2\phi ^r \,dV_t\,dt \\&\quad \le - \left. \left( \int |\nabla F|^2\phi ^re^{-\lambda F} \,dV_t \right) \right| _a^b +\int _a^b\int |\square ^*\phi ^r|Fe^{-\lambda F} \,dV_t\,dt \\&\qquad +\lambda \int _a^b\int \left( (\lambda +2{\bar{\tau }}^{-1})F+2r^{-\frac{1}{2}}F^{\frac{1}{2}}\right) Fe^{-\lambda F} \,dV_t\,dt. \end{aligned}$$

Now we let \(r \rightarrow \infty \) and the conclusion follows from Lemmas 1 and 2. \(\square \)

Theorem 7

On the Ricci flow spacetime \(M \times (-\infty , 1)\) induced by a Ricci shrinker (Mgf), there exists a positive heat kernel function H(xtys) for all \(x, y \in M\) and \(s, t \in (-\infty , 1)\) with \(x \ne y\) and \(s<t\). It satisfies

$$\begin{aligned}&\square _{x,t} H(x,t,y,s) {:}{=}\left( \partial _t - \varDelta _{x} \right) H(x,t,y,s)=0, \end{aligned}$$
(52)
$$\begin{aligned}&\square _{y,s}^{*} H(x,t,y,s) {:}{=}\left( -\partial _{s} -\varDelta _{y}+R(y,s) \right) H(x,t,y,s)=0, \end{aligned}$$
(53)
$$\begin{aligned}&\lim _{t \rightarrow s^{+}} H(x,t,y,s)=\delta _{y}, \end{aligned}$$
(54)
$$\begin{aligned}&\lim _{s \rightarrow t^{-}} H(x,t,y,s)=\delta _{x}. \end{aligned}$$
(55)

Furthermore, the heat kernel H satisfies the semigroup property

$$\begin{aligned} H(x,t,y,s)=\int H(x,t,z,\rho )H(z,\rho ,y,s)\, dV_{\rho }(z), \quad \forall \; x, y \in M, \; \rho \in (s,t) \subset (-\infty , 1), \end{aligned}$$
(56)

and the following integral relationships

$$\begin{aligned}&\int H(x,t,y,s)\,dV_t(x) \le 1, \end{aligned}$$
(57)
$$\begin{aligned}&\int H(x,t,y,s)\,dV_s(y) =1. \end{aligned}$$
(58)

Proof

We shall divide the proof of Theorem 7 into four steps.

Step 1. Existence of a heat kernel function H solving heat equation and conjugate heat equation.

Fix a compact interval \(I=[a,b] \subset (-\infty ,1)\) and a compact set \(\varOmega \subset M\) with smooth boundary, there exists a Dirichlet heat kernel. The proof can be found in in [19, Chapter 24, Sect. 5]. Regarding \((-\infty ,1)\) as the union of \([-2^{k}, 1-2^{-k}]\), it is easy to see that the Dirichlet heat kernel actually exists on \(\varOmega \times (-\infty , 1)\). Now we let \(\left\{ \varOmega _i \right\} \) be an exhaustion of M by relatively compact domains with smooth boundary such that \(\overline{\varOmega _i}\subset \varOmega _{i+1}\). Let \(H_i(x,t,y,s)\) be the Dirichlet heat kernel of \((\overline{\varOmega _i},g)\). Then the following properties hold.

$$\begin{aligned} \partial _tH_i(x,t,y,s)&=\varDelta _{x,t} H_i(x,t,y,s), \end{aligned}$$
(59)
$$\begin{aligned} \partial _sH_i(x,t,y,s)&=-\varDelta _{y,s}H_i(x,t,y,s)+R(y,s)H_i(x,t,y,s); \end{aligned}$$
(60)
$$\begin{aligned} \lim _{ t \searrow s}H_i(x,t,y,s)&=\delta _y, \end{aligned}$$
(61)
$$\begin{aligned} \lim _{ s \nearrow t}H_i(x,t,y,s)&=\delta _x. \end{aligned}$$
(62)

Let \(\mathbf {n}\) be the outward normal vector of \(\partial \varOmega _i\), then the positivity of \(H_i\) implies that \(\frac{\partial H_i}{\partial \mathbf {n}} \le 0\). Since \(R \ge 0\) on Ricci shrinkers, direct computation shows that

$$\begin{aligned} \partial _t\int _{\varOmega _i}H_i(x,t,y,s)\,dV_t(x)=\int _{\varOmega _i} \left( \varDelta _{x,t} -R \right) H_i(x,t,y,s) \,dV_t(x) \le \int _{\partial \varOmega _i} \frac{\partial H_i}{\partial \mathbf {n}}\,d\sigma _t(x) \le 0. \end{aligned}$$
(63)

Hence from (61), we have

$$\begin{aligned} \int _{\varOmega _i}H_i(x,t,y,s)\,dV_t(x) \le 1. \end{aligned}$$
(64)

Similarly, we have

$$\begin{aligned} \partial _s\int _{\varOmega _i}H_i(x,t,y,s)\,dV_s(y)&=-\int _{\varOmega _i}\varDelta _{y,s} H_i(x,t,y,s) \,dV_s(y) =-\int _{\partial \varOmega _i}\frac{\partial H_i}{\partial \mathbf {n}}\,d\sigma _s(y) \ge 0, \end{aligned}$$
(65)

which implies that

$$\begin{aligned} \int _{\varOmega _i}H_i(x,t,y,s)\,dV_s(y) \le 1. \end{aligned}$$
(66)

As \(H_i>0\) on \(\varOmega _i \times (-\infty ,1)\), it follows from the classical maximum principle that

$$\begin{aligned} 0\le H_i \le H_{i+1} \end{aligned}$$
(67)

on \(\varOmega _i \times \varOmega _i \times (-\infty , 1)\). Now we define the heat kernel on \(M \times (-\infty , 1)\) by

$$\begin{aligned} H(x,t,y,s) {:}{=}\lim _{i \rightarrow \infty }H_i(x,t,y,s). \end{aligned}$$
(68)

From the well-known mean value theorem (cf. Theorem 25.2 in [19]) the interior regularity estimates for the heat equation and conjugate heat equation, it follows from (64) and (66) that \(H_i\) is uniformly bounded when st are fixed. Threfore, H exists as a smooth function. Its positivity is guranteed by (67). The regularity estimates also imply that the convergence from \(H_i\) to H is locally smooth. In particular, we can take limit of (59) and (60) to obtain that H solves heat equation and conjugate heat equation on \(M \times (-\infty , 1)\). In other words, (52) and (53) are satisfied.

Step 2 The heat kernel is a fundamental solution of heat equation and conjugate heat equation.

Let \(\phi \) be a smooth function on M with compact support K. For fixed y and s, we have

$$\begin{aligned}&\left| \partial _t\int _{\varOmega _i}H_i(x,t,y,s)\phi (x)\,dV_t(x)\right| \nonumber \\&\quad =\left| \int _{\varOmega _i} (\varDelta _{x,t}-R) H_i(x,t,y,s)\phi (x)\,dV_t(x)\right| \nonumber \\&\quad \le \left| \int _{\varOmega _i} H_i(x,t,y,s)\varDelta \phi (x)\,dV_t(x)\right| +\left| \int _{\varOmega _i}RH_i(x,t,y,s)\phi (x) \,dV_t(x)\right| \nonumber \\&\quad \le C \left| \int _{\varOmega _i}H_i(x,t,y,s)\,dV_t(x)\right| \le C, \end{aligned}$$
(69)

where C is independent of \(H_i\). Notice that the last two inequalities hold since we just need to restrict the integral on K, and for a fixed s, when t is close to s, the metrics are uniformly equivalent on \(K \times [s,t]\). Combining (61) with (69), we obtain

$$\begin{aligned} \left| \int _{\varOmega _i}H_i(x,t,y,s)\phi (x)\,dV_t(x)-\phi (y)\right| \le C(t-s). \end{aligned}$$
(70)

Since \(\phi \) has compact support, it is clear that

$$\begin{aligned} \lim _{i \rightarrow \infty } \int _{\varOmega _i} H_i(x,t,y,s)\phi (x)\,dV_t(x)=\int H(x,t,y,s) \phi (x) dV_{t}(x). \end{aligned}$$

Plugging the above equation into (70) yields that

$$\begin{aligned} \left| \int H(x,t,y,s)\phi (x)\,dV_t(x)-\phi (y)\right| \le C(t-s), \end{aligned}$$

which means that

$$\begin{aligned} \lim _{ t \rightarrow s^{+}} \int H(x,t,y,s)\phi (x)\,dV_t(x)=\phi (y). \end{aligned}$$

By the arbitrary choice of \(\phi \), we obtain (54). Therefore, H is a fundamental solution of the heat equation. Similary, we can use the limit argument to derive (55) and claim that H is a fundamental solution of the conjugate heat equation.

Step 3 The heat kernel satisfies the semigroup property.

From its construction, \(H_i\) satisfies the semigroup property:

$$\begin{aligned} H_i(x,t,y,s)=\int _{\varOmega _i}H_i(x,t,z,\rho )H_i(z,\rho ,y,s)\, dV_{\rho }(z), \quad \forall \; x, y \in \varOmega _i, \; \rho \in (s,t) \subset (-\infty , 1). \end{aligned}$$
(71)

For each compact set \(K \subset M\), it is clear that \(K \subset \varOmega _i\) for large i. By the positivity of each \(H_i\), we have

$$\begin{aligned} H(x,t,y,s)&=\lim _{i \rightarrow \infty } H_i(x,t,y,s)=\lim _{i \rightarrow \infty } \int _{\varOmega _i}H_i(x,t,z,\rho )H_i(z,\rho ,y,s)\, dV_{\rho }(z)\\&\ge \lim _{i \rightarrow \infty } \int _{K} H_i(x,t,z,\rho )H_i(z,\rho ,y,s)\, dV_{\rho }(z)=\int _{K} H(x,t,z,\rho )H(z,\rho ,y,s)\, dV_{\rho }(z). \end{aligned}$$

By the arbitrary choice of \(K \subset M\), the above inequality implies that

$$\begin{aligned} H(x,t,y,s) \ge \int H(x,t,z,\rho )H(z,\rho ,y,s)\, dV_{\rho }(z). \end{aligned}$$
(72)

By (67), (68) and the positivity of H, we have

$$\begin{aligned} H_i(x,t,y,s)&=\int _{\varOmega _i}H_i(x,t,z,\rho )H_i(z,\rho ,y,s)\, dV_{\rho }(z) \le \int _{\varOmega _i} H(x,t,z,\rho )H(z,\rho ,y,s) dV_{\rho }(z)\\&<\int H(x,t,z,\rho ) H(z,\rho ,y,s) dV_{\rho }(z), \end{aligned}$$

whose limit form is

$$\begin{aligned} H(x,t,y,s) \le \int H(x,t,z,\rho ) H(z,\rho ,y,s) dV_{\rho }(z). \end{aligned}$$
(73)

Therefore, the semigroup property (56) follows from the combination of (72) and (73).

Step 4 The integral relationships (57) and (58) are satisfied.

On each compact set \(K \subset M\), since \(K \subset \varOmega _i\) for large i and each \(H_i\) is positive on \(\varOmega _i\), we have

$$\begin{aligned} \int _{K} H(x,t,y,s) dV_{t}(x) =\lim _{i \rightarrow \infty } \int _{K} H_i(x,t,y,s) dV_{t}(x) \le \lim _{i \rightarrow \infty } \int _{\varOmega _i} H_i(x,t,y,s) dV_{t}(x) \le 1, \end{aligned}$$

where (64) is applied in the last step. The arbitrary choice of K then yields that

$$\begin{aligned} \int H(x,t,y,s) dV_{t}(x) \le 1, \end{aligned}$$

which is nothing but (57). Similar reasoning can pass (66) to obtain

$$\begin{aligned} \int _{K} H(x,t,y,s) dV_{s}(y) \le 1, \end{aligned}$$
(74)

where the inequality will be improved to equality (58) in the following argument. In fact, let \(\phi ^{r}\) be the cutoff function defined in (36). For any fixed x and t, it follows from the cutoff function estimate (40) that

$$\begin{aligned}&\left| \partial _s\int H(x,t,y,s)\phi ^r(y,s)\,dV_s(y)\right| \\&\quad =\left| \int H(x,t,y,s)\square _{y,s}\phi ^r(y,s)\, dV_s(y)\right| \le Cr^{-1}\int H(x,t,y,s) \, dV_s(y). \end{aligned}$$

Plugging (74) into the above inequality, we obtain

$$\begin{aligned} \left| \partial _s\int H(x,t,y,s)\phi ^r(y,s)\,dV_s(y)\right| \le Cr^{-1}. \end{aligned}$$

When r is large, x is covered by the support of \(\phi ^r\) at the time t. Using (55), the above inequality implies that

$$\begin{aligned} \left| \int H(x,t,y,s)\phi ^r(y,s)\,dV_s(y)-1\right| \le Cr^{-1}(t-s). \end{aligned}$$

Since r could be arbitrarily large in the above inequality, we obtain (58) by letting \(r \rightarrow \infty \). \(\square \)

Lemma 5

Suppose \([a,b] \subset (-\infty , 1)\) and \(u_{a}\) is a bounded function on the time slice (Mg(a)). Then

$$\begin{aligned} u(x,t) {:}{=}\int H(x,t,y,a)u_a(y) dV_{a}(y), \quad \forall \; t \in [a,b] \end{aligned}$$
(75)

is the unique bounded heat solution with initial value \(u_a\).

Proof

Clearly, u is a well-defined heat solution with the initial value \(u_a\). Suppose \(\tilde{u}\) is another heat solution with initial value \(u_a\). Then \(u-\tilde{u}\) is a bounded heat solution with initial value 0. Therefore, we can apply maximum principle Theorem 6 on \(\pm (u-\tilde{u})\) to obtain that

$$\begin{aligned} u -\tilde{u} \equiv 0 \quad \text {on} \; M \times [a,b]. \end{aligned}$$

In other words, \(\tilde{u} \equiv u\) and the uniqueness is proved. \(\square \)

Corollary 1

Suppose \(u_a\) is a smooth, bounded, integrable function on (Mg(a)). Let u be the unique bounded heat solution on \(M \times [a,b]\) starting from \(u_a\). Then we have

$$\begin{aligned} \sup _{M} |\nabla u(\cdot , b)| \le \sup _{M} |\nabla u(\cdot , a)|. \end{aligned}$$
(76)

Proof

Fix \(r>>1\) and multiply both sides of \(\square u=0\) by \(u (\phi ^r)^2\) and integrating on \(M \times [a,b]\), we obtain

$$\begin{aligned} \left. \left( \frac{1}{2}\int u^2(\phi ^r)^2 \,dV_t \right) \right| _a^b-\int _a^b \int u^2 \phi ^r\phi ^r_t \,dV_t \,dt=\int _a^b \int \left\{ -|\nabla (u \phi ^r)|^2+|\nabla \phi ^r|^2u^2 \right\} dV_t \,dt. \end{aligned}$$
(77)

By Lemma 5 we know

$$\begin{aligned} u =\int H(x,t,y,a)u_a(y) dV_{a}. \end{aligned}$$

Then it follows from (57) that u is bounded and integrable. Consequently, \(u^2\) is integrable. It follows from (37) and (38) that by letting \(r \rightarrow \infty \), we obtain from (77)

$$\begin{aligned} \int _a^b \int |\nabla u|^2\,dV_t \,dt \le -\left. \left( \frac{1}{2} \int u^2 dV_t \right) \right| _a^b+C\bar{\tau }^{-1}\int _a^b \int u^2\,dV_t \,dt <\infty . \end{aligned}$$

Therefore, the assumption of Theorem 6 is satisfied. Since \(\square |\nabla u|^2=-2|\text {Hess}\,u|^2 \le 0\), following the maximum principle, we arrive at (76). \(\square \)

Proposition 1

Suppose u is a heat solution and w is a conjugate heat solution on \(M \times [a,b]\) for \([a,b] \subset (-\infty , 1)\) such that

$$\begin{aligned} \sup _{t \in [a,b]} \int |w|\,dV_t+\sup _{M \times [a,b]} |u| \le C <\infty . \end{aligned}$$

Then we have

$$\begin{aligned} \int uw \,dV_{b} = \int uw \,dV_{a}. \end{aligned}$$
(78)

Proof

Fix \(r>>1\). We calculate

$$\begin{aligned} \partial _t\int wu\phi ^r \,dV_t&=\int \left\{ w \square (u\phi ^{r}) - (u \phi ^{r}) \square ^{*} w \right\} \,dV_t=\int w \square \left( u \phi ^{r} \right) \,dV_t \nonumber \\&=\int w\left\{ u \square \phi ^{r} + \phi ^{r} \square u - 2\langle \nabla u, \nabla \phi ^{r}\rangle \right\} \,dV_t \nonumber \\&=\int w\left\{ u \square \phi ^{r} - 2\langle \nabla u, \nabla \phi ^{r}\rangle \right\} \,dV_t. \end{aligned}$$
(79)

Note that \(|\nabla u|\le C\) by Corollary 1. Plugging the cutoff function estimates (37) and (40) into the above inequality, we obtain

$$\begin{aligned} \left| \left. \left( \int wu \phi ^{r} \,dV_t \right) \right| _{a}^{b} \right| \le C(r^{-1}+r^{-\frac{1}{2}}). \end{aligned}$$

Taking \(r \rightarrow \infty \), the right hand side of the above inequality tends to zero, the left hand side converges to

$$\begin{aligned} \int wu \,dV_b - \int wu \,dV_a, \end{aligned}$$

since u is bounded and w is integrable. Consequently, we arrive at (78). \(\square \)

Lemma 6

Suppose \([a,b] \subset (-\infty , 1)\) and \(w_{b}\) is an integrable function on the time slice (Mg(b)). Then

$$\begin{aligned} w(y,s) {:}{=}\int H(x,b,y,s)w_b(x) \, dV_{b}(x) \end{aligned}$$
(80)

is the unique conjugate heat solution with initial value \(w_{b}\) such that

$$\begin{aligned} \sup _{t \in [a,b]} \int |w|\,dV_t<\infty . \end{aligned}$$
(81)

Proof

Fix a time \(a_0 \in [a,b]\) and let h be an arbitrary smooth function with compact support. Then we solve the heat equation starting from h to obtain a unique bounded function u as

$$\begin{aligned} u(x,t)=\int H(x,t,y,a)h(y) \,dV_{a}(y). \end{aligned}$$

Since w is given by (80), it follows from (58) that w satisfies (81). Suppose \(\tilde{w}\) is another conjugate heat solution starting from \(w_b\) satisfying (81). Then we can apply Lemma 5 to the couple of u and \(\tilde{w}-w\) to obtain that for any \(t \in [a_0,b]\),

$$\begin{aligned} \int \left( \tilde{w}(x,t)-w(x,t) \right) u(x,t) \,dV_{t}(x)=0. \end{aligned}$$

In particular,

$$\begin{aligned} \int \left( \tilde{w}(x,a_0)-w(x,a_0) \right) h(x) \,dV_{a_0}(x)=0. \end{aligned}$$

By the arbitrary choice of h, we obtain \(\tilde{w}(\cdot , a_0)-w(\cdot , a_0) \equiv 0\). Then by the arbitrary choice of \(a_0\), we see that

$$\begin{aligned} \tilde{w}(\cdot , t) \equiv w(\cdot , t), \quad \; \forall \; t \in [a,b]. \end{aligned}$$

Therefore, the uniqueness is proved. \(\square \)

Lemma 7

Suppose w is a bounded function on \(M \times [a,b]\) satisfying \(\square ^* w \le 0\) and (81). Then we have

$$\begin{aligned} \sup _{M} w(\cdot , a) \le \sup _{M} w(\cdot , b). \end{aligned}$$
(82)

Proof

Without loss of generality, by adding a constant, we may assume that \(\displaystyle \sup _{M} w(\cdot , b)=0\). Then it suffices to show that

$$\begin{aligned} \sup _{M} w(\cdot , a) \le 0. \end{aligned}$$
(83)

At the time slice \(t=a\), we choose an arbitrary nonnegative smooth function h with compact support. Then we solve the forward heat solution starting from h and denote the function by u. It is clear that \(u \ge 0\). Similar to the proof of Proposition 1, we obtain that

$$\begin{aligned} \int w(x,a)h(x) \,dV_{a}(x) \le \int w(x,b) u(x,b) \,dV_{b}(x) \le 0, \end{aligned}$$

since at time \(t=b\) we have \(u \ge 0\) and \(w \le 0\). Therefore, the inequality (83) follows from the arbitrary choice of h. \(\square \)

Theorem 8

(Bounded heat solution) Suppose \(t_0 \in (-\infty , 1)\) and h is a bounded function on the time-slice \((M, g(t_0))\). On \(M \times (t_0, 1)\), starting from h, there is a unique heat solution u which is bounded on each compact time-interval of \([t_0,1)\). The solution is

$$\begin{aligned} u(x,t) = \int H(x,t,y,t_0) h(y) \,dV_{t_0}(y), \quad \forall \; x \in M, \; t \in (t_0, 1). \end{aligned}$$
(84)

Similarly, for any bouned integrable function h, starting from h there is a unique conjugate heat solution w which is bounded and integrable uniformly on each compact time interval of \((-\infty , t_0]\). The solution is

$$\begin{aligned} w(x,t)=\int H(y,t_0,x,t) h(y) \, dV_{t_0}(y), \quad \forall \; x \in M, \; t \in (-\infty , t_0). \end{aligned}$$
(85)

Theorem 9

(Maximum principle of bounded functions) Suppose u is a bounded super-heat-solution, i.e., \(\square u \le 0\) on \(M \times [a,b]\). Then

$$\begin{aligned} \sup _{M} u(\cdot , b) \le \sup _{M} u(\cdot , a). \end{aligned}$$
(86)

Similarly, if w is a bounded super-conjugate-heat-solution, i.e., \(\square ^* w \le 0\) on \(M \times [a,b]\) satisfying (81). Then

$$\begin{aligned} \sup _{M} w(\cdot , b) \ge \sup _{M} w(\cdot , a). \end{aligned}$$
(87)

From (27) and (28) from previous section, on the space-time \(M \times (-\infty , 1)\), there are standard heat solution and conjugate heat solutions \(F+\frac{n}{2}t\) and \(\bar{v}=(4\pi (1-t))^{-\frac{n}{2}} e^{-f}\). We can apply Theorems 6 and 9 to compare other supersolutions or subsolutions with \(F+\frac{n}{2}t\) and \(\bar{v}=(4\pi (1-t))^{-\frac{n}{2}} e^{-f}\). In particular, we have the following Lemma.

Lemma 8

Given a smooth function \(\phi \) with compact support on a Ricci shrinker \((M^n,g,f)\). For any \(b<1\), let \(w(x,t)=\int H(y,b,x,t)\phi (y) \,dV_b(y)\) be the bounded solution of conjugate heat equation with \(w(\cdot ,b)=\phi \). Then there exists a constant \(C>0\) such that for \(t \le b\)

$$\begin{aligned} w(x,t) \le C\bar{v}(x,t)=C\frac{e^{-f(x,t)}}{(4\pi \bar{\tau })^{n/2}}. \end{aligned}$$
(88)

Lemma 8 tells us that starting from a compact supported function, the solution of the conjugate heat equation is at least exponentially decaying.

4 Monotonicity of Perelman’s entropy

Recall that on any compact Riemannian manifold \((M^n,g)\), Perelman’s \(\mathcal W\) entropy [46] is defined as

$$\begin{aligned} \mathcal W(g,\phi ,\tau )=\int \left( \tau (|\nabla \phi |^2+R)+\phi -n \right) \frac{e^{-\phi }}{(4 \pi \tau )^{n/2}}\,dV \end{aligned}$$
(89)

for \(\phi \) a smooth function and \(\tau >0\). Let \(u^2=\frac{e^{-\phi }}{(4\pi \tau )^{n/2}}\), we can rewrite above functional as

$$\begin{aligned} \overline{\mathcal W}(g,u,\tau )=\int \tau (4|\nabla u|^2+Ru^2)-u^2\log u^2 \,dV-\left( n+\frac{n}{2}\log (4\pi \tau )\right) \int u^2 \,dV. \end{aligned}$$
(90)

For a general Ricci shrinker \((M^n,g,f)\), we define the \(\varvec{\mu }\)-functional as

(91)

where

$$\begin{aligned} W_{*}^{1,2}(M)=\left\{ u \left| \int |\nabla u|^2\,dV<\infty , \right. \int u^2 \,dV=1 \,\text {and}\, \int d^2(p,\cdot )u^2\,dV<\infty \right\} . \end{aligned}$$
(92)

The last integral condition \(\int d^2(p,\cdot )u^2\,dV<\infty \) is imposed for two reasons. First, it follows from Lemma 1 and (20) that

$$\begin{aligned} \int Ru^2\,dV<\infty . \end{aligned}$$
(93)

Second, the term \(\int u^2 \log u^2\,dV\) in the definition of \(\overline{\mathcal W}(g,u,\tau )\) is well defined. Indeed, if we consider the rescaled measure \(d\tilde{V} {:}{=}e^{-d^2(p,x)}V\), then it follows from the volume estimate Lemma 2 that \(\tilde{V}(M)\) is finite. Given a \(u \in W_{*}^{1,2}\), we set \(A {:}{=}\{x\in M\mid u(x)<1\}\) and \(\tilde{u} {:}{=}\chi _Au\), where \(\chi _A\) is the characteristic function of the set A. Then it is clear that \(\int d^2(p,x)\tilde{u}^2(x)\,dV<\infty \). By a direct calculation,

$$\begin{aligned} \int \tilde{u}^2\log \tilde{u}^2\,dV=\int \hat{u}^2\log \hat{u}^2\,d\tilde{V}-\int d^2(p,\cdot )\tilde{u}^2\,dV, \end{aligned}$$
(94)

where \(\hat{u}^2=\tilde{u}^2 e^{d^2(p,\cdot )}\). By Jensen’s inequality, we obtain

$$\begin{aligned} \int \hat{u}^2\log \hat{u}^2\,d\tilde{V} \ge \left( \int \hat{u}^2\,d\tilde{V} \right) \log \left( \frac{1}{\tilde{V}(M)}\int \hat{u}^2\,d\tilde{V} \right) >-\infty \end{aligned}$$

since \(\int \hat{u}^2\,d\tilde{V}=\int \tilde{u}^2\,dV \in [0,1]\). Therefore it follows from (94) that

$$\begin{aligned} \int \tilde{u}^2\log \tilde{u}^2\,dV>-\infty . \end{aligned}$$

In other words, it implies that for any \(u \in W_{*}^{1,2}(M)\), the negative part of \(u^2\log u^2\) is integrable and \(\overline{\mathcal W}(g,u,\tau ) \in [-\infty ,+\infty )\). In fact, it will be proved later, see Proposition 15 that \(\overline{\mathcal W}(g,u,\tau )\) cannot be \(-\infty \).

Remark 1

The space \(W_{*}^{1,2}(M)\) can be regarded as a collection of probability measure v such that

  1. (i)

    \(v=\rho V\), that is, v is absolutely continuous with respect to the volume form V.

  2. (ii)

    v has finite moment of second order (\(v \in P_2(M)\)), that is, for any point \(q \in M\),

    $$\begin{aligned} \int d^2(q,\cdot )\,dv < \infty . \end{aligned}$$
  3. (iii)

    The Fisher information

    $$\begin{aligned} F(\rho ) {:}{=}4\int |\nabla \sqrt{\rho }|^2\,dV<\infty . \end{aligned}$$

Now we show that for any Ricci shrinker, we can always restrict the infimum on all smooth functions with compact support.

Proposition 2

For any Ricci shrinker \((M^n,g,f)\),

$$\begin{aligned} \varvec{\mu }(g,\tau )= \inf \left\{ \overline{\mathcal W}(g,u,\tau ) \left| \, u \in C_0^{\infty }(M) \, \, \text {and} \, \int u^2 \,dV=1 \right. \right\} . \end{aligned}$$
(95)

Proof

For any function \(u \in W_{*}^{1,2}(M)\) such that \(\overline{\mathcal W}(g,u,\tau )\) is finite, we define a positive constant

$$\begin{aligned} c_r^2=\int u^2(\phi ^r)^2 \,dV. \end{aligned}$$

It is clear from the definition that \(c_r \le 1\) and \(\lim _{r \rightarrow \infty }c_r=1\). From direct computations,

$$\begin{aligned}&\overline{\mathcal W}(g,c_r^{-1}u\phi ^r,\tau ) \\&\quad =\int c_r^{-2}\tau (4|\nabla (u\phi ^r)|^2+R(u\phi ^r)^2)-(c_r^{-1}u\phi ^r)^2\log (c_r^{-1}u\phi ^r)^2 \,dV-n-\frac{n}{2}\log (4\pi \tau ) \\&\quad =\int 4\tau c_r^{-2}\left( (\phi ^r)^2|\nabla u|^2+|\nabla \phi ^r|^2u^2+2u\phi ^r\langle \nabla u, \nabla \phi ^r \rangle \right) +c_r^{-2}\tau R(u\phi ^r)^2 \, dV \\&\qquad -\int (c_r^{-1}\phi ^r)^2u^2\log u^2+(c_r^{-1}\phi ^r)^2 \log {(\phi ^r)^2}u^2\, dV+\log c_r^2-n-\frac{n}{2}\log (4\pi \tau ). \end{aligned}$$

Now by the definition of \(W_{*}^{1,2}\) and the dominated convergence theorem,

$$\begin{aligned}&\lim _{r \rightarrow \infty }\overline{\mathcal W}(g,c_r^{-1}u\phi ^r,\tau )-\overline{\mathcal W}(g,u,\tau ) \\&\quad =-\lim _{r \rightarrow \infty }\int \left( 1-(c_r^{-1}\phi ^r)^2\right) u^2\log u^2 \,dV. \end{aligned}$$

Since \(u^2\log u^2\) is absolutely integrable, by the dominated convergence theorem,

$$\begin{aligned} \lim _{r \rightarrow \infty }\int \left( 1-(c_r^{-1}\phi ^r)^2\right) u^2\log u^2 \,dV=0 \end{aligned}$$

and hence

$$\begin{aligned}&\lim _{r \rightarrow \infty }\overline{\mathcal W}(g,c_r^{-1}u\phi ^r,\tau )=\overline{\mathcal W}(g,u,\tau ). \end{aligned}$$

Similarly, if \(\overline{\mathcal W}(g,u,\tau )=-\infty \), then

$$\begin{aligned} \lim _{r \rightarrow \infty }\overline{\mathcal W}(g,c_r^{-1}u\phi ^r,\tau )=-\infty . \end{aligned}$$

For a fixed r, it is not hard to choose a sequence of smooth functions \(u_s\) with compact support by the usual smoothing process such that

$$\begin{aligned}&\lim _{s \rightarrow \infty }\overline{\mathcal W}(g,u_s,\tau )=\overline{\mathcal W}(g,c_r^{-1}u\phi ^r,\tau ). \end{aligned}$$

\(\square \)

Now we prove the celebrated monotonicity theorem of Perelman on Ricci shrinkers.

Theorem 10

For any Ricci shrinker \((M^n,g,f)\) and \(\tau >0\),

$$\begin{aligned} \varvec{\mu }(g(t),\tau -t) \end{aligned}$$
(96)

is increasing for \(t < \min \{1,\tau \}\).

Proof

We fix a time \(t_1 < \min \{1,\tau \}\) and an nonnegative smooth function \(\sqrt{\bar{w}}\) with compact support such that \(\int \bar{w}\,dV_{t_1}=1\). By defining

$$\begin{aligned} w(x,t)=\int H(y,t_1,x,t)\bar{w}(y)\,dV_{t_1}(y), \end{aligned}$$
(97)

it is straightforward to check that

$$\begin{aligned} \int w(x,t) \, dV_t(x)=\iint H(y,t_1,x,t)\bar{w}(y)\,dV_t(x)\,dV_{t_1}(y)=\int \bar{w}(y)dV_{t_1}(y)=1, \end{aligned}$$

where we have used stochastic completeness (58) for the last equality.

Lemma 9

For any time \(t_0 < t_1\),

$$\begin{aligned} 4\int _{t_0}^{t_1}\int |\nabla \sqrt{w}|^2\,dV_t\,dt =\int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\,dV_t\,dt <\infty . \end{aligned}$$
(98)

Proof of Lemma 9:

By direct computations,

$$\begin{aligned}&\int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\phi ^r\,dV_t\,dt \nonumber \\&\quad =\int _{t_0}^{t_1}\int \langle \nabla (\log w), \nabla w \rangle \phi ^r\,dV_t\,dt \nonumber \\&\quad = -\int _{t_0}^{t_1}\int (\log w) \varDelta w \phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int \log w \langle \nabla w, \nabla \phi ^r \rangle \,dV_t\,dt \nonumber \\&\quad =I+II. \end{aligned}$$
(99)

We estimate I first.

$$\begin{aligned} I {:}{=}&-\int _{t_0}^{t_1}\int (\log w) \varDelta w \phi ^r\,dV_t\,dt \\&=\int _{t_0}^{t_1}\int (\log w) w_t \phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int (\log w) Rw \phi ^r\,dV_t\,dt \\&=\left. \left( \int (\log w) w \phi ^r\,dV_t\right) \right| _{t_0}^{t_1}-\int _{t_0}^{t_1}\int (\log w)_t w \phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int (\log w) w \phi ^r_t\,dV_t\,dt \\&\quad +\int _{t_0}^{t_1}\int (\log w) Rw \phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int (\log w) Rw \phi ^r\,dV_t\,dt \\&=\left. \left( \int (\log w) w \phi ^r\,dV_t\right) \right| _{t_0}^{t_1}-\int _{t_0}^{t_1}\int w_t \phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int (\log w) w \phi ^r_t\,dV_t\,dt \\&=\left. \left( \int (\log w) w \phi ^r\,dV_t\right) \right| _{t_0}^{t_1}-\left. \left( \int w \phi ^r\,dV_t\right) \right| _{t_0}^{t_1} \\&\quad +\int _{t_0}^{t_1}\int w \phi ^r_t\,dV_t\,dt-\int _{t_0}^{t_1}\int w R\phi ^r\,dV_t\,dt-\int _{t_0}^{t_1}\int (\log w) w \phi ^r_t\,dV_t\,dt \end{aligned}$$

Now it is easy to show that all integrals in I are bounded. Indeed, from Lemma 8, there exists a constant C such that

$$\begin{aligned} w(x,t) \le Ce^{-f(x,t)} \end{aligned}$$

on \(M \times [t_0,t_1]\), where C depends only on \(t_1\), \(t_2\) and the upper bound of \(w(\cdot , t_1)\).

Therefore for \(t \in [t_0,t_1]\)

$$\begin{aligned} \int |w(\log w)| \, dV_t \le C\int w^{1/2}+w^2\, dV_t \le C\int e^{-f/2}+e^{-f}\, dV_t \le C. \end{aligned}$$

Moreover, by using (20),

$$\begin{aligned} \int _{t_0}^{t_1}\int w R\,dV_t\,dt \le C\int _{t_0}^{t_1}\int fe^{-f}\,dV_t\,dt \le C. \end{aligned}$$

Now we estimate II in (99).

$$\begin{aligned} |II|&\le \left| \int _{t_0}^{t_1}\int \log w \langle \nabla w, \nabla \phi ^r \rangle \,dV_t\,dt \right| \nonumber \\&\le \int _{t_0}^{t_1}\int |\log w||\nabla w||\nabla \phi ^r|\,dV_t\,dt \nonumber \\&=\int _{t_0}^{t_1}\int |\log w|\frac{|\nabla w|}{\sqrt{w}}\frac{|\nabla \phi ^r|}{\sqrt{\phi ^r}}\sqrt{w}\sqrt{\phi ^r}\,dV_t\,dt \nonumber \\&\le \frac{1}{2}\int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\phi ^r\,dV_t\,dt+\frac{1}{2}\int _{t_0}^{t_1}\int w|\log w|^2\frac{|\nabla \phi ^r|^2}{\phi ^r}\,dV_t\,dt. \end{aligned}$$
(100)

By our construction of \(\phi ^r\), \(\frac{|\nabla \phi ^r|^2}{\phi ^r}\) is uniformly bounded. Reasoning as before,

$$\begin{aligned} \int _{t_0}^{t_1}\int w|\log w|^2\frac{|\nabla \phi ^r|^2}{\phi ^r}\,dV_t\,dt \le C\iint w^{1/2}+w^2\, dV_t\,dt \le C\iint e^{-f/2}+e^{-f}\, dV_t\,dt \le C. \end{aligned}$$

Now it is easy to see from (99) and (100) that

$$\begin{aligned} \int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\phi ^r\,dV_t\,dt \le C, \end{aligned}$$

where C depends only on \(t_0\), \(t_1\) and the upper bound of \(w(\cdot ,t_1)\). By taking \(r \rightarrow \infty \), we have proved Lemma 9.

Now we define the function \(\phi \) as

$$\begin{aligned} w(x,t)=\frac{e^{-\phi }}{(4\pi (\tau -t))^{n/2}}. \end{aligned}$$

By direct computations, see Theorem 9.1 of [46], that if we set

$$\begin{aligned} v=\left( (\tau -t)(2\varDelta \phi -|\nabla \phi |^2+R)+\phi -n\right) w, \end{aligned}$$

then for \(t<\tau \),

$$\begin{aligned} \square ^* v=-2(\tau -t)\left| Rc+\text {Hess}\,\phi -\frac{g}{2(\tau -t)}\right| ^2w \le 0, \end{aligned}$$
(101)

that is, v is a subsolution of the conjugate heat equation.

We set \(\tau _1=\tau _1(t)=\tau -t\) for simplicity. By the definition,

$$\begin{aligned} v=\tau _1\left( -2\varDelta w+\frac{|\nabla w|^2}{w}+Rw\right) -w\log w-\left( n+\frac{n}{2}\log (4\pi \tau _1)\right) w. \end{aligned}$$
(102)

Now we multiply both sides of (101) by \(\phi ^r\) so that

$$\begin{aligned} \int _{t_0}^{t_1}\int v_t \phi ^r\,dV_t\,dt \ge -\int _{t_0}^{t_1}\int \varDelta v \phi ^r\,dV_t\,dt+\int _{t_0}^{t_1}\int Rv\phi ^r\,dV_t\,dt. \end{aligned}$$
(103)

The left side of (103) is

$$\begin{aligned} \int _{t_0}^{t_1}\int v_t \phi ^r\,dV_t\,dt = -\int _{t_0}^{t_1}\int v \phi _t^r\,dV_t\,dt +\int _{t_0}^{t_1}\int Rv\phi ^r\,dV_t\,dt+\left. \left( \int v\phi ^r\,dV_t\right) \right| _{t_0}^{t_1}. \end{aligned}$$
(104)

The right side of (103) is

$$\begin{aligned} -\int _{t_0}^{t_1}\int \varDelta v \phi ^r\,dV_t\,dt+\int _{t_0}^{t_1}\int Rv\phi ^r\,dV_t\,dt=-\int _{t_0}^{t_1}\int v\varDelta \phi ^r\,dV_t\,dt+\int _{t_0}^{t_1}\int R\phi ^r\,dV_t\,dt \end{aligned}$$
(105)

Therefore, we have

$$\begin{aligned} \left. \left( \int v\phi ^r\,dV_t \right) \right| _{t_0}^{t_1} \ge \int _{t_0}^{t_1}\int v \square \phi ^r\,dV_t\,dt. \end{aligned}$$
(106)

Now it is important to use the exact expression of \(\square \phi ^r\), that is,

$$\begin{aligned} \square \phi ^r=-nr^{-1}{\eta }'/2-r^{-2}{\eta }''|\nabla F|^2. \end{aligned}$$
(107)

We consider the first term of v and prove the following lemma. \(\square \)

Lemma 10

$$\begin{aligned} \lim _{r \rightarrow \infty }\int _{t_0}^{t_1}\int \varDelta w \square \phi ^r\,dV_t\,dt=0. \end{aligned}$$
(108)

Proof of Lemma 10:

From (107), we have

$$\begin{aligned} \int _{t_0}^{t_1}\int \varDelta w \square \phi ^r\,dV_t\,dt&=-\frac{n}{2r}\int _{t_0}^{t_1}\int \varDelta w\eta '\,dV_t\,dt-r^{-2}\int _{t_0}^{t_1}\int \varDelta w {\eta }'' |\nabla F|^2 \,dV_t\,dt \\&= I+II. \end{aligned}$$

Now

$$\begin{aligned} |I|&= \left| -\frac{n}{2r}\int _{t_0}^{t_1}\int \varDelta w\eta '\,dV_t\,dt \right| =\left| \frac{n}{2r}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla {\eta }' \rangle \,dV_t\,dt \right| \nonumber \\&= \left| \frac{n}{2r^2}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla F\rangle {\eta }''\,dV_t\,dt \right| \le \frac{n}{2r^2}\int _{t_0}^{t_1}\int |\nabla w||\nabla F||{\eta }''|\,dV_t\,dt \nonumber \\&=\frac{n}{2r^2}\int _{t_0}^{t_1}\int \frac{|\nabla w|}{\sqrt{w}}|\nabla F||{\eta }''|\sqrt{w}\,dV_t\,dt \nonumber \\&\le \frac{n}{2r^2}\left( \int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\,dV_t\,dt \right) ^{1/2}\left( \int _{t_0}^{t_1}\int |\nabla F|^2|{\eta }''|^2w\,dV_t\,dt \right) ^{1/2}. \end{aligned}$$
(109)

Now the first integral of (109) is bounded by (98) while the second

$$\begin{aligned} \int _{t_0}^{t_1}\int |\nabla F|^2|{\eta }''|^2w\,dV_t\,dt \le C\int _{t_0}^{t_1}\int Fw\,dV_t\,dt \le C\int _{t_0}^{t_1}\int Fe^{-f}\,dV_t\,dt \le C \end{aligned}$$
(110)

where the last constant C depends only on \(t_0\), \(t_1\) and the upper bound of \(w(\cdot ,t_1)\).

It is immediate that from (109) by taking \(r \rightarrow \infty \) that \(\lim _{r \rightarrow \infty } I=0\).

We continue to estimate II.

$$\begin{aligned} |II|&= \left| -r^{-2}\int _{t_0}^{t_1}\int \varDelta w {\eta }'' |\nabla F|^2 \,dV_t\,dt \right| \\&\le \left| r^{-2}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla {\eta }'' \rangle |\nabla F|^2\,dV_t\,dt \right| +\left| r^{-2}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla |\nabla F|^2\rangle {\eta }''\,dV_t\,dt \right| \\&=III+IV. \end{aligned}$$

Now we have

$$\begin{aligned} III&=\left| r^{-2}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla {\eta }'' \rangle |\nabla F|^2\,dV_t\,dt \right| \le r^{-3}\int _{t_0}^{t_1}\int |\nabla w||\nabla F|^3 |{\eta }'''|\,dV_t\,dt \\&\le Cr^{-3}\int _{t_0}^{t_1}\int \frac{|\nabla w|}{\sqrt{w}}|\nabla F|^3\sqrt{w}\,dV_t\,dt \\&\le Cr^{-3}\left( \int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\,dV_t\,dt \right) ^{1/2}\left( \int _{t_0}^{t_1}\int |\nabla F|^6w\,dV_t\,dt \right) ^{1/2} \le C \end{aligned}$$

since

$$\begin{aligned} \int _{t_0}^{t_1}\int |\nabla F|^6w\,dV_t\,dt \le \int _{t_0}^{t_1}\int F^3e^{-f}\,dV_t\,dt \le C. \end{aligned}$$

Therefore \(\lim _{r \rightarrow \infty } III=0\).

Similarly,

$$\begin{aligned} IV&=\left| r^{-2}\int _{t_0}^{t_1}\int \langle \nabla w,\nabla |\nabla F|^2\rangle {\eta }''\,dV_t\,dt \right| \le Cr^{-2}\int _{t_0}^{t_1}\int |\nabla w||\nabla F||\text {Hess} F||{\eta }''|\,dV_t\,dt \\&\le Cr^{-3/2}\int _{t_0}^{t_1}\int |\nabla w||\text {Hess} F|\,dV_t\,dt=Cr^{-3/2}\int _{t_0}^{t_1}\int \frac{|\nabla w|}{\sqrt{w}}|\text {Hess} F|\sqrt{w}\,dV_t\,dt \\&\le Cr^{-3/2}\left( \int _{t_0}^{t_1}\int \frac{|\nabla w|^2}{w}\,dV_t\,dt \right) ^{1/2}\left( \int _{t_0}^{t_1}\int |\text {Hess}F|^2w\,dV_t\,dt \right) ^{1/2}. \end{aligned}$$

Now from Lemma 4 the last integral is bounded since \(w \le Ce^{-f}\), so \(\lim _{r \rightarrow \infty } IV=0\). Therefore, Lemma 10 is proved.

We can estimate the integral of \(v \square \phi ^r\).

From the expression of v in (102), we have

$$\begin{aligned}&\int _{t_0}^{t_1}\int v \square \phi ^r\,dV_t\,dt \\&\quad =\int _{t_0}^{t_1}\int \left( \tau _1 (-2\varDelta w+\frac{|\nabla w|^2}{w}+Rw)-w\log w-\left( n+\frac{n}{2}\log (4\pi \tau _1)\right) w\right) \square \phi ^r dV_t\,dt. \end{aligned}$$

Since we have \(|\square \phi ^r|\le Cr^{-1}\) from (40) and all terms except the first above have bounded integral on spacetime, it is easy to show, by taking into account of the claim, that

$$\begin{aligned} \lim _{r \rightarrow \infty } \int _{t_0}^{t_1}\int v \square \phi ^r\,dV_t\,dt=0. \end{aligned}$$
(111)

Now from (106),

$$\begin{aligned} \lim _{r \rightarrow \infty } \int v\phi ^r\,dV_{t_1} \ge \lim _{r \rightarrow \infty } \int v\phi ^r\,dV_{t_0}. \end{aligned}$$
(112)

Since we choose \(\sqrt{w}(\cdot ,t_1)\) to be a smooth function with compact support, it is immediate that

$$\begin{aligned} \lim _{r \rightarrow \infty } \int v\phi ^r\,dV_{t_1}=\overline{\mathcal W}(g(t_1),\sqrt{w}(\cdot ,t_1),\tau -t_1). \end{aligned}$$
(113)

\(\square \)

Lemma 11

$$\begin{aligned} \sqrt{w}(\cdot ,t_0)\in W_{*}^{1,2} \end{aligned}$$

and

$$\begin{aligned} \lim _{r \rightarrow \infty }\int \varDelta w\phi ^r \,dV_{t_0} =0. \end{aligned}$$
(114)

Proof of Lemma 11:

From the definition of v,

$$\begin{aligned}&\lim _{r \rightarrow \infty } \int v\phi ^r\,dV_{t_0}\\&\quad =\lim _{r \rightarrow \infty }\int \left( (\tau _1 (-2\varDelta w+\frac{|\nabla w|^2}{w}+Rw)-w\log w-(n+\frac{n}{2}\log (4\pi \tau _1))w\right) \phi ^r \,dV_{t_0}. \end{aligned}$$

All terms except for the first two in the above integral are absolutely integrable, due to \(w \le Ce^{-f}\) and \(R \le \tau ^{-2}F\).

Combining with (112), we conclude that

$$\begin{aligned} \lim _{r \rightarrow \infty }\int \left( -2\varDelta w+\frac{|\nabla w|^2}{w}\right) \phi ^r \,dV_{t_0} \end{aligned}$$

is bounded above.

Then we have

$$\begin{aligned}&\lim _{r \rightarrow \infty }\int \left( -2\varDelta w+\frac{|\nabla w|^2}{w}\right) \phi ^r \,dV_{t_0} \nonumber \\&\quad =\lim _{r \rightarrow \infty }\int 2\langle \nabla w,\nabla \phi ^r \rangle +\frac{|\nabla w|^2}{w}\phi ^r \,dV_{t_0} \nonumber \\&\quad \ge \lim _{r \rightarrow \infty }\int - \frac{|\nabla w|^2}{2w}\phi ^r-2\frac{|\nabla \phi ^r|^2}{\phi ^r}w+\frac{|\nabla w|^2}{w}\phi ^r \,dV_{t_0} \nonumber \\&\quad = \frac{1}{2}\lim _{r \rightarrow \infty }\int \frac{|\nabla w|^2}{w}\phi ^r \,dV_{t_0}, \end{aligned}$$
(115)

where we have used

$$\begin{aligned} \lim _{r \rightarrow \infty }\int \frac{|\nabla \phi ^r|^2}{\phi ^r}w \,dV_{t_0}=0. \end{aligned}$$

To prove (114), for any \(\epsilon >0\),

$$\begin{aligned} \lim _{r \rightarrow \infty }\left| \int \varDelta w\phi ^r \,dV_{t_0}\right|&=\lim _{r \rightarrow \infty }\left| \int \langle \nabla w,\nabla \phi ^r \rangle \,dV_{t_0}\right| \\&\le \lim _{r \rightarrow \infty } \int \epsilon \frac{|\nabla w|^2}{w}\phi ^r+\epsilon ^{-1}\frac{|\nabla \phi ^r|^2}{4\phi ^r}w \,dV_{t_0} \\&=\epsilon \int \frac{|\nabla w|^2}{w}\,dV_{t_0} \end{aligned}$$

By taking \(\epsilon \rightarrow 0\), we conclude that (114) holds. Therefore, the proof of Lemma 11 is complete.

Therefore,

$$\begin{aligned} \lim _{r \rightarrow \infty }\int v\phi ^r\,dV_{t_0}=\overline{\mathcal W}(g(t_0),\sqrt{w}(\cdot ,t_0),\tau -t_0). \end{aligned}$$

In summary, we have shown from (112) that

$$\begin{aligned} \overline{\mathcal W}(g(t_1),\sqrt{w}(\cdot ,t_1),\tau -t_1) \ge \overline{\mathcal W}(g(t_0),\sqrt{w}(\cdot ,t_0),\tau -t_0) \ge \varvec{\mu }(g(t_0),\tau -t_0). \end{aligned}$$

Since \(\tau \), \(t_0\), \(t_1\) and \(\sqrt{w}(\cdot ,t_1)\) are arbitrary, the proof of Theorem 10 is complete. \(\square \)

Corollary 2

On a Ricci shrinker \((M^n,g,f)\), the functional \(\varvec{\mu }(g,\tau )\) is decreasing for \( 0<\tau <1\) and increasing for \(\tau >1\).

Proof

The same argument appeared in Step 1, Proposition 9.5 of [34]. We repeat the argument here for the convenience of the readers.

For a fixed constant \(\tau _0>1\), from Theorem 10,

$$\begin{aligned} \varvec{\mu }(g(t),\tau _0-t)=\varvec{\mu }((1-t)(\psi ^t)^*g,\tau _0-t)=\varvec{\mu }\left( g,\frac{\tau _0-t}{1-t} \right) \end{aligned}$$

is increasing for \(t <1\). Now as t goes from 0 to 1, \(\frac{\tau _0-t}{1-t}\) goes from \(\tau _0\) to \(\infty \). As \(\tau _0>1\) is arbitrary, we have proved that \(\varvec{\mu }(g,\tau )\) is increasing for all \(\tau >1\). Similarly, for any \(\tau _0<1\), as t goes from 0 to \(\tau _0\), \(\frac{\tau _0-t}{1-t}\) goes from \(\tau _0\) to 0. Therefore, \(\varvec{\mu }(g,\tau )\) is decreasing for all \(\tau <1\).

\(\square \)

5 Optimal logarithmic Sobolev constant—part I

For any Ricci shrinker \((M^n,g,f)\) with the normalization (2), we define

$$\begin{aligned} \varvec{\mu }=\varvec{\mu }(g){:}{=}\log \int \frac{e^{-f}}{(4\pi )^{n/2}}\, dV. \end{aligned}$$
(116)

It follows from a direct calculation that \(e^{\varvec{\mu }}\) is comparable to the volume of the unit ball B(p, 1).

Lemma 12

(cf. Lemma 2.5 of  [34]) For any Ricci shrinker \((M^n,g,f)\), there exists a constant \(C=C(n)>1\) such that

$$\begin{aligned} C^{-1}e^{\varvec{\mu }} \le |B(p,1)| \le Ce^{\varvec{\mu }}. \end{aligned}$$

Next we recall from [1] some standard definitions and properties of the space which satisfies the curvature-dimension estimate.

Definition 1

A Riemannian manifold (Mgv), equipped with a reference measure \(v=e^{-W}V\) where \(W \in C^2\) and V is the standard volume form, satisfies the \(CD(K,\infty )\) condition if the generalized Ricci tensor

$$\begin{aligned} Ric_W {:}{=}Ric+\text {Hess}\,W \ge Kg. \end{aligned}$$

In particular, on a Ricci shrinker \((M^n,g,f)\), if we define

$$\begin{aligned} f_0&=f+\varvec{\mu }+\frac{n}{2}\log (4\pi ), \end{aligned}$$
(117)
$$\begin{aligned} v_0&=e^{-f_0}V \end{aligned}$$
(118)

then \(v_0\) is a probability measure and \((M,g,v_0) \in CD(\frac{1}{2},\infty )\). Then the following celebrated theorem of Bakry–Émery can be applied on Ricci shrinkers.

Theorem 11

(Bakry–Émery theorem [2]) For any Riemannian manifold (Mgv) satisfying the \(CD(K,\infty )\) condition for some \(K>0\), the following logarithmic Sobolev inequality holds

$$\begin{aligned} \int \rho \log \rho \,dv \le \frac{1}{2K}\int \frac{|\nabla \rho |^2}{\rho } \,dv, \end{aligned}$$
(119)

where v and \(\rho \, v\) are probability measures which have finite moments of second order and \(\rho \) is locally Lipschitz.

The original proof by Bakry and Émery is complete for compact manifolds. A proof using the optimal transport by Lott and Villani for the general case can be found in [39, Corollary 6.12], see also [52, Theorem 21.2]. For the self-containedness, we give a proof of the Bakry–Émery theorem for Ricci shrinkers.

Theorem 12

For any Ricci shrinker \((M^n,g,f)\) and any nonnegative function \(\rho \) such that \(\sqrt{\rho }\in W^{1,2}(M,v_0)\) and \(\int d^2(p,\cdot )\rho \,dv_0 <\infty \),

$$\begin{aligned} \int \rho \log \rho \,dv_0-\left( \int \rho \,dv_0\right) \log {\left( \int \rho \, dv_0\right) } \le \int \frac{|\nabla \rho |^2}{\rho }\,dv_0. \end{aligned}$$

If the equality holds, then either \(\rho \) is a constant or \((M^n,g)\) splits off a \(\mathbb {R}\) factor.

Before we prove Theorem 12, we prove the following two lemmas.

Lemma 13

For any smooth function u(tx) on \(M \times [0,T]\) such that

$$\begin{aligned} \square _f u{:}{=}(\partial _t-\varDelta _f)u \le 0, \end{aligned}$$

and for some constant \(a>0\),

$$\begin{aligned} \int _0^T\int u^2(t,x)e^{-ad^2(p,x)}\,dv_0dt<\infty , \end{aligned}$$

if \(u(\cdot ,0) \le c\), then \(u \le c\) on \(M \times [0,T]\).

Proof

The proof follows from [35, Theorem 15.2] verbatim by using \(\varDelta _f\) and the measure \(v_0\) instead of \(\varDelta \) and the volume form V. \(\square \)

We define a new familiy of cutoff functions by setting

$$\begin{aligned} \overline{\phi }^r {:}{=}\eta \left( \frac{f}{r} \right) , \end{aligned}$$

where \(\eta \) is the same function as in (36) and f is the potential function at time 0. A direct calculation shows that

$$\begin{aligned} \varDelta _f \overline{\phi }^r=r^{-2}\eta ''|\nabla f|^2+r^{-1}\eta ' \varDelta _f f=r^{-2}\eta ''|\nabla f|^2+r^{-1}\eta '(\frac{n}{2}-f). \end{aligned}$$

Then it is clear that \(\varDelta _f \overline{\phi }^r\) is supported on \(\{f \ge r\}\) and there exists a constant \(C=C(n)\) such that

$$\begin{aligned} |\varDelta _f \overline{\phi }^r | \le C. \end{aligned}$$
(120)

Lemma 14

For any smooth bounded function u on M,

$$\begin{aligned} \lim _{r \rightarrow \infty }\int (\varDelta _f u) \overline{\phi }^r\,dv_0=0. \end{aligned}$$

Proof

From the integration by parts,

$$\begin{aligned} \lim _{r \rightarrow \infty } \int (\varDelta _f u)\overline{\phi }^r \,dv_0=\lim _{r \rightarrow \infty } \int u(\varDelta _f \overline{\phi }^r) \,dv_0=0, \end{aligned}$$

where the last equality holds since u is bounded and \(v_0\) is a probability measure. \(\square \)

Proof of Theorem 12:

We only prove the inequality for \(\rho _0\) such that \(\sqrt{\rho }_0\) is a compactly supported smooth function and the general case follows from approximations as in Proposition 95. In addition, we assume that \(\int \rho _0\,dv_0=1\).

Given such \(\rho _0\), we consider the heat flow with respect to the measure \(v_0\), that is,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\partial _t \rho =\varDelta _f \rho ,\\ &{}\rho (0,\cdot )=\rho _0. \end{array}\right. } \end{aligned}$$

It is clear that there exists a constant C such that \(\rho \le C\) on \(M \times [0,\infty )\). Now we set

$$\begin{aligned} E(t) {:}{=}\left( \int \rho \log \rho \,dv_0 \right) (t). \end{aligned}$$

By direct computations

$$\begin{aligned} \partial _t \int \rho (\log \rho ) \overline{\phi }^r \,dv_0&=\int \rho _t(\log \rho +1) \overline{\phi }^r \,dv_0\\&=\int \varDelta _f \rho (\log \rho +1) \overline{\phi }^r \,dv_0 \\&=\int -\frac{|\nabla \rho |^2}{\rho } \overline{\phi }^r+\varDelta _f(\rho \log \rho )\overline{\phi }^r\,dv_0. \end{aligned}$$

Therefore, for any \(T>0\),

$$\begin{aligned}&\left( \int \rho (\log \rho ) \overline{\phi }^r \,dv_0 \right) (T)-\left( \int \rho (\log \rho ) \overline{\phi }^r \,dv_0\right) (0) \\&\quad = \int _0^T \int -\frac{|\nabla \rho |^2}{\rho } \overline{\phi }^r+\varDelta _f(\rho \log \rho )\overline{\phi }^r\,dv_0dt. \end{aligned}$$

It follows from Lemma 14 that

$$\begin{aligned} \int _0^T \int \frac{|\nabla \rho |^2}{\rho }\,dv_0dt <\infty \end{aligned}$$
(121)

and

$$\begin{aligned} E(T)-E(0)=- \int _0^T \int \frac{|\nabla \rho |^2}{\rho }\,dv_0dt. \end{aligned}$$
(122)

We compute

$$\begin{aligned} \partial _t \int \frac{|\nabla \rho |^2}{\rho } \overline{\phi }^r\,dv_0 =\int \square _f \left( \frac{|\nabla \rho |^2}{\rho }\right) \overline{\phi }^r+ \varDelta _f \left( \frac{|\nabla \rho |^2}{\rho }\right) \overline{\phi }^r\,dv_0. \end{aligned}$$
(123)

From Bochner’s formula,

$$\begin{aligned} \partial _t |\nabla \rho |^2&= 2\langle \nabla \varDelta _f \rho ,\nabla \rho \rangle \\&= \varDelta _f |\nabla \rho |^2-2|\text {Hess}\,\rho |^2-2(Rc+\text {Hess}\,f)(\nabla \rho ,\nabla \rho ) \\&=\varDelta _f |\nabla \rho |^2-2|\text {Hess}\,\rho |^2-|\nabla \rho |^2, \end{aligned}$$

where we have used the Ricci shrinker equation for the last equality.

Therefore,

$$\begin{aligned} \square _f |\nabla \rho |^2=-2|\text {Hess}\,\rho |^2-|\nabla \rho |^2. \end{aligned}$$
(124)

A direct calculation shows that

$$\begin{aligned} \square _f \frac{|\nabla \rho |^2}{\rho }=-\frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2 -\frac{|\nabla \rho |^2}{\rho }. \end{aligned}$$
(125)

It follows from (124) and Lemma 13 that there exists a constant \(C>0\) such that

$$\begin{aligned} \frac{|\nabla \rho |^2}{\rho } \le C. \end{aligned}$$
(126)

Therefore, by (123) and Lemma 14, for any \(T>S>0\),

$$\begin{aligned}&\left( \int \frac{|\nabla \rho |^2}{\rho } \,dv_0 \right) (T)-\left( \int \frac{|\nabla \rho |^2}{\rho } \,dv_0\right) (S) \nonumber \\&\quad = \int _S^T \int -\frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2 -\frac{|\nabla \rho |^2}{\rho }\,dv_0dt. \end{aligned}$$
(127)

It follows from (122) that for any \(t \ge 0\),

$$\begin{aligned} E'(t)=-\left( \int \frac{|\nabla \rho |^2}{\rho } \,dv_0 \right) (t) \le 0 \end{aligned}$$
(128)

Moreover, for any \(t >s \ge 0\), it follows from (127) that

$$\begin{aligned} -E'(t)+E'(s) \le \int _s^t E'(z)\,dz \le 0. \end{aligned}$$
(129)

Then it is easy to see from (129) that

$$\begin{aligned} E'(t) \ge E'(0)e^{-t}. \end{aligned}$$
(130)

Now we claim that \(E(t) \rightarrow 0\) if \(t \rightarrow \infty \). Since E(t) is decreasing by (128), we only need to prove the claim by considering a sequence \(t_i \rightarrow \infty \). We define \(u_i=\sqrt{\rho (t_i,\cdot )}\), then

$$\begin{aligned} \int u_i^2 \,dv_0=1 \end{aligned}$$
(131)

and by (130),

$$\begin{aligned} \int |\nabla u_i|^2 \,dv_0 \rightarrow 0. \end{aligned}$$
(132)

Then by taking a subsequence, we claim that \(u_i\) converges to \(u_{\infty }\) weakly in \(W^{1,2}(M,v_0)\). It is clear from (131) and (132) that \(u_{\infty } \equiv 1\). Since we can assume that \(u_i\) converges to 1 almost everywhere,

$$\begin{aligned} \lim _{i \rightarrow \infty } \int u_i^2 \log u_i^2\,dv_0 =0 \end{aligned}$$
(133)

by the dominated convergence theorem. Therefore, \(E(t) \rightarrow 0\) if \(t \rightarrow \infty \).

It follows from (122) and (130) that

$$\begin{aligned} \int \rho _0\log \rho _0\,dv_0=E(0)=-\int _0^{\infty }E'(t)\,dt \le -E'(0) \int _0^{\infty }e^{-t}\,dt=\int \frac{|\nabla \rho _0|^2}{\rho _0} \,dv_0. \end{aligned}$$

If the equality holds and \(\rho _0\) is not a constant, it follows from (127) that

$$\begin{aligned} \text {Hess}(\log \rho )=\frac{1}{\rho }\left( \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho } \right) =0. \end{aligned}$$

Therefore, \((M^n,g)\) splits off a \(\mathbb {R}\) factor.

In summary, the proof of Theorem 12 is complete.

Using the Bakry–Émery theorem, Carrillo and Ni have proved in [10] the following result. \(\square \)

Proposition 3

(Carrillo–Ni [10]) For any Ricci shrinker \((M^n,g,f)\), we have

$$\begin{aligned} \overline{\mathcal {W}} (g,e^{-\frac{f_0}{2}},1)=\varvec{\varvec{\mu }}(g,1)=\varvec{\varvec{\mu }}, \end{aligned}$$
(134)

where \(f_0\) is the normalization of f defined in (117).

Proof

We shall follow the argument of Carrillo and Ni. The proof is given for the self-containedness.

For any Ricci shrinker \((M^n,g,f)\) and any smooth function u on M with compact support such that \(\int u^2 \,dV=1\), we define \(w=u^2e^{f_0}\). Then it is clear that both \(v_0\) and \(wv_0\) belong to \(P_2(M)\) from the estimates of f and dV.

It follows from Theorem 12 that

$$\begin{aligned} \int w \log w \,dv_0 \le \int \frac{|\nabla w|^2}{w} \,dv_0. \end{aligned}$$
(135)

By rewriting (135) in terms of u, we have

$$\begin{aligned} \int u^2 \log u^2\,dV+\int f_0u^2\,dV \le \int 4|\nabla u|^2+|\nabla f_0|^2u^2+4\langle \nabla u,\nabla f_0\rangle u\,dV. \end{aligned}$$
(136)

It follows from the integration by parts for the last term that (136) becomes

$$\begin{aligned} \int u^2 \log u^2\,dV+\varvec{\mu }+\frac{n}{2}\log (4\pi ) \le \int 4|\nabla u|^2+u^2(|\nabla f|^2-2\varDelta f-f)\,dV. \end{aligned}$$
(137)

It follows from the \(|\nabla f|^2+R=f\) and \(\varDelta f+R=\frac{n}{2}\) that \(|\nabla f|^2-2\varDelta f-f=R-f\). Therefore, by (137) that

$$\begin{aligned} \overline{\mathcal {W}} (g,u,1)=\int \left\{ 4|\nabla u|^2+Ru^2-u^2\log u^2 \right\} dV-n-\frac{n}{2}\log (4\pi )\ge \varvec{\mu }. \end{aligned}$$

By the arbitrary choice of u, the above inequality means that

$$\begin{aligned} \varvec{\mu }(g,1) \ge \varvec{\mu }. \end{aligned}$$
(138)

On the other hand, if we set \(u_1=e^{-\frac{f_0}{2}}\), it follows from direct calculation that

$$\begin{aligned} \overline{\mathcal {W}} (g,u_1,1) =\int \left( |\nabla f|^2+R+f-n \right) \,e^{-f_0}dV+\varvec{\mu }. \end{aligned}$$

Recall that \(R+|\nabla f|^2=f\) and \(R+\varDelta f=\frac{n}{2}\) on a Ricci shrinker. So the above equation can be simplified as

$$\begin{aligned} \overline{\mathcal {W}} (g,u_1,1)-\varvec{\mu }=\int \left( 2f-n \right) \,e^{-f_0}dV=-2\int (\varDelta _{f} f) e^{-f_0} dV =-2\int (\varDelta _{f_0} f) e^{-f_0} dV=0. \end{aligned}$$

Then it follows from definition that

$$\begin{aligned} \varvec{\mu }(g,1) \le \overline{\mathcal {W}} (g,u_1,1)=\varvec{\mu }. \end{aligned}$$
(139)

Therefore, (134) follows from the combination of (138) and (139). \(\square \)

Corollary 3

For any Ricci shrinker \((M^n,g,f)\), if there exist more than one minimizer \(u \in W^{1,2}_{*}\) for \(\overline{\mathcal {W}} (g,u,1)\), then (Mg) must split off a \(\mathbb {R}\) factor.

Proof

If u is a minimizer other than \(e^{-\frac{f_0}{2}}\), then the same proof as Proposition 3 shows that

$$\begin{aligned} \int w \log w \,dv_0 = \int \frac{|\nabla w|^2}{w} \,dv_0, \end{aligned}$$

where \(w=u^2e^{f_0}\). Then the conclusion follows from the equality case of Theorem 12. \(\square \)

Proposition 3 indicates that \(\varvec{\mu }\) is the optimal log-Sobolev constant for \((M^n,g,f)\) on scale 1. We shall improve (134) by showing that \(\varvec{\mu }\) is in fact the optimal log-Sobolev constant for all scales. Note that the same result has already been proved for compact Ricci shrinkers in Proposition 9.5 of [34].

Proposition 4

For any Ricci shrinker \((M^n,g,f)\), we have

$$\begin{aligned} \varvec{\nu }(g) {:}{=}\inf _{\tau >0}\varvec{\mu }(g,\tau )=\varvec{\mu }. \end{aligned}$$
(140)

We first show two important intermediate steps before we prove Proposition 4.

Lemma 15

For each \(\tau \in (0,1)\), we have

$$\begin{aligned} \varvec{\mu }(g, \tau ) \ge \varvec{\mu }=\varvec{\mu }(g,1). \end{aligned}$$
(141)

Proof

Fix \(\eta _0 \in (0,1)\). Let w be a nonnegative, compactly supported smooth function satisfying the normalization condition \(\int w \,dV=1\). We now regard w as a smooth function on the time slice \(t=0\) and solve the conjugate heat equation \(\square ^* w=0\). Then w is a smooth function on the space-time \(M \times (-\infty , 0)\). It follows from Lemma 8 that there exists a constant \(C>0\) such that

$$\begin{aligned} w(x,t) \le C (4\pi (1-t))^{-\frac{n}{2}} e^{-f(x,t)}, \quad \forall \; x \in M, \; t \in (-\infty , 0]. \end{aligned}$$
(142)

By the diffeomorphism invariance of the \(\overline{\mathcal {W}}\)-functional, it is easy to see that

$$\begin{aligned} \overline{\mathcal {W}}\left( g(t), \sqrt{w(\cdot , t)}, \eta _0-t \right) =\overline{\mathcal {W}}\left( (1-t) (\psi ^{t})^{*} g, \sqrt{w(\cdot , t)}, \eta _0-t \right) =\overline{\mathcal {W}}\left( g, u(\cdot , t), \theta (t) \right) \end{aligned}$$
(143)

where we have used the notation

$$\begin{aligned}&u(\cdot , t) {:}{=}(1-t)^{\frac{n}{4}}\sqrt{((\psi ^{t})^{-1})^{*} w(\cdot , t)}, \end{aligned}$$
(144)
$$\begin{aligned}&\theta (t) {:}{=}\frac{\eta _0-t}{1-t}. \end{aligned}$$
(145)

Notice that \(\int u^2 \,dV \equiv 1\) according to our construction. It follows from definition and direct calculations that

$$\begin{aligned}&\overline{\mathcal {W}}\left( g, u(\cdot , t), \theta (t) \right) \nonumber \\&\quad =\int \left\{ \theta \left( 4|\nabla u|^2 + Ru^2 \right) -u^2 \log u^2 \right\} dV -n-\frac{n}{2} \log (4\pi \theta ) \nonumber \\&\quad = \theta \left\{ \int \left\{ \left( 4|\nabla u|^2 + Ru^2 \right) -u^2 \log u^2 \right\} dV -n-\frac{n}{2} \log (4\pi ) \right\} \nonumber \\&\qquad +(\theta -1) \left\{ \int u^2 \log u^2 dV + n+\frac{n}{2} \log (4\pi ) \right\} -\frac{n}{2} \log \theta \nonumber \\&\quad \ge \theta \varvec{\mu }(g,1) + (\theta -1) \left\{ \int u^2 \log u^2 dV +n +\frac{n}{2} \log (4\pi ) \right\} -\frac{n}{2} \log \theta . \end{aligned}$$
(146)

By (144), the inequality (142) can be understood as

$$\begin{aligned} u^2(x,t) \le C e^{-f(x,0)} \end{aligned}$$

for some constant C indepenent of t. Consequently, as \(f \ge 0\), we obtain

$$\begin{aligned} \int u^2 \log u^2 dV \le \int \left\{ -f+\log C \right\} u^2 dV \le \log C - \int f \cdot u^2 dV \le \log C. \end{aligned}$$

Note that \(\theta (t)<1\) when \(t<0\). Plugging the above inequality into (146), and noting that

$$\begin{aligned} \overline{\mathcal {W}}\left( g(0), \sqrt{w(\cdot , 0)}, \eta _0 \right) \ge \overline{\mathcal {W}}\left( g(t), \sqrt{w(\cdot , t)}, \eta _0-t \right) , \quad \forall \; t \in (-\infty , 0), \end{aligned}$$

we can use (143) to obtain

$$\begin{aligned} \overline{\mathcal {W}}\left( g(0), \sqrt{w(\cdot , 0)}, \eta _0 \right) \ge \theta \varvec{\mu }(g,1) + (\theta -1)\left\{ \log C +n+\frac{n}{2} \log (4\pi ) \right\} -\frac{n}{2} \log \theta . \end{aligned}$$

From (145), it is clear that \(\displaystyle \lim _{t \rightarrow -\infty } \theta (t)=1\). On the right hand side of the above inequlaity, letting \(t \rightarrow -\infty \), we arrive at

$$\begin{aligned} \overline{\mathcal {W}}\left( g(0), \sqrt{w(\cdot , 0)}, \eta _0 \right) \ge \varvec{\mu }(g, 1). \end{aligned}$$

Since \(w(\cdot , 0)\) could be arbitrary smooth nonnegative function satisfying the normalization condition, and \(g=g(0)\), in light of (95), it is clear that (141) follows from the above inequality. \(\square \)

Lemma 16

For each \(\tau \in (1, \infty )\), we have

$$\begin{aligned} \varvec{\mu }(g, \tau ) \ge \varvec{\mu }=\varvec{\mu }(g,1). \end{aligned}$$
(147)

Proof

For any \(u \in W_{*}^{1,2}\) and \(\tau >1\),

$$\begin{aligned} \overline{\mathcal W}(g,u,\tau )&=\int \left\{ \tau (4|\nabla u|^2+Ru^2)-u^2\log u^2 \right\} dV-n-\frac{n}{2}\log (4\pi \tau ) \\&\ge \int \left\{ (4|\nabla u|^2+Ru^2)-u^2\log u^2 \right\} dV-n-\frac{n}{2}\log (4\pi \tau ) \\&\ge \,\varvec{\mu }(g, 1)-\frac{n}{2}\log \tau =\varvec{\mu } -\frac{n}{2}\log \tau . \end{aligned}$$

By the arbitrary choice of \(u \in W_{*}^{1,2}\), it follows that

$$\begin{aligned} \varvec{\mu }(g, \tau )\ge \varvec{\mu }-\frac{n}{2}\log \tau . \end{aligned}$$

Let \(\tau \rightarrow 1^{+}\), we obtain that

$$\begin{aligned} \liminf _{\tau \rightarrow 1^{+}} \varvec{\mu }(g, \tau ) \ge \varvec{\mu }. \end{aligned}$$

By Corollary 2, we know that \(\varvec{\mu }(g, \tau )\) is an increasing function of \(\tau \) for \(\tau \in (1, \infty )\). Then it is clear that (147) follows directly from the above inequality. \(\square \)

Proof of Proposition 4:

It follows from the combination of Lemmas 15 and  16. \(\square \)

Lemma 17

Suppose (Mg) is a complete Riemannian manifold with Sobolev constant \(C_{RS}\). Namely, for each smooth function u with compact support, we have

$$\begin{aligned} \left( \int u^{\frac{2n}{n-2}} dV \right) ^{\frac{n-2}{n}} \le C_{RS} \int \left\{ 4|\nabla u|^2 + Ru^2 \right\} dV. \end{aligned}$$
(148)

Then for each positive \(\tau \), the following estimates hold for any \(u \in W_{*}^{1,2}\),

$$\begin{aligned} e^{-\frac{2E}{n}} \le \tau \int \left\{ 4|\nabla u|^2 + Ru^2 \right\} dV \le \max \left\{ n^2, 2E \right\} , \end{aligned}$$
(149)

where

$$\begin{aligned} E = \overline{\mathcal W}(g,u,\tau ) + \frac{n}{2} \log (4\pi e^2 C_{RS}). \end{aligned}$$
(150)

Proof

By Jensen’s inequality, we know that

$$\begin{aligned} \int u^2\log u^2 \,dV =\frac{n-2}{2}\int u^2\log u^{\frac{4}{n-2}} \,dV \le \frac{n-2}{2}\log \left( \int u^{\frac{2n}{n-2}} \,dV \right) . \end{aligned}$$

Plugging the Sobolev inequality (148) into the above inequality yields that

$$\begin{aligned} \int u^2\log u^2 \,dV \le \frac{n}{2}\log C_{RS} + \frac{n}{2} \log \int \left\{ 4|\nabla u|^2 + Ru^2 \right\} dV. \end{aligned}$$
(151)

It follows that

$$\begin{aligned} \overline{\mathcal W}(g,u,\tau )&\ge \int \left\{ \tau (4|\nabla u|^2+Ru^2)-u^2\log u^2 \right\} dV-n-\frac{n}{2}\log (4\pi \tau )\\&\ge \int \tau \left\{ 4|\nabla u|^2+Ru^2 \right\} dV -\frac{n}{2} \log \int \tau \left\{ 4|\nabla u|^2 + Ru^2 \right\} dV -n-\frac{n}{2} \log (4\pi C_{RS}). \end{aligned}$$

Let \(x=\int \left( 4|\nabla u|^2+Ru^2 \right) dV\). The above inequality can be rewritten as

$$\begin{aligned} \tau x -\frac{n}{2} \log (\tau x) \le \overline{\mathcal W}(g,u,\tau ) +n+\frac{n}{2} \log (4\pi C_{RS})=E. \end{aligned}$$
(152)

Since \(\tau x>0\), it follows from (152) that

$$\begin{aligned} \tau x \ge e^{-\frac{2}{n}E}. \end{aligned}$$
(153)

On the other hand, it is clear that

$$\begin{aligned} s-\frac{n}{2} \log s \ge \frac{s}{2} \quad \text {on} \; [n^2, \infty ). \end{aligned}$$
(154)

Suppose \(\tau x \ge n^2\), then the combination of (152) and (154) implies that \(\tau x \le 2E\). Consequently, we always have

$$\begin{aligned} \tau x \le \max \left\{ n^2, 2E \right\} . \end{aligned}$$
(155)

Clearly, (149) follows from the combination of (153) and (155). \(\square \)

Corollary 4

(Sobolev inequality) Let \(\Big \{(M^n,g(t)),\, t\in (-\infty ,1)\Big \}\) be the Ricci flow solution of a Ricci shrinker \((M^n,p,g,f)\), there exists a constant \(C=C(n)\) such that at any time \(t<1\),

$$\begin{aligned} \left( \int u^{\frac{2n}{n-2}}\,dV_t \right) ^{\frac{n-2}{n}} \le Ce^{-\frac{2 \varvec{\mu }}{n}} \int \left\{ 4|\nabla u|^2+Ru^2 \right\} dV_t \end{aligned}$$
(156)

for any smooth function u with compact support.

Proof

We consider the Schrödinger operator \(H=-2\varDelta +\frac{R}{2}\) and the quadratic forms \(Q(u){:}{=}\int (Hu)u\,dV_t\) with its corresponding Markov semigroup \(\{e^{-Hs},\,s\ge 0\}\). Since \(\varvec{\mu }(g(t),\tau ) =\varvec{\mu }(g,\frac{\tau }{1-t})\ge \varvec{\mu }\), we have

$$\begin{aligned} \int u^2 \log u\,dV_t \le \tau Q(u)+\beta (\tau ) \end{aligned}$$

for any \(\int u^2\,dV_t=1\), where \(\beta (\tau )=-\frac{n}{2}-\frac{n}{4}\log (4\pi \tau )-\varvec{\mu }\). Then it follows from [21, Corollary 2.2.8] that for any \(s>0\),

$$\begin{aligned} \Vert e^{-Hs}\Vert _{\infty ,2} \le e^{M(s)}\le Cs^{-\frac{n}{4}}e^{-\frac{\varvec{\mu }}{2}}, \end{aligned}$$
(157)

where \(M(s){:}{=}\frac{1}{s}\int _0^s \beta (\tau )\,d\tau \). Now we use the same argument as in [21, Theorem 2.4.2] to derive the Sobolev inequaltiy. It follows from (157) that for any \(u \in L^2\),

$$\begin{aligned} \Vert e^{-Hs}u\Vert _{\infty } \le Cs^{-\frac{n}{4}}e^{-\frac{\varvec{\mu }}{2}}\Vert u\Vert _2. \end{aligned}$$
(158)

Since \(e^{-Hs}\) is self-adjoint, by taking the conjugation of (158) we obtain

$$\begin{aligned} \Vert e^{-Hs}u\Vert _2 \le Cs^{-\frac{n}{4}}e^{-\frac{\varvec{\mu }}{2}}\Vert u\Vert _1. \end{aligned}$$
(159)

Therefore, for any \(s>0\),

$$\begin{aligned} \Vert e^{-Hs}u\Vert _{\infty } \le Cs^{-\frac{n}{4}}e^{-\frac{\varvec{\mu }}{2}}\Vert e^{-\frac{Hs}{2}}u\Vert _2 \le Cs^{-\frac{n}{2}}e^{-\varvec{\mu }}\Vert u\Vert _1. \end{aligned}$$
(160)

Combining (160) with the fact that \(e^{-Hs}\) is a contraction on \(L^{\infty }\), it follows from the Riesz-Thorin interpolation that for any \(q \in [1,\infty ]\).

$$\begin{aligned} \Vert e^{-Hs}u\Vert _{\infty } \le Cs^{-\frac{n}{2q}}e^{-\frac{\varvec{\mu }}{q}}\Vert u\Vert _q. \end{aligned}$$
(161)

We now write

$$\begin{aligned} H^{-\frac{1}{2}}u=a+b \end{aligned}$$

where

$$\begin{aligned} a&=\varGamma ^{-1}(1/2) \int _0^T s^{-\frac{1}{2}}e^{-Hs}u\,ds,\\ b&=\varGamma ^{-1}(1/2) \int _T^{\infty } s^{-\frac{1}{2}}e^{-Hs}u\,ds. \end{aligned}$$

It follows from (161) that

$$\begin{aligned} \Vert b\Vert _{\infty } \le C\varGamma ^{-1}(1/2) \int _T^{\infty } s^{-\frac{1}{2}-\frac{n}{2q}}e^{-\frac{\varvec{\mu }}{q}}\Vert u\Vert _q\,ds=ce^{-\frac{\varvec{\mu }}{q}}\Vert u\Vert _qT^{\frac{1}{2}-\frac{n}{2q}} \end{aligned}$$

for some constant \(c=c(n)\). Given \(\lambda >0\), we define \(T>0\) by \(\frac{\lambda }{2}=ce^{-\frac{\varvec{\mu }}{q}}\Vert u\Vert _qT^{\frac{1}{2}-\frac{n}{2q}}\). It is clear that

$$\begin{aligned} |\{x:\,|H^{-\frac{1}{2}}u(x)|\ge \lambda \}|\le |\{x:\,|a(x)|\ge \lambda /2\}| \le 2^q\lambda ^{-q}\Vert a\Vert _{q}^q \le C\lambda ^{-q}T^{\frac{q}{2}}\Vert u\Vert _q^q, \end{aligned}$$

since \(e^{-Hs}\) is a contraction on \(L^q\). For any \(1<q<n\), we set \(\frac{1}{r}=\frac{1}{q}-\frac{1}{n}\), then it follows from our choice of \(\lambda \) that

$$\begin{aligned} |\{x:\,|H^{-\frac{1}{2}}u(x)|\ge \lambda \}| \le Ce^{-\frac{\varvec{\mu } q}{n-q}}\lambda ^{-r} \Vert u\Vert _q^r. \end{aligned}$$

In other words,

$$\begin{aligned} \Vert H^{-\frac{1}{2}}u\Vert _{r,w} \le Ce^{-\frac{\varvec{\mu } q}{r(n-q)}}\Vert u\Vert _q \end{aligned}$$
(162)

where \(\Vert \cdot \Vert _{r,w}\) denotes the weak \(L^r\) space. Therefore, it follows from the Marcinkiewicz interpolation theorem that

$$\begin{aligned} \Vert H^{-\frac{1}{2}}u\Vert _{p} \le Ce^{-\frac{2\varvec{\mu } }{p(n-2)}}\Vert u\Vert _2=Ce^{-\frac{\varvec{\mu } }{n}}\Vert u\Vert _2, \end{aligned}$$
(163)

where \(\frac{1}{p}=\frac{1}{2}-\frac{1}{n}\). Therefore, (156) is a direct consequence. \(\square \)

Remark 2

It follows from the above corollary that the Yamabe invariant of \((M^n,g,f) \)

$$\begin{aligned} Y([g]) {:}{=}\inf _{u \in C_0^{\infty }(M)}\frac{\int \frac{4(n-1)}{n-2}|\nabla u|^2+Ru^2 \,dV}{\left( \int u^{\frac{2n}{n-2}}\,dV\right) ^{\frac{n-2}{n}}} >0. \end{aligned}$$
(164)

Here Y depends only on the conformal class of g. Hence it implies some connections between a Ricci shrinker and its conformal class. Note that it is shown in [63] that each Ricci shrinker has a conformal metric such that its Ricci curvature has local bound depending only on the dimension. This fact plays a key role in [34].

Proposition 5

On a Ricci shrinker \((M^n,g,f)\), the functional \(\varvec{\mu }(g,\tau )\) is a continuous function of \(\tau \in (0, \infty )\).

Proof

Fix \(\tau _0 \in (0, \infty )\). We need to show both the upper semi-continuity and the lower semi-continuity as \(\tau _0\).

The upper-semicontinuity is more or less standard. Fix \(u \in W_{*}^{1,2}\), we have

$$\begin{aligned} \limsup _{\tau \rightarrow \tau _0}\varvec{\mu }(g,\tau )&\le \limsup _{\tau \rightarrow \tau _0}\overline{\mathcal W}(g,u,\tau ) \\&=\limsup _{\tau \rightarrow \tau _0}\int \tau (4|\nabla u|^2+Ru^2)-u^2\log u^2 \,dV-n-\frac{n}{2}\log (4\pi \tau )\\&=\int \tau _0(4|\nabla u|^2+Ru^2)-u^2\log u^2 \,dV-n-\frac{n}{2}\log (4\pi \tau _0)\\&=\overline{\mathcal W}(g,u,\tau _0). \end{aligned}$$

By taking the infimum among all qualified u’s, we have

$$\begin{aligned} \limsup _{\tau \rightarrow \tau _0}\varvec{\mu }(g,\tau )\le \varvec{\mu }(g,\tau _0). \end{aligned}$$
(165)

Hence \(\varvec{\mu }(g,\tau )\) is upper semicontinuous.

The lower semicontinuity relies on the estimate (149) in Lemma 17. Actually, for arbitrary \(u \in W_{*}^{1,2}\) satisfying the normalization condition, direct calculation shows that

$$\begin{aligned} \overline{\mathcal W}(g,u,\tau )&=\overline{\mathcal W}(g,u,\tau _0) + (\tau -\tau _0)\int \left\{ 4|\nabla u|^2+Ru^2 \right\} dV-\frac{n}{2}\log \left( \frac{\tau }{\tau _0} \right) \nonumber \\&\ge \varvec{\mu }(g, \tau _0) -|\tau -\tau _0|\int \left\{ 4|\nabla u|^2+Ru^2 \right\} dV-\frac{n}{2}\log \left( \frac{\tau }{\tau _0} \right) . \end{aligned}$$
(166)

For any \(\tau _i \rightarrow \tau _0\), we choose \(u_i \in W_{*}^{1,2}\) such that

$$\begin{aligned} \overline{\mathcal W}(g,u_i,\tau _i)-\varvec{\mu }(g,\tau _i)<i^{-1}. \end{aligned}$$
(167)

Together with (165), this implies that

$$\begin{aligned} \limsup _{i \rightarrow \infty } \overline{\mathcal W}(g,u_i,\tau _i) =\limsup _{i \rightarrow \infty } \varvec{\mu }(g,\tau _i) \le \varvec{\mu }(g, \tau _0). \end{aligned}$$
(168)

By Corollary 4, the Sobolev constant on each Ricci shrinker is finite. It follows from (149) and (150) that \(\int (4|\nabla u_i|^2+Ru_i^2)\,dV\) is uniformly bounded. In (166), replacing u by \(u_i\) and letting \(i \rightarrow \infty \), we obtain

$$\begin{aligned} \liminf _{i \rightarrow \infty } \overline{\mathcal W}(g,u_i,\tau _i) \ge \varvec{\mu }(g, \tau _0). \end{aligned}$$

Combining the above inequality with (167), we obtain that

$$\begin{aligned} \liminf _{i \rightarrow \infty } \varvec{\mu }(g,\tau _i) \ge \varvec{\mu }(g, \tau _0), \end{aligned}$$

which is the lower semi-continuity at \(\tau _0\). The continuity of \(\varvec{\mu }(g,\tau )\) with respect to \(\tau \) at \(\tau _0\) follows from the combination of the above inequality and (165). \(\square \)

6 Optimal logarithmic Sobolev constant—part II

We first prove the log-Sobolev inequality for the conjugate heat kernel following [28]. The proof in [28] is for spacetime with bounded geometry. Since we do not impose any curvature restriction here, more should be done due to the integration by parts.

Theorem 13

For any Ricci shrinker \((M^n,g,f)\) with its heat kernel H(xtys),

$$\begin{aligned} \int \rho \log \rho \,dv_s-\left( \int \rho \,dv_s\right) \log {\left( \int \rho \, dv_s\right) } \le (t-s) \int \frac{|\nabla \rho |^2}{\rho }\,dv_s. \end{aligned}$$

Here \(dv_s(y)=H(x,t,y,s)dV_s(y)\) for any \(x\in M\) and \(s<t<1\) and \(\rho \) is any nonnegative function such that \(\sqrt{\rho }\in W^{1,2}(M,v_s)\) and \(\int d^2(p,\cdot )\rho \,dv_s <\infty \). If the equality holds, then either \(\rho \) is a constant or \((M^n,g)\) splits off a \(\mathbb {R}\) factor.

Proof

By a similar approximation process as in Sect. 4, we only need to prove the inequality for any \(\rho \) such that \(\sqrt{\rho }\) is a compactly supported smooth function. Without loss of generality, we assume \(s=0\) and fix \(T>0\) and \(q \in M\). Moreover, we set \(w(x,t)=H(q,T,x,t)\), \(dv=w(y,0)\,dV_0(y)\) and \(\rho (x,t)\) is the bounded solution of the heat equation starting from \(\rho (x)\). In the proof, we denote \(\rho (x,t)\) by \(\rho \) with the time t implicitly understood. We also assume that \(\rho \) is uniformly bounded by 1 on \(M \times [0,T]\).

It is clear from the definition of w that

$$\begin{aligned} \lim _{t \nearrow T} \int \rho (\log \rho ) w\phi ^r \,dV_t=\rho (q,T) \log \rho (q,T)=\left( \int \rho \,dv\right) \log {\left( \int \rho \, dv\right) } \end{aligned}$$

and

$$\begin{aligned} \int \rho (\log \rho ) w\phi ^r \,dV_0=\int \rho \log \rho \phi ^r \,dv. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \int \rho \log \rho \phi ^r \,dv-\left( \int \rho \,dv\right) \log {\left( \int \rho \, dv\right) } = \int _0^T -\partial _t \int \rho (\log \rho ) w\phi ^r \,dV_t \,dt. \end{aligned}$$

By direct computations

$$\begin{aligned}&-\partial _t \int \rho (\log \rho ) w\phi ^r \,dV_t \nonumber \\&\quad =-\int \rho _t(\log \rho +1)w\phi ^r+\rho (\log \rho )w_t \phi ^r+\rho (\log \rho )w\phi ^r_t-R\rho (\log \rho )w\phi ^r \,dV_t \nonumber \\&\quad =-\int \varDelta \rho (\log \rho +1)w\phi ^r-\rho (\log \rho )\varDelta w \phi ^r+\rho (\log \rho )w\phi ^r_t \,dV_t \nonumber \\&\quad =\int \frac{|\nabla \rho |^2}{\rho }w\phi ^r+2w\langle \nabla (\rho \log \rho ),\nabla \phi ^r \rangle -\rho (\log \rho )w \square \phi ^r\,dV_t. \end{aligned}$$
(169)

Similarly,

$$\begin{aligned} \partial _t \int \frac{|\nabla \rho |^2}{\rho }w\phi ^r\,dV_t =\int \square \left( \frac{|\nabla \rho |^2}{\rho }\right) w\phi ^r-2w\left\langle \nabla \left( \frac{|\nabla \rho |^2}{\rho }\right) ,\nabla \phi ^r \right\rangle +\frac{|\nabla \rho |^2}{\rho }w \square \phi ^r\,dV_t. \end{aligned}$$

Since

$$\begin{aligned} \square \frac{|\nabla \rho |^2}{\rho }=-\frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2 \end{aligned}$$
(170)

we have for any \(s \in [0,T]\),

$$\begin{aligned}&\int \frac{|\nabla \rho |^2}{\rho }w\phi ^r\,dV_s \nonumber \\&\quad = \int \frac{|\nabla \rho |^2}{\rho } \phi ^r \,dv-\int _0^s\int 2w\langle \nabla \left( \frac{|\nabla \rho |^2}{\rho }\right) ,\nabla \phi ^r \rangle \,dV_t \,dt \nonumber \\&\qquad +\int _0^s\int \frac{|\nabla \rho |^2}{\rho }w \square \phi ^r\,dV_t \,dt- \int _0^s\int \frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2w\phi ^r \,dV_t \,dt. \end{aligned}$$
(171)

With (169) and (171), we have proved so far that if r is sufficiently large,

$$\begin{aligned}&\int \rho \log \rho \,dv-\left( \int \rho \,dv\right) \log {\left( \int \rho \, dv\right) }- T\int \frac{|\nabla \rho |^2}{\rho } \,dv \\&\quad =\int _0^T\int 2w\langle \nabla (\rho \log \rho ),\nabla \phi ^r \rangle \,dV_t\,dt-\int _0^T\int \rho (\log \rho )w \square \phi ^r\,dV_t\,dt \\&\qquad +\int _0^T\int _0^s\int \frac{|\nabla \rho |^2}{\rho }w \square \phi ^r\,dV_t dtds-\int _0^T\int _0^s\int 2w\langle \nabla \left( \frac{|\nabla \rho |^2}{\rho }\right) ,\nabla \phi ^r \rangle \, dV_t dtds \\&\qquad - \int _0^T\int _0^s\int \frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2w\phi ^r \,dV_t \,dt\,ds\\&\quad =I+II+III+IV+V, \end{aligned}$$

where

$$\begin{aligned} I&= \int _0^T\int 2w\langle \nabla (\rho \log \rho ),\nabla \phi ^r \rangle \,dV_t\,dt,\\ II&= \int _0^T\int -\rho (\log \rho )w \square \phi ^r\,dV_t\,dt, \\ III&= \int _0^T\int _0^s\int \frac{|\nabla \rho |^2}{\rho }w \square \phi ^r\,dV_t dtds, \\ IV&=\int _0^T\int _0^s\int -2w\langle \nabla \left( \frac{|\nabla \rho |^2}{\rho }\right) ,\nabla \phi ^r \rangle \,dV_t dtds, \\ V&=\int _0^T\int _0^s\int -\frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2w\phi ^r \,dV_t \,dtds. \end{aligned}$$

It remains to show that when \(r \rightarrow \infty \) the sum is less or equal to 0.

We first notice that as \(\rho \) is smooth with compact support, by using (170) and the maximum principle,

$$\begin{aligned} \frac{|\nabla \rho |^2}{\rho } \le C. \end{aligned}$$

Here the assumption in Theorem 6 can be checked as (98).

Now we have for the first term I

$$\begin{aligned} \lim _{r \rightarrow \infty }|I| \le \lim _{r \rightarrow \infty } 2\int _0^T\int w|\nabla \rho |(1+|\log \rho |)|\nabla \phi ^r| \,dV_t\,dt \le \lim _{r \rightarrow \infty } Cr^{-1/2}=0. \end{aligned}$$

For the second term II,

$$\begin{aligned} \lim _{r \rightarrow \infty }|II| \le \lim _{r \rightarrow \infty } \int _0^T\int w\rho |\log \rho | |\square \phi ^r| \,dV_t\,dt \le \lim _{r \rightarrow \infty } Cr^{-1}=0. \end{aligned}$$

Similarly for the third term III,

$$\begin{aligned} \lim _{r \rightarrow \infty }|III| \le \lim _{r \rightarrow \infty } \int _0^T\int _0^s\int \frac{|\nabla \rho |^2}{\rho }w |\square \phi ^r|\,dV_t dtds \le \lim _{r \rightarrow \infty } Cr^{-1}=0. \end{aligned}$$

The fourth term IV is more involved, by computation we have

$$\begin{aligned} \nabla \frac{|\nabla \rho |^2}{\rho }=2 \frac{\langle \text {Hess}\,\rho ,\nabla \rho \rangle }{\rho } -\frac{|\nabla \rho |^2}{\rho ^2}\nabla \rho =2 \frac{\langle \text {Hess}\,\rho -\rho ^{-1}d\rho \otimes d\rho ,\nabla \rho \rangle }{\rho }+\frac{|\nabla \rho |^2}{\rho ^2}\nabla \rho . \end{aligned}$$
(172)

From (172), we have

$$\begin{aligned} |IV|&\le \int _0^T\int _0^s\int 2w|\nabla \phi ^r| \left( 2\frac{|h||\nabla \rho |}{\rho }+\frac{|\nabla \rho |^3}{\rho ^2} \right) \,dV_t dtds \\&\le \int _0^T\int _0^s\int 2\epsilon \frac{|h|^2}{\rho }w\phi ^r+2\epsilon ^{-1}\frac{|\nabla \rho |^2}{\rho }\frac{|\nabla \phi ^r|^2}{\phi ^r}w+2w|\nabla \phi ^r| \frac{|\nabla \rho |^3}{\rho ^2} \,dV_t dtds \\&\le -\epsilon V+C\epsilon ^{-1}r^{-1}+2r^{-1/2}\int _0^T\int _0^s\int w\frac{|\nabla \rho |^3}{\rho ^2} \,dV_t dtds \end{aligned}$$

where we denote \(\text {Hess}\,\rho -\rho ^{-1}d\rho \otimes d\rho \) by h and \(\epsilon \in (0,1)\).

To deal with the last integral, we notice from Lemma 18 that

$$\begin{aligned} \frac{|\nabla \rho |^3}{\rho ^2} =\frac{|\nabla \rho |^{3/2}}{\rho ^{3/2}} \frac{|\nabla \rho |^{3/2}}{\rho ^{3/4}} \rho ^{1/4} \le \frac{C}{t^{3/4}}\left( \rho ^{1/6} \log {\frac{M}{\rho }} \right) ^{3/2} \le \frac{C}{t^{3/4}} \end{aligned}$$

and hence

$$\begin{aligned} \int _0^s\int w\frac{|\nabla \rho |^3}{\rho ^2} \,dV_t dt \le C\int _0^s t^{-3/4} \,dt \le C. \end{aligned}$$

Therefore, \(\lim _{r \rightarrow \infty }V\) is finite and \(\lim _{r \rightarrow \infty }|IV| \le -\epsilon \left( \lim _{r \rightarrow \infty }V\right) \). By taking \(\epsilon \rightarrow 0\), we obtain that \(\lim _{r \rightarrow \infty }|IV|=0\) and hence

$$\begin{aligned}&\int \rho \log \rho \,dv-\left( \int \rho \,dv\right) \log {\left( \int \rho \, dv\right) }- T\int \frac{|\nabla \rho |^2}{\rho } \,dv \end{aligned}$$
(173)
$$\begin{aligned}&\quad =-\int _0^T\int _0^s\int \frac{2}{\rho }\left| \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho }\right| ^2w \,dV_t \,dt\,ds \le 0. \end{aligned}$$
(174)

If the equality holds and \(\rho \) is not a constant, it follows from (173) that

$$\begin{aligned} \text {Hess}(\log \rho )=\frac{1}{\rho }\left( \text {Hess}\,\rho -\frac{d\rho \otimes d\rho }{\rho } \right) =0. \end{aligned}$$

Therefore, \((M^n,g)\) splits off a \(\mathbb {R}\) factor. \(\square \)

For fixed xt and s, Theorem 13 implies that the probability measure \(dv_s(y)=H(x,t,y,s)dV_s(y)\) satisfies the log-Sobolev inequality with the constant \(\frac{1}{2(t-s)}\). It is a standard fact that log-Sobolev condition implies the Talagrand’s inequality and equivalently, the Gaussian concentration, see [52, Theorems 22.17, 22.10]. In particular we have the following theorem, see also [28, Theorem 1.13].

Theorem 14

(Gaussian concentration) For any Ricci shrinker \((M^n,g,f)\) with its heat kernel H(xtys) and reference measure \(dv_s(y)=H(x,t,y,s)dV_s(y)\) and any \(\sigma >0\)

$$\begin{aligned} v_s(A)v_s^{\frac{1}{\sigma }}(B)\le \exp {\left( -\frac{r^2}{4(1+\sigma )(t-s)}\right) } \end{aligned}$$

where A and B are two sets on M such that \(d_s(A,B) \ge r>0\).

Proof

From Theorem (13), we have for any probability measure \(\rho dv_s\),

$$\begin{aligned} \int \rho \log \rho \,dv_s \le (t-s)\int \frac{|\nabla \rho |^2}{\rho } \,dv_s. \end{aligned}$$
(175)

By a further approximation, we can assume (175) holds for any locally Lipschitz \(\rho \). Now it follows from [52, Theorem 22.17] that \(dv_s\) satisfies the \(T_2\) Talagrand inequality, that is,

$$\begin{aligned} W_2(\eta , v_s) \le \sqrt{4(t-s)}\left( \int \rho \log \rho \,dv_s\right) ^{1/2} \end{aligned}$$
(176)

for any measure \(\eta \in P_2(M)\), where \(W_2\) is the Wasserstein distance of second order. For any two sets A and B on M such that \(d_s(A,B) \ge r>0\). We set \(\eta =\frac{1_A}{v_s(A)}v_s\) and \(v=\frac{1_B}{v_s(B)}v_s\). Then on the one hand,

$$\begin{aligned} W_2(\eta ,v)&\le W_2(\eta ,v_s)+W_2(v,v_s) \\&\le \sqrt{4(t-s)}\left( \left( \int \frac{1_A}{v_s(A)} \log \frac{1_A}{v_s(A)} \,dv_s\right) ^{1/2}+\left( \int \frac{1_B}{v_s(B)} \log \frac{1_B}{v_s(B)}\,dv_s\right) ^{1/2}\right) \\&= \sqrt{4(t-s)}\left( \left( \log {\frac{1}{v_s(A)}}\right) ^{1/2}+\left( \log {\frac{1}{v_s(B)}}\right) ^{1/2}\right) \end{aligned}$$

and hence

$$\begin{aligned} W^2_2(\eta ,v)&\le 4(t-s)\left( \left( \log {\frac{1}{v_s(A)}}\right) ^{1/2}+\left( \log {\frac{1}{v_s(B)}}\right) ^{1/2}\right) ^2 \\&\le 4(t-s)\left( (1+\sigma )\log {\frac{1}{v_s(A)}}+(1+\sigma ^{-1})\log {\frac{1}{v_s(B)}}\right) . \end{aligned}$$

On the other hand, it follows from the definition of \(W_2\) that

$$\begin{aligned} W^2_2(\eta ,v) =\int d^2_s(x,y)\,d\pi (x,y) \ge r^2, \end{aligned}$$

where \(\pi \) is the optimal transport between \(\eta \) and v.

Therefore by computation

$$\begin{aligned} v_s(A)v_s^{\frac{1}{\sigma }}(B)\le \exp {\left( -\frac{r^2}{4(1+\sigma )(t-s)}\right) }. \end{aligned}$$

\(\square \)

In fact, with the Gaussian concentration, we can prove that \(v_s\) has finite square-exponential moment.

Corollary 5

For any Ricci shrinker \((M^n,g,f)\) with its heat kernel H(xtys) and reference measure \(dv_s(y)=H(x,t,y,s)dV_s(y)\), if \(a< \frac{1}{4(t-s)}\), then

$$\begin{aligned} \int e^{ad_s^2(p,x)}\,dv_s < \infty . \end{aligned}$$

Proof

We choose a constant \(\sigma >0\) such that \(a < \frac{1}{4(1+\sigma )(t-s)} \). It follows from Theorem 14 that for any integer \(k \ge 2\),

$$\begin{aligned} v_s(M \backslash B_s(p,k))\le \exp {\left( -\frac{(k-1)^2}{4(1+\sigma )(t-s)}\right) }. \end{aligned}$$

Hence

$$\begin{aligned} \int e^{ad_s^2(p,x)}\,dv_s&\le C\left( 1+\sum _{k=2}^{\infty }\int _{B_s(p,k+1)\backslash B_s(p,k)}e^{ad_s^2(p,x)}\,dv_s\right) \\&\le C +C\sum _{k=2}^{\infty }(k+1)^n\exp {\left( a(k+1)^2-\frac{(k-1)^2}{4(1+\sigma )(t-s)}\right) } \end{aligned}$$

where we have used Lemma 2. Since \(a< \frac{1}{4(1+\sigma )(t-s)}\), it is easy to show that the last sum is finite. \(\square \)

7 Heat kernel estimates

We first prove a pointwise upper bound for the heat kernel H. The idea of the proof is from [21, Chapter 2], see also [61].

Theorem 15

(Ultracontractivity) For any Ricci shrinker \((M^n,g,f)\),

$$\begin{aligned} H(x,t,y,s) \le \frac{e^{-\varvec{\mu }}}{(4\pi (t-s))^{\frac{n}{2}}}. \end{aligned}$$

Proof

We fix \(x \in M\) and two constants \(s<T<1\). For notational simplicity, we assume that \(\tau =T-t\) and \(\partial _{\tau }=-\partial _t\). We also fix a function \(p(\tau )=\frac{T-s}{T-s-\tau }\) for \(\tau \in [0,T-s)\). For any nonnegative smooth function h with compact support we define

$$\begin{aligned} w(y,\tau )=\int H(x,T,y,T-\tau )h(x)\,dV_T(x), \end{aligned}$$
(177)

then \(\square ^* w=0\).

Now we compute,

$$\begin{aligned} \partial _{\tau }||w\phi ^r||_{p({\tau })}&=\partial _{\tau }\left( \int (w\phi ^r)^{p({\tau })}\,dV_{T-\tau }\right) ^{\frac{1}{p({\tau })}} \\&=-\frac{p'({\tau })}{p^2({\tau })}||w\phi ^r||_{p({\tau })} \log \left( \int (w\phi ^r)^{p({\tau })} \,dV_{T-\tau }\right) \\&\quad +\frac{1}{p({\tau })}\left( \int (w\phi ^r)^{p({\tau })}dV_{T-\tau }\right) ^{\frac{1}{p({\tau })}-1}\left( \int (w\phi ^r)^{p({\tau })}(\log w\phi ^r) p'({\tau }) dV_{T-\tau } \right) \\&\quad +\frac{1}{p({\tau })}\left( \int (w\phi ^r)^{p({\tau })}dV_{T-\tau }\right) ^{\frac{1}{p({\tau })}-1}\left( \int p({\tau })(w\phi ^r)^{p({\tau })-1}(w\phi ^r)_{\tau }+R(w\phi ^r)^{p(\tau )} dV_{T-\tau } \right) . \end{aligned}$$

If we multiply both sides above by \(p^2({\tau })||w\phi ^r||^{p({\tau })}_{p({\tau })}\) and use the fact

$$\begin{aligned} (w\phi ^r)_{\tau }=\varDelta w \phi ^r-Rw\phi ^r+w\phi ^r_{\tau }=\varDelta (w\phi ^r)-Rw\phi ^r-(\square _{\tau }\phi ^r)w-2\langle \nabla w,\nabla \phi ^r\rangle , \end{aligned}$$
(178)

then we have

$$\begin{aligned}&p^2({\tau })||w\phi ^r||^{p({\tau })}_{p({\tau })}\partial _{\tau }||w\phi ^r||_{p({\tau })} \nonumber \\&\quad =-p'({\tau })||w\phi ^r||^{p({\tau })+1}_{p({\tau })} \log \left( \int (w\phi ^r)^{p({\tau })} \,dV_{T-\tau }\right) \nonumber \\&\qquad +p({\tau })p'({\tau })||w\phi ^r||_{p({\tau })}\int (w\phi ^r)^{p({\tau })}\log (w\phi ^r) \,dV_{T-\tau } \nonumber \\&\qquad -p^2({\tau })(p({\tau })-1)||w\phi ^r||_{p({\tau })}\int (w\phi ^r)^{p({\tau })-2}|\nabla (w\phi ^r)|^2 \,dV_{T-\tau } \nonumber \\&\qquad -(p({\tau })-1) ||w\phi ^r||_{p({\tau })}\int R(w\phi ^r)^{p({\tau })} \,dV_{T-\tau }+X, \end{aligned}$$
(179)

where

$$\begin{aligned} X=p^2({\tau })||w\phi ^r||_{p({\tau })}\int (w\phi ^r)^{p({\tau })-1}\left( -(\square _{\tau }\phi ^r)w-2\langle \nabla w,\nabla \phi ^r\rangle \right) \,dV_{T-\tau }. \end{aligned}$$

Now we divide both sides of (179) by \(||w\phi ^r||_{p({\tau })}\), then

$$\begin{aligned}&p^2({\tau })||w\phi ^r||^{p({\tau })}_{p({\tau })}\partial _{\tau }\log ||w\phi ^r||_{p({\tau })}\nonumber \\&\quad =-p'({\tau })||w\phi ^r||^{p({\tau })}_{p({\tau })} \log \left( \int (w\phi ^r)^{p({\tau })} \,dV_{T-\tau }\right) \nonumber \\&\qquad +p({\tau })p'({\tau })\int (w\phi ^r)^{p({\tau })}\log (w\phi ^r) \,dV_{T-\tau } \nonumber \\&\qquad -4(p({\tau })-1)\int |\nabla (w\phi ^r)^{\frac{p({\tau })}{2}}|^2 \,dV_{T-\tau } \nonumber \\&\qquad -(p({\tau })-1) \int R(w\phi ^r)^{p({\tau })} \,dV_{T-\tau }+Y, \end{aligned}$$
(180)

where

$$\begin{aligned} Y=p^2({\tau })\int (w\phi ^r)^{p({\tau })-1}\left( -(\square _{\tau }\phi ^r)w-2\langle \nabla w,\nabla \phi ^r\rangle \right) \,dV_{T-\tau }. \end{aligned}$$

We denote \(v=(w\phi ^r)^{\frac{p({\tau })}{2}}/||(w\phi ^r)^{\frac{p({\tau })}{2}}||_2\) so that \(||v||_2=1\). Now by direct computations,

$$\begin{aligned} v^2\log v^2=p({\tau })v^2\log (w\phi ^r)-2v^2\log ||(w\phi ^r)^{\frac{p({\tau })}{2}}||_2. \end{aligned}$$

So (180) becomes

$$\begin{aligned}&p^2({\tau })\partial _{\tau }\log ||w\phi ^r||_{p({\tau })}\\&\quad = p'({\tau })\int v^2\log v^2 \,dV_{T-\tau }-4(p({\tau })-1)\int |\nabla v|^2 \,dV_{T-\tau }-(p({\tau })-1)\int Rv^2\,dV_{T-\tau }+Z \end{aligned}$$

where

$$\begin{aligned} Z=\frac{p^2({\tau })}{||w\phi ^r||^{p({\tau })}_{p({\tau })}}\int (w\phi ^r)^{p({\tau })-1}\left( -(\square _{\tau }\phi ^r)w-2\langle \nabla w,\nabla \phi ^r\rangle \right) \,dV_{T-\tau }. \end{aligned}$$

Now we obtain

$$\begin{aligned} p^2({\tau })\partial _{\tau }\log ||w\phi ^r||_{p({\tau })}&= p'({\tau })\left( \int v^2\log v^2 \,dV_{T-\tau }-\frac{p({\tau })-1}{p'({\tau })}\int 4|\nabla v|^2+ Rv^2\,dV_{T-\tau }\right) +Z. \end{aligned}$$
(181)

Since \(\frac{p({\tau })-1}{p'({\tau })}=\frac{\tau (T-s-\tau )}{T-s}>0\), we have from (181)

$$\begin{aligned} p^2({\tau })\partial _{\tau }\log ||w\phi ^r||_{p({\tau })}\le p'({\tau })\left( -\varvec{\mu }-n-\frac{n}{2}\log (4\pi (p({\tau })-1)/p'({\tau }))\right) +Z. \end{aligned}$$
(182)

Now we divide both sides by \(p^2(\tau )\), we have

$$\begin{aligned} \partial _{\tau }\log ||w\phi ^r||_{p({\tau })}\le \frac{p'({\tau })}{p^2(\tau )}\left( -\varvec{\mu }-n-\frac{n}{2}\log (4\pi )-\frac{n}{2}\log \left( \frac{p({\tau })-1)}{p'({\tau })}\right) \right) +U(\tau ), \end{aligned}$$
(183)

where

$$\begin{aligned} U(\tau )=\frac{1}{||w\phi ^r||^{p({\tau })}_{p({\tau })}}\int (w\phi ^r)^{p({\tau })-1}\left( -(\square _{\tau }\phi ^r)w-2\langle \nabla w,\nabla \phi ^r\rangle \right) \,dV_{T-\tau }. \end{aligned}$$

Now we integrate both sides of (183) and estimate the two terms of right side separately.

For a number \(L < T-s\), we integrate (183) from 0 to L so that

$$\begin{aligned}&\log ||w\phi ^r||_{p(L)}-\log ||w\phi ^r||_1 \nonumber \\&\quad \le \int _0^L \frac{p'({\tau })}{p^2(\tau )}\left( -\varvec{\mu }-n-\frac{n}{2}\log (4\pi )-\frac{n}{2}\log \left( \frac{p({\tau })-1)}{p'({\tau })}\right) \right) \, d\tau +\int _0^LU(\tau )\,d\tau \nonumber \\&\quad = I(L)+II(L). \end{aligned}$$
(184)

By direct computations,

$$\begin{aligned} I(T-s)&= \int _0^{T-s} \frac{p'({\tau })}{p^2(\tau )}\left( -\varvec{\mu }-n-\frac{n}{2}\log (4\pi )-\frac{n}{2}\log \left( \frac{p({\tau })-1)}{p'({\tau })}\right) \right) \, d\tau \nonumber \\&=-\frac{n}{2}\log (T-s)-\varvec{\mu }-\frac{n}{2}\log (4\pi ). \end{aligned}$$
(185)

Now we consider the term \(U(\tau )\).

$$\begin{aligned} |U(\tau )| \le \frac{1}{||w\phi ^r||^{p({\tau })}_{p({\tau })}}\int w^{p({\tau })}|\square _{\tau }\phi ^r|+2|\nabla w||\nabla \phi ^r|\,dV_{T-\tau }. \end{aligned}$$
(186)

Since we construct w through a smooth function with compact support,

$$\begin{aligned} w \le Ce^{-f} \end{aligned}$$

for a constant C uniformly on \(M \times [T-s-L,T-s]\). On the other hand, by Lemma 11\(\sqrt{w} \in W_{*}^{1,2}\) for any \(\tau >0\), in particular

$$\begin{aligned} \int \frac{|\nabla w|^2}{w} \,dV_{T-\tau } < \infty . \end{aligned}$$

Now the second term in (186) can be estimated as

$$\begin{aligned} \int |\nabla w||\nabla \phi ^r| \, dV_{T-\tau }=\int \frac{|\nabla w|}{\sqrt{w}}|\nabla \phi ^r|\sqrt{w} \, dV_{T-\tau } \le \left( \int \frac{|\nabla w|^2}{ w} \, dV_{T-\tau } \right) ^{\frac{1}{2}} \left( \int |\nabla \phi ^r|^2 w \, dV_{T-\tau } \right) ^{\frac{1}{2}}. \end{aligned}$$

For any fixed L, it is easy to say \(U(\tau )\) is uniformly bounded for any \(\tau \in [T-s-L,T-s]\) and \(r \ge 1\). By taking \(r \rightarrow \infty \) in (184), from the dominated convergence theorem,

$$\begin{aligned} \log ||w||_{p(L)}-\log ||w||_1 \le I(L). \end{aligned}$$

Now by taking \(L \rightarrow T-s\) we have

$$\begin{aligned} \log ||w||_{\infty }-\log ||w||_1 \le -\frac{n}{2}\log (T-s)-\varvec{\mu }-\frac{n}{2}\log (4\pi ). \end{aligned}$$

Therefore,

$$\begin{aligned} \int H(x,T,y,s)h(x)\,dV_T(x)&\le \frac{e^{-\varvec{\mu }}}{(4\pi (T-s))^{n/2}} \iint H(x,T,y,s)h(x)\,dV_T(x)\,dV_s(y) \\&=\frac{e^{-\varvec{\mu }}}{(4\pi (T-s))^{n/2}} \int h(x)\,dV_T(x). \end{aligned}$$

Since h(x) can be any smooth function with compact support, we derive that

$$\begin{aligned} H(x,T,y,s) \le \frac{e^{-\varvec{\mu }}}{(4\pi (T-s))^{n/2}}. \end{aligned}$$

\(\square \)

Now we derive the lower bound of H. Recall that the reduced distance between (xt) and (ys) are defined as

$$\begin{aligned} l_{(x,t)}(y,s)=\frac{1}{2\sqrt{t-s}} \inf \left\{ \mathcal L(\gamma ): \, \gamma :[s,t] \rightarrow M \, \text {between } (x,t)\text { and } (y,s)\right\} , \end{aligned}$$
(187)

where

$$\begin{aligned} \mathcal L(\gamma )=\int _s^t \sqrt{t-z} \left( |\gamma '(z)|_{z}^2+R(\gamma (z),z)\right) \,dz. \end{aligned}$$
(188)

Now we have the following important estimate, see Corollary 9.5 of [46]. The proof is motivated by [16, Proposition 1].

Theorem 16

For any Ricci shrinker \((M^n,g,f)\),

$$\begin{aligned} H(x,t,y,s) \ge \frac{e^{-l_{(x,t)}(y,s)}}{(4\pi (t-s))^{\frac{n}{2}}}. \end{aligned}$$

Proof

We set

$$\begin{aligned} L(x,t,y,s)= \frac{e^{-l_{(x,t)}(y,s)}}{(4\pi (t-s))^{\frac{n}{2}}}. \end{aligned}$$
(189)

It follows from the definition of \(l_{(x,t)}(y,s)\), see [46] and [56], that

$$\begin{aligned} -\partial _s L(x,t,y,s) \le \varDelta _{y,s} L(x,t,y,s)-R(y,s)L(x,t,y,s) \end{aligned}$$
(190)

and

$$\begin{aligned} \lim _{s \nearrow t}L(x,t,y,s)= \delta _x. \end{aligned}$$
(191)

For any \(x,y \in M\), \(s<T\) and small \(\epsilon >0\) we have

$$\begin{aligned}&\int L(x,T,z,T-\epsilon )H(z,T-\epsilon ,y,s)\phi ^r(z,T-\epsilon )\,dV_{T-\epsilon }(z)\nonumber \\&\qquad -\int L(x,T,z,s+\epsilon )H(z,s+\epsilon ,y,s)\phi ^r(z,s+\epsilon )\,dV_{s+\epsilon }(z) \nonumber \\&\quad =\int _{s+\epsilon }^{T-\epsilon } \partial _t\left( \int L(x,T,z,t)H(z,t,y,s)\phi ^r(z,t)\, dV_t\right) \,dt \nonumber \\&\quad =\int _{s+\epsilon }^{T-\epsilon } \int L_tH\phi ^r \, dV_t \,dt+\int _{s+\epsilon }^{T-\epsilon } \int LH_t\phi ^r \,dV_t \,dt \nonumber \\&\qquad +\int _{s+\epsilon }^{T-\epsilon } \int LH\phi ^r_t\, dV_t\,dt-\int _{s+\epsilon }^{T-\epsilon } \int LH\phi ^rR\, dV_t \,dt \nonumber \\&\quad \ge -\int _{s+\epsilon }^{T-\epsilon } \int \varDelta LH\phi ^r dV_t dt+\int _{s+\epsilon }^{T-\epsilon } \int L\varDelta H\phi ^r dV_tdt +\int _{s+\epsilon }^{T-\epsilon } \int LH\phi _t^r dV_tdt. \end{aligned}$$
(192)

Here and after we omit all zt for notational simplicity.

By the integration by parts, we have

$$\begin{aligned} -\int _{s+\epsilon }^{T-\epsilon } \int \varDelta LH\phi ^r\, dV_t\,dt=-\int _{s+\epsilon }^{T-\epsilon } \int L\left( \varDelta H\phi ^r+H\varDelta \phi ^r+2\langle \nabla H,\nabla \phi ^r \rangle \right) \, dV_t\,dt \end{aligned}$$
(193)

Therefore,

$$\begin{aligned}&\int L(x,T,z,T-\epsilon )H(z,T-\epsilon ,y,s)\phi ^r(z,T-\epsilon )\,dV_{T-\epsilon }(z)\nonumber \\&\qquad -\int L(x,T,z,s+\epsilon )H(z,s+\epsilon ,y,s)\phi ^r(z,s+\epsilon )\,dV_{s+\epsilon }(z) \nonumber \\&\quad \ge \int _{s+\epsilon }^{T-\epsilon } \int LH \square \phi ^r-2L\langle \nabla H,\nabla \phi ^r \rangle \, dV_t\,dt. \end{aligned}$$
(194)

Now we multiply both sides of \(\square H=0\) by \((\phi ^r)^2H\) and do the integration.

$$\begin{aligned} \int _{s+\epsilon }^{T-\epsilon }\int |\nabla (\phi ^rH)|^2 \,dV_t\,dt&\le \int _{s+\epsilon }^{T-\epsilon }\int |\nabla \phi ^r|^2H^2 \,dV_t\,dt+ \int _{s+\epsilon }^{T-\epsilon }\int \frac{H^2}{2}(\phi ^r)^2_t \,dV_t\,dt \\&\quad -\left. \left( \int \frac{H^2}{2}(\phi ^r)^2 \,dV_t \right) \right| _{s+\epsilon }^{T-\epsilon }. \end{aligned}$$

It is immediate by taking \(r \rightarrow \infty \) that

$$\begin{aligned} \int _{s+\epsilon }^{T-\epsilon }\int |\nabla H|^2 \,dV_t\,dt <\infty . \end{aligned}$$
(195)

For fixed \(\epsilon \), we have

$$\begin{aligned}&\left| \int _{s+\epsilon }^{T-\epsilon } \int LH \square \phi ^r-2L\langle \nabla H,\nabla \phi ^r \rangle \, dV_t\,dt \right| \nonumber \\&\quad \le \int _{s+\epsilon }^{T-\epsilon } \int LH |\square \phi ^r|+2L|\nabla H||\nabla \phi ^r|\, dV_t\,dt=I+II. \end{aligned}$$
(196)

For the first term,

$$\begin{aligned} \lim _{r\rightarrow \infty }I=\lim _{r\rightarrow \infty }\int _{s+\epsilon }^{T-\epsilon } \int LH |\square \phi ^r|\, dV_t\,dt=0 \end{aligned}$$

since L is uniformly bounded on \(M \times [s+\epsilon ,T-\epsilon ]\) and H is integrable.

For the second term,

$$\begin{aligned} II=\int _{s+\epsilon }^{T-\epsilon } \int L|\nabla H||\nabla \phi ^r|\, dV_t\,dt \le 2\left( \int _{s+\epsilon }^{T-\epsilon } \int L^2|\nabla \phi ^r|^2\, dV_t\,dt\right) ^{\frac{1}{2}}\left( \int _{s+\epsilon }^{T-\epsilon } \int |\nabla H|^2\, dV_t\,dt\right) ^{\frac{1}{2}}. \end{aligned}$$

Now we claim

$$\begin{aligned} \int _{s+\epsilon }^{T-\epsilon } \int L^2\, dV_t\,dt<\infty . \end{aligned}$$
(197)

Indeed, it follows from [58, Eq. (3.3)] that for any \(t \in [s+\epsilon ,T-\epsilon ]\),

$$\begin{aligned} l_{(x,T)}(z,t) \ge \sqrt{\frac{1-t}{T-t}}f(z,t)-\sqrt{\frac{1-T}{T-t}}f(x,T) \end{aligned}$$

and hence

$$\begin{aligned} L(x,T,z,t) \le \eta _1 e^{-\eta _2 F(z,t)}, \end{aligned}$$

where

$$\begin{aligned} \eta _1= \frac{\exp \left( \sqrt{\frac{1-T}{\epsilon }}f(x,T)\right) }{(4\pi \epsilon )^{\frac{n}{2}}} \quad \text {and} \quad \eta _2=\frac{1}{\sqrt{(T-s-\epsilon )(1-s-\epsilon )}}. \end{aligned}$$

Therefore, it is clear from Lemmas 1 and 2 that the claim (197) holds.

It is immediate from (195) that

$$\begin{aligned} \lim _{r \rightarrow \infty }II \le \lim _{r\rightarrow \infty }2\left( \int _{s+\epsilon }^{T-\epsilon } \int L^2|\nabla \phi ^r|^2\, dV_t\,dt\right) ^{\frac{1}{2}}\left( \int _{s+\epsilon }^{T-\epsilon } \int |\nabla H|^2\, dV_t\,dt\right) ^{\frac{1}{2}}=0. \end{aligned}$$
(198)

Now it follows from (194) that by taking \(r \rightarrow \infty \),

$$\begin{aligned} \int L(x,T,z,T-\epsilon )H(z,T-\epsilon ,y,s)\,dV_{T-\epsilon }(z)\ge \int L(x,T,z,s+\epsilon )H(z,s+\epsilon ,y,s)\,dV_{s+\epsilon }(z). \end{aligned}$$
(199)

As \(\epsilon \rightarrow 0\), both \(H(z,T-\epsilon ,y,s)\) and \(L(x,T,z,s+\epsilon )\) are uniformly bounded (in terms of z). We conclude from the definition of \(\delta \) function that by taking \(\epsilon \rightarrow 0\)

$$\begin{aligned} H(x,T,y,s)\ge L(x,T,y,s). \end{aligned}$$

\(\square \)

We also need the following gradient estimate from [60].

Lemma 18

For any Ricci shrinker \((M^n,g,f)\), suppose u is a positive bounded solution of the heat equation \(\square u=0\) on \(M \times [0,T]\), then

$$\begin{aligned} \frac{|\nabla u|}{u} \le \sqrt{\frac{1}{t}}\sqrt{\log \frac{\varLambda }{u}} \end{aligned}$$

where \(\varLambda =\max _{M \times [0,T]}u\).

Proof

From a direction computation

$$\begin{aligned} \square \left( t\frac{|\nabla u|^2}{u}-u\log \frac{\varLambda }{u}\right) =-\frac{2}{u}\left| \text {Hess}\,u-\frac{du \otimes du}{u}\right| ^2 \le 0. \end{aligned}$$

Now the theorem follows from Theorem 6 if

$$\begin{aligned} \int _0^T\int \frac{|\nabla u|^2}{u}e^{-2f}\,dV_t\,dt <\infty . \end{aligned}$$

Notice that this follows the same proof as Lemma 9. \(\square \)

Now we have the following corollary of Lemma 18, see [60, Eq. (3.44)].

Corollary 6

With the same conditions as Lemma 18, for any \(\sigma >0\),

$$\begin{aligned} u(y,t) \le \varLambda ^{\frac{\sigma }{1+\sigma }}u(x,t)^{\frac{1}{1+\sigma }}\exp \left( \frac{d_t^2(x,y)}{4\sigma t} \right) . \end{aligned}$$
(200)

Proof

We rewrite Lemma 18 as

$$\begin{aligned} \left| \nabla \sqrt{\log {\frac{\varLambda }{u}}}\right| \le \frac{1}{2\sqrt{t}}, \end{aligned}$$

and hence

$$\begin{aligned} \sqrt{\log {\frac{\varLambda }{u(x,t)}}}&\le \sqrt{\log {\frac{\varLambda }{u(y,t)}}}+\frac{d_t(x,y)}{2\sqrt{t}}. \end{aligned}$$

By squaring both sides above, we have

$$\begin{aligned} \log {\frac{\varLambda }{u(x,t)}}&\le \left( \sqrt{\log {\frac{\varLambda }{u(y,t)}}}+\frac{d_t(x,y)}{2\sqrt{t}} \right) ^2 \\&\le (1+\sigma ) \log {\frac{\varLambda }{u(y,t)}}+\frac{1+\sigma }{\sigma }\frac{d_t^2(x,y)}{4t}. \end{aligned}$$

Then the conclusion follows immediately. \(\square \)

We now prove the pointwise lower bound of the heat kernel H.

Theorem 17

For any Ricci shrinker \((M^n,g,f)\), \(0<\delta <1\), \(D>1\) and \(0<\epsilon <4\), there exists a constant \(C=C(n,\delta ,D)>0\) such that

$$\begin{aligned} H(x,t,y,s) \ge \frac{C^{\frac{4}{\epsilon }}e^{\varvec{\mu }(\frac{4}{\epsilon }-1)}}{(4\pi (t-s))^{n/2}} \exp {\left( -\frac{d_t^2(x,y)}{(4-\epsilon )(t-s)}\right) } \end{aligned}$$

for any \(t \in [-\delta ^{-1},1-\delta ]\) and \(d_t(p,y)+\sqrt{t-s}\le D\).

Proof

From Theorem 16,

$$\begin{aligned} H(y,t,y,s) \ge \frac{e^{-l_{(y,t)}(y,s)}}{(4\pi (t-s))^{\frac{n}{2}}}. \end{aligned}$$
(201)

By the definition of l and \(\partial _z f(y,z)=|\nabla f|^2 \ge 0\),

$$\begin{aligned} l_{(y,t)}(y,s)&\le \frac{1}{2\sqrt{t-s}}\int _s^t \sqrt{t-z} R(y,z)\,dz \nonumber \\&\le \frac{1}{2\sqrt{t-s}}\int _s^t \frac{\sqrt{t-z}}{1-z} f(y,z)\,dz \nonumber \\&\le \frac{f(y,t)}{2\sqrt{t-s}}\int _s^t \frac{\sqrt{t-z}}{1-z}\,dz \le \frac{(t-s)}{3(1-t)^2}F(y,t) \end{aligned}$$
(202)

and hence

$$\begin{aligned} H(y,t,y,s) \ge \frac{C}{(4\pi (t-s))^{n/2}} \end{aligned}$$

for some constant \(C=C(n,\delta ,D)>0\).

By using (200) for the heat kernel on \(M \times [\frac{t+s}{2},t]\), we obtain

$$\begin{aligned} H(y,t,y,s) \le e^{-\varvec{\mu } \frac{\sigma }{1+\sigma }} (4\pi (t-s))^{-\frac{n}{2}\frac{\sigma }{1+\sigma }} H^{\frac{1}{1+\sigma }}(x,t,y,s)\exp {\left( \frac{d_t^2(x,y)}{4\sigma (t-s)}\right) } \end{aligned}$$

where we have used the result in Theorem 15 for the upper bound.

Therefore,

$$\begin{aligned} H(x,t,y,s) \ge \frac{C^{1+\sigma }e^{\varvec{\mu } \sigma }}{(4\pi (t-s))^{n/2}} \exp {\left( -\frac{1+\sigma }{\sigma }\frac{d_t^2(x,y)}{4(t-s)}\right) }. \end{aligned}$$

The conclusion follows by choosing \(\sigma =4/\epsilon -1\). \(\square \)

Remark 3

From the proof a more precise bound is, for any \(0<\epsilon <4\),

$$\begin{aligned} H(x,t,y,s) \ge \frac{e^{\varvec{\mu }(\frac{4}{\epsilon }-1)}}{(4\pi (t-s))^{n/2}} \exp {\left( -\frac{d_t^2(x,y)}{(4-\epsilon )(t-s)}-\frac{4(t-s)}{3(1-t)^2\epsilon }F(y,t)\right) }. \end{aligned}$$
(203)

In order to further estimate the upper bound of H, it is crucial to compare distance functions from different time slices. We first prove the second order estimate of the heat equation soluton on Ricci shrinkers, see [3, Lemma 3.1].

Lemma 19

Let \((M^n,g(t)),\, t\in [0,1)\) be the Ricci flow solution of a Ricci shrinker and let u be a postive solution to the heat equation \(\square u=0\) and \(u \le \varLambda \) on \(M \times [0,T]\). Then there exists a constant \(C =C(n)\) such that

$$\begin{aligned} |\varDelta u|+\frac{|\nabla u|^2}{u}-\varLambda R \le \frac{C\varLambda }{t}. \end{aligned}$$
(204)

Proof

By rescaling, we assume that \(\varLambda =1\). Let \(L_1=-\varDelta u+\frac{|\nabla u|^2}{u}-R\), then it follows from [3, Eqs. (3.3), (3.4)] that

$$\begin{aligned} \square L_1 \le -\frac{1}{n}L_1^2+\frac{1}{e^2 t^2}. \end{aligned}$$
(205)

From (205) we have

$$\begin{aligned} \square (L_1\phi ^r)&=\phi ^r\square L_1+L_1 \square \phi ^r-2\langle \nabla \phi ^r,\nabla L_1 \rangle \nonumber \\&\le \phi ^r( -\frac{1}{n}L_1^2+\frac{1}{e^2 t^2})+L_1 \square \phi ^r-2\langle \nabla \phi ^r,\nabla L_1 \rangle \nonumber \\&=\phi ^r( -\frac{1}{n}L_1^2+\frac{1}{e^2 t^2})+L_1 \square \phi ^r-2\frac{\langle \nabla (L_1\phi ^r),\nabla \phi ^r \rangle }{\phi ^r}+2\frac{L_1|\nabla \phi ^r|^2}{\phi ^r}. \end{aligned}$$
(206)

Now at the maximum point of \(L_1\phi ^r\), we have

$$\begin{aligned} -\frac{1}{n}(L_1\phi ^r)^2+(\phi ^re^{-1}t^{-1})^2+(L_1\phi ^r)\left( \square \phi ^r+2\frac{|\nabla \phi ^r|^2}{\phi ^r}\right) \ge 0, \end{aligned}$$
(207)

so we obtain

$$\begin{aligned} L_1\phi ^r \le n\left( \square \phi ^r+2\frac{|\nabla \phi ^r|^2}{\phi ^r}\right) +\sqrt{n}\phi ^re^{-1}t^{-1} \le C(n)(r^{-1}+t^{-1}). \end{aligned}$$
(208)

By taking \(r \rightarrow \infty \), we have \(L_1 \le C(n)t^{-1}\). Now if we set \(L_2=\varDelta u+\frac{|\nabla u|^2}{u}-R\), then similarly

$$\begin{aligned} \square L_2 \le -\frac{1}{2n}L_2^2+\frac{1+\frac{4}{n}}{e^2 t^2}. \end{aligned}$$
(209)

Therefore by the same method, we prove that \(L_2 \le C(n)t^{-1}\).

Now the proof is complete. \(\square \)

By applying the above lemma to the heat kernel, we immediately have from Theorem 15 that

Lemma 20

For any Ricci shrinker \((M^n,g,f)\), there exists a constant \(C =C(n)\) such that

$$\begin{aligned} |\partial _tH(x,t,y,s)|=|\varDelta _xH(x,t,y,s)| \le C\frac{e^{-\varvec{\mu }}}{(t-s)^{\frac{n}{2}}}\left( R(x,t)+\frac{1}{t-s}\right) \end{aligned}$$
(210)

for any \(s<t<1\).

Now we can prove the local distance distorsion on Ricci shrinkers. Notice that a similar estimate has been obtained on compact manifolds, see [3, Theorem 1.1].

Theorem 18

(Local distance distorsion estimate) For any Ricci shrinker \((M^n,p,g,f) \in \mathcal M_n(A)\), \(0<\delta <1\) and \(D>1\), there exists a constant \(Y=Y(n,A,\delta ,D)>1\) such that for any two points q and z in M with \(d_t(p,q) \le D\) and \(d_t(q,z)=r \le D\),

$$\begin{aligned} Y^{-1}d_s(q,z) \le d_t(q,z) \le Y d_s(q,z) \end{aligned}$$

for any \(t \in [-\delta ^{-1},1-\delta -r^2]\) and \(s \in [t-Y^{-1} r^2,t+Y^{-1}r^2]\).

Proof

In the proof, all constants \(C_i\) and \(c_i\) depend on \(n,A,\delta \) and D. Fix a time \(T \in [-\delta ^{-1},1-\delta -r^2]\), a point q with \(d_T(p,q) \le D\) and \(r \le D\), we set \(w(x,t)=H(x,t,q,T-r^2)\). It follows from Theorem 17 that \(w(y,T) \ge C_1 r^{-n}\) for any y with \(d_T(q,y) \le r\). For any \(y \in B_T(q,r)\), we have from Lemma 20 that

$$\begin{aligned} |\partial _t w(y,t)| \le C_2r^{-n}(R(y,t)+r^{-2}) \end{aligned}$$
(211)

for \(t \in [T-r^2/2,T+r^2]\). Since \(d_T(p,y) \le d_T(p,q)+d_T(q,y) \le 2D\), it is clear from Lemma 1 that \(F(y,T) \le c_1\). Moreover, it follows from (22) and (24) that

$$\begin{aligned} |\partial _t F(y,t)|=|(1-t)R(y,t)| \le \frac{F(y,t)}{1-t} \le c_2 F(y,t). \end{aligned}$$

Therefore, it is clear that for any \(t \in [T-r^2/2,T+r^2]\), \(F(y,t) \le c_3\) and hence \(R(y,t) \le c_4\) from (24). Since \(r \le D\), we have from (211)

$$\begin{aligned} |\partial _t w(y,t)| \le C_3r^{-n-2}. \end{aligned}$$
(212)

Now we set \(c_5=C_1(2C_3)^{-1}\), it follows from \(w(q,T) \ge C_1r^{-n}\) and (212) that \(w(y,t) \ge \frac{C_1}{2}r^{-n}\) on \(B_T(q,r) \times [T-c_5r^2,T+c_5r^2]\). On the one hand, it follows from Corollary 6 that \(w \ge C_4r^{-n}\) on \(B_t(y,r) \times \{t\}\). On the other hand, by Lemma 1, F and hence R is bounded on \(B_t(y,r) \times \{t\}\), we conclude from Theorem 23 that

$$\begin{aligned} |B_t(y,r)|_t \ge C_5r^n. \end{aligned}$$

For any point z with \(d_T(q,z)=r\), we consider a geodesic \(\gamma \) connecting q and z. We claim that for any \(t \in [T-c_5r^2,T+c_5r^2]\), \(d_t(q,z) \le C_6 r\), where \(C_6=8(C_4C_5)^{-1}\). Otherwise, we take a maximal set \(\{y_i\}_{i=1}^N \subset \gamma \) such that \(B_t(y_i,r)\) are mutually disjoint. In particular, it implies that \(\{B_t(y_i,2r)\}\) covers \(\gamma \). Then it is easy to see \(C_6 r \le 4Nr\) and hence \(N \ge \frac{C_6}{4}\). However, it follows from (57) that

$$\begin{aligned} 1 \ge \int w \,dV_t \ge \sum _{i=1}^N \int _{B_t(y_i,r)}w\,dV_t \ge \sum _{i=1}^N C_4r^{-n}|B_t(y_i,r)|_t \ge NC_4C_5 \ge 2, \end{aligned}$$

which is a contradiction. Now we set \(c_6=c_5(2C_6)^{-2}\) and claim that \(d_t(y,z) \ge (2C_6)^{-1}r\) for any \(t \in [T-c_6r^2,T+c_6r^2]\). Otherwise, we can find a time \(t \in [T-c_6r^2,T+c_6r^2]\) such that \(d_t(y,z)=(2C_6)^{-1}r\). Since \(c_6r^2 = c_5(2C_6)^{-2}r^2\), the argument before shows that \(r=d_T(q,z) \le C_6 d_t(q,z)=r/2\) and this is impossible.

Therefore, by choosing \(Y=\max \{c_6^{-1},2C_6\}\), the conclusion follows. \(\square \)

Now we prove that H has the exponential decay in the integral sense.

Theorem 19

For any Ricci shrinker \((M^n,p,g,f)\in \mathcal M_n(A)\), \(0<\delta <1\), \(D>1\) and \(\epsilon >0\), there exists a constant \(C=C(n,A,\delta ,D,\epsilon )>1\) such that

$$\begin{aligned} \int _{M \backslash B_s(x,r\sqrt{t-s})} H(x,t,y,s)\,dV_s(y) \le C\exp {\left( -\frac{(r-1)^2}{4(1+\epsilon )}\right) } \end{aligned}$$

for any point \(x \in M\), \(t \in [-\delta ^{-1},1-\delta ]\), \(d_t(p,x)+\sqrt{t-s}\le D\) and \(r \ge 1\).

Proof

It follows from Theorem (14) with \(\sigma =\epsilon \) that

$$\begin{aligned}&\left( \int _{B_s(x,\sqrt{t-s})} H(x,t,y,s)\,dV_s(y)\right) ^{\frac{1}{\epsilon }} \left( \int _{M \backslash B_s(x,r\sqrt{t-s})} H(x,t,y,s)\,dV_s(y)\right) \nonumber \\&\quad \le \exp {\left( -\frac{(r-1)^2}{4(1+\epsilon )}\right) } \end{aligned}$$
(213)

for any \(r \ge 1\). So we only need to prove the first integral to be bounded below.

Theorem 18 shows that there exists a constant \(Y=Y(n,A,\delta ,D)>1\) such that for any y with \(d_s(x,y) \le \sqrt{t-s}\), we have \(d_t(x,y) \le Y\sqrt{t-s}\). Therefore, it follows from Theorem 17 that

$$\begin{aligned} H(x,t,y,s) \ge C(t-s)^{-n/2} \end{aligned}$$

for any y with \(d_s(x,y) \le \sqrt{t-s}\).

It implies that

$$\begin{aligned} \int _{B_s(x,\sqrt{t-s})} H(x,t,y,s)\,dV_s(y) \ge C(t-s)^{-n/2}|B_s(x,\sqrt{t-s})|_s \ge C \end{aligned}$$

where we have used the fact that R is locally bounded. \(\square \)

As we have proved that all distance functions to the base point p are comparable, we prove the following weaker upper bound.

Theorem 20

For any Ricci shrinker \((M^n,p,g,f)\in \mathcal M_n(A)\), \(x \in M\) and \(s<t<1\), there exist constants \(C=C(n,A,x,t,s)>1\) and \(c=c(n,A,x,t,s)>0\) such that

$$\begin{aligned} H(x,t,y,s) \le Ce^{-cd_0^2(p,y)}. \end{aligned}$$

Proof

Fix \(s<t<1\) and x and we require that all constants in the proof depend on nxst and A. Notice that since s and t are fixed, f is comparable to \(d_0^2(p,\cdot )\) by Lemma 1.

For an \(\epsilon >0\) to be chosen later, we have from the semigroup property

$$\begin{aligned} H(x,t,y,s)&=\int H(x,t,z,l)H(z,l,y,s) \,dV_{l}(z)\\&=\int _{d_0(p,z) \ge \epsilon d_0(p,y) } H(x,t,z,l)H(z,l,y,s) \,dV_{l}(z) \\&\quad +\int _{d_0(p,z) \le \epsilon d_0(p,y)} H(x,t,z,l)H(z,l,y,s) \,dV_{l}(z)=I+II \end{aligned}$$

where \(l=\frac{s+t}{2}\).

Now from Theorem 19

$$\begin{aligned} I&=\int _{d_0(p,z) \ge \epsilon d_0(p,y) } H(x,t,z,l)H(z,l,y,s) \,dV_{l}(z) \nonumber \\&\le C_1 \int _{d_0(p,z) \ge \epsilon d_0(p,y) } H(x,t,z,l) \,dV_{l}(z) \le C_2e^{-c_1 \epsilon ^2d_0^2(p,y)}. \end{aligned}$$
(214)

Note that here we can always assume that \(\epsilon d_0(p,y)\) is large.

We choose \(\phi \) which is identical 1 on \(B_l(p,c_2\epsilon d_0(p,y))\) and supported on \(B_l(p,2c_2\epsilon d_0(p,y))\) where we choose \(c_2\) that \(B_0(p,\epsilon d_0(p,y)) \subset B_l(p,c_2\epsilon d_0(p,y))\).

If we set \(w=\frac{e^{-f}}{(4\pi \tau )^{n/2}}\), there are \(c_3\) and \(c_4\) that for any \(z \in M\)

$$\begin{aligned} c_3e^{c_4\epsilon ^2d_0^2(p,y)}w(z,l) \ge \phi (z). \end{aligned}$$
(215)

Now, we have

$$\begin{aligned} II&= \int _{B_0(p,\epsilon d_0(p,y))} H(x,t,z,l)H(z,l,y,s) \,dV_{l}(z) \le c_5\int _{B_l(p,c_2\epsilon d_0(p,y))} H(z,l,y,s)\,dV_l(z) \\&\le c_5\int H(z,l,y,s)\phi (z)\,dV_l(z) \le c_6e^{c_4\epsilon ^2d_0^2(p,y)}w(y,s), \end{aligned}$$

where \(c_6=c_5c_3\) and the last inequality follows from Lemma 8. Indeed, if we consider \(m(u,s):=\int H(z,l,u,s) \phi (z)\,dV_l(z)\), then it follows from (215) and Lemma 8 that

$$\begin{aligned} m(u,s) \le c_3e^{c_4\epsilon ^2d_0^2(p,y)}w(u,s) \end{aligned}$$
(216)

for any \(u \in M\) and \(s \le l\). In particular, (216) holds if \(u=y\).

By the definition of w,

$$\begin{aligned} w(y,s) \le c_7e^{-c_8d_0^2(p,y)}. \end{aligned}$$

Hence,

$$\begin{aligned} II \le c_9e^{-(c_8-c_4 \epsilon ^2)d_0^2(p,y)}. \end{aligned}$$
(217)

If we choose \(\epsilon =\sqrt{\frac{c_8}{2c_4}}\), it follows from (214) and (217) that

$$\begin{aligned} H(x,t,y,s) \le Ce^{-cd_0^2(p,y)}. \end{aligned}$$

\(\square \)

8 Differential Harnack inequality on Ricci shrinkers

In this subsection, we prove that Perelman’s differential Harnack inequality holds on Ricci shrinkers.

For any Ricci shrinker \((M^n,g,f)\), we fix a point \(q \in M\) and a time \(T <1\). Moreover, we set

$$\begin{aligned} w(x,t)=H(q,T,x,t)=\frac{e^{-b(x,t)}}{(4\pi (T-t))^{n/2}} \end{aligned}$$
(218)

and \(\tau =T-t\).

We first prove

Lemma 21

For any r such that \(\phi ^r=1\) on an open neighborhood of (qT),

$$\begin{aligned} \lim _{t \nearrow T} \int bw\phi ^r \,dV_t = \frac{n}{2}. \end{aligned}$$
(219)

Proof

We set \(K_r=\text {supp}\, \phi ^r \bigcap M \times [T-1,T]\). Since we only care about the integral on the compact set \(K_r\) when t is sufficiently close to T, we can assume that the distances on different time slices from t to T are uniformly comparable. Now all constants C’s in the rest of the proof depend on \(q,T,\varvec{\mu }\) and the geometry on \(K_r\). In particular, they are independent of \(\tau \).

Now we have from Theorem 19 that

$$\begin{aligned} \int _{d_t(q,x)\ge 2A\sqrt{\tau }} w(x,t)\,dV_t \le Ce^{-A^2/2}. \end{aligned}$$
(220)

Moreover, from Theorem 17,

$$\begin{aligned} b(x,t) \le C\left( 1+\frac{d_t^2(q,x)}{\tau }\right) \end{aligned}$$
(221)

for \((x,t) \in K_r\).

Now we set \(d_t=d_t(q,x)\), then for any \(A \ge 1\), we have

$$\begin{aligned} \int _{K_r \cap \{d_t\ge 2A\sqrt{\tau }\}} bw\,dV_t&\le C\int _{K_r \cap \{d_t \ge 2A\sqrt{\tau }\}}w+\tau ^{-1}d^2_tw\,dV_t \le Ce^{-A^2/2}\\&\quad +C\tau ^{-1}\int _{K_r \cap \{d_t \ge 2A\sqrt{\tau }\}}d^2_tw\,dV_t. \end{aligned}$$

Now we have

$$\begin{aligned} \int _{K_r \cap \{d_t \ge 2A\sqrt{\tau }\}}d^2_tw\,dV_t&= \sum _{k=1}^{\infty }\int _{K_r \cap \{2^kA \sqrt{\tau } \le d_t \le 2^{k+1}A\sqrt{\tau }\}}d^2_tw\,dV_t \\&\le \sum _{k=1}^{\infty }2^{2k+2}A^2\tau \int _{K_r \cap \{2^kA \sqrt{\tau } \le d_t \le 2^{k+1}A\sqrt{\tau }\}}w\,dV_t \\&\le \sum _{k=1}^{\infty }2^{2k+2}A^2e^{-2^{2k-3}A^2}\tau . \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} \int _{K_r \cap \{d_t\ge 2A\sqrt{\tau }\}} bw\phi ^r \,dV_t \le \int _{K_r \cap \{d_t\ge 2A\sqrt{\tau }\}} bw\,dV_t \le \eta (A) \end{aligned}$$
(222)

where \(\eta (A) \rightarrow 0\) if \(A \rightarrow +\infty \).

In addition, it follows from Theorem 15 that \( b(x,t) \ge \varvec{\mu }\) and hence

$$\begin{aligned} \int _{K_r \cap \{d_t\ge 2A\sqrt{\tau }\}} bw\phi ^r \,dV_t \ge \varvec{\mu }\int _{K_r \cap \{d_t\ge 2A\sqrt{\tau }\}} w \,dV_t \ge -Ce^{-A^2/2} \end{aligned}$$
(223)

where the last inequality is from (220).

The inequalities (222) and (223) indicates that the integral \(\int bw\phi ^r\,dV_t\) is concentrated on the scale \(\sqrt{\tau }\).

We take a sequence \(\tau _i \rightarrow 0\) and set \(g_i(t)=\tau _i^{-1}g(T-\tau _it)\) and \(w_i(\cdot ,t)=\tau _i^{n/2}w(\cdot ,T-\tau _it)\). Then we have

$$\begin{aligned} \partial _t w_i=\varDelta _i w_i-R_iw_i, \end{aligned}$$

where \(\varDelta _i\) and \(R_i\) are with respect to \(g_i\).

Since \(g_i\) is a blow-up sequence for the metric g and \(K_r\) has bounded geometry, it is easy to show that \((M,g_i,q)\) subconverges to \((\mathbb {R}^n,g_E,0)\) and \(w_i\) converges a positive smooth function \(w_{\infty }\) on \(\mathbb {R}^n \times (0,\infty )\) such that

$$\begin{aligned} \partial _t w_{\infty }=\varDelta _{g_E} w_{\infty }. \end{aligned}$$

Now we can show as (55) that \(w_{\infty }\) is in fact a fundamental solution of the heat equation on the Euclidean space. Moreover it is easy to see by Fatou’s inequality that

$$\begin{aligned} \int w_{\infty } \,dx \le 1 \end{aligned}$$

for any time \(t>1\). Now it follows from [22, Corollary 9.6] that \(w_{\infty }\) is the heat kernel based at 0, that is,

$$\begin{aligned} w_{\infty }(x,t)=\frac{e^{-\frac{|x|^2}{4t}}}{(4\pi t)^{n/2}}. \end{aligned}$$

From the smooth convergence,

$$\begin{aligned} \lim _{i \rightarrow \infty }\int _{K_r \cap \{d_{T-\tau _i}\le 2A\sqrt{\tau _i}\}} bw\,dV_{T-\tau _i} =\int _{|x|\le 2A} \frac{|x|^2}{4}\frac{e^{-\frac{|x|^2}{4}}}{(4\pi )^{n/2}}\,dx. \end{aligned}$$
(224)

By direct computations,

$$\begin{aligned} \int \frac{|x|^2}{4}\frac{e^{-\frac{|x|^2}{4}}}{(4\pi )^{n/2}}\,dx=\frac{n}{2}. \end{aligned}$$

Therefore, it is straightforward from (222), (223) and the fact that \(\phi ^r\) is equal to 1 on a neighborhood of (qT) that

$$\begin{aligned} \lim _{t \nearrow T} \int bw\phi ^r \,dV_t = \frac{n}{2}. \end{aligned}$$

\(\square \)

Remark 4

The same proof of Lemma 21 shows that if u is a bounded smooth function on \(M \times [T-1,T]\), then

$$\begin{aligned} \lim _{t \nearrow T} \int bwu\phi ^r \,dV_t = \frac{n}{2}u(q,T). \end{aligned}$$
(225)

Now we set \(d=d_T(q,\cdot )\), it follows from (203) that

$$\begin{aligned} H(q,T,x,t) \ge C \frac{e^{-c_1\frac{d^2}{\tau }-c_2\tau F}}{\tau ^{\frac{n}{2}}}. \end{aligned}$$
(226)

In terms of b, we have

$$\begin{aligned} b(x,t) \le c_1\frac{d^2}{\tau }+c_2\tau F(x,T)+c_3 \end{aligned}$$
(227)

We denote \(K^r_t=\{r \le F(\cdot , t) \le 2r\}\), then we have

Lemma 22

There exist constants \(C_0\) and \(C_1\) which depend only on \(\varvec{\mu },q\) and T such that

$$\begin{aligned} \int _{T-1}^T \int _{K_t^r} |b|w \,dV_t\,dt \le C_0 \end{aligned}$$
(228)

for any \(r \ge C_1\).

Proof

From Lemma 1, there exists \(C_1>0\) such that for any \(x \in K_t^r\) where \(t \in [T-1,T]\),

$$\begin{aligned} \frac{1}{5}d_t^2(p,x) \le F(x,t) \le d_t^2(p,x) \end{aligned}$$

if \(r \ge C_1\).

It follows from (227) that \(|b|\le -\varvec{\mu }+c_1\frac{d^2}{\tau }+c_2\). So we only need to estimate

$$\begin{aligned} \int _{T-1}^T \int _{K_t^r} d^2w \,dV_t\,dt. \end{aligned}$$
(229)

Now it follows from the definition of \(\phi ^r\) that \(K_t^r \subset \{c_4r \le d^2 \le c_5r\}\) if \(C_1\) is sufficiently large, therefore

$$\begin{aligned} \int _{T-1}^T \int _{K_t^r} d^2w \,dV_t\,dt&\le C\int _{T-1}^Tr \int _{K_t^r} w \,dV_t\,dt \le C\int _{T-1}^Tr e^{-\frac{c_3r}{\tau }} \,dt \le C_0. \end{aligned}$$
(230)

Note that here we have used (220). \(\square \)

Now we have the following spacetime integral estimate.

Lemma 23

$$\begin{aligned} \int _{T-1}^{T-\epsilon } \int (|\nabla b|^2+R)w\,dV_t\,dt \le C\log {\epsilon ^{-1}}, \end{aligned}$$
(231)

where C depends only on \(\varvec{\mu },n,q\) and T.

Proof

From the evolution equation

$$\begin{aligned} \partial _t w=-\varDelta w+Rw, \end{aligned}$$
(232)

we immediately have

$$\begin{aligned} \partial _t b=-\varDelta b+|\nabla b|^2-R+\frac{n}{2\tau }. \end{aligned}$$
(233)

From an elementary computation,

$$\begin{aligned}&\partial _t \int wb\phi ^r \,dV_t \nonumber \\&\quad = \int \left( b_tw\phi ^r+bw_t\phi ^r+bw\phi ^r_t-bw\phi ^rR \right) \,dV_t \nonumber \\&\quad =\int \left( -\varDelta b+|\nabla b|^2-R+\frac{n}{2\tau }\right) w\phi ^r-b\varDelta w \phi ^r+bw\phi ^r_t\,dV_t \nonumber \\&\quad =\int \langle \nabla b,\nabla (w\phi ^r) \rangle +\langle \nabla w,\nabla (b\phi ^r) \rangle +|\nabla b|^2w\phi ^r-Rw\phi ^r+bw\phi ^r_t+\frac{n}{2\tau }w\phi ^r \,dV_t \nonumber \\&\quad =\int \langle \nabla b,\nabla \phi ^r\rangle w+\langle \nabla w,\nabla \phi ^r \rangle b-(|\nabla b|^2+R)w\phi ^r+bw\phi ^r_t+\frac{n}{2\tau }w\phi ^r \,dV_t, \end{aligned}$$
(234)

where we have used \(\nabla w=-w\nabla b\).

On the one hand we have,

$$\begin{aligned} \int \langle \nabla b,\nabla \phi ^r\rangle w \,dV_t&\le \int |\nabla b||\nabla \phi ^r| w \,dV_t \nonumber \\&\le \frac{1}{4}\int |\nabla b|^2w\phi ^r \,dV_t+\int \frac{|\nabla \phi ^r|^2}{\phi ^r} w \,dV_t. \end{aligned}$$
(235)

On the other hand

$$\begin{aligned} \int \langle \nabla w,\nabla \phi ^r\rangle b \,dV_t&\le \int |\nabla w||\nabla \phi ^r| b \,dV_t =\int |\nabla b||\nabla \phi ^r| wb \,dV_t \nonumber \\&\le \frac{1}{4}\int |\nabla b|^2w\phi ^r \,dV_t+\int \frac{|\nabla \phi ^r|^2}{\phi ^r} b^2w \,dV_t. \end{aligned}$$
(236)

Now (234) becomes

$$\begin{aligned} \partial _t \int wb\phi ^r \,dV_t \le -\frac{1}{2}\int (|\nabla b|^2+R)w\phi ^r \,dV_t+X+\frac{n}{2\tau } \end{aligned}$$
(237)

where

$$\begin{aligned} X=\int bw\phi ^r _t-\frac{|\nabla \phi ^r|^2}{\phi ^r} w-\frac{|\nabla \phi ^r|^2}{\phi ^r} b^2w \,dV_t. \end{aligned}$$

Integrate (237) from \(T-1\) to \(T-\epsilon \), we have

$$\begin{aligned} \frac{1}{2}\int _{T-1}^{T-\epsilon } \int (|\nabla b|^2+R)w\phi ^{r}\,dV_t\,dt \le \left. \left( \int wb\phi ^r\,dV_t\right) \right| _{T-\epsilon }^{T-1}+\frac{n}{2}\log {\epsilon ^{-1}}+Y \end{aligned}$$
(238)

where

$$\begin{aligned} Y=\int _{T-1}^{T-\epsilon }\int bw\phi ^r _t-\frac{|\nabla \phi ^r|^2}{\phi ^r} w-\frac{|\nabla \phi ^r|^2}{\phi ^r} b^2w \,dV_t\,dt. \end{aligned}$$

At the time \(T-1\), since \(b=-\log w-\frac{n}{2}\log {4\pi }\), we have

$$\begin{aligned} \int wb\phi ^r\,dV_{T-1}=\int w\left( -\log w-\frac{n}{2}\log {4\pi }\right) \phi ^r \,dV_{T-1} \le C \end{aligned}$$
(239)

where the last inequality can be seen from Theorem 20.

Moverover,

$$\begin{aligned} \int wb\phi ^r\,dV_{T-\epsilon } \ge \varvec{\mu } \int w\phi ^r \,dV_{T-1} \ge \varvec{\mu }. \end{aligned}$$
(240)

Now it follows from Theorem 20 and Lemma 2 that

$$\begin{aligned} \lim _{r \rightarrow \infty }|Y|=0. \end{aligned}$$
(241)

So if we let \(r \rightarrow \infty \) in (238), the proof is complete. \(\square \)

From Lemma 23, we have

Lemma 24

There exist a sequence \(\tau _i \rightarrow 0\) and a constant \(C>0\) such that

$$\begin{aligned} \tau _i\int (|\nabla b|^2+R)w\,dV_{T-\tau _i} \le C. \end{aligned}$$
(242)

Proof

If the conclusion does not hold, we can find a function \(C(\tau )\) such that \(\lim _{\tau \rightarrow 0}C(\tau )=+\infty \) and

$$\begin{aligned} \int (|\nabla b|^2+R)w\,dV_{T-\tau } \ge \frac{C(\tau )}{\tau }. \end{aligned}$$
(243)

But it obviously contradicts Lemma 23 if \(\epsilon \) is sufficiently small. \(\square \)

Lemma 25

For any \(\theta >0\),

$$\begin{aligned} \int _{T-1}^{T} \int \tau ^{\theta }(|\nabla b|^2+R)w\,dV_t\,dt < \infty . \end{aligned}$$
(244)

Proof

It follows from Lemma 23 that

$$\begin{aligned}&\int _{T-1}^{T} \int \tau ^{\theta }(|\nabla b|^2+R)w\,dV_t\,dt \\&\quad =\sum _{k=0}^{\infty } \int _{T-2^{-k}}^{T-2^{-k-1}} \int \tau ^{\theta }(|\nabla b|^2+R)w\,dV_t\,dt \\&\quad \le \sum _{k=0}^{\infty } 2^{-\theta k} \int _{T-1}^{T-2^{-k-1}} \int (|\nabla b|^2+R)w\,dV_t\,dt\\&\quad \le \sum _{k=0}^{\infty } 2^{-\theta k}\log {2^{-k-1}} <\infty . \end{aligned}$$

\(\square \)

Now we fix a nonnegative function u on the time slice \(T-1\) such that \(\sqrt{u}\) smooth and compactly supported. We denote by the same u as its heat equation solution.

Then we have

Lemma 26

There exists a constant \(C>0\) such that

$$\begin{aligned} \frac{|\nabla u|^2}{u} \le C \end{aligned}$$

on \(M \times [T-1,T]\).

Proof

The conclusion follows directly from

$$\begin{aligned} \square \frac{|\nabla u|^2}{u}=-\frac{2}{u}\left| \text {Hess}\,u-\frac{du \otimes du}{u}\right| ^2 \end{aligned}$$

and Theorem 6. Note that the assumption in Theorem 6 can be checked similarly as Lemma 9\(\square \)

We also need the following lemma, whose proof is similar to Lemma 4.

Lemma 27

There exists a constant \(C>0\) such that

$$\begin{aligned} \int _{T-1}^T \int |\text {Hess}\,F|^2 w \,dV_t\,dt \le C. \end{aligned}$$
(245)

Proof

From the evolution equation \(\square |\nabla F|^2=-2|\text {Hess}\,F|^2\), we have

$$\begin{aligned} \partial _t \int |\nabla F|^2 w\phi ^r \,dV_t=\int (\varDelta |\nabla F|^2-2|\text {Hess}\,F|^2 )w\phi ^r-|\nabla F|^2 \varDelta w\phi ^r+|\nabla F|^2w \phi ^r_t \, dV_t. \end{aligned}$$

Integrate above from \(T-1\) to T, we get

$$\begin{aligned}&\int _{T-1}^T\int 2|\text {Hess}\,F|^2 w\phi ^r \,dV_t \,dt \\&\quad \le \int _{T-1}^T\int -2\langle \nabla |\nabla F|^2, \nabla \phi ^r\rangle w +|\nabla F|^2w \square \phi ^r\,dV_t \,dt-\left. \left( \int |\nabla F|^2 w\phi ^r \,dV_t\right) \right| _{T-1}^{T}\\&\quad \le \int _{T-1}^T\int |\text {Hess}\,F|^2 w\phi ^r+4|\nabla F|^2\frac{|\nabla \phi ^r|^2}{\phi ^r}w+|\nabla F|^2w \square \phi ^r\,dV_t \,dt -\left. \left( \int |\nabla F|^2 w\phi ^r \,dV_t\right) \right| _{T-1}^{T}. \end{aligned}$$

From (37) and (40), there exists a constant C independent of r such that

$$\begin{aligned} |\nabla F|^2\frac{|\nabla \phi ^r|^2}{\phi ^r}+|\nabla F|^2 |\square \phi ^r| \le C. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{T-1}^T\int |\text {Hess}\,F|^2 w\phi ^r \,dV_t \,dt \le C. \end{aligned}$$

Now the lemma follows by taking \(r \rightarrow \infty \). \(\square \)

With the same proof, we have

Lemma 28

There exists a constant \(C>0\) such that

$$\begin{aligned} \int _{T-1}^T \int |\text {Hess}\,u|^2 w \,dV_t\,dt \le C. \end{aligned}$$

As before, we set

$$\begin{aligned} v=\left( \tau (2\varDelta b-|\nabla b|^2+R)+b-n\right) w, \end{aligned}$$

and therefore

$$\begin{aligned} \partial _tv=-\varDelta v+Rv+2\tau \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2w. \end{aligned}$$
(246)

Now we prove

Lemma 29

There exist a sequence \(\tau _i \rightarrow 0\) and a constant \(C>0\) independent of r and i such that

$$\begin{aligned} \int vu\phi ^r \,dV_{T-\tau _i} \le C. \end{aligned}$$

Proof

From integration by parts, we have

$$\begin{aligned} \int vu\phi ^r \,dV_t&=\int \left( \tau (2\varDelta b-|\nabla b|^2+R)+b-n\right) wu\phi ^r\,dV_t \\&=\int -2\tau \langle \nabla b,\nabla w\rangle u\phi ^r-2\tau \langle \nabla b, \nabla u \rangle w\phi ^r-2\tau \langle \nabla b, \nabla \phi ^r \rangle wu\,dV_t \\&\quad +\int \left( \tau (R-|\nabla b|^2)+b-n\right) wu\phi ^r\,dV_t \\&=\int \left( \tau (|\nabla b|^2+R)+b-n\right) wu\phi ^r-2\tau \langle \nabla b, \nabla u \rangle w\phi ^r-2\tau \langle \nabla b, \nabla \phi ^r \rangle wu\,dV_t. \end{aligned}$$

In addition,

$$\begin{aligned}&\int -2\tau \langle \nabla b, \nabla u \rangle w\phi ^r-2\tau \langle \nabla b, \nabla \phi ^r \rangle wu\,dV_t\\&\quad \le \int 2\tau |\nabla b||\nabla u|w\phi ^r+2\tau |\nabla b||\nabla \phi ^r|wu\,dV_t \\&\quad \le \tau \int 2|\nabla b|^2wu\phi ^r+\frac{|\nabla u|^2}{u}w\phi ^r+\frac{|\nabla \phi ^r|^2}{\phi ^r}wu\,dV_t. \end{aligned}$$

Now the conclusion follows immediately from Lemmas 21, 24 and 26. \(\square \)

We are now ready to estimate the squared term in (246).

Lemma 30

$$\begin{aligned} \int _{T-1}^T \int \tau \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu \,dV_t\,dt <\infty . \end{aligned}$$

Proof

We denote \(A= 2\tau \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2w\). By computations,

$$\begin{aligned} \partial _t\int vu\phi ^r \,dV_t&= \int v_tu\phi ^r+vu_t\phi ^r+uv\phi ^r_t-Ruv\phi ^r\,dV_t \\&= \int -\varDelta vu\phi ^r+Au\phi ^r+\varDelta u v \phi ^r+uv\phi ^r_t\,dV_t \\&= \int uv\square \phi ^r-2v\langle \nabla \phi ^r,\nabla u \rangle +Au\phi ^r\,dV_t. \end{aligned}$$

Now we have

$$\begin{aligned}&\int uv\square \phi ^r\,dV_t \nonumber \\&\quad =\int \left( \tau (2\varDelta b-|\nabla b|^2+R)+b-n\right) wu\square \phi ^r\,dV_t \nonumber \\&\quad = \int -2\tau \langle \nabla b,\nabla w\rangle u\square \phi ^r-2\tau \langle \nabla b,\nabla u\rangle w\square \phi ^r-2\tau \langle \nabla b,\nabla \square \phi ^r\rangle uw \,dV_t \nonumber \\&\qquad +\int \left( \tau (R-|\nabla b|^2)+b-n\right) wu\square \phi ^r\,dV_t \nonumber \\&\quad =\int \left( \tau (|\nabla b|^2+R)+b-n\right) wu\square \phi ^r\,dV_t \nonumber \\&\qquad -2\tau \int \langle \nabla b,\nabla u\rangle w\square \phi ^r+ \langle \nabla b,\nabla \square \phi ^r\rangle uw \,dV_t. \end{aligned}$$
(247)

For the last integral,

$$\begin{aligned} 2\int \langle \nabla b,\nabla u\rangle w\square \phi ^r \,dV_t \le 2\int | \nabla b||\nabla u|w|\square \phi ^r| \,dV_t\le \int |\nabla b|^2wu|\square \phi ^r|+\frac{|\nabla u|^2}{u}w|\square \phi ^r| \,dV_t \end{aligned}$$

and

$$\begin{aligned} 2\int \langle \nabla b,\nabla \square \phi ^r\rangle uw \,dV_t \le 2\int | \nabla b||\nabla \square \phi ^r|uw \,dV_t \le \int _{K^r_t} |\nabla b|^2wu+|\nabla \square \phi ^r|^2uw \,dV_t. \end{aligned}$$

By the explicit expression \(\square \phi ^r=-nr^{-1}\eta '/2-r^{-2}\eta ''|\nabla F|^2\), we have

$$\begin{aligned} |\nabla \square \phi ^r|&=\left| -nr^{-2}\nabla F\eta ''/2-r^{-3}\eta '''\nabla F|\nabla F|^2-2r^{-2}\eta ''\text {Hess}\,F(\nabla F)\right| \\&\le Cr^{-2}|\nabla F|\left( 1+|\text {Hess}\,F|+r^{-1}|\nabla F|^2 \right) . \end{aligned}$$

In addition,

$$\begin{aligned}&\int v\langle \nabla \phi ^r,\nabla u \rangle \,dV_t \\&\quad =\int \left( \tau (2\varDelta b-|\nabla b|^2+R)+b-n\right) w\langle \nabla \phi ^r,\nabla u \rangle \,dV_t \\&\quad = \int -2\tau \langle \nabla b,\nabla w\rangle \langle \nabla \phi ^r,\nabla u \rangle -2\tau \text {Hess}\,\phi ^r(\nabla b,\nabla u)w-2\tau \text {Hess}\,u(\nabla b,\nabla \phi ^r)w \,dV_t \\&\qquad +\int \left( \tau (R-|\nabla b|^2)+b-n\right) w\langle \nabla \phi ^r,\nabla u \rangle \,dV_t \\&\quad =\int \left( \tau (|\nabla b|^2+R)+b-n\right) w\langle \nabla \phi ^r,\nabla u \rangle \,dV_t -2\tau \text {Hess}\,\phi ^r(\nabla b,\nabla u)w\\&\quad -2\tau \text {Hess}\,u(\nabla b,\nabla \phi ^r)w \,dV_t. \end{aligned}$$

To estimate the last two terms, since \(|\nabla u|\) is uniformly bounded,

$$\begin{aligned} \int \text {Hess}\,\phi ^r(\nabla b,\nabla u)w \,dV_t \le \int |\text {Hess}\,\phi ^r| | \nabla b||\nabla u| w\,dV_t \le C\int _{K^r_t} |\nabla b|^2w+ |\text {Hess}\,\phi ^r|^2w \,dV_t. \end{aligned}$$

Note that we have

$$\begin{aligned} |\text {Hess}\,\phi ^r|=\left| r^{-2}\eta '' F_iF_j+r^{-1}\eta ' \text {Hess}\,F\right| \le Cr^{-1}\left( |\text {Hess}\,F|+r^{-1}|\nabla F|^2\right) \end{aligned}$$

and

$$\begin{aligned} \int \text {Hess}\,u(\nabla b,\nabla \phi ^r)w \,dV_t \le \int |\text {Hess}\,u| | \nabla b||\nabla \phi ^r| w\,dV_t \le Cr^{-\frac{1}{2}}\int _{K^r_t} |\nabla b|^2w+ |\text {Hess}\,u|^2w \,dV_t. \end{aligned}$$

Now we integrate (247) from \(T-1\) to \(T-\tau _i\),

$$\begin{aligned}&\int _{T-1}^{T-\tau _i} \int Au\phi ^r\,dV_t \\&\quad \le \left. \left( \int vu\phi ^r \,dV_t\right) \right| _{T-1}^{T-\tau _i}+\int _{T-1}^{T-\tau _i}\int \left( \tau (|\nabla b|^2+R)+b-n\right) w(2\langle \nabla \phi ^r,\nabla u \rangle -u\square \phi ^r)\,dV_t\,dt \\&\qquad +\int _{T-1}^{T-\tau _i}\tau \int _{K^r_t} |\nabla b|^2wu|\square \phi ^r|+\frac{|\nabla u|^2}{u}w|\square \phi ^r| + |\nabla b|^2wu\,dV_t\,dt \\&\qquad +Cr^{-4}\int _{T-1}^{T-\tau _i}\tau \int _{K^r_t}|\nabla F|^2\left( 1+|\text {Hess}\,F|^2+r^{-2}|\nabla F|^4 \right) uw\,dV_t\,dt \\&\qquad +C\int _{T-1}^{T-\tau _i}\tau \int _{K^r_t} |\nabla b|^2w+|\text {Hess}\,u|^2w\,dV_t\,dt \\&\qquad +Cr^{-2}\int _{T-1}^{T-\tau _i}\tau \int _{K^r_t}\left( |\text {Hess}\,F|^2+r^{-2}|\nabla F|^4 \right) w\,dV_t\,dt. \end{aligned}$$

Therefore,

$$\begin{aligned}&\int _{T-1}^{T-\tau _i} \int Au\phi ^r\,dV_t \nonumber \\&\quad \le \left. \left( \int vu\phi ^r \,dV_t\right) \right| _{T-1}^{T-\tau _i}+Cr^{-\frac{1}{2}}\int _{T-1}^{T}\int _{K^r_t} \left( \tau (|\nabla b|^2+R)+|b|+n\right) w\,dV_t\,dt \nonumber \\&\qquad +\int _{T-1}^{T}\tau \int _{K^r_t} Cr^{-1}(|\nabla b|^2w+w) + |\nabla b|^2w\,dV_t\,dt \nonumber \\&\qquad +Cr^{-2}\int _{T-1}^{T}\tau \int _{K^r_t}\left( 1+|\text {Hess}\,F|^2 \right) uw\,dV_t\,dt \nonumber \\&\qquad +C\int _{T-1}^{T}\tau \int _{K^r_t} |\nabla b|^2w+|\text {Hess}\,u|^2w\,dV_t\,dt \end{aligned}$$
(248)

For a fixed i, from Theorem 20, Lemmas 4, 25, 26 and 27, we have by taking \(r \rightarrow \infty \) that

$$\begin{aligned} \int _{T-1}^{T-\tau _i} \int Au \,dV_t =\lim _{r \rightarrow \infty }\int _{T-1}^{T-\tau _i} \int Au\phi ^r \,dV_t =\lim _{r \rightarrow \infty }\left. \left( \int vu\phi ^r \,dV_t\right) \right| _{T-1}^{T-\tau _i} \le C, \end{aligned}$$
(249)

where the last inequality follows from Lemma 29.

Now the lemma follows from (249) by taking \(i \rightarrow \infty \). \(\square \)

A consequence of Lemma 30 is

Lemma 31

There exists a sequence \(\tau _j \rightarrow 0\) such that

$$\begin{aligned} \lim _{j \rightarrow \infty } \int \tau _j^2\left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu+\tau _j^{\frac{3}{2}}|\nabla b|^2w \,dV_t =0. \end{aligned}$$

Proof

It follows from Lemmas 30 and 25 that

$$\begin{aligned} \int _{T-1}^{T} \int \tau \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu+\tau ^{\frac{1}{2}}|\nabla b|^2w \,dV_t < \infty . \end{aligned}$$

Now the conclusion is obvious. \(\square \)

Note that the sequence \(\tau _j\) may not be the same sequence \(\tau _i\) in Lemma 24.

Finally, we can prove Perelman’s differential Harnack inequality.

Theorem 21

$$\begin{aligned} \tau (2\varDelta b-|\nabla b|^2+R)+b-n \le 0. \end{aligned}$$

Proof

As \(T-1\) can be any time \(S<T\), we just need to prove \(v \le 0\) on \(T-1\).

For the chosen \(\tau _j\) obtained in Lemma 31, we have

$$\begin{aligned}&\int vu\phi ^r \,dV_{T-\tau _j} \nonumber \\&\quad =\int \left( \tau _j(2\varDelta b-|\nabla b|^2+R)+b-n\right) wu\phi ^r\,dV_{T-\tau _j} \nonumber \\&\quad =\int \tau _j(\varDelta b+R-\frac{n}{2\tau _j})wu\phi ^r-\tau _j\langle \nabla b,\nabla u \rangle w\phi ^r \,dV_{T-\tau _j} \nonumber \\&\qquad +\int -\tau _j\langle \nabla b,\nabla \phi ^r \rangle uw+(b-\frac{n}{2})wu\phi ^r\,dV_{T-\tau _j}. \end{aligned}$$
(250)

On the one hand,

$$\begin{aligned}&\int \tau _j(\varDelta b+R-\frac{n}{2\tau _j})wu\phi ^r \,dV_{T-\tau _j} \nonumber \\&\quad \le \tau _j \left( \int \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu \,dV_{T-\tau _j} \right) ^{\frac{1}{2}}\left( \int wu \, dV_{T-\tau _j} \right) ^{\frac{1}{2}} \nonumber \\&\quad \le C\left( \int \tau _j^2\left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu \,dV_{T-\tau _j} \right) ^{\frac{1}{2}}. \end{aligned}$$
(251)

On the other hand,

$$\begin{aligned}&\int -\tau _j\langle \nabla b,\nabla u \rangle w\phi ^r-\tau _j\langle \nabla b,\nabla \phi ^r \rangle uw \, dV_{T-\tau _j} \nonumber \\&\quad \le C\tau _j\int |\nabla b|w \,dV_{T-\tau _j} \le C\left( \int \tau _j^{\frac{3}{2}}|\nabla b|^2 \,dV_{T-\tau _j} \right) ^{\frac{1}{2}}\left( \int \tau _j^{\frac{1}{2}}w \, dV_{T-\tau _j} \right) ^{\frac{1}{2}} \nonumber \\&\quad =C\tau _j^{\frac{1}{4}}\left( \int \tau _j^{\frac{3}{2}}|\nabla b|^2 \,dV_{T-\tau _j} \right) ^{\frac{1}{2}}. \end{aligned}$$
(252)

In addition, it follows from Lemma 21 and Remark 4 that

$$\begin{aligned} \int (b-\frac{n}{2})wu\phi ^r\,dV_{T-\tau _j}=0. \end{aligned}$$
(253)

Combining (251), (252) and (253), it follows immediately from Lemmas 31 and 21 that

$$\begin{aligned} \lim _{j \rightarrow \infty }\int vu\phi ^r \,dV_{T-\tau _j}=0. \end{aligned}$$

Now we consider (248), with \(\tau _i\) replaced by \(\tau _j\), and let \(j \rightarrow \infty \).

$$\begin{aligned}&\int vu\phi ^r \,dV_{T-1} \\&\quad \le Cr^{-\frac{1}{2}}\int _{T-1}^{T}\int _{K^r_t} \left( \tau (|\nabla b|^2+R)+|b|+n\right) w\,dV_t\,dt \\&\qquad +\int _{T-1}^{T}\tau \int _{K^r_t} Cr^{-1}(|\nabla b|^2w+w) + |\nabla b|^2w\,dV_t\,dt \\&\qquad +Cr^{-2}\int _{T-1}^{T}\tau \int _{K^r_t}\left( 1+|\text {Hess}\,F|^2 \right) uw\,dV_t\,dt \\&\qquad +C\int _{T-1}^{T}\tau \int _{K^r_t} |\nabla b|^2w+|\text {Hess}\,u|^2w\,dV_t\,dt. \end{aligned}$$

It is easy to see all integrals above converge to zero if \(r \rightarrow \infty \), by Lemmas 22, 26, 27 and 28. Therefore,

$$\begin{aligned} \int vu \,dV_{T-1} \le 0. \end{aligned}$$

By the arbitrary choice of u at \(T-1\), we have proved that \(v \le 0\). \(\square \)

Remark 5

Note that as in Perelman’s paper [46], Theorem 16 is a corollary of Theorem 21. Our proof of Theorem 21 is different from most literature, for instance [11, 44], in that we do not need a pointwise gradient estimate of the conjugate heat kernel, see [44, Lemma 2.2].

Remark 6

The proof of Theorem 21 shows the following identity. For any \(S<T<1\),

$$\begin{aligned} \int vu \,dV_S=-\int _{S}^T \int 2\tau \left| Rc+\text {Hess}\,b-\frac{g}{2\tau }\right| ^2wu \,dV_t\,dt. \end{aligned}$$

9 The no-local-collapsing theorems

We need to use the local entropy in [53]. Let us first recall some notations. Let \(\varOmega \) be a domain in M. Then we define (cf. (91) and (92) and Sect. 2 of [53]):

$$\begin{aligned}&\varvec{\mu }(\varOmega , g, \tau ) {:}{=}\inf \left\{ \overline{\mathcal {W}}(g,u,\tau ) \left| u \in \mathcal W_{*}^{1,2}(M), \; u \; \text {is supported on} \; \varOmega \right. \right\} , \end{aligned}$$
(254)
$$\begin{aligned}&\varvec{\nu }(\varOmega , g, \tau ) {:}{=}\inf _{s \in (0, \tau )} \varvec{\mu }(\varOmega , g, s). \end{aligned}$$
(255)

When the meaning is clear in the context, the metric g may be dropped. Note that if \(\varOmega \) does not appear, it means the default set is M. We shall exploit the argument in [53] to obtain volume ratio estimate.

Theorem 22

Suppose \((M^n, g, f)\) is a Ricci shrinker and \(B=B(x,r) \subset M\) is a geodesic ball with \(R \le \varLambda \), then we have

$$\begin{aligned} r^{-n} |B| \ge c \cdot e^{\varvec{\mu } -\varLambda r^2}, \end{aligned}$$
(256)

for some \(c=c(n)>0\). If \(r \in (0, 1)\), then (256) can be improved to

$$\begin{aligned} r^{-n} |B| \ge c \cdot e^{\varvec{\mu }(g, r^2) -\varLambda r^2}. \end{aligned}$$
(257)

Proof

We first show (256). By Theorem 3.3 of [53], we know that

$$\begin{aligned} r^{-n} |B| \ge c(n) e^{\varvec{\nu }(B, r^2)} e^{-\varLambda r^2}, \end{aligned}$$
(258)

where \(\varvec{\nu }(B,r^2)\) is the local \(\varvec{\nu }\)-functional of B on the scale \(r^2\). Since (Mg) is a Ricci shrinker, it follows from (6) in Theorem 1 that

$$\begin{aligned} \varvec{\nu }(B, r^2) \ge \varvec{\nu }(M, r^2)=\inf _{\tau \in (0,r^2)} \varvec{\mu }(M, g, \tau ) \ge \varvec{\mu }. \end{aligned}$$
(259)

If \(r \in (0, 1)\), then \(r^2 \in (0, 1)\). By the monotonicity in Theorem 1, the above inequality can be written as

$$\begin{aligned} \varvec{\nu }(B, r^2) \ge \varvec{\mu }(g, r^2). \end{aligned}$$
(260)

Therefore, we obtain (256) and (257), after we plugging (259) and (260) into (258) respectively. \(\square \)

Theorem 23

Suppose \((M^n, g, f)\) is a Ricci shrinker and \(B=B(q,r) \subset M\) is a geodesic ball with \(R \le \varLambda \), then we have

$$\begin{aligned} r^{-n}|B| \ge c \cdot e^{\varvec{\mu }}(1+\varLambda r^2)^{-\frac{n}{2}}. \end{aligned}$$
(261)

Proof

Choose \(\rho _0 \in [0,r]\) such that \(\displaystyle \inf _{s \in [0, r]}s^{-n}|B(q,s)|\) is achieved at \(\rho _0\). There are two cases \(\rho _0=0\) and \(\rho _0>0\), which we shall discuss separately.

Case 1 \(\rho _0=0\).

In this case, we have

$$\begin{aligned} |B(q,r)|\ge \omega _n r^n, \end{aligned}$$
(262)

where \(\omega _n\) is the volume of the unit Euclidean ball. Actually, it is not hard to observe that

$$\begin{aligned} \varvec{\mu } \le 0. \end{aligned}$$
(263)

Let \(\tau \rightarrow 0^{+}\), it is clear that \((M^n,p,\tau ^{-1}g)\) converges to \((\mathbb {R}^n,0,g_E)\) in the Cheeger–Gromov sense. By Lemma 3.2 of  [36], we have

$$\begin{aligned} \limsup _{\tau \rightarrow 0^+}\varvec{\mu }(g,\tau )=\limsup _{\tau \rightarrow 0^+}\varvec{\mu }(\tau ^{-1}g,1) \le \varvec{\mu }(g_E,1)=0. \end{aligned}$$
(264)

As \(\varvec{\mu }(g,\tau )\) is decreasing on (0, 1) by Lemma 15, then (263) follows from the above inequality. Consequently, (261) follows from the combination of (262) and (263).

Case 2 \(\rho _0>0\).

We choose a nonincreasing smooth function \(\eta \) on \(\mathbb {R}\) such that \(\eta =1\) on \((-\infty ,1/2]\) and 0 on \([1,\infty )\). We also define \(u(x)=\eta (\frac{d(q,x)}{\rho _0})\). From (156) in Corollary 4, we obtain

$$\begin{aligned} |B(q,\rho _0/2)|^{\frac{n-2}{n}}&\le Ce^{-\frac{2 \varvec{\mu }}{n}}\int \left\{ 4|\nabla u|^2+Ru^2 \right\} dV \\&\le Ce^{-\frac{2 \varvec{\mu }}{n}}\left( \rho _0^{-2}|B(q,r)|+\int Ru^2 \,dV \right) \\&\le Ce^{-\frac{2 \varvec{\mu }}{n}} \rho _0^{-2}(1+\varLambda r^2)|B(q,\rho _0)| \end{aligned}$$

where the last inequality follows from \(R \le \varLambda \le \varLambda r^2 \rho _0^{-2}\). According to the choice of \(\rho _0\), we obtain

$$\begin{aligned} |B(q,\rho _0/2)| \ge 2^{-n}|B(q,\rho _0)|. \end{aligned}$$

Combining the previous two steps yields that

$$\begin{aligned} |B(q,\rho _0)| \ge 2^n |B(q,\rho _0/2)| \ge Ce^{\varvec{\mu }}(1+\varLambda r^2)^{-\frac{n}{2}}\rho _0^n. \end{aligned}$$

Recall that \(r^{-n}|B(q,r)| \ge \rho _0^{-n} |B(q,\rho _0)|\) by our choice of \(\rho _0\). Therefore, (261) follows directly from the above inequality. \(\square \)

Remark 7

Theorem 23 indicates that any Ricci shrinker is \(\kappa \)-noncollapsed for some positive constant \(\kappa \) which depends only on the dimension n and the lower bound of \(\varvec{\mu }\).

Note that Theorem 22 is based on the Logarithmic Sobolev inequality, and Theorem 23 relies on the Sobolev inequality. Each of Theorems 22 and 23 has its own advantage and will be used in the remainder of the section. Bascially, Theorem 22 is sharper when r is very small and Theorem 23 is more accurate in the situation when \(\varLambda r^2\) is large.

Using the Sobolev constant estimate in Corollary 4, we can further improve Theorem 6.1 of  [41] stating that for any noncompact Ricci shrinker, the volume increases at least linearly.

Proposition 6

For any noncompact Ricci shrinker \((M^n,p,g,f)\), there exist big positive constant \(r_0=r_0(n)\) and small positive constant \(\epsilon _0=\epsilon _0(n)\) such that

$$\begin{aligned} |B(p,r)| \ge \epsilon _0 e^{\varvec{\mu }} r, \quad \forall \; r \ge r_0. \end{aligned}$$
(265)

Proof

Similar to the proof of Lemma 2, we follow the notation of [41] to denote

$$\begin{aligned}&\rho {:}{=}2\sqrt{f}, \quad D(r) {:}{=}\{x\in M \mid \rho \le r\}, \quad A(s,r) {:}{=}D(r)\backslash D(s);\\&V(r) {:}{=}|D(r)|, \quad \chi (r) {:}{=}\int _{D(r)}R \,dV. \end{aligned}$$

From Lemma 1, V(r) is almost the volume of geodesic ball B(pr), with the advantage that the estimate of V(r) is relatively easier than the estimate of |B(pr)|. Actually, by Eqs. (6.24) and (6.25) of [41], we know that

$$\begin{aligned}&V(t+1) \le 2V(t), \end{aligned}$$
(266)
$$\begin{aligned}&V(t+1)-V(t) \le C_1\frac{V(t)}{t}, \end{aligned}$$
(267)

whenever \(t \ge C_1\) for some dimensional constant \(C_1=C_1(n)\). Now we define

$$\begin{aligned} r_0 {:}{=}\max \{100n,10C_1\}. \end{aligned}$$
(268)

Therefore, in order to prove (265), it suffices to show that

$$\begin{aligned} V(r) \ge \epsilon _0 e^{\varvec{\mu }} r, \quad \forall \; r \ge r_0, \end{aligned}$$
(269)

where \(\epsilon _0=\epsilon _0(n)\) will be determined later.

We shall prove (269) by a contradiction argument. If (269) were wrong, then there exists an \(r \ge 2r_0\) such that \(V(r) \le \epsilon _0 e^{\varvec{\mu }}r\) for \(\epsilon _0\) to be determined later, we claim that

$$\begin{aligned} V(t_m)\le 2\epsilon _0e^{\varvec{\mu }}t_m, \quad t_m=r+m, \quad \forall m \in \mathbb N. \end{aligned}$$
(270)

Indeed, by our assumption the case \(m=0\) is true. We assume that the conclusion is true for all \(m=0,1,2,\ldots ,k\) and proceed to show it holds for \(m=k+1\).

For any \(t \ge r_0\), we define

$$\begin{aligned} u(x) {:}{=}{\left\{ \begin{array}{ll} 1 \quad \quad &{}\text {on} \quad A(t,t+1),\\ t+2-\rho (x) \quad \quad &{}\text {on} \quad A(t+1,t+2),\\ \rho (x)-(t-1) \quad \quad &{}\text {on} \quad A(t-1,t),\\ 0 \quad \quad &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Let \(t=t_m\) and plug the above u into the Sobolev inequality (156). We obtain

$$\begin{aligned} |A(t_m,t_{m+1})|^{\frac{n-2}{n}} \le C_3e^{-\frac{2\varvec{\mu }}{n}} \left( |A(t_{m-1},t_m)|+|A(t_{m+1},t_{m+2})|+\chi (t_{m+2})-\chi (t_{m-1}) \right) \end{aligned}$$
(271)

for some \(C_3=C_3(n)\). For \(0\le m\le k\), it follows from our induction assumption and (267) that

$$\begin{aligned} |A(t_m,t_{m+1})|=V(t_{m+1})-V(t_m) \le C_1\frac{V(t_m)}{t_m} \le 2C_1\epsilon _0e^{\varvec{\mu }}. \end{aligned}$$
(272)

Summing (271) from \(m=0\) to \(m=k\), we have

$$\begin{aligned} \sum _{m=0}^k|A(t_m,t_{m+1})|^{\frac{n-2}{n}}&\le C_3e^{-\frac{2\varvec{\mu }}{n}} \sum _{m=0}^k\left( |A(t_{m-1},t_m)|+|A(t_{m+1},t_{m+2})|+\chi (t_{m+2})-\chi (t_{m-1}) \right) \\&\le 3C_3e^{-\frac{2\varvec{\mu }}{n}} \left( |A(t_{-1},t_{k+2})|+\chi (t_{k+2})\right) . \end{aligned}$$

Recall that \(\chi (t) \le \frac{n}{2}V(t)\) by (3.4) of [9]. Plugging this fact into the above inequality yields that

$$\begin{aligned} \sum _{m=0}^k|A(t_m,t_{m+1})|^{\frac{n-2}{n}} \le 3C_3e^{-\frac{2\varvec{\mu }}{n}} \left( V(t_{k+2})+\frac{n}{2}V(t_{k+2})\right) \le C_4e^{-\frac{2\varvec{\mu }}{n}} V(t_{k+1}), \end{aligned}$$
(273)

where \(C_4=(6+3n)C_3=C_4(n)\). Now we choose

$$\begin{aligned} \epsilon _0 {:}{=}(2C_1)^{-1}(2C_4)^{-\frac{n}{2}}. \end{aligned}$$
(274)

Clearly, \(\epsilon _0=\epsilon _0(n)\). Then it follows from (272) that

$$\begin{aligned} 2C_4e^{-\frac{2\varvec{\mu }}{n}}|A(t_m,t_{m+1})| \le |A(t_m,t_{m+1})|^{\frac{n-2}{n}}, \quad \forall m \in \{ 1, 2, \ldots , k\}. \end{aligned}$$

It is clear from (273) that

$$\begin{aligned} 2C_4e^{-\frac{2\varvec{\mu }}{n}}(V(t_{k+1})-V(r)) \le C_4e^{-\frac{2\varvec{\mu }}{n}} V(t_{k+1}) \end{aligned}$$

and hence

$$\begin{aligned} V(t_{k+1}) \le 2V(r) \le 2\epsilon _0e^{\varvec{\mu }}r \le 2\epsilon _0e^{\varvec{\mu }}t_{k+1}. \end{aligned}$$

Therefore, the induction is complete and (270) is proved. By the arbitrary choice of m, the total volume of the Ricci shrinker is finite, which contradicts Lemma 6.2 of [41](See also Theorem 3.1 of [7] by Cao–Zhu). Therefore, the proof of (269) is established by this contradiction. Consequently, (265) holds by Lemma 1. Note that \(r_0\) and \(\epsilon _0\) are defined in (268) and (274). Both of them can be calculated explicitly. \(\square \)

Remark 8

In [41, Theorem 6.1], the authors have obtained a weaker lower bound

$$\begin{aligned} |B(p,r)| \ge Ce^{c \varvec{\mu }}r \end{aligned}$$

for two constants \(C>0\) and \(c>1\) depending only on n.

We are now ready to prove the improved no-local-collapsing, i.e., Theorem 2.

Proof of Theorem 2

It follows from Lemma 2 and Proposition 6 that

$$\begin{aligned} \epsilon _0 r \le |B(p,r)|e^{-\varvec{\mu }} \le C r^n. \end{aligned}$$

By Lemma 12, we know \(e^{\varvec{\mu }}|B(p,1)|^{-1}\) is uniformly bounded from above and from below. Multiplying each term of the above inequality by \(e^{\varvec{\mu }}|B(p,1)|^{-1}\) and adjusting C if necessary, we arrive at

$$\begin{aligned} \frac{1}{C}r \le \frac{|B(p,r)|}{|B(p,q)|} \le Cr^n, \end{aligned}$$

which is nothing but (9a). We proceed to prove (9b). Recall that \(q \in \partial B(p,r)\) and \(\rho \in (0, r^{-1})\) for some \(r>1\). Triangle inequality implies that

$$\begin{aligned} B(q,\rho ) \subset B(p, 2r). \end{aligned}$$

It follows from Lemma 1 that \(f \le C r^2\) for some \(C=C(n)\) on B(p, 2r). Since \(R+|\nabla f|^2=f\) and \(R \ge 0\), it follows that \(R \le Cr^2\) on B(p, 2r). In particular, we have \(R\rho ^2 \le Rr^{-2} \le C(n)\) on \(B(q,\rho )\). Consequently, we can apply Theorem 23 on the ball \(B(q,\rho )\) to obtain (9b). \(\square \)

10 The pseudolocality theorems

In this section, we prove the pseudo-locality theorems on Ricci shrinker and discuss their applications.

Based on the Harnack estimate, following a classical point-picking, or maximum principle argument, we are able to obtain the following pseudo-locality theorem.

Theorem 24

There exist positive numbers \(\epsilon _0=\epsilon _0(n)\) and \(\delta _0=\delta _0(n)\) with the following properties.

Let \(\{(M^n, g(t)), -\infty< t < 1\}\) be the Ricci flow induced from a Ricci shrinker \((M^n, p, g)\). Suppose \(t_0 \in (-\infty , 1)\) and \(B_{g(t_0)}(x,r) \subset M\) is a geodesic ball satisfying

$$\begin{aligned} \varvec{\nu } (B_{g(t_0)}(x,r), g(t_0), r^2) > -\delta _0. \end{aligned}$$
(275)

Then for each \(t \in (t_0, \min \{t_0+\epsilon _0^2 r^2, 1\})\) and \(y \in B_{g(t)}(x, 0.5r)\), we have

figure c

The statement in Theorem 24 is a slight improvement of Theorem 10.1 of [46]. The basic idea of the proof is already contained in Propositions 3.1 and 3.2 of Tian–Wang [51]. Note that the isoperimetric constant estimate in Peleman’s statement is only used to (cf. Lemma 3.5 of [53]) estimate the local entropy (i.e., (254) and (255)) \(\varvec{\nu } (B_{g(t_0)}(x,r), g(t_0), r^2)\). The statement (275) seems to be more straightforward. The conclusion (276) follows from a standard point-picking argument, whenever the differential Harnack estimate, i.e., Theorem 21 holds. More details can be found in [32, Sect. 30], [11, Sect. 8],  [19, Chapter 21], or [54].

As Ricci shrinker Ricci flows are self-similar, we can improve the estimate (276) by the following property.

Theorem 25

Suppose \((M^n, p, g, f)\) is a Ricci shrinker, \(B=B(q,r) \subset M\) is a geodesic ball satisfying

$$\begin{aligned} \varvec{\nu } (B, g, r^2) > -\delta _0. \end{aligned}$$
(275)

Then we have

$$\begin{aligned} \sup _{x \in B(q, 0.5\epsilon _0 r)} |Rm|(x) \le \max \{ 1, \epsilon _0 D r\} \cdot (\epsilon _0 r)^{-2}, \end{aligned}$$
(276)

where \(D=d(p,q)+\sqrt{2n}\).

Proof of Theorem 25

We fix \(\xi \le \epsilon _0\) a small positive number, whose value will be determined later (i.e., (281)). We set

$$\begin{aligned} t {:}{=}-(\xi r)^2, \quad \tilde{q}=(\psi ^{t})^{-1}(q), \quad D {:}{=}d(p,q)+\sqrt{2n}, \end{aligned}$$
(277)

where \(\psi ^{s}\) is the diffeormorphism (i.e., (15)) generated by \(\frac{\nabla f}{1-s}\).

Claim

By choosing \(\xi \) properly, we have

$$\begin{aligned} d(q, \tilde{q}) \le \frac{ \epsilon _0 r}{2}. \end{aligned}$$
(278)

By (15) and (2), along the flow line \(\psi ^s(\tilde{q})\) where s goes from t to 0, we compute

$$\begin{aligned} d(q,\tilde{q}) \le \int _t^0 \frac{|\nabla f|(\psi ^s(\tilde{q}))}{1-s}\,ds \le \int _t^0 \frac{\sqrt{f(\psi ^s(\tilde{q}))}}{1-s}\,ds. \end{aligned}$$
(279)

From the definition of \(\psi ^s\), we have

$$\begin{aligned} \frac{d}{ds} f(\psi ^s(\tilde{q}))=\frac{|\nabla f|^2(\psi ^s(\tilde{q}))}{1-s} \le \frac{f(\psi ^s(\tilde{q}))}{1-s}. \end{aligned}$$

For each \(s \ge t=-(\xi r)^2\), the integration of the above inequality yields that

$$\begin{aligned} f(\psi ^s(\tilde{q})) \le \frac{1-t}{1-s}f(\psi ^t(\tilde{q}))=\frac{1-t}{1-s}f(q) \le \frac{1-t}{1-s} \cdot \frac{D^2}{4}, \end{aligned}$$

where we applied (30) in the last step. Therefore, it follows from (279) that

$$\begin{aligned} d(q,\tilde{q})&\le D\sqrt{1-t}\int _t^0 \frac{1}{2}(1-s)^{-3/2}\,ds=D \left( \sqrt{1-t}-1\right) . \end{aligned}$$

Plugging the fact that \(t=-(\xi r)^2\) into the above inequality, we arrive at

$$\begin{aligned} d(q,\tilde{q}) \le D \left( \sqrt{1+(\xi r)^2}-1\right) . \end{aligned}$$
(280)

Now we define \(\xi \) as follows.

$$\begin{aligned} \xi {:}{=}{\left\{ \begin{array}{ll} \epsilon _0, &{} \text {if} \; Dr \le \epsilon _0^{-1};\\ \sqrt{\frac{\epsilon _0}{Dr}} &{} \text {if} \; Dr > \epsilon _0^{-1}. \end{array}\right. } \end{aligned}$$
(281)

Therefore, if \(Dr \le \epsilon _0^{-1}\), it follows from (280) that

$$\begin{aligned} d(q,\tilde{q})\le D \left( \sqrt{1+(\epsilon _0 r)^2}-1\right) \le \frac{D(\epsilon _0r)^2}{2}\le \frac{\epsilon _0 r}{2}. \end{aligned}$$

If \(Dr >\epsilon _0^{-1}\), it also follows from (280) that

$$\begin{aligned} d(q,\tilde{q})\le D \left( \sqrt{1+(\xi r)^2}-1\right) \le \frac{D}{2} \cdot (\xi r)^2=\frac{\epsilon _0 r}{2}. \end{aligned}$$

Therefore, no matter what the value of r is, we always have (278). The proof of the Claim is complete.

We proceed to prove (276). Since \(g(t)=(1-t)(\psi ^t)^*g\), it is clear that

$$\begin{aligned} \psi ^t\left( B_t \left( \tilde{q},\sqrt{1-t}\,r \right) \right) =B(q,r). \end{aligned}$$

It follows from the scaling property of \(\varvec{\nu }\) that

$$\begin{aligned} \varvec{\nu } \left( B_{g(t)} \left( \tilde{q},\sqrt{1-t}\,r \right) , g(t), (1-t)r^2 \right) =\varvec{\nu } (B, g, r^2)>-\delta _0. \end{aligned}$$

Therefore, we can apply Theorem 24. For each \(s \in (t, \min \{t+(\epsilon _0r)^2, 1\}]\) and \(x \in B_{g(s)}(\tilde{q}, 0.5 r)\), we have

figure d

In particular, we can choose \(s=0\). Since \(g=g(0)\), for each \(x \in B(\tilde{q}, 0.5r)\), we obtain

figure e

Note that \(B(q, 0.5 \epsilon _0 r) \in B(\tilde{q}, \epsilon _0 r) \subset B(\tilde{q}, 0.5r)\) by (278). Plugging (281) into (285a), we obtain (276). \(\square \)

Now we apply Theorems 24 and 25 to study the geometric properties of (Mg) in terms of \(\varvec{\mu }\). In particular, we are ready to finish the proof of Theorem 3.

Proof of Theorem 3

We divide the proof into several steps.

Step 1 The gap property (10) holds.

It suffices to show that \(\varvec{\mu } \ge -\delta _0\) implies that (Mg) is isometric to the Euclidean space.

Following directly from its definition, as \(B(x,r) \subset M\), it is clear that

$$\begin{aligned} \varvec{\nu } (B(x,r), g, r^2) \ge \varvec{\nu } (M, g, r^2) =\varvec{\nu } (g, r^2). \end{aligned}$$

Combining the above inequality with the optimal Logarithmic Sobolev inequality, we obtain

$$\begin{aligned} \varvec{\nu } (B(x,r), g, r^2) \ge \varvec{\mu }. \end{aligned}$$
(286)

Therefore, if \(\varvec{\mu } \ge -\delta _0\), then each ball B(xr) will satisfy the condition (275). By choosing \(r>>D\), we can apply (276) to obtain that

$$\begin{aligned} |Rm|(x) \le \epsilon _0 D r \cdot (\epsilon _0 r)^{-2}=D \epsilon _0^{-1} r^{-1}. \end{aligned}$$

Let \(r \rightarrow \infty \), we obtain that \(|Rm|(x) \equiv 0\). By the arbitrary choice of x, we obtain that \(|Rm| \equiv 0\). In particular, \(Rc \equiv 0\). Then the Ricci shrinker equation implies that \(f_{ij}=\frac{g_{ij}}{2}\). Therefore, (Mg) is isometric to a metric cone which is also a smooth manifold. This forces that (Mg) is isometric to the standard Euclidean space \((\mathbb {R}^n, g_{E})\). Thus, the proof of (10) is complete.

Step 2 The inequality (12) and (13) imply the curvature and injectivity radius bound (14).

Recall that (10) means \(\varvec{\mu }(g, 1)<-\delta _0\). If (12) holds, by continuity and monotonicity of \(\varvec{\mu }(g, \tau )\), it is clear that there exists some \(\tau \in (0,1)\) such that

$$\begin{aligned} \varvec{\mu }(g,\tau )=-\delta _0. \end{aligned}$$

Then the \(\tau _0\) in (14) is well defined. Namely, \(\tau _0\) is the largest \(\tau \in (0, 1)\) such that the above equality holds. It follows from the definition of \(\tau _0\) and \(\varvec{\nu }\) that

$$\begin{aligned} \varvec{\nu }(g, \tau _0) = \varvec{\mu }(g, \tau _0)=-\delta _0. \end{aligned}$$
(287)

For each ball \(B_{g(0)}(x,r) \subset M\), we know \( \varvec{\nu }(B_{g(0)}(x,r), g, \tau _0) \ge -\delta _0\). In particular, we can choose \(r=\sqrt{\tau _0}\). Now we apply Theorem 24 on the time slice \(t_0=0\), with scale \(\sqrt{\tau _0}\), to obtain that

$$\begin{aligned} |Rm|(x,t) \le t^{-1}, \quad \forall \; x \in M, \quad \forall \; t \in (0, \epsilon _0^2 r^2]. \end{aligned}$$

In particular, we have

$$\begin{aligned} \sup _{x \in M} |Rm|(x, \epsilon _0^2 \tau _0) \le \epsilon _0^{-2} \tau _0^{-1}. \end{aligned}$$

Up to rescaling, since \(g(0)=g\), we arrive atl

$$\begin{aligned} \sup _{x \in M} |Rm|_{g}(x) \le \epsilon _0^{-2} \tau _0^{-1} (1-\epsilon _0^2 \tau _0)=\epsilon _0^{-2} \tau _0^{-1}-1< C(n) \tau _0^{-1}, \end{aligned}$$

which is nothing but (14a). Plugging (287) into (257) of Theorem 22, we obtain that each geodesic ball \(B(\cdot , \sqrt{\tau _0})\) has volume bounded below by \(c(n) \tau _0^{\frac{n}{2}}\). Therefore, the injectivity radius estimate of Cheeger–Gromov–Taylor [13] applies and we arrive at (14b). The proof of (14) is complete.

Step 3 The bounded geometry estimate (14) implies the equality (11), i.e., \(\displaystyle \lim _{\tau \rightarrow 0^+}\varvec{\mu }(g,\tau ) =0\).

We shall argue in the way similar to that in Theorem 1.1 of [62], with more details on the regularity estimate.

Assume otherwise that there exists a sequence \(\tau _i \rightarrow 0^+\) such that

$$\begin{aligned} \lim _{i \rightarrow \infty }\varvec{\mu }(g,\tau _i)=\varvec{\mu }_{\infty }<0. \end{aligned}$$
(288)

If we set \(g_i=\tau _i^{-1}g\), then all metrics \(g_i\) have uniformly bounded geometry. More precisely, there exist positive constants K and \(v_0\) such that

figure f

Notice that for any i, there exists a large domain

$$\begin{aligned} B_i {:}{=}\left\{ x \left| 2\sqrt{f} \le r_i \right. \right\} \end{aligned}$$
(290)

for some large \(r_i>>1\) such that

$$\begin{aligned} \varvec{\mu }(B_i, g_i,1)-\varvec{\mu }(g_i,1)=\varvec{\mu }(B_i, g_i,1)-\varvec{\mu }(g,\tau _i)<i^{-1}. \end{aligned}$$
(291)

The geometry bound (289) actually implies higher order derivatives of curvatures and \(\sqrt{f}\) are also uniformly bounded (cf. Sect. 4 of [34]). Therefore, it is not hard to see that \(\partial B_i\) is smooth. All the covariant derivatives of second fundamental forms of \(\partial B_i\) are bounded independent of i.

It follows from [48] that a minimizer \(u_i\) of \(\varvec{\mu }(B_i, g_i,1)\) exists. More precisely, \(u_i \in W_0^{1,2}(B_i)\) is a positive smooth function on \(B_i\) satisfying the normalization condition

$$\begin{aligned} \int _{B_i} u_i^2\,dV_i=1 \end{aligned}$$
(292)

and solve the Dirichlet problem

figure g

Here \(dV_i\), \(\varDelta _i\) and \(R_i\) denote the volume form, Laplacian operator and scalar curvature with respect to \(g_i\) respectively. The number \(\lambda _i\) is defined by

$$\begin{aligned} \lambda _i {:}{=}n+\frac{n}{2}\log (4\pi )+\varvec{\mu }(B_i, g_i,1). \end{aligned}$$

Recall that \(\displaystyle \lim _{\tau \rightarrow 0^{+}} \varvec{\mu }(g,\tau ) \le 0\) by (264). Then it follows from (291) that \(\lambda _i\) is uniformly bounded. Since curvature is uniformly bounded, the classical \(L^2\)-Sobolev constant of \((B_i, g_i)\) is uniformly bounded. In light of (293), the Moser iteration then implies \({ \left\| u_i \right\| }_{C^0}\) is uniformly bounded, see [62, Lemma 2.1(a)] or the proof of Proposition 3.1 of [51]. Then it follows from [23, Corollary 8.36] that \(\Vert u_i\Vert _{C^{1,\frac{1}{2}}(\bar{B}_i)}\) are uniformly bounded. Since all \(\partial B_i\) have uniformly higher regularities, the bootstrapping, see [23, Theorem 6.19], shows that \(\Vert u_i\Vert _{C^{k,\frac{1}{2}}(\bar{B}_i)}\) are uniformly bounded for any \(k \ge 2\).

Let \(q_i\) be a point where \(u_i\) achieves maximum value in \(B_i\). By (293), at \(q_i\) we have

$$\begin{aligned} R_iu_i-2u_i\log u_i-\lambda _i u_i \le 0, \end{aligned}$$

whence we derive

$$\begin{aligned} u_i(q_i) \ge \exp \left( \frac{R_i-\lambda _i}{2} \right) \ge c_0 \end{aligned}$$
(294)

for some uniform constant \(c_0\).

In light of (289) and the discussion below (291), we know that \((M^n, q_i, g_i)\) subconverges to Euclidean space \((\mathbb {R}^n, 0, g_{E})\) in \(C^{\infty }\)-Cheeger–Gromov topology. The set \(B_i\) converges to a limit set \(B_{\infty }\). If \(d(q_i, \partial B_i) \rightarrow \infty \), then \(B_{\infty }=\mathbb {R}^n\). Otherwise, by the estimate of second fundamental form and its covariant derivatives, \(\partial B_i\) converge to a smooth \((n-1)\)-dimensional set \(\partial B_{\infty }\). In light of the uniform bound of \({ \left\| u_i \right\| }_{C^{k,\frac{1}{2}}}\) and the uniform regularity of \(\partial B_i\), by taking subsequence if necessary, we can assume that \(u_i\) converges in smooth topology to a smooth function \(u_{\infty } \in C^{\infty }(\bar{B}_{\infty })\). Furthermore, \(u_{\infty } \equiv 0\) on \(\partial B_{\infty }\).

In view of (294), the convergence process implies that

$$\begin{aligned} 0<c^2 = \int _{B_{\infty }} u_{\infty }^2 dV_{\infty } \le 1. \end{aligned}$$
(295)

Furthermore, we have on \(B_{\infty }\) that

$$\begin{aligned} -4\varDelta _{g_E} u_{\infty }-2u_{\infty }\log u_{\infty }-\lambda _{\infty } u_{\infty } =0, \end{aligned}$$
(296)

where \(\lambda _{\infty }=n+\frac{n}{2}\log (4\pi )+\varvec{\mu }_{\infty }\). Let \(\tilde{u}=c^{-1}u_{\infty }\). Then \(\int _{B_{\infty }} \tilde{u}^2 dV_{\infty }=1\). The above equation becomes

$$\begin{aligned} -4\varDelta _{g_E} \tilde{u}-2\tilde{u} \log \tilde{u} -\left( n+\frac{n}{2}\log (4\pi )+\varvec{\mu }_{\infty }+2\log c\right) \tilde{u}=0. \end{aligned}$$

Since \(c \in (0, 1)\) by (295) and \(\varvec{\mu }_{\infty }<0\) by (288), then an integration by parts shows that

$$\begin{aligned} \varvec{\mu }(g_E,1) \le \overline{\mathcal {W}}(g_{E}, \tilde{u}, 1)=\varvec{\mu }_{\infty }+2\log c<0, \end{aligned}$$

which is a contradiction. So we finish the proof of Step 3.

Step 4 The three properties are equivalent.

By Step 2, it is clear that \((c) \Rightarrow (a)\). Then Step 3 means that \((a) \Rightarrow (b)\). It is obvious that \((b) \Rightarrow (c)\). Therefore, we obtained the equivalence of properties (a), (b) and (c) in Theorem 3. The proof of the Theorem is complete. \(\square \)

Corollary 7

There exists a small positive number \(\epsilon =\epsilon (n)>0\) such that for any nonflat Ricci shrinker \((M^n, p, g, f)\), we have

$$\begin{aligned} d_{PGH} \left\{ (M^n,p,g), (\mathbb {R}^n, 0,g_{E})\right\} >\epsilon . \end{aligned}$$
(297)

Proof

We argue by contradiction.

If (297) were wrong, then we can have a sequence of nonflat Ricci shrinkers \((M_i, p_i, g_i)\) such that

$$\begin{aligned} d_{PGH} \left( (M_i, p_i, g_i), \left( \mathbb {R}^n,0, g_{E} \right) \right) \rightarrow 0. \end{aligned}$$

By Proposition 5.8 of [34], it is clear that \(\varvec{\mu }_i = \varvec{\mu } (M_i, p_i, g_i)\) is uniformly bounded from below. Using Theorem 1.1 of [34], the above convergence can be improved to be in the \(C^{\infty }\)-Cheeger–Gromov sense

$$\begin{aligned} (M_i, p_i, g_i) \longrightarrow \left( \mathbb {R}^n, 0, g_{E}\right) . \end{aligned}$$

It is not hard to see that \(\varvec{\mu }\) is continuous with respect to the above convergence (cf. Theorem 1.2(c) of [34]). Therefore, we have

$$\begin{aligned} \varvec{\mu }_i = \varvec{\mu } (M_i, p_i, g_i) \rightarrow \varvec{\mu } (\mathbb {R}^n, 0, g_{E}) =0. \end{aligned}$$

It follows that \(\varvec{\mu }_i >-\delta _0\) for large i. Therefore, each \((M_i, g_i)\) is isometric to Euclidean space by Theorem 3. This contradicts our choice of \((M_i, g_i)\). The proof of (297) is established by this contradiction. \(\square \)

Corollary 8

Let \((M^n,g,f)\) be a Ricci shrinker and let \(q \in M\) be a point such that

$$\begin{aligned} \nu (B(q,\epsilon _0^{-1}),g,\epsilon _0^{-2})>-\delta _0. \end{aligned}$$

Then there exist a positive constant \(C=C(n)\) such that

$$\begin{aligned} |Rm|(\psi ^t(x)) \le CD(1-t)\le CD\frac{f(x)}{f(\psi ^t(x))} \end{aligned}$$

for any \(x \in B(q,\frac{1}{2}e^{-CD}D^{-\frac{1}{2}})\) and \(t \in [0,1)\), where \(D=d(p,q)+\sqrt{2n}\).

Proof

By the assumption, it follows from Theorem 24 by choosing \(r=\epsilon _0^{-1}\) that

$$\begin{aligned} |Rm|(x,t) \le \frac{1}{t} \end{aligned}$$
(298)

for any \(t \in (0,1)\) and \(d_{g(t)}(q,x) \le \frac{1}{2}\epsilon _0^{-1}\). In addition, from Theorem 25 we have

$$\begin{aligned} |Rm|(x) \le D \end{aligned}$$
(299)

for any \(x \in B(q,\frac{1}{2})\). From (298), (299) and [15, Theorem 3.1] that there exist a positive constant \(C=C(n)\) such that for any \(x \in B_t(q,\frac{1}{2}D^{-\frac{1}{2}})\),

$$\begin{aligned} |Rm|(x,t) \le CD. \end{aligned}$$
(300)

From (300), it is easy to see by comparing the distances that

$$\begin{aligned} B \left( q, \frac{1}{2}e^{-CD}D^{-\frac{1}{2}} \right) \subset B_t \left( q,\frac{1}{2}D^{-\frac{1}{2}} \right) \end{aligned}$$
(301)

for any \(t \in [0,1)\).

Therefore, for any \(x \in B(q,\frac{1}{2}e^{-CD}D^{-\frac{1}{2}})\),

$$\begin{aligned} |Rm|(\psi ^t(x))=(1-t)|Rm|(x,t)\le CD(1-t). \end{aligned}$$
(302)

Along the flow line of \(\psi ^t(x)\),

$$\begin{aligned} \frac{d}{dt}f(\psi ^t(x))=\frac{|\nabla f|^2(\psi ^t(x))}{1-t}\le \frac{f(\psi ^t(x))}{1-t}, \end{aligned}$$
(303)

and hence by solving the corresponding ODE,

$$\begin{aligned} f(\psi ^t(x)) \le \frac{f(x)}{1-t}. \end{aligned}$$
(304)

Combining (302) and (304), the conclusion follows. \(\square \)

Since f is almost \(\frac{d^2}{4}\) by Lemma 1, Corollary 8 shows that the curvature is quadratically decaying along the flow line. Next we prove that if there exists a tubular neighborhoold of some level set of f whose isoperimetric constant is almost Euclidean, then globally the curvature is quadratically decaying.

Corollary 9

For any Ricc shrinker \((M^n,g,f)\), if there exists an \(a>0\) such that for any \(x \in f^{-1}(a)\),

$$\begin{aligned} \nu (B(x,\epsilon _0^{-1}),g,\epsilon _0^{-2})>-\delta _0, \end{aligned}$$

then the curvature is quadratically decaying and each end has a unique smooth tangent cone at infinity.

Proof

We can assume that (Mf) is nonflat, otherwise there is nothing to prove. Now we reparametrize \(\psi ^t\) by defining for any \(s \in (-\infty ,\infty )\)

$$\begin{aligned} \tilde{\psi }^s=\psi ^{1-e^{-s}}. \end{aligned}$$

It is clear from the definition of \(\psi ^t\) that

$$\begin{aligned} \frac{d}{ds}\tilde{\psi }^s(x)=\nabla f(\tilde{\psi }^s(x)). \end{aligned}$$

In other words, \(\tilde{\psi }^s\) is the one-parameter group of diffeomorphisms generated by \(\nabla f\). Now we set

$$\begin{aligned} \epsilon _1=\epsilon _1(a,n)=\frac{1}{2}e^{-CD_1}D_1^{-\frac{1}{2}}, \end{aligned}$$

where \(D_1=2\sqrt{a}+5n+\sqrt{2n}+4\).

We claim that any \(x \in T_{\epsilon _1}(f^{-1}(a)){:}{=}\bigcup _{q\in f^{-1}(a)}B(q,\epsilon _1)\) is not a stationary point of \(\tilde{\psi }^s\). Otherwise, it follows from Corollary 8 that

$$\begin{aligned} |Rm|(x)=|Rm|(\tilde{\psi }^s(x))\le CD_1e^{-s} \end{aligned}$$

for any \(s \ge 0\). However, when \(s \rightarrow \infty \), \(|Rm|(x)=0\) and this contradicts our nonflatness assumption.

Now we choose \(c<a<d\) such that for any \(x \in \partial T_{\frac{\epsilon _1}{2}}(f^{-1}(a))\), either \(f(x) \le c\) or \(f(x) \ge d\). By continuity, there exists a positive constant \(\epsilon \ll \epsilon _1\) such that for any \(x \in T_{\epsilon }(f^{-1}(a))\), \(f(x) \in (c+\epsilon ,d-\epsilon )\). We set \(U {:}{=}T_{\epsilon }(f^{-1}(a))\) and claim that for any \(y \in U\), there exists an \(x \in f^{-1}(a)\) such that \(\tilde{\psi }^{s}(x)=y\) for some s. If \(f(y)=a\), then the claim is obvious. If \(f(y) <a\), we consider the flow line \(\tilde{\psi }^{s}(y)\) for \(s \ge 0\). Notice that by the definition of \(\tilde{\psi }^{s}\),

$$\begin{aligned} \frac{d}{ds}f(\tilde{\psi }^{s}(y))=|\nabla f|^2(\tilde{\psi }^{s}) \ge 0. \end{aligned}$$

Therefore, by the local compactness and our previous no stationary argument, the flow will continue and along the flow f is strictly increasing as long as \(\tilde{\psi }^{s}(y)\) stays in \(T_{\epsilon _1}(f^{-1}(a))\). We set \(s_0\) to be the first time such that \(\tilde{\psi }^{s}(y)\) reaches \(\partial T_{\frac{\epsilon _1}{2}}(f^{-1}(a))\). In particular, \(f(\tilde{\psi }^{s}(y)) \le c\) or \(f(\tilde{\psi }^{s}(y)) \ge d\). Since \(f(y) \in (c+\epsilon ,d-\epsilon )\), it must be \(f(\tilde{\psi }^{s}(y)) \ge d\). As \(f(y)<a<f(\tilde{\psi }^{s_0}(y))\), there exists an \(s \in (0,s_0)\) such that \(f(\tilde{\psi }^{s}(y))=a\) by continuity. Therefore, if we set \(x=\tilde{\psi }^{s}(y) \in f^{-1}(a)\), then \(\tilde{\psi }^{-s}(x)=y\) and the claim follows. Similarly, for the case \(f(y)>a\), the claim is also true.

Next we prove that for any y such that \(f(y)>a\), there exists an \(x \in U\) such that \(\tilde{\psi }^{s}(x)=y\) for some s. Fix such y, we choose any curve \(\{\gamma (z):\,z\in [0,1]\}\) such that \(\gamma (0)=p\) and \(\gamma (1)=y\). In particular, since p is the minimum point of f, there exists a \(z_0 \in [0,1)\) such that \(\gamma (z_0) \in f^{-1}(a)\) and for all \(z \in (z_0,1]\), \(f(\gamma (z))>a\). Now we define \(I \subset [z_0,1]\) such that \(z \in I\) if and only if there exists an \(x \in U\) such that \(\tilde{\psi }^{s}(x)=\gamma (z)\) for some s. In particular, I is not empty as \(z_0 \in I\). It is clear that I is open, since U is open. Now we prove the closedness of I. For a sequence \(z_i \in I\) such that \(z_i \rightarrow z_{\infty } \in [z_0,1]\), \(f(z_i)>a\) if i is sufficiently large. By our definition of I and the claim with its proof, there exists \(x_i \in f^{-1}(a)\) and \(s_i >0\) such that \(\tilde{\psi }^{s_i}(x_i)=\gamma (z_i)\). Note that \(s_i\) must be bounded. Indeed, by Corollary 8,

$$\begin{aligned} |Rm|(\gamma (z_i))=|Rm|(\tilde{\psi }^{s_i}(x_i))\le CD_1e^{-s_i}. \end{aligned}$$

If \(s_i \rightarrow \infty \), then it forces \(|Rm|(\gamma (z_{\infty }))=0\) and this is a contradiction. By compactness and taking the subsequence, there exist \(x_{\infty } \in f^{-1}(a)\) and \(s_{\infty } \ge 0\) such that \(x_i \rightarrow x_{\infty }\) and \(s_i \rightarrow s_{\infty }\). By continuity, \(\tilde{\psi }^{s_{\infty }}(x_i)=\gamma (z_{\infty })\). To summarize, \(I=[z_0,1]\) and in particular, \(\tilde{\psi }^{s}(x)=\gamma (1)=y\) for some \(x \in U\) and \(s \in \mathbb {R}\). By the claim again, we have proved that for any y with \(f(y) \ge a\), there exists an \(x \in f^{-1}(a)\) such that \(\psi ^s(x)=y\) for some \(s \ge 0\).

Therefore, for any point y outside the compact set \(\{f \le a\}\), it follows from Corollary 8 that

$$\begin{aligned} |Rm|(y) \le \frac{CD_1a}{f(y)} \le C\frac{ \max \{1,a^{\frac{3}{2}}\}}{f(y)}. \end{aligned}$$
(305)

See Figure 4 for intuition in the case \(a>1\).

Fig. 4
figure 4

The quadratic decay of curvature

In other words, the curvature is quadratically decaying. Since a Ricci shrinker can be regarded as an ancient Ricci flow, it follows from Shi’s local estimates [50] that

$$\begin{aligned} |\nabla ^kRm|(y) \le \frac{C_k}{d^{k+2}(p,y)} \end{aligned}$$

for all \(k=1,2,\ldots \). It follows immediately that any tangent cone at infinity must be smooth. Finally, the uniqueness follows from [17, Theorem 2], see also [33, Lemma A.3]. \(\square \)

Remark 9

The proof of Corollary 9 shows that the manifold \(\{x\in M \mid \,f(x) \ge a\}\) is diffeomorphic to \(f^{-1}(a) \times [0,1)\).

11 Strong maximum principle for curvature operator

The purpose of this section is to prove Theorem 4. We remind the readers that all constants C’s in this section depend only on the dimension n.

We first show an \(L^2\)-integral estimate of Riemannian curvature.

Theorem 26

Suppose \((M^n,p,g,f)\) is a Ricci shrinker satisfying \(\varvec{\mu } \ge -A\), and \(\lambda \) is a positive number. Then we have

$$\begin{aligned} \int |Rm|^2e^{-\lambda f}\,dV \le I \end{aligned}$$
(306)

for some \(I=I(n,A,\lambda )<\infty \).

Theorem 26 is the consequence of the improved no-local-collapsing theorem (i.e., Theorem 2), the local conformal transformation technique (cf. Sect. 3 of [34]), and the curvature estimate of Jiang–Naber (i.e.,  [31]).

Lemma 32

For any Ricci shrinker \((M^n,p,g,f)\) and any constant \(D >100n\), we have

$$\begin{aligned} \int _{A(D,2D)} |Rc|^2e^{-f}\,dV \le Ce^{\varvec{\mu }}D^{n+2}e^{-D^2/5} \end{aligned}$$
(307)

where A(D, 2D) is the annulus \(B(p,2D) \backslash B(p, D)\).

Proof

Fix a cutoff function \(\psi \) on \(\mathbb {R}\) such that \(\psi =1\) on [1, 2] and \(\psi =0\) outside \([\frac{1}{2},3]\). By defining \(\eta (x)=\psi (\frac{d(p,x)}{D})\), we compute

$$\begin{aligned} \int \eta ^2|Rc|^2e^{-f} \,dV&=\int \eta ^2\langle \frac{g}{2}-\text {Hess}\,f,Rc \rangle e^{-f} \,dV \\&= \int \left( \frac{1}{2}\eta ^2 R+2\eta Rc(\nabla \eta ,\nabla f) \right) e^{-f} \,dV \\&\le \int \left( \frac{1}{2}\eta ^2 R+\frac{1}{2}\eta ^2|Rc|^2+2|\nabla \eta |^2|\nabla f|^2 \right) e^{-f} \,dV \end{aligned}$$

where for the second line we have used \(\text {div}(Rc\,e^{-f})=0\). Consequently, by Lemmas 1 and 2, we have

$$\begin{aligned} \int \eta ^2|Rc|^2e^{-f} \,dV&\le \int \left( \eta ^2 R+4|\nabla \eta |^2|\nabla f|^2\right) e^{-f} \,dV \le C\int _{A(D/2,3D)} fe^{-f}\,dV. \end{aligned}$$

Plugging the estimates in Lemmas 1 and 2 into the above inequality, we arrive at (307). \(\square \)

In the proof of Lemma 32, if we choose \(\psi \) such that \(\psi =1\) on \((-\infty ,1]\) and \(\psi =0\) on \([2,\infty )\), then a similar argument shows the following Lemma.

Lemma 33

For any Ricci shrinker \((M^n,p,g,f)\), we have

$$\begin{aligned} \int |Rc|^2e^{-f} \,dV \le Ce^{\varvec{\mu }}. \end{aligned}$$
(308)

The details of the proof of Lemma 33 is almost identical to that of Lemma 32. So we leave it to interested readers. Note that Lemma 33 provides an explicit upper bound of [40, Theorem 1.1]. Starting from Lemmas 32 and 33, we are ready to prove Theorem 26.

Proof of Theorem 26:

We only prove the case when \(\lambda =1\). The general case is similar and is left to interested readers.

For any point \(q \in M\) such that \(d(p,q)=D>100n\), we set \(r=\frac{1}{D}\), \(\bar{f}=f-f(q)\), then under the conformal transformation \(\bar{g}{:}{=}e^{-\frac{2\bar{f}}{n-2}}g\), we have

(309)
(310)

where the proof and the definition of the Kulkarni–Nomizu product can be found in [4, Theorem 1.165]. It follows from [34, Lemma 3.5] that

$$\begin{aligned} B_{\bar{g}}\left( q, e^{-\frac{1}{n-2}} r \right) \subset B(q, r) \subset B_{\bar{g}}\left( q, e^{\frac{1}{n-2}}r \right) . \end{aligned}$$
(311)

Therefore, by the same proof as in [34, Lemma 3.7], we have

$$\begin{aligned} |\bar{f}| \le C \quad \text {and} \quad { \left| \overline{Rc} \right| }_{\bar{g}} \le CD^2 \quad \text {on} \quad B_{\bar{g}}\left( q, e^{\frac{1}{n-2}}r \right) . \end{aligned}$$
(312)

Since \(R \le CD^2\) on B(qr), it follows from Theorem 23 that \(|B(q,r)| \ge Ce^{\varvec{\mu }}r^n\) and hence

$$\begin{aligned} \left| B_{\bar{g}}\left( q, e^{\frac{1}{n-2}}r \right) \right| _{\bar{g}} \ge Ce^{\varvec{\mu }}r^n. \end{aligned}$$
(313)

One can also use Theorem 2 to obtain the above estimate directly.

By defining \(\tilde{g}{:}{=}r^{-2}\bar{g}\), we have \(|\widetilde{Rc}|_{\tilde{g}} \le C\) on \(B_{\tilde{g}}(q,e^{\frac{1}{n-2}})\) and \(|B_{\tilde{g}}(q,e^{\frac{1}{n-2}})|_{\tilde{g}} \ge Ce^{\varvec{\mu }}\). By shrinking balls to its half size if necessary, it follows from [31, Theorem 1.6] that

$$\begin{aligned} r^{4-n}\int _{B_{\bar{g}}(q,e^{\frac{1}{n-2}}r)} |\overline{Rm}|^2 \,dV_{\bar{g}}=\int _{B_{\tilde{g}}(q,e^{\frac{1}{n-2}})} |\widetilde{Rm}|^2 \,dV_{\tilde{g}} \le I_0 \end{aligned}$$
(314)

for some constant \(I_0=I_0(n,A)\).

From (310), we have on B(qr),

$$\begin{aligned} |Rm|^2 \le C\left( |\overline{Rm}|^2+|\nabla f|^4+|Rc|^2 \right) \le C\left( |\overline{Rm}|^2+f^2+|Rc|^2 \right) . \end{aligned}$$

Therefore, we have

$$\begin{aligned}&\int _{B(q,r)} |Rm|^2 e^{-f}\,dV\\&\quad \le C \left( \int _{B_{\bar{g}}(q,e^{\frac{1}{n-2}}r)} |\overline{Rm}|^2 e^{-f}\,dV_{\bar{g}}+\int _{B(q,r)} f^2 e^{-f}\,dV+\int _{B(q,r)} |Rc|^2e^{-f}\,dV \right) \\&\quad \le Ce^{-\frac{D^2}{5}} \left( D^{4-n}I_0+D^{n+2}e^{\varvec{\mu }} \right) \end{aligned}$$

where we have used Lemma 32 and (314). Consequently, there exists \(I_1=I_1(n,A)\) such that

$$\begin{aligned} \int _{B(q,r)} |Rm|^2 e^{-f}\,dV \le I_1D^{n+2}e^{-\frac{D^2}{5}} . \end{aligned}$$
(315)

For any constant \(D >100n\), we apply Vitali’s lemma for the covering \(\{B(q,\frac{1}{4D})\}_{q\in A(D,2D)}\). If we assume that \(\{B(q_i,\frac{1}{4D})\}_{1 \le i \le k}\) is a maximal collection of mutually disjoint sets, then \(\{B(q_i,\frac{1}{2D})\}_{1 \le i \le k}\) cover A(D, 2D). It is clear from definition that

$$\begin{aligned} \sum _{i=1}^k \left| B\left( q_i,\frac{1}{4D} \right) \right| \le |A(D,2D)| \le |B(p,2D)|. \end{aligned}$$

By Lemma 2 and (313), we obtain \(k \le CD^{2n}\). Combining (315) with the above inequality implies that

$$\begin{aligned} \int _{A(D,2D)} |Rm|^2 e^{-f}\,dV \le \sum _{i=1}^k \int _{B(q_i,\frac{1}{2D})} |Rm|^2 e^{-f}\,dV \le kI_1D^{n+2}e^{-\frac{D^2}{5}} \le CI_1D^{3n+2}e^{-\frac{D^2}{5}}. \end{aligned}$$
(316)

Similarly, by exploiting Lemma 33, we have

$$\begin{aligned} \int _{B(p,D_0)} |Rm|^2 e^{-f}\,dV \le I_2 \end{aligned}$$
(317)

where \(D_0=100n\) and \(I_2=I_2(n,A)\).

Now we set \(D_i=2^i D_0\) and decompose the integral as

$$\begin{aligned} \int |Rm|^2 e^{-f}\,dV=\int _{B(p,D_0)} |Rm|^2 e^{-f}\,dV+\sum _{i \ge 0}\int _{A(D_i,2D_i)} |Rm|^2 e^{-f}\,dV. \end{aligned}$$

Plugging (316) and (317) into the above equation, we arrive at

$$\begin{aligned} \int |Rm|^2 e^{-f}\,dV&\le I_2+\sum _{i \ge 0}CI_1D_i^{3n+2}e^{-\frac{D_i^2}{5}}=I_2+ CI_1 \sum _{i \ge 0} 2^{i(3n+2)}D_0^{3n+2}e^{-\frac{4^i D_0^2}{5}} {:}{=}I. \end{aligned}$$

Since both \(I_1\) and \(I_2\) depend only on n and A, it is clear that I relies only on n and A and we arrive at (306). The proof of Theorem 26 is complete. \(\square \)

From (306) and [40, Theorem 1.2], a direct corollary of Theorem 26 is the following estimate.

Corollary 10

For any Ricci shrinker \((M^n, g, f)\in \mathcal M_n(A)\), there exists a constant \(I=I(n,A)<\infty \) such that

$$\begin{aligned} \int |\nabla Rc|^2 e^{-f}\,dV = \int |div(Rm)|^2 e^{-f}\,dV \le I. \end{aligned}$$

Theorem 26 is an important step for verifying maximum principle on curvature operators. The curvature operator on two-forms are defined as \(\mathcal {R}: \varLambda ^2 \rightarrow \varLambda ^2: \mathcal R(e^i \wedge e^j,e^k \wedge e^l)=R_{ijkl}\). The two-form \(e^i \wedge e^j {:}{=}e^i \otimes e^j-e^j\otimes e^i\) and the inner product on \(\varLambda ^2\) is defined as \(\langle A,B\rangle {:}{=}-\frac{1}{2}tr(AB)\) for \(A,B\in \varLambda ^2=\mathfrak {so}(n)\). In other words, for \(w=\frac{1}{2}\sum _{i,j}w_{ij}e^i \wedge e^j\), we have

$$\begin{aligned} \mathcal R(w)_{ij}=\frac{1}{2}R_{ijkl}w_{kl}. \end{aligned}$$

In the setting of Ricci shrinker \((M^n,g,f)\), the following equation (see  [24]) holds:

$$\begin{aligned} \varDelta _f \mathcal R=\mathcal R-2Q(\mathcal R). \end{aligned}$$

Here \(Q(\mathcal R) {:}{=}\mathcal R^2+\mathcal R^{\#}\) and \(\mathcal R^{\#}\) is defined as

$$\begin{aligned} \mathcal R^{\#}(u,v)=-\frac{1}{2}tr(ad_u \,\mathcal R\, ad_v\, \mathcal R) \end{aligned}$$

for any \(u,v \in \varLambda ^2\). If we choose an orthonormal basis \(\{\phi _i\}\) of \(\varLambda ^2\), then

$$\begin{aligned} \mathcal R^{\#}(u,v)=-\frac{1}{2}\sum _{i,j}\langle [\mathcal R(\phi _i),\phi _j],u \rangle \langle [\mathcal R(\phi _j),\phi _i],v \rangle . \end{aligned}$$

If we assume \(\lambda _1 \le \lambda _2 \le \cdots \) are all eigenvalues of \(\mathcal R\) on \(\varLambda ^2\), then we have the following rigidity theorem.

Theorem 27

There exists a constant \(\epsilon =\epsilon (n)>0\) such that for any Ricci shrinkers \((M^n,g,f)\), if \(\lambda _2 \ge -\epsilon \dfrac{\lambda _1^2}{|R-2\lambda _1|}\), then \(\lambda _1 \ge 0\). Consequently, \((M^n,g)\) is isometric to a quotient of \(N^k \times \mathbb {R}^{n-k}\) for some \(0 \le k \le n\), where \(N^k\) is a closed symmetric space.

Proof

It suffices to prove \(\lambda _1 \ge 0\). Namely, \((M^n,g)\) has nonnegative curvature operator. The further conclusion follows from [42, Corollary 4].

We fix a point q and assume that \(\phi _1\) is an eigenvector of \(\lambda _1\). Extending \(\phi _1\) by parallel transport on a small neighborhood of q, we have

$$\begin{aligned} \varDelta _f \mathcal R(\phi _1,\phi _1)=\mathcal R(\phi _1,\phi _1)-2Q(\mathcal R)(\phi _1,\phi _1). \end{aligned}$$

Therefore if we assume that \(\phi _i\) are eigenvectors of \(\lambda _i\), then in the barrier sense,

$$\begin{aligned} \varDelta _f \lambda _1&\le \lambda _1-\left( 2\lambda _1^2-\sum _{i,j}\langle [\mathcal R(\phi _i),\phi _j],\phi _1 \rangle \langle [\mathcal R(\phi _j),\phi _i],\phi _1 \rangle \right) \nonumber \\&=\lambda _1-\left( 2\lambda _1^2+\sum _{i,j}C_{ij}^2 \lambda _i\lambda _j \right) \end{aligned}$$
(318)

where \(C_{i,j}=\langle [\phi _i,\phi _j],\phi _1 \rangle \). Notice that \(C_{i,j}=0\) if \(i=1\) or \(j=1\).

We claim that \(|C_{i,j}| \le 2\). Indeed, if we assume that \(\phi _i,\phi _j\) and \(\phi _1\) are represented by the antisymmetric matrices AB and C respectively, then \(C_{i,j}=-\frac{1}{2}tr((AB-BA)C)=-tr(ABC)\). By choosing a basis such that \(A_{2k-1,2k}=a_k=-A_{2k,2k-1}\) for \(k \le [n/2]\) and 0 otherwise, we have

$$\begin{aligned} |tr(ABC)|&\le \sum _{k,l}|a_k||B_{2k,l}C_{l,2k-1}-B_{2k-1,l}C_{l,2k}| \\&\le \frac{1}{2}\sum _{k,l} (B^2_{2k,l}+C^2_{l,2k-1}+B^2_{2k-1,l}+C^2_{l,2k}) \\&\le \frac{1}{2}(|B|^2+|C|^2)=2. \end{aligned}$$

Here we have used the fact that \(|A|^2=|B|^2=|C|^2=2\).

Next we prove that if \(\epsilon \) is properly chosen, then we have

$$\begin{aligned} P {:}{=}2\lambda _1^2+\sum _{i,j}C_{ij}^2 \lambda _i\lambda _j \ge 0. \end{aligned}$$

From the definition of \(\lambda _i\), we notice that \(\sum \lambda _i=R/2\). Therefore, we fix \(\lambda _1\) and \(\lambda _2\) and minimize P under the restriction \(\sum \lambda _i=R/2\). We can assume that \(\lambda _2<0\), otherwise \(P \ge 0\) from its definition. We also set \(c_n=n(n-1)/2\) and assume that \(\lambda _1 \le \lambda _2 \le \cdots \le \lambda _{s+1}\) are all eigenvalues smaller than 0. Therefore,

$$\begin{aligned} P\ge P_1 {:}{=}2\lambda _1^2+2\sum _{\begin{array}{c} 2\le i\le s+1 \\ s+2 \le j \le c_n \end{array}}C_{ij}^2 \lambda _i\lambda _j. \end{aligned}$$

It is easy to show that \(P_1\) is minimized when \(\lambda _2=\lambda _3=\cdots =\lambda _{s+1}\) and \(\lambda _{s+2}=\cdots =\lambda _{c_n}\). It follows that

$$\begin{aligned} \frac{P_1}{2}\ge&\lambda _1^2+\sum _{\begin{array}{c} 2\le i\le s+1 \\ s+2 \le j \le c_n \end{array}}\frac{1}{c_n-s-1}C^2_{i,j}\lambda _2(R/2-\lambda _1-s\lambda _2) \\ \ge&\lambda _1^2+4s\lambda _2(R/2-\lambda _1-s\lambda _2). \end{aligned}$$

By solving the above quadratic inequality, we obtain that \(P_1\) and hence P are nonnegative if

$$\begin{aligned} \lambda _2 \ge \frac{\frac{R}{2}-\lambda _1-\sqrt{(\frac{R}{2}-\lambda _1)^2+\lambda _1^2}}{2s}. \end{aligned}$$

If we choose \(\epsilon =\frac{1}{(1+\sqrt{2})(c_n-2)}\), then it is clear that for any \(1\le s \le c_n-2\),

$$\begin{aligned} \lambda _2 \ge -\epsilon \frac{\lambda _1^2}{R-2\lambda _1} \ge \frac{\frac{R}{2}-\lambda _1-\sqrt{(\frac{R}{2}-\lambda _1)^2+\lambda _1^2}}{2(c_n-2)}\ge \frac{\frac{R}{2}-\lambda _1-\sqrt{(\frac{R}{2}-\lambda _1)^2+\lambda _1^2}}{2s}. \end{aligned}$$

Therefore, from (318) we obtain \(\varDelta _f \lambda _1 \le \lambda _1\). Since \(\lambda _1 \in L^2(e^{-f}\,dV)\) by (306), then it follows from [47, Theorem 4.4] that \(\lambda _1 \ge 0\). \(\square \)

We conclude this section by the proof of Theorem 4.

Proof of Theorem 4:

Since \(\lambda _2 \ge 0\), we can apply Theorem 27 to obtain \(\lambda _1 \ge 0\). Therefore, \(M^n\) is a finite quotient of \(N^k \times \mathbb {R}^{n-k}\). Note that only the case \(k=n\) is possible. For otherwise the second smallest eigenvalue must be 0. Since \(N^n\) is a compact Einstein manifold such that the curvature operator is 2-positive, it follows from [5] that its universal covering must be \(S^n\). \(\square \)