Elliptic gradient estimates for a weighted heat equation and applications

Abstract

We obtain two elliptic gradient estimates for positive solutions to the $f$-heat equation on a complete smooth metric measure space with only Bakry–Émery Ricci tensor bounded below. One is a local sharp Souplet–Zhang’s type and the other is a global Hamilton’s type. As applications, we prove parabolic Liouville theorems for ancient solutions satisfying some growth restriction near infinity. In particular the Liouville results are suitable for the gradient shrinking or steady Ricci solitons. The estimates of derivation of the $f$-heat kernel are also obtained.

1 Introduction

Let $(M,g)$ be an $n$-dimensional complete Riemannian manifold, and let $f$ be a smooth function on $M$. Then the triple $(M,g,e^{-f}dv)$ is called a complete smooth metric measure space, where $dv$ is the volume element of $g$, and $e^{-f}dv$ is called the weighted measure. On $(M,g,e^{-f}dv)$, the $m$-Bakry–Émery Ricci tensor [3] is defined by

$$\begin{aligned} Ric_f^m:=Ric+\nabla ^2 f-\frac{1}{m}df\otimes df \end{aligned}$$

for some constant $m>0$, where $Ric$ is the Ricci tensor of the manifold and $\nabla ^2$ is the Hessian with respect to the metric $g$. When $m<\infty$, the Bochner formula for $Ric_f^m$ can be regarded as the classical Bochner formula for the Ricci tensor of an $(n+m)$-dimensional manifold, i.e.,

$$\begin{aligned} \frac{1}{2}\Delta _f|\nabla u|^2= & {} |\nabla ^2 u|^2+\langle \nabla \Delta _f u, \nabla u\rangle +Ric_f^m(\nabla u, \nabla u) +\frac{1}{m}|\langle \nabla f,\nabla u\rangle |^2\nonumber \\\ge & {} \frac{(\Delta _f u)^2}{m+n}+\langle \nabla \Delta _f u, \nabla u\rangle +Ric_f^m(\nabla u, \nabla u). \end{aligned}$$

(1.1)

Therefore many geometric results for manifolds with Ricci tensor bounded below can be possibly extended to smooth metric measure spaces with $m$-Bakry–Émery Ricci tensor bounded below, see for example [4, 16, 27]. When $m=\infty$, we have the Bakry–Émery Ricci tensor

$$\begin{aligned} Ric_f:=\lim _{m\rightarrow \infty }Ric_f^m=Ric+\nabla ^2 f. \end{aligned}$$

This tensor is related to the gradient Ricci soliton

$$\begin{aligned} Ric_f=\lambda g, \end{aligned}$$

where $\lambda$ is a real constant. A Ricci soliton is called shrinking, steady, or expanding, if the constant $\lambda$ is positive, zero, or negative, respectively. Ricci solitons play an important role in studying the singularities of the Ricci flow [12]. We refer the readers to [7] for a recent survey on this subject. Our interest in the Bakry–Émery Ricci tensor is largely due to gradient Ricci solitons. For the case of $Ric_f$, due to the lack of the extra term $\frac{1}{m}|\langle \nabla f,\nabla u\rangle |^2$ in the Bochner formula (1.1), one can derive only local $f$-mean curvature comparison (see [28]), which highly rely on the potential function $f$, and this makes it much more difficult to investigate smooth metric measure spaces with $Ric_f$ bounded below. However when $f$ is assumed to some restriction, there still have been many work in this direction. We refer the readers to [18, 20–22, 28] for nice details.

The $f$-Laplacian $\Delta _f$ on $(M,g,e^{-f}dv)$, which is self-adjoint with respect to the weighted measure, is defined by

$$\begin{aligned} \Delta _f:=\Delta -\nabla f\cdot \nabla . \end{aligned}$$

A function $u$ is called $f$-harmonic if $\Delta _f u=0$. The $f$-heat equation (or weighted heat equation) is defined as

$$\begin{aligned} \frac{\partial u}{\partial t}=\Delta _f u. \end{aligned}$$

(1.2)

We let $H(x,y,t)$ denote the $f$-heat kernel. That is, for each $y\in M$, $H(x,y,t)=u(x,t)$ is the minimal positive solution to the $f$-heat equation satisfying $\lim _{t\rightarrow 0}u(x,t)=\delta _{f,y}(x)$, where $\delta _{f,y}(x)$ is the weighted delta function defined by

$$\begin{aligned} \int _M\phi (x)\delta _{f,y}(x)e^{-f}dv=\phi (y) \end{aligned}$$

for $\phi \in C_0^{\infty }(M)$.

On smooth metric measure spaces, various Liouville-type theorems for the $f$-harmonic function were obtained, see for example Cao and Zhou [8], Li [16], Munteanu and Sesum [19], Pigola et al. [24], Wei and Wylie [28], Wu [29] and Wu and Wu [30, 31]. In particular, some Liouville-type results were directly derived by Yau’s gradient estimate [32]. For example, Brighton [5] proved a elliptic gradient estimate for positive $f$-harmonic functions by applying Yau’s idea to function $u^\epsilon$ $(0<\epsilon <1)$ instead of $\ln u$, and hence obtained a Liouville theorem for positive bounded $f$-harmonic functions when $Ric_f\ge 0$. Munteanu and Wang [20] refined Brighton’s argument and proved that any positive $f$-harmonic function of sub-exponential growth on smooth metric measure spaces with $Ric_f\ge 0$ must be constant. They further applied the De Giorgi–Nash–Moser theory to get a sharp gradient estimate and a Liouville-type result for positive $f$-harmonic functions. Recently, Li [17] applied probabilistic arguments to derive a Hamilton’s estimate for positive bounded solutions to the $f$-heat equation when $Ric_f\ge -K$ for some constant $K\ge 0$, using which he gave an alternative proof of Brighton’s Liouville theorem.

In this paper we will study gradient estimates for positive solutions to the $f$-heat equation on a complete smooth metric measure space $(M,g,e^{-f}dv)$ with only the Bakry–Émery Ricci tensor $Ric_f$ bounded below. According to [20, 21], there seems essential obstacles to obtain Li–Yau gradient estimates for positive solutions to the $f$-heat equation when $Ric_f$ is bounded below, even assuming growth assumption on $f$. However, we observe that, without any assumption on $f$, we can prove local sharp elliptic (Souplet–Zhang’s) gradient estimates and global elliptic (Hamilton’s) gradient estimates of positive solutions to the $f$-heat equation. As applications, we will derive some Liouville-type theorems for some growth condition solutions to the $f$-heat equation. The estimates of derivation of the $f$-heat kernel are also discussed.

Our first result gives a local sharp gradient estimate for the positive solutions to the $f$-heat equation. The important point here is that our gradient estimate does not depend on any assumption on $f$.

Theorem 1.1

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space. Fix $x_0\in M$ and $R\ge 2$. Assume that $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$ in $B(x_0,R)$. If $0<u(x,t)\le D$ for some constant $D$, is a smooth solution to $f$-heat equation (1.2) in $Q_{R,T}\equiv B(x_0,R)\times [t_0-T,t_0]\subset M\times (-\infty ,\infty )$, where $t_0\in \mathbb {R}$ and $T>0$, then there exists a constant $c(n)$ such that

$$\begin{aligned} |\nabla \ln u|\le c(n)\left( \sqrt{\frac{1+|\alpha |}{R}}+\frac{1}{\sqrt{t-t_0+T}}+\sqrt{K}\right) \left( 1+\ln \frac{D}{u}\right) \end{aligned}$$

(1.3)

in $Q_{R/2, T}$ with $t\ne t_0-T$. Here $\alpha :=\max _{\{x|d(x,x_0)=1\}}\Delta _f\,r(x)$, where $r(x)$ is the distance to $x$ from the fixed point $x_0\in M$.

The motivation of our theorem comes from Souplet–Zhang [26], where they discovered a localized elliptic Cheng–Yau estimate of heat equation, by inserting a necessary logarithmic correction term. In our case, the proof not only depends on Souplet–Zhang’s arguments [26], but also relies on some Brighton’s proof trick [5] and the $f$-Laplacian comparison theorem of Wei and Wylie [28].

Arnaudon et al. [1] sketchily proved Souplet–Zhang type estimate for a general equation $\partial _tu=\Delta u+Zu$ for a smooth vector field $Z$. However, to the author, an important issue seems not to be elaborate, especially to the usage of generalized Laplacian comparison theorem.

The following example confirms that Theorem 1.1 is sharp when $f$ has a linear growth on smooth metric measure spaces.

Example 1.2

For any $a, b>0$, consider

$$\begin{aligned} u=e^{ax+(a^2+ab)t}\quad \mathrm {and}\quad f=-bx. \end{aligned}$$

Then $u$ is a positive solution to the $f$-heat equation in $Q:=(0,4)\times [1,2]\subset \mathbb {R}\times (-\infty ,\infty )$. We compute that $|\alpha |=b$, $D:=\sup _{Q}u=e^{4a+2(a^2+ab)}$,

$$\begin{aligned} \frac{|\nabla u|}{u}(2,2)=a\quad \mathrm {and}\quad \ln \left( \frac{D}{u}\right) (2,2)=2a. \end{aligned}$$

At the space-time point $(x,t)=(2,2)$, the left hand side of (1.3) is $a$, while the right hand side of (1.3) with $x_0=2$, $t_0=2$, $R=2$ and $T=1$ equals to $c(n)(\frac{\sqrt{1+b}}{2}+1)(1+2a)$. Therefore they are equivalent when $a$ is sufficient large compared with the fixed $b$ and $c(n)$.

An interesting application of Theorem 1.1 is a parabolic Liouville theorem under certain growth restriction near infinity when $Ric_f\ge 0$. This result is similar to the case of heat equation on a manifold with nonnegative Ricci tensor, obtained by Souplet and Zhang [26].

Theorem 1.3

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space with $Ric_f\ge 0$.

• (1) If $u$ is a positive ancient solution to the $f$-heat equation (that is, a solution defined in all space and negative time) such that $u(x,t)=e^{o\,(r^{1/2}(x)+|t|^{1/4}\,)}$ near infinity, then $u$ is constant.

• (2) If $u$ is an ancient solution to the $f$-heat equation such that $u(x,t)=o\,(r^{1/2}(x)+|t|^{1/4}\,)$ near infinity, then $u$ is constant.

Any complete noncompact shrinking or steady gradient Ricci soliton satisfies the assumption of Theorem 1.3. Hence,

Corollary 1.4

Let $(M,g,e^{-f}dv)$ be a complete noncompact gradient shrinking or steady Ricci soliton.

• (1) If $u$ is a positive ancient solution to the $f$-heat equation such that $u(x,t)=e^{o\,(r^{1/2}(x)+|t|^{1/4}\,)}$ near infinity, then $u$ is constant.

• (2) If $u$ is an ancient solution to the $f$-heat equation such that $u(x,t)=o\,(r^{1/2}(x)+|t|^{1/4}\,)$ near infinity, then $u$ is constant.

Remark 1.5

If $f$ is also bounded, by a different $f$-Laplacian comparison theorem of Wei and Wylie [28], we can give stronger elliptic gradient estimates and Liouville theorems for the $f$-heat equation, see Theorems 3.1 and 3.2 in Sect. 3.

Our second objective is to derive a global Hamilton’s gradient estimate for positive bounded solutions to the $f$-heat equation on smooth metric measure spaces with $Ric_f$ bounded below.

Theorem 1.6

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space with $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$. If $0<u(x,t)\le D$ for some constant $D$ is a solution to the $f$-heat equation in $M\times [0,T]$ for $0<T<\infty$, then

$$\begin{aligned} t|\nabla \ln u|^2\le (1+2(n-1)Kt)\ln \left( \frac{D}{u}\right) \end{aligned}$$

for all $x\in M$ and $t\in [0,T]$.

Li [17] also proved Theorem 1.6 by a probabilistic method. Our proof is geometric, which combines the arguments of Kotschwar [14] and Brighton [5], including a Bernstein-type estimate of Shi [25], the $f$-Laplacian comparison theorem of Wei and Wylie [28] and a weighted Karp and Li [13] maximum principle. When $f$ is constant, Theorem 1.6 returns to the Kotschwar’s result [14].

If we further assume that $f$ is bounded, we can apply Theorem 1.6 and two-side Gaussian bounds on the $f$-heat kernel [29] to obtain a estimate on the derivation of the $f$-heat kernel.

Theorem 1.7

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space with $Ric_f\ge 0$. If $|f|(x)\le A$ for some constant $A$ in $M$, then there exists a constant $c=c(n, A)$ such that

$$\begin{aligned} |\nabla \ln H(x,y,t)|^2\le \frac{c}{t}\left( 1+\frac{d^2(x,y)}{t}\right) \end{aligned}$$

for all $x, y\in M$ in $M$ and $t>0$.

Theorems 1.1 and 1.6 are elliptic gradient estimates, which are proved by using the maximum principle. Historically, the gradient estimate technique is originated by Yau [32] (see also Cheng–Yau [9]), who first proved a gradient estimate for the Laplacian equation via the maximum principle. This technique was further developed by Li and Yau [15] for the heat equation. Then Hamilton [11] gave an elliptic type gradient estimate for the heat equation. But this type of estimate is a global result which requires the heat equation defined on closed manifolds. Later Souplet and Zhang [26] proved a localized elliptic Cheng–Yau estimate for the heat equation by adding a necessary logarithmic correction term. In [14], Kotschwar applied the Karp–Li’s maximum principle [13] and a Bernstein-type estimate to extend Hamilton’s result to the complete non-compact case.

This paper is organized as follows. In Sect. 2, we first give a lemma and then apply this lemma and cut-off function technique to prove Theorem 1.1. In Sect. 3, we will apply Theorem 1.1 to prove Theorems 1.3. When $f$ is bounded, we give another local elliptic gradient estimate for the $f$-heat equation. In Sect. 4, we will discuss a Hamilton’s gradient estimate of the $f$-heat equation and give a geometric proof of Theorem 1.6. Finally we apply this estimate to prove Theorem 1.7.

2 Souplet–Zhang’s gradient estimate

In this section, we will apply the arguments of Souplet and Zhang [26] and Brighton’s idea [5] to prove Theorem 1.1. Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space. Consider the $f$-heat equation

$$\begin{aligned} u_t=\Delta u-\nabla f\cdot \nabla u. \end{aligned}$$

(2.1)

Let $x_0\in M$ and $R>0$. Assume that $0<u(x,t)\le D$ for some constant $D$ is a smooth solution to (2.1) in $Q_{R,T}\equiv B(x_0,R)\times [t_0-T,t_0]\subset M\times (-\infty ,\infty )$, $t_0\in R$. It is clear that the gradient estimate in Theorem 1.1 is invariant under the scaling $u\rightarrow u/D$. Therefore we can assume that $0<u\le 1$. Define a smooth function

$$\begin{aligned} h(x,t):=\ln u(x,t) \end{aligned}$$

in $Q_{R,T}$. By (2.1), then

$$\begin{aligned} \left( \Delta _f-\frac{\partial }{\partial t}\right) h+|\nabla h|^2=0. \end{aligned}$$

(2.2)

Obviously, $h\le 0$, since $0<u\le 1$. From above, we have the following lemma, which is a mild generalization of [26].

Lemma 2.1

Let $(M,g,e^{-f}dv)$ be a complete smooth metric measure space. Fix $x_0\in M$ and $R>0$. Assume that $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$ in $B(x_0,R)$. Let $h(x,t)$ is a smooth function defined in $Q_{R,T}$ satisfying equation (2.2). Then the function

$$\begin{aligned} \omega :=\left| \nabla \ln (1-h)\right| ^2=\frac{|\nabla h|^2}{(1-h)^2} \end{aligned}$$

(2.3)

for all $(x,t)$ in $Q_{R,T}$, satisfies

$$\begin{aligned} \left( \Delta _f-\frac{\partial }{\partial t}\right) \omega \ge \frac{2h}{1-h}\left\langle \nabla h,\nabla \omega \right\rangle +2(1-h)\omega ^2-2(n-1)K\omega . \end{aligned}$$

Proof

By (2.3), we compute that

$$\begin{aligned} \omega _j=\frac{2h_i h_{ij}}{(1-h)^2}+\frac{2h^2_i h_j}{(1-h)^3} \end{aligned}$$

(2.4)

and

$$\begin{aligned} \Delta \omega =\frac{2|h_{ij}|^2}{(1-h)^2}+\frac{2h_i h_{ijj}}{(1-h)^2}+\frac{8h_ih_j h_{ij}}{(1-h)^3}+\frac{2h^2_i h_{jj}}{(1-h)^3}+\frac{6h^2_i h^2_j}{(1-h)^4}. \end{aligned}$$

Hence

$$\begin{aligned} \Delta _f\,\omega= & {} \Delta \omega -\langle \nabla f,\nabla \omega \rangle \nonumber \\= & {} \frac{2|h_{ij}|^2}{(1-h)^2}+\frac{2h_i h_{ijj}}{(1-h)^2}+\frac{8h_ih_j h_{ij}}{(1-h)^3} +\frac{2h^2_i h_{jj}}{(1-h)^3}+\frac{6h^4_i}{(1-h)^4}\nonumber \\&-\frac{2h_{ij}h_if_j}{(1-h)^2} -\frac{2h^2_ih_jf_j}{(1-h)^3}\nonumber \\= & {} \frac{2|h_{ij}|^2}{(1-h)^2}+\frac{2h_i (\Delta _f h)_i}{(1-h)^2}+\frac{2(R_{ij}+f_{ij})h_ih_j}{(1-h)^2} +\frac{8h_ih_jh_{ij}}{(1-h)^3}\nonumber \\&+\frac{6h^4_i}{(1-h)^4}+\frac{2h^2_i\cdot \Delta _fh}{(1-h)^3}, \end{aligned}$$

(2.5)

where $h_i:=\nabla _i h$ and $h_{ijj}:=\nabla _j\nabla _j\nabla _i h$, etc. By (2.2) and (2.3), we also have

$$\begin{aligned} \omega _t= & {} \frac{2\nabla _ih\cdot \nabla _i\left( \Delta _fh+|\nabla h|^2\right) }{(1-h)^2}+\frac{2|\nabla h|^2\left( \Delta _fh+|\nabla h|^2\right) }{(1-h)^3}\nonumber \\= & {} \frac{2\nabla h \nabla \Delta _fh}{(1-h)^2} +\frac{4h_ih_jh_{ij}}{(1-h)^2}+\frac{2h^2_i\Delta _f h}{(1-h)^3} +\frac{2|\nabla h|^4}{(1-h)^3}. \end{aligned}$$

(2.6)

Combining (2.5) and (2.6) yields

$$\begin{aligned} \left( \Delta _f-\frac{\partial }{\partial t}\right) \omega= & {} \frac{2|h_{ij}|^2}{(1-h)^2} +\frac{2(R_{ij}+f_{ij})h_ih_j}{(1-h)^2}+\frac{8h_ih_j h_{ij}}{(1-h)^3}+\frac{6h^4_i}{(1-h)^4}\nonumber \\&-\frac{4h_ih_jh_{ij}}{(1-h)^2}-\frac{2h^4_i}{(1-h)^3}. \end{aligned}$$

(2.7)

Since $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$,

$$\begin{aligned} (R_{ij}+f_{ij})f_if_j\ge -(n-1)K|\nabla f|^2. \end{aligned}$$

(2.8)

By (2.4), we obtain

$$\begin{aligned} \omega _jh_j=\frac{2h_ih_j h_{ij}}{(1-h)^2}+\frac{2h^2_i h^2_j}{(1-h)^3}, \end{aligned}$$

which implies

$$\begin{aligned} 0=-2\omega _jh_j+\frac{4h_ih_j h_{ij}}{(1-h)^2}+\frac{4h^4_i}{(1-h)^3}, \end{aligned}$$

(2.9)

and

$$\begin{aligned} 0=\frac{1}{1-h}\left( 2\omega _jh_j-\frac{4h^4_i}{(1-h)^3}\right) -\frac{4h_ih_jh_{ij}}{(1-h)^3}. \end{aligned}$$

(2.10)

Substituting (2.8) into (2.7) and then adding (2.7) with (2.9) and (2.10), we get

$$\begin{aligned} \left( \Delta _f-\frac{\partial }{\partial t}\right) \omega\ge & {} \frac{2|h_{ij}|^2}{(1-h)^2} +\frac{4h_ih_jh_{ij}}{(1-h)^3}+\frac{2h^4_i}{(1-h)^4}\nonumber \\&+\frac{2h}{1-h}h_i\omega _i+\frac{2h^4_i}{(1-h)^3} -\frac{2(n-1)K|\nabla h|^2}{(1-h)^2}. \end{aligned}$$

(2.11)

Since $1-h\ge 1$, we further have

$$\begin{aligned} \frac{2|h_{ij}|^2}{(1-h)^2}+\frac{4h_ih_j h_{ij}}{(1-h)^3}+\frac{2h^4_i}{(1-h)^4}\ge 0. \end{aligned}$$

This, together with (2.11), proves the desired estimate.$\square$

In the rest of this section, we will apply Lemma 2.1 and the localization technique of Souplet–Zhang [26] to give a elliptic type gradient estimate for the positive smooth solutions to the $f$-heat equation (1.2).

We first introduce a well-known cut-off function originated by Li and Yau [15] (see also [26] and [2]) as follows.

Lemma 2.2

Fix $t_0\in \mathbb {R}$ and $T>0$. For any given $\tau \in (t_0-T,t_0]$, there exists a smooth function ${\bar{\psi }}:[0,\infty )\times [t_0-T,t_0]\rightarrow \mathbb R$ satisfying following propositions:

• (1)
  $$\begin{aligned} 0\le {\bar{\psi }}(r,t)\le 1 \end{aligned}$$

  in $[0,R]\times [t_0-T,t_0]$, and it is supported in a subset of $[0,R]\times [t_0-T,t_0]$.

• (2)
  $$\begin{aligned} {\bar{\psi }}(r,t)=1\quad \mathrm {and}\quad \frac{\partial {\bar{\psi }}}{\partial r}(r,t)=0 \end{aligned}$$

  in $[0,R/2]\times [\tau ,t_0]$ and $[0,R/2]\times [t_0-T,t_0]$, respectively.

• (3)
  $$\begin{aligned} \left| \frac{\partial {\bar{\psi }}}{\partial t}\right| \le \frac{C{\bar{\psi }}^{\frac{1}{2}}}{\tau -(t_0-T)} \end{aligned}$$

  in $[0,\infty )\times [t_0-T,t_0]$ for some $C>0$, and ${\bar{\psi }}(r,t_0-T)=0$ for all $r\in [0,\infty )$.

• (4)
  $$\begin{aligned} -\frac{C_\epsilon {\bar{\psi }}^\epsilon }{R}\le \frac{\partial {\bar{\psi }}}{\partial r}\le 0 \quad \mathrm {and}\quad \left| \frac{\partial ^2{\bar{\psi }}}{\partial r^2}\right| \le \frac{C_\epsilon {\bar{\psi }}^\epsilon }{R^2} \end{aligned}$$

  in $[0,\infty )\times [t_0-T,t_0]$ for every $\epsilon \in (0,1)$ with some constant $C_\epsilon$ depending on $\epsilon$.

Then we apply Lemma 2.1 and 2.2 to prove Theorem 1.1 via the maximum principle in a local space-time supported set. We remind the readers that our proof is a little different from [15] and [26]. Here, we mainly follow the arguments of Bailesteanua et al. [2], combining some proof trick of Brighton [5].

Proof of Theorem 1.1

Pick a number $\tau \in (t_0-T,t_0]$ and fix a cutoff function ${\bar{\psi }}(r,t)$ satisfying those conditions of Lemma 2.2. Briefly, we will show that (1.3) holds at the space-time point $(x,\tau )$ for all $x$ such that $d(x,x_0)<R/2$. Since $\tau$ is arbitrary, the conclusion then follows.

Now we give a detailed discussion. Introduce a cutoff function $\psi :M\times [t_0-T,t_0]\rightarrow \mathbb R$, such that

$$\begin{aligned} \psi =\bar{\psi }(d(x,x_0),t)\equiv \psi (r,t). \end{aligned}$$

It is easy to see that $\psi (x,t)$ is be a smooth cut-off function supported in $Q_{R,T}$. Our aim is to estimate $\left( \Delta _f-\frac{\partial }{\partial t}\right) (\psi \omega )$ and carefully analyze the result at a space-time point where the function $\psi \omega$ attains its maximum.

By Lemma 2.1, a straight forward calculation yields

$$\begin{aligned}&\left( \Delta _f-\frac{\partial }{\partial t}\right) (\psi \omega )-\left( \frac{2h}{1-h}\nabla h+2\frac{\nabla \psi }{\psi }\right) \cdot \nabla (\psi \omega )\nonumber \\&\quad \ge 2\psi (1-h)\omega ^2-\left( \frac{2h}{1-h}\nabla h\cdot \nabla \psi \right) \omega -2\frac{|\nabla \psi |^2}{\psi }\omega \nonumber \\&\qquad +(\Delta _f\psi )\omega -\psi _t\omega -2(n-1)K\psi \omega . \end{aligned}$$

(2.12)

Let $(x_1,t_1)$ be a maximum space-time point for the function $\psi \omega$ in the closed set

$$\begin{aligned} \left\{ (x,t)\in M\times [t_0-T,\tau ]\,|d(x,x_0)\le R\right\} . \end{aligned}$$

We may assume $(\psi \omega )(x_1,t_1)>0$; otherwise, $\omega (x,\tau )\le 0$ and (1.3) naturally holds at $(x,\tau )$ whenever $d(x, x_0)<\frac{R}{2}$. Here $t_1\ne t_0-T$, since we assume $(\psi \omega )(x_1,t_1)>0$. We may also assume that function $\psi (x,t)$ is smooth at $(x_1,t_1)$ due to the standard Calabi’s argument [6]. Since $(x_1,t_1)$ is a maximum space-time point, at this point we have

$$\begin{aligned} \Delta _f(\psi \omega )\le 0,\quad (\psi \omega )_t\ge 0 \quad \mathrm {and}\quad \nabla (\psi \omega )=0. \end{aligned}$$

Using the above estimates at $(x_1,t_1)$, (2.12) can be simplified as

$$\begin{aligned} 2\psi (1-h)\omega ^2\le & {} \left( \frac{2h}{1-h}\nabla h\cdot \nabla \psi +2\frac{|\nabla \psi |^2}{\psi }\right) \omega \nonumber \\&-(\Delta _f\psi )\omega +\psi _t\omega +2(n-1)K\psi \omega \end{aligned}$$

(2.13)

at $(x_1,t_1)$.

The rest of this part we will make use of (2.13) to give our desired gradient estimate. If $x_1\in B(x_0,1)$, then $\psi$ is constant in space direction in $B(x_0,R/2)$ by our assumption, where $R\ge 2$. So (2.13) yields

$$\begin{aligned} \omega \le \left( \frac{1}{2}\frac{\psi _t}{\psi }+(n-1)K\right) \le \frac{C}{\tau -(t_0-T)}+(n-1)K \end{aligned}$$

at $(x_1,t_1)$, where we used $1-h\ge 1$ and proposition (3) in Lemma 2.2. Since $\psi (x,\tau )=1$ when $d(x,x_0)<R/2$ by the proposition (2) in Lemma 2.2, the above estimate indeed gives that

$$\begin{aligned} \omega (x,\tau )= & {} (\psi \omega )(x,\tau )\\\le & {} (\psi \omega )(x_1,t_1)\\\le & {} \omega (x_1,t_1)\\\le & {} \frac{C}{\tau -t_0+T}+(n-1)K \end{aligned}$$

for all $x\in M$ such that $d(x,x_0)<R/2$. By the definition of $w(x,\tau )$ and the fact that $\tau \in (t_0-T,t_0]$ was chosen arbitrarily, we in fact prove that

$$\begin{aligned} \frac{|\nabla h|}{(1-h)}(x,t)\le \frac{C}{\sqrt{t-t_0+T}}+\sqrt{(n-1)K} \end{aligned}$$

for all $(x,t)\in Q_{R/2,T}\equiv B(x_0,R/2)\times [t_0-T,t_0]$ with $t\ne t_0-T$. Then (1.3) follows since $h=\ln (u/D)$ with $D$ scaled to $1$.

Thus we may assume $x_1\not \in B(x_0,1)$. At this time, since $Ric_f\ge -(n-1)K$ and $r(x_1,x_0)\ge 1$ in $B(x_0,R)$, we have the $f$-Laplacian comparison theorem (Theorem 3.1 in [28])

$$\begin{aligned} \Delta _f\,r(x_1)\le \alpha +(n-1)K(R-1), \end{aligned}$$

(2.14)

where $\alpha :=\max _{\{x|d(x,x_0)=1\}}\Delta _f\,r(x)$. This comparison theorem will be used later. Below we will carefully estimate upper bounds for each term of the right-hand side of (2.13), similar to the arguments of Souplet–Zhang ([26], pp. 1050–1051). Note that the Young’s inequality

$$\begin{aligned} ab\le \frac{a^p}{p}+\frac{b^q}{q},\quad \forall \,\,\, p,q>0 \,\,\,\mathrm {with}\,\,\, \frac{1}{p}+\frac{1}{q}=1 \end{aligned}$$

will be repeatedly used in the following. We let $c$ denote a constant depending only on $n$ whose value may change from line to line.

First, we have the estimates of first term of the right hand side of (2.13):

$$\begin{aligned} \left( \frac{2h}{1-h}\nabla h\cdot \nabla \psi \right) \omega\le & {} 2|h|\cdot |\nabla \psi |\cdot \omega ^{3/2}\nonumber \\= & {} 2\left[ \psi (1-h)\omega ^2\right] ^{3/4} \cdot \frac{|h|\cdot |\nabla \psi |}{[\psi (1-h)]^{3/4}}\nonumber \\\le & {} \psi (1-h)\omega ^2+c \frac{(h|\nabla \psi |)^4}{[\psi (1-h)]^3}\nonumber \\\le & {} \psi (1-h)\omega ^2+c\frac{h^4}{R^4(1-h)^3}. \end{aligned}$$

(2.15)

For the second term of the right hand side of (2.13), we have

$$\begin{aligned} 2\frac{|\nabla \psi |^2}{\psi }\omega= & {} 2\psi ^{1/2}\omega \cdot \frac{|\nabla \psi |^2}{\psi ^{3/2}}\nonumber \\\le & {} \frac{1}{8}\psi \omega ^2+c \left( \frac{|\nabla \psi |^2}{\psi ^{3/2}}\right) ^2\nonumber \\\le & {} \frac{1}{8}\psi \omega ^2+\frac{c}{R^4}. \end{aligned}$$

(2.16)

For the third term of the right hand side of (2.13), since $\psi$ is a radial function, then at $(x_1,t_1)$, using (2.14) we have

$$\begin{aligned} -(\Delta _f\psi )\omega= & {} -\left[ (\partial _r\psi )\Delta _fr+(\partial ^2_r\psi )\cdot |\nabla r|^2\right] \omega \nonumber \\\le & {} -\left[ \partial _r\psi \left( \alpha +(n-1)K(R-1)\right) +\partial ^2_r\psi \right] \omega \nonumber \\\le & {} \left[ |\partial ^2_r\psi |+\left( |\alpha |+(n-1)K(R-1)\right) |\partial _r\psi |\right] \omega \nonumber \\= & {} \psi ^{1/2}\omega \frac{|\partial ^2_r\psi |}{\psi ^{1/2}} +\left( |\alpha |+(n-1)K(R-1)\right) \psi ^{1/2}\omega \frac{|\partial _r\psi |}{\psi ^{1/2}}\nonumber \\\le & {} \frac{1}{8}\psi \omega ^2+c \left[ \left( \frac{|\partial ^2_r\psi |}{\psi ^{1/2}}\right) ^2 +\left( \frac{|\alpha |\cdot |\partial _r\psi |}{\psi ^{1/2}}\right) ^2 +\left( \frac{K(R-1)|\partial _r\psi |}{\psi ^{1/2}}\right) ^2\right] \nonumber \\\le & {} \frac{1}{8}\psi \omega ^2+\frac{c}{R^4}+\frac{c\alpha ^2}{R^2} +cK^2, \end{aligned}$$

(2.17)

where in the last inequlity we used proposition (4) in Lemma 2.2.

Then we estimate the fourth term:

$$\begin{aligned} |\psi _t|\omega= & {} \psi ^{1/2}\omega \frac{|\psi _t|}{\psi ^{1/2}}\nonumber \\\le & {} \frac{1}{8}\left( \psi ^{1/2}\omega \right) ^2+c \left( \frac{|\psi _t|}{\psi ^{1/2}}\right) ^2\nonumber \\\le & {} \frac{1}{8}\psi \omega ^2+\frac{c}{(\tau -t_0+T)^2}. \end{aligned}$$

(2.18)

Finally, we estimate the last term:

$$\begin{aligned} 2(n-1)K\psi \omega \le \frac{1}{8}\psi \omega ^2+cK^2. \end{aligned}$$

(2.19)

Now substituting (2.15)–(2.19) into the right hand side of (2.13), we have that

$$\begin{aligned} 2\psi (1-h)\omega ^2\le & {} \psi (1-h)\omega ^2+\frac{ch^4}{R^4(1-h)^3} +\frac{1}{2}\psi \omega ^2+\frac{c}{R^4}\nonumber \\&+\frac{c\alpha ^2}{R^2}+\frac{c}{(\tau -t_0+T)^2}+cK^2 \end{aligned}$$

(2.20)

at $(x_1,t_1)$. Recall that $1-h\ge 1$, then (2.20) implies

$$\begin{aligned} \psi \omega ^2\le \frac{ch^4}{R^4(1-h)^4}+\frac{1}{2}\psi \omega ^2 +\frac{c}{R^4}+\frac{c\alpha ^2}{R^2}+\frac{c}{(\tau -t_0+T)^2}+cK^2 \end{aligned}$$

at $(x_1,t_1)$. Moreover, since $\frac{h^4}{(1-h)^4}\le 1$, the above inequality implies that

$$\begin{aligned} (\psi ^2\omega ^2)(x_1,t_1)\le & {} (\psi \omega ^2)(x_1,t_1)\\\le & {} \frac{c}{R^4}+\frac{c\alpha ^2}{R^2}+\frac{c}{(\tau -t_0+T)^2}+cK^2. \end{aligned}$$

Since $\psi (x,\tau )=1$ when $d(x,x_0)<R/2$ by the proposition (2) in Lemma 2.2, from the above estimate, we have

$$\begin{aligned} \omega (x,\tau )= & {} (\psi \omega )(x,\tau )\nonumber \\\le & {} (\psi \omega )(x_1,t_1)\nonumber \\\le & {} \frac{c}{R^2}+\frac{c|\alpha |}{R}+\frac{c}{\tau -t_0+T}+cK \end{aligned}$$

for all $x\in M$ such that $d(x,x_0)<R/2$. By the definition of $w(x,\tau )$ and the fact that $\tau \in (t_0-T,t_0]$ was chosen arbitrarily, we in fact show that

$$\begin{aligned} \frac{|\nabla h|}{(1-h)}(x,t)\le \frac{c}{R}+\frac{c\sqrt{|\alpha |}}{\sqrt{R}}+\frac{c}{\sqrt{t-t_0+T}}+c\sqrt{K} \end{aligned}$$

for all $(x,t)\in Q_{R/2,T}\equiv B(x_0,R/2)\times [t_0-T,t_0]$ with $t\ne t_0-T$. We have finished the proof of theorem since $h=\ln (u/D)$ with $D$ scaled to $1$ and $R\ge 2$.$\square$

Theorem 1.1 immediately implies a global gradient estimate for positive solutions to the $f$-heat equation.

Proposition 2.3

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space with $Ric_f\ge 0$. If $u(x,t)$ is a positive smooth solution to the $f$-heat equation (1.2) in $M\times (0,\infty )$, then there exist constants $c_1(n)$ and $c_2(n)$ such that

$$\begin{aligned} \frac{|\nabla u(x,t)|}{u(x,t)}\le c_1\frac{1+\sqrt{|\alpha |}}{t^{1/4}}\left( 1+e^{c_2A}+\ln \frac{u(x,2t)}{u(x,t)}\right) \end{aligned}$$

for $x\in M$ and $t>4$, where $A:=\sup _{y\in B(x,\sqrt{t})}f(y)$.

Proof of Proposition 2.3

We will apply (1.3) on the cube $Q_{\sqrt{t},t}:= B(x,\sqrt{t})\times [t/2, t]$. Since $Ric_f\ge 0$, by the parabolic Moser’s Harnack inequality [31], we have

$$\begin{aligned} D:=\sup _{(x,t)\in Q_{\sqrt{t},t}}u(x,t)\le \exp \{c_1e^{c_2A}\}u(x,2t), \end{aligned}$$

where $A:=\sup _{y\in B(x,\sqrt{t})}f(y)$. Substituting this into (1.3) finishes the proof.$\square$

3 Proof of Theorem 1.3

In this section, following the arguments of Souplet–Zhang in [26], we will prove Theorem 1.3. When $Ric_f\ge 0$ and $f$ is bounded, we apply a different mean value comparison to derive another form of gradient estimate and Liouville theorem of the $f$-heat equation.

Proof of Theorem 1.3

(1). By our theorem assumption, the function $u+1$ satisfies $\ln (u+1)=o(r^{1/2}(x)+|t|^{1/4})$ near infinity. Fixing any space time point $(x_0, t_0)$ and using Theorem 1.1 for $u+1$ in the set $B(x_0, R)\times [t_0-R^2, t_0]$, we have

$$\begin{aligned} \frac{|\nabla u(x_0,t_0)|}{u(x_0,t_0)+1} \le \frac{C(n,\alpha )}{\sqrt{R}}(1+o(\sqrt{R}\,)) \end{aligned}$$

for $R>2$. Letting $R\rightarrow \infty$, we have that $|\nabla u(x_0, t_0)|=0$. Since $(x_0, t_0)$ is arbitrary, then $u$ is a constant.

(2). For any space time point $(x_0, t_0)$, we let $D_R:=\sup _{Q_{\sqrt{R}, R}}|u|$. Consider the function $U=u+2D_{2R}$. Clearly we have

$$\begin{aligned} D_{2R}\le U(x,t)\le 3D_{2R} \end{aligned}$$

whenever $(x,t)\in Q_{2\sqrt{R},4R}$. Applying Theorem 1.1 to $U$, we conclude that

$$\begin{aligned} \frac{|\nabla u(x_0, t_0)|}{u(x_0, t_0)+2D_{2R}} \le \frac{c(n,\alpha )}{\sqrt{R}} \end{aligned}$$

for $R\ge 2$. Since $D_{2R}=o(\sqrt{R})$ by our assumption, the conclusion follows by letting $R$ to $\infty$.$\square$

When $Ric_f\ge 0$ and $f$ is locally bounded, we can prove another gradient estimate of $f$-heat equation. Its proof is only a weighted version of the case of Souplet and Zhang [26] with main difference is that we use the $f$-Laplacian comparison theorem of Wei and Wylie (Theorem 1.1(b) in [28]) instead of the classical Laplacian comparison theorem. Hence we only provide the result without the proof.

Theorem 3.1

Let $(M^n,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space. Fix $x_0\in M$ and $R>0$. Assume that $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$ in $B(x_0,R)$ and

$$\begin{aligned} A=A(R):=\sup _{x\in B(x_0,R)}|f(x)|. \end{aligned}$$

Suppose that $u(x,t)$ is a positive smooth solution to the $f$-heat equation (1.2) in $Q_{R,T}\equiv B(x_0,R)\times [t_0-T,t_0]\subset M\times (-\infty ,\infty )$, where $t_0\in \mathbb {R}$ and $T>0$. Then there exists a dimensional constant $c(n)$ such that

$$\begin{aligned} |\nabla \ln u|\le c(n)\left( \frac{1+A}{R}+\frac{1}{\sqrt{t-t_0+T}}+\sqrt{K}\right) \left( 1+ \ln \frac{D}{u}\right) \end{aligned}$$

in $Q_{R/2, T}$ with $t\ne t_0-T$, where $D:=\sup _{(x,t)\in Q_{R,T}}u(x,t)$.

As an application of Theorem 3.1, we can derive another Liouville theorem of the $f$-heat equation. Since its proof is also similar to Souplet and Zhang’s case [26], we omit the proof.

Theorem 3.2

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space with $Ric_f\ge 0$ and bounded $f$.

• (1) If $u$ is a positive ancient solution to the $f$-heat equation (that is, a solution defined in all space and negative time) such that $u(x,t)=e^{o\,(r(x)+|t|^{1/2}\,)}$ near infinity, then $u$ is constant.

• (2) If $u$ is an ancient solution to the $f$-heat equation such that $u(x,t)=o\,(r(x)+|t|^{1/2}\,)$ near infinity, then $u$ is constant.

4 Hamilton’s gradient estimate

In this section, inspired by the work of Kotschwar [14] and Brighton [5], we will prove a global Hamilton’s gradient estimate for $f$-heat equation when Bakry–Émery Ricci tensor is bounded below. First, we give a local estimate for $f$-heat equation, which is important for deriving the Hamilton’s gradient estimate.

Theorem 4.1

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space. Fix $x_0\in M$ and $R\ge 2$. Assume that $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$ in $B(x_0,R)$. If $0<u(x,t)\le D$ for some constant $D$, is a smooth solution to the $f$-heat equation (1.2) in $B(x_0,R)\times [0,T]$, where $T>0$, then there exists absolute constants $C_5$ and $C_6$ such that

$$\begin{aligned} t|\nabla u|^2\le C_5D^2\left[ 1+C_6T\left( (n-1)K+\frac{\alpha }{R}+\frac{1}{R^2}\right) \right] \end{aligned}$$

(4.1)

in $B(x_0,R/2)\times [0,T]$, where $\alpha :=\max _{\{x|d(x,x_0)=1\}}\Delta _f\,r(x)$ and $r(x)=d(x,x_0)$.

To prove Theorem 4.1, we employ a technique of Shi [25] from the estimation of derivatives of curvature under the Ricci flow (see also [10]). Define

$$\begin{aligned} F(x,t):=(4D^2+u^2)|\nabla u|^2 \end{aligned}$$

for $t>0$. Then

Lemma 4.2

There exist positive absolute constants $C_1$ and $C_2$ such that

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) F \le C_1(n-1)KF-\frac{C_2}{D^4}F^2. \end{aligned}$$

Proof of Lemma 4.2

Direct computation yields

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) u^2 =-2|\nabla u|^2. \end{aligned}$$

and

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) |\nabla u|^2=-2|\nabla ^2 u|^2-2Ric_f(\nabla u, \nabla u). \end{aligned}$$

Since $Ric_f\ge -(n-1)K$, then

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) F\le & {} (4D^2+u^2)\left[ -2|\nabla ^2 u|^2+2(n-1)K|\nabla u|^2\right] \nonumber \\&-2|\nabla u|^4-8u(\nabla \nabla u)(\nabla u,\nabla u)\nonumber \\\le & {} -10u^2|\nabla ^2 u|^2 + 10(n-1)D^2K|\nabla u|^2- 2|\nabla u|^4\nonumber \\&-8u(\nabla \nabla u)(\nabla u,\nabla u). \end{aligned}$$

Since

$$\begin{aligned} 8|u||\nabla \nabla u||\nabla u|^2 \le 10u^2|\nabla ^2 u|^2 + \frac{8}{5}|\nabla u|^4, \end{aligned}$$

and $4D^2 |\nabla u|^2\le F \le 5D^2|\nabla u|^2$, we have that

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) F&\le 10(n-1)D^2K|\nabla u|^2 -\frac{2}{5}|\nabla u|^4\\&\le \frac{5}{2}(n-1)KF -\frac{2F^2}{625D^4}. \end{aligned}$$

This finishes the proof of lemma.$\square$

As the Li–Yau argument in [15], we then take a $C^2$ cut-off function $\bar{\psi }$ defined on $[0,\infty )$ such that $\bar{\psi }(s)=1$ for $s\in [0,1/2]$, $\bar{\psi }(s)=0$ for $s\in [1,\infty )$, and $0\le \bar{\psi }(s)\le 1$. And $\bar{\psi }$ satisfies

$$\begin{aligned} -C_1\le \frac{\bar{\psi }'(s)}{\bar{\psi }^{1/2}(s)}\le 0\quad \mathrm {and}\quad \bar{\psi }''(s)\ge -C_2 \end{aligned}$$

for some positive absolute constants $C_1$ and $C_2$. Denote by $r(x)$ the distance function from the fixed point $x_0\in M$ to $x$. Set $\psi (x):=\bar{\psi }\left( \frac{r(x)}{R}\right)$. Using an argument of Calabi [6] (see also Cheng-Yau [9]), we can assume $\psi (x) \in C^2(M)$ with support in $B(x_0,R)$. Direct calculation shows that

$$\begin{aligned} \frac{|\nabla \psi |^2}{\psi }\le \frac{C_3}{R^2} \end{aligned}$$

(4.2)

for some absolute constant $C_3>0$, and

$$\begin{aligned} \Delta _f\psi =\frac{\bar{\psi }' \Delta _f\,r}{R}+\frac{\bar{\psi }''|\nabla r|^2}{R^2}. \end{aligned}$$

(4.3)

in $B(x_0,R)$.

Now, applying Lemma 4.2 and the above cut-off function, we give a proof of Theorem 4.1.

Proof of Theorem 4.1

Define $F$ as in Lemma 4.2. On $\mathrm {supp}(\psi )\times [0,T]$, by Lemma 4.2, we have

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) (t\psi F)= \psi F + t\psi \left( \frac{\partial }{\partial t}-\Delta _f\right) F - tF\Delta _f\,\psi -2t\left\langle \nabla \psi , \nabla F\right\rangle . \end{aligned}$$

(4.4)

If $t\psi F$ is not identically zero (i.e., if $u$ is not constant on $\mathrm {supp}(\psi )$), then $t\psi F$ attains a positive maximum at $(x_1,t_1)\in B(x_0,R)\times (0,T]$. In the following we will evaluate (4.4) at $(x_1,t_1)$ and solving for the $t\psi F$ term. At this point, we notice that

$$\begin{aligned} \nabla (t\psi F)=0\quad \mathrm {and}\quad \left( \frac{\partial }{\partial t}-\Delta _f\right) (t\psi F) \ge 0. \end{aligned}$$

Now we note that if $x_1\in B(x_0,1)$, then since $\psi =1$ on $B(x_0,R/2)$, where $R\ge 2$. So (4.4) yields

$$\begin{aligned} \frac{C_2}{D^4}t_1 F \le 1+C_1(n-1)KT \end{aligned}$$

at $(x_1,t_1)$. At this time, $(x_1,t_1)$ is also the maximum point of $t F$ in $B(x_0,R/2)\times (0,T]$. Hence, for any $(x,t)\in B(x_0,R/2)\times [0,T]$,

$$\begin{aligned} tF(x,t)\le t_1 F(x_1,t_1)\le \frac{D^4}{C_2}\left[ 1+ C_1(n-1)KT\right] . \end{aligned}$$

Also notice that $|\nabla u|^2\le \frac{F}{4 D^2}$, and the theorem follows.

Thus we may assume $x_1\not \in B(x_0,1)$. At this time, since $Ric_f\ge -(n-1)K$ and $r(x_1,x_0)\ge 1$ in $B(x_0,R)$, the $f$-Laplacian comparison theorem (Theorem 3.1 in [28]) implies

$$\begin{aligned} \Delta _f\,r(x_1)\le \alpha +(n-1)K(R-1), \end{aligned}$$

where $\alpha :=\max _{\{x|d(x,x_0)=1\}}\Delta _f\,r(x)$. Hence at $x_1$, (4.3) implies

$$\begin{aligned} \Delta _f \psi \ge -\frac{C_1}{R}\left[ \alpha +(n-1)K(R-1)\right] -\frac{C_2}{R^2} \end{aligned}$$

(4.5)

for some positive constants $C_1$ and $C_2$. Now by Lemma 4.2, (4.2) and (4.5), we have

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) (t\psi F)&= \psi F + t\psi \left( \frac{\partial }{\partial t}-\Delta _f\right) F - tF\Delta _f\,\psi -2t\left\langle \nabla \psi , \nabla F\right\rangle \nonumber \\&= \left( \psi + 2t\frac{|\nabla \psi |^2}{\psi } - t\Delta _f\, \psi \right) F + t\psi \left( \frac{\partial }{\partial t}-\Delta _f\right) F\\&\quad -2\left\langle \nabla (t\psi F), \frac{\nabla \psi }{\psi }\right\rangle \\&\le \left[ \left( 1+C_1(n-1)Kt\right) \psi +\frac{C_4t}{R^2}+\frac{C_1\alpha t}{R} + C_1(n-1)K\frac{R-1}{R}t\right] F\\&\quad -\frac{C_2}{D^4}t\psi F^2-2\left\langle \nabla (t\psi F), \frac{\nabla \psi }{\psi }\right\rangle . \end{aligned}$$

Noticing that at the point $(x_1,t_1)$,

$$\begin{aligned} \nabla (t\psi F)=0\quad \mathrm {and}\quad \left( \frac{\partial }{\partial t}-\Delta _f\right) (t\psi F) \ge 0, \end{aligned}$$

so we have

$$\begin{aligned} \frac{C_2}{D^4}t_1\psi F^2 \le \left[ 1+2C_1(n-1)KT +\frac{C_4}{R^2}T+\frac{C_1\alpha }{R} T\right] F. \end{aligned}$$

Consequently, for any $(x,t)\in B(x_0,R/2)\times [0,T]$,

$$\begin{aligned} tF(x,t)&= tF(x,t)\psi (x)\\&\le t_1 F(x_1,t_1)\psi (x_1) \\&\le \frac{D^4}{C_2}\left[ 1+2C_1(n-1)KT+\frac{C_4}{R^2}T+\frac{C_1\alpha }{R} T\right] . \end{aligned}$$

But $|\nabla u|^2\le \frac{F}{4 D^2}$, and the theorem follows.$\square$

Theorem 4.1 has the following corollaries.

Corollary 4.3

If $Ric_f\ge 0$ and $0<u\le D$ in $B(x_0,R)\times [0,T]$, $R\ge 2$, then there exists a absolute constant $C_5$ such that

$$\begin{aligned} t|\nabla u|^2\le C_5 D^2\left( 1+\frac{\alpha T}{R}+\frac{T}{R^2}\right) \end{aligned}$$

in $B(x_0,R/2)\times [0,T]$, where $\alpha :=\max _{\{x|d(x,x_0)=1\}}\Delta _f\,r(x)$. In particular, if $0<u\le D$ in defined in $M\times [0,T]$, then

$$\begin{aligned} t|\nabla u|^2\le C_5D^2 \end{aligned}$$

in $M\times [0,T]$.

Corollary 4.4

Suppose the solution $0<u\le D$ defined in $M\times [0,T]$. If $Ric_f\ge -(n-1)K$ for some constant $K\ge 0$ in $M$, then there exists a absolute constant $C_5$ such that

$$\begin{aligned} t|\nabla u|^2\le C_5D^2\left( 1+(n-1)KT\right) \end{aligned}$$

in $M\times [0,T]$.

To prove Theorem 1.6, we also need the following lemma, which is a mild extension of the manifold case due originally to Karp and Li [13] (see also Ni and Tam [23])

Lemma 4.5

Let $(M,g,e^{-f}dv)$ be an $n$-dimensional complete smooth metric measure space. Suppose $h(x,t)$ is a smooth function in $M\times [0,T]$, $0<T<\infty$, such that

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) h(x,t)\le 0 \quad \mathrm {whenever}\quad h(x,t)\le 0. \end{aligned}$$

Let $h_+(x,t):=\max \{h(x,t),0\}$ and $r(x)$ be the distance to $x$ from some fixed $x_0\in M$. Assume that

$$\begin{aligned} \int ^T_0\int _Me^{-\alpha r^2(x)}h^2_+(x,t)e^{-f}dvdt<\infty \end{aligned}$$

(4.6)

for some constant $\alpha >0$. If $h(x,0)\le 0$ for all $x\in M$, then $h(x,t)\le 0$ for all $(x,t)\in M\times [0,T]$.

Finally, we apply the above gradient estimates and the maximum principle to give a global Hamilton’s gradient estimate of $f$-heat equation in complete noncompact smooth metric measure spaces.

Proof of Theorem 1.6

Let $u_\epsilon =u+\epsilon$ for $\epsilon >0$. Then we have a solution, which satisfies $\epsilon <u_\epsilon \le D_\epsilon$, where $D_\epsilon :=D+\epsilon$. Once the estimate has been proved for $u_\epsilon$, the theorem immediately follows by letting $\epsilon \rightarrow 0$.

Similar to the arguments in [11], we let a function

$$\begin{aligned} P(x,t):=\varphi (t)\frac{|\nabla u_\epsilon |^2}{u_\epsilon } -u_\epsilon \ln \left( \frac{D_\epsilon }{u_\epsilon }\right) , \end{aligned}$$

where $\varphi (t):=\frac{t}{1+2(n-1)Kt}$, which satisfies

$$\begin{aligned} \left( \frac{\partial }{\partial t}-\Delta _f\right) P(x,t)\le 0 \end{aligned}$$

and

$$\begin{aligned} P(x,0)=-u_\epsilon \ln (D_\epsilon /u_\epsilon )\le 0. \end{aligned}$$

By our assumptions on $u_\epsilon$, we also have

$$\begin{aligned} P_+(x,t) \le \frac{t}{\epsilon }|\nabla u_\epsilon |^2. \end{aligned}$$

If $Ric_f\ge -(n-1)K$ for some constant $K>0$, then using Corollary 4.4, for any $x\in B(x_0,R)$, and $R>0$, we have

$$\begin{aligned}&\int _0^T\int _{B(x_0,R)}e^{-\alpha r^2(x)}P_{+}^2(x,t)\,e^{-f}dvdt \le \frac{1}{\epsilon ^2}\int _0^T\int _{B(x_0,R)}e^{-\alpha r^2(x)} (t|\nabla u_\epsilon |^2)^2e^{-f}dvdt\\&\quad \le \frac{C_5^2D_\epsilon ^4}{\epsilon ^2} \left( 1+(n-1)KT\right) ^2\int _0^T\int _{M}\,e^{-\alpha r^2(x)}\,e^{-f}dv dt. \end{aligned}$$

It follows from the Bishop volume comparison theorem due to Wei-Wylie (see Theorem 4.1 in [27]) that the rightmost integral in the inequality above is finite as long as $\alpha$ is large enough. Hence

$$\begin{aligned} \int _0^T\int _{M}\,e^{-\alpha r^2(x)}P_{+}^2(x,t)\,e^{-f}dvdt&\le \liminf _{R\rightarrow \infty }\int _0^T\int _{B(x_0,R)}\,e^{-\alpha r^2(x)}P_{+}^2(x,t)\,e^{-f}dv dt\\&<\infty , \end{aligned}$$

and we conclude that $P(x,t)\le 0$ for all $t\le T$.$\square$

In the rest of this section, we will use the arguments of Kotschwar [14] and prove Theorem 1.7.

Proof of Theorem 1.7

Let $H(x,y,t)$ be the heat kernel of $f$-heat equation. Since $Ric_f\ge 0$ and $|f|\le A$ for some constant $A$. By Theorem 1.2 in [29], there exist constants $c_1, c_2, c_3, c_4>0$ depending only on $n$ and $A$ such that

$$\begin{aligned} \frac{c_2}{V_f(B_y(\sqrt{t}))} \exp \left( -c_1\frac{d^2(x,y)}{t}\right) \le H(x, y, t) \le \frac{c_4}{V_f(B_y(\sqrt{t}))} \exp \left( -c_3\frac{d^2(x,y)}{t}\right) \end{aligned}$$

(4.7)

for all $x, y \in M$ and $t>0$. We would like to point out that the above $f$-heat kernel estimate is global. The reason is that our curvature conditions make sure the global $f$-volume doubling property and the global $f$-Neumann Poincaré inequality, which are equivalent to the global Gaussian bounds on the $f$-heat kernel (4.7).

For any $t>0$ and $y\in M$, we set $u(x,s):=H(x,y,s+t/2)$, and then $u$ is a smooth, positive solution to the heat equation on $[0,T)$. By (4.7), we have

$$\begin{aligned} \frac{c_2}{V_f(B_y(\sqrt{s+t/2}))}\exp \left( -c_1\frac{d^2(x,y)}{s+t/2}\right) \le u(x,s)\le \frac{c_4}{V_f(B_y(\sqrt{s+t/2}))} \end{aligned}$$

(4.8)

for all $x,y\in M$, and $s\ge 0$. Letting

$$\begin{aligned} D:=\frac{c_4}{V_f(B_y(\sqrt{t/2}))}, \end{aligned}$$

then the latter part of inequality (4.8) implies $u\le D$ for all $x$ and $s$. Since our theorem assumptions imply the global $f$-volume doubling property (see (2.2) in [29]), then there exists a positive constant $c_5:=c_5(n,A)$ such that

$$\begin{aligned} V_f(B_y(\sqrt{s+t/2}))\le V_f(B_y(\sqrt{t})) \le c_5 V_f(B_y(\sqrt{t/2})) \end{aligned}$$

for all $0\le s\le t/2$. Thus, by the front part of inequality (4.8) and Theorem 1.6, we have

$$\begin{aligned} s|\nabla \ln u|^2\le \ln \left( \frac{D}{u}\right) \le \ln \left( \frac{c_4c_5}{c_2}\right) +c_1\frac{d^2(x,y)}{(s+t/2)} \end{aligned}$$

on $M\times [0,t/2]$. Choosing at $s=t/2$, from above, the desired result follows.$\square$