An $L^1$-Estimate for Certain Spectral Multipliers Associated with the Ornstein–Uhlenbeck Operator

Abstract

We study a class of spectral multipliers $\phi (L)$ for the Ornstein–Uhlenbeck operator L arising from the Gaussian measure on $\mathbb {R}^n$ and find a sufficient condition for integrability of $\phi (L)f$ in terms of the admissible conical square function and a maximal function.

1 Introduction

On the Euclidean space $\mathbb {R}^n$, the Ornstein–Uhlenbeck operator

$$\begin{aligned} L = -\frac{1}{2} \Delta + x\cdot \nabla \end{aligned}$$

is associated with the Gaussian measure

$$\begin{aligned} d\gamma (x) = \pi ^{-n/2} e^{-|x|^2}dx \end{aligned}$$

by the Dirichlet form

$$\begin{aligned} \int _{\mathbb {R}^n} (Lf) g \, d\gamma = \frac{1}{2} \int _{\mathbb {R}^n} \nabla f \cdot \nabla g \, d\gamma , \quad f,g\in C^\infty _0(\mathbb {R}^n) . \end{aligned}$$

It has discrete spectrum $\sigma (L) = \{ 0,1,2,\ldots \}$ on $L^2(\gamma )$, and an orthonormal basis of eigenfunctions is given by Hermite polynomials $h_\beta$, $\beta \in \mathbb {N}^n$, so that $Lh_\beta = |\beta | h_\beta$. Moreover, it generates a (positive) diffusion semigroup on $L^2(\gamma )$ which can be expressed as

$$\begin{aligned} e^{-tL}f(x) = \int _{\mathbb {R}^n} M_t(x,y) f(y) \, d\gamma (y) , \quad t>0 , \end{aligned}$$

by means of the Mehler kernel (see [12])

$$\begin{aligned} M_t(x,y) = \frac{1}{(1-e^{-2t})^{n/2}} \exp \left( -\frac{e^{-t}}{1-e^{-2t}} |x-y|^2 \right) \exp \left( \frac{e^{-t}}{1+e^{-t}} (|x|^2 + |y|^2) \right) . \end{aligned}$$

The semigroup is contractive on $L^p(\gamma )$ for each $1\le p \le \infty$, and acts conservatively so that $e^{-tL}1 = 1$. Therefore, Stein’s theory [13] is applicable in studying the boundedness of spectral multipliers $\phi (L)$ defined as $\phi (L)h_\beta = \phi (|\beta |) h_\beta$ for $\beta \in \mathbb {N}^n$. More precisely, [13, Corollary 3, p. 121] guarantees $L^p(\gamma )$-boundedness with $1 < p < \infty$, for any spectral multiplier of ‘Laplace transform type’, i.e. of the form

$$\begin{aligned} \phi (\lambda ) = \lambda \int _0^\infty \Phi (t) e^{-t\lambda } \, dt , \quad \lambda \ge 0, \end{aligned}$$

where $\Phi : (0,\infty ) \rightarrow \mathbb {C}$ is bounded. In particular, the imaginary powers $L^{i\tau }$, $\tau \in \mathbb {R}$, arising from $\Phi (t) = t^{-i\tau } / \Gamma (1-i\tau )$ are bounded on $L^p(\gamma )$. The general theory was improved by Cowling [3] who showed that, for a given $p\in (1,\infty )$, $\phi (L)$ is bounded on $L^p(\gamma )$ as soon as $\phi$ extends analytically to a sector of angle greater than $\pi |1/p - 1/2|$. (See also the more recent development [2].)

The $L^1$-theory in the Gaussian setting is quite problematic. Although finite linear combinations of Hermite polynomials are dense in $L^1(\gamma )$, the spectral projections onto their eigenspaces are not $L^1(\gamma )$-bounded. Moreover, $tLe^{-tL}$ is bounded (uniformly) on $L^p(\gamma )$ only when $1 < p < \infty$ (see [7, Chap. 5]).

A Gaussian weak (1, 1)-type estimate for spectral multipliers of Laplace transform type was established by García-Cuerva et al. [5]. Moreover, in [4] they showed that requiring analyticity of $\phi$ in a sector of angle smaller than $\arcsin |1-2/p|$ will not alone suffice for boundedness of $\phi (L)$ on $L^p(\gamma )$. Observing that $\arcsin |1-2/p| \rightarrow \pi /2$ as $p\rightarrow 1$ is in line with the fact that the spectrum of L on $L^1(\gamma )$ is the (closed) right half-plane. Furthermore, $L^1(\gamma )$-boundedness of dilation invariant spectral multipliers for L was characterised in [6, Theorem 3.5(ii)].

The main obstruction in developing a metric theory of Hardy spaces in the Gaussian setting arises from the fact that the rapidly decaying measure $\gamma$ is non-doubling, that is, for every $t>0$

$$\begin{aligned} \frac{\gamma (B(x,2t))}{\gamma (B(x,t))} \longrightarrow \infty , \quad as \quad |x|\rightarrow \infty . \end{aligned}$$

Mauceri and Meda overcame this problem in [10] and developed an atomic theory for a Gaussian Hardy space which relies of the fact that the Gaussian measure behaves well locally with respect to the admissibility function

$$\begin{aligned} m(x) = \min (1 , |x|^{-1}) , \quad x\in \mathbb {R}^n . \end{aligned}$$

Indeed, $\gamma$ is doubling on families of ‘admissible’ Euclidean balls

$$\begin{aligned} \mathscr {B}_\alpha = \{ B(x,t) : 0 < t \le \alpha m(x) \} , \quad \alpha \ge 1, \end{aligned}$$

in the sense that for all $\lambda \ge 2$ we have

$$\begin{aligned} \gamma (\lambda B) \le e^{4\lambda ^2\alpha ^2} \gamma (B) , \quad B\in \mathscr {B}_\alpha . \end{aligned}$$

(1)

Other natural objects that are suitable for defining Hardy spaces, namely maximal functions and square functions, were studied in the Gaussian setting by Maas et al. In [8] they considered (a version¹ of) the admissible conical square function

$$\begin{aligned} Sf(x) = \left( \int _0^{2m(x)} \frac{1}{\gamma (B(x,t))} \int _{B(x,t)} |t^2L e^{-t^2L}f(y)|^2 \, d\gamma (y) \, \frac{dt}{t} \right) ^{1/2} , \quad x\in \mathbb {R}^n, \end{aligned}$$

and showed that it is controlled by a non-tangential semigroup maximal function. The converse inequality was presented in [11] along with a proof that the Riesz transform satisfies $\Vert \nabla L^{-1/2} f \Vert _1 \lesssim \Vert Sf \Vert _1 + \Vert f \Vert _1$. The benefit of conical objects (as opposed to vertical ones) is the applicability of tent space theory, which in the Gaussian setting was initiated in [9] and further developed by Amenta and Kemppainen [1].

The aim of this paper is to examine the decomposition method presented in [11] and to see what kind of $L^1$-estimates one can obtain for spectral multipliers $\phi (L)f$ in terms of the admissible conical square function Sf and other relevant objects. The hope is that these considerations will help in developing a fully satisfactory theory of Gaussian Hardy spaces.

Theorem

Let

$$\begin{aligned} \phi (\lambda ) = \int _0^\infty \Phi (t) (t\lambda )^2 e^{-t\lambda } \, \frac{dt}{t} , \quad \lambda \ge 0, \end{aligned}$$

where $\Phi : (0,\infty ) \rightarrow \mathbb {C}$ is twice continuously differentiable and satisfies

$$\begin{aligned} \sup _{0<t<\infty } (|\Phi (t)| + t|\Phi '(t)| + t^2|\Phi ''(t)|) + \int _1^\infty (|\Phi '(t)| + t|\Phi ''(t)|) \, dt < \infty . \end{aligned}$$

Then, for all $f\in L^1(\gamma )$, we have

$$\begin{aligned} \Vert \phi (L)f \Vert _1 \lesssim \Vert Sf \Vert _1 + \Vert f \Vert _1 + \Vert (1+\log _+ |\cdot |) \, Mf \Vert _1 , \end{aligned}$$

where

$$\begin{aligned} Mf(x) = \sup _{\varepsilon m(x)^2 < t \le 1} |e^{-tL}f(x)|, \quad x\in \mathbb {R}^n , \end{aligned}$$

and $\varepsilon > 0$ does not depend on f.

Remarks

Several remarks are in order:

• (1) The term $\Vert (1+\log _+ |\cdot |) \, Mf \Vert _1$ is highly undesirable for two reasons. Firstly, the maximal operator M is of a non-admissible kind in the sense that it is not restricted to times $t\lesssim m(\cdot )$. Secondly, the weight factor $(1+\log _+ |\cdot |)$, which arises from the admissibility function m, seems problematic. However, it is difficult to see how the appearance of the term could be avoided. Notice, nevertheless, that $\Vert (1+\log _+ |\cdot |) \, Mf \Vert _1$ is finite at least if $f\in L^p(\gamma )$ with $1 < p < \infty$.

• (2) The operators in the theorem above are special kind of Laplace type multipliers;

  $$\begin{aligned} \phi (\lambda ) = \int _0^\infty \Phi (t) (t\lambda )^2 e^{-t\lambda } \, \frac{dt}{t} = \lambda \int _0^\infty (\Phi (t) + t\Phi '(t)) e^{-t\lambda } \, dt , \quad \lambda \ge 0, \end{aligned}$$

  and therefore bounded on $L^p(\gamma )$ when $1 < p < \infty$. Note that if, in addition, we had

  $$\begin{aligned} \int _0^1 (|\Phi '(t)| + t|\Phi ''(t)|) \, dt < \infty , \end{aligned}$$

  then $\phi (L)$ would be bounded even on $L^1(\gamma )$. Indeed, using integration by parts we have

  $$\begin{aligned} \phi (L)f = -\Phi (0)f + \int _0^\infty (2\Phi '(t) + t\Phi ''(t)) e^{-tL}f \, dt \end{aligned}$$

  so that $\Vert e^{-tL}f \Vert _1 \le \Vert f \Vert _1$ implies

  $$\begin{aligned} \Vert \phi (L)f \Vert _1 \lesssim \left( |\Phi (0)| + \int _0^\infty (|\Phi '(t)| + t|\Phi ''(t)|) \, dt \right) \Vert f \Vert _1 . \end{aligned}$$

• (3) An example of a multiplier satisfying the conditions of the theorem is the localized imaginary power arising from $\Phi (t) = t^{i\tau } \chi (t)$, where $\tau \in \mathbb {R}$ and $\chi$ is a smooth cutoff with, say, $1_{(0,1]} \le \chi \le 1_{(0,2]}$. Observe that for $0 < t \le 1$ we have $|\Phi '(t)| \eqsim t^{-1}$ and $|\Phi ''(t)| \eqsim t^{-2}$ so that

  $$\begin{aligned} \int _0^1 (|\Phi '(t)| + t|\Phi ''(t)|) \, dt = \infty . \end{aligned}$$

2 Proof of the Theorem

Strategy The proof of the theorem follows the decomposition method from [11]. Let us begin by introducing a discretized version of the admissibility function

$$\begin{aligned} \widetilde{m}(x) = {\left\{ \begin{array}{ll} 1, &{}|x|<1, \\ 2^{-k}, &{}2^{k-1} \le |x| < 2^k , \quad k\ge 1, \end{array}\right. } \end{aligned}$$

and write $\widetilde{\mathscr {B}}_\alpha$ for the associated family of admissible balls. From $\widetilde{m} \le m \le 2\widetilde{m}$ it follows that $\widetilde{\mathscr {B}}_\alpha \subset \mathscr {B}_\alpha \subset \widetilde{\mathscr {B}}_{2\alpha }$. This discretization is relevant for Proposition 7.

We define the Gaussian tent space adapted to this new admissibility function as the space $\mathfrak {t}^1(\gamma )$ of functions u on the admissible region $D = \{ (y,t)\in \mathbb {R}^n \times (0,\infty ) : 0 < t < \widetilde{m}(y) \}$ for which

$$\begin{aligned} \Vert u \Vert _{\mathfrak {t}^1(\gamma )} = \int _{\mathbb {R}^n} \left( \iint _{\Gamma (x)} |u(y,t)|^2 \, \frac{d\gamma (y) \, dt}{t\gamma (B(y,t))} \right) ^{1/2} d\gamma (x) < \infty . \end{aligned}$$

Here $\Gamma (x) = \{ (y,t)\in D : |y-x| < t \}$ is an admissible cone at $x\in \mathbb {R}^n$.

The main theorem of [1] guarantees that every $u\in \mathfrak {t}^1(\gamma )$ admits a decomposition into ‘atoms’ $a_k$ so that

$$\begin{aligned} u = \sum _k \lambda _k a_k , \quad with \quad \sum _k |\lambda _k| \eqsim \Vert u \Vert _{\mathfrak {t}^1(\gamma )} . \end{aligned}$$

Recall that atom is a function a on D associated with a ball $B\in \widetilde{\mathscr {B}}_5$ for which $supp \, a \subset B\times (0,r_B)$ and

$$\begin{aligned} \left( \int _0^{r_B} \Vert a(\cdot , t) \Vert _2^2 \, \frac{dt}{t} \right) ^{1/2} \le \gamma (B)^{-1/2} . \end{aligned}$$

For such a function, $\Vert a \Vert _{\mathfrak {t}^1(\gamma )} \lesssim 1$.

Let then $\phi$ and $\Phi$ be as in Theorem and let f be a polynomial. For any $\delta , \delta ' > 0$ and $\kappa \ge 1$ we can decompose $\phi (L)f$ into three parts as follows:

$$\begin{aligned} \phi (L)f&= c_{\delta ,\delta '} \int _0^\infty \Phi ((\delta '+\delta )t^2) (t^2L)^2 e^{-(\delta '+\delta )t^2L} f \, \frac{dt}{t} \\&= c_{\delta ,\delta '} \left( \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2L e^{-\delta ' t^2L} u(\cdot , t) \, \frac{dt}{t} \right. \\&\left. \quad +\, \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2L e^{-\delta ' t^2L} (1_{D^c}(\cdot , t) t^2L e^{-\delta t^2L}f) \, \frac{dt}{t} \right. \\&\left. \quad +\,\int _{\widetilde{m}(\cdot )/\kappa }^\infty \widetilde{\Phi }(t^2) (t^2L)^2 e^{-(\delta '+\delta )t^2L}f \, \frac{dt}{t} \right) \\&=: c_{\delta ,\delta '} ( \pi _1 u + \pi _2 f + \pi _3 f ), \end{aligned}$$

where $u(\cdot , t) = 1_D(\cdot , t) t^2L e^{-\delta t^2L}f$ and $\widetilde{\Phi } (t) = \Phi ((\delta '+\delta )t)$.

Now

$$\begin{aligned} \Vert \phi (L)f \Vert _1 \le |c_{\delta ,\delta '} | (\Vert \pi _1 u \Vert _1 + \Vert \pi _2 f \Vert _1 + \Vert \pi _3 f \Vert _1) , \end{aligned}$$

and the proof consists of estimating these three terms separately for sufficiently small $\delta > \delta ' > 0$ and large enough $\kappa \ge 1$.

Analysis of the three parts Proposition 2 deals with

$$\begin{aligned} \pi _1 u = \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2L e^{-\delta ' t^2L} u(\cdot , t) \, \frac{dt}{t} \end{aligned}$$

and relies on the following $L^2$–$L^2$-off diagonal estimate (cf. [11, Proposition 4.2] and [14]).

Lemma 1

There exists a constant $c_{od} > 0$ such that for $j=0,1$ we have

$$\begin{aligned} \Vert 1_{E'} (tL)^j e^{-tL} 1_E \Vert _{2\rightarrow 2} \lesssim \exp \left( -\frac{d(E,E')^2}{c_{od}t}\right) , \quad t>0, \end{aligned}$$

whenever $E,E'\subset \mathbb {R}^n$.

Proposition 2

Let $\kappa \ge 1$ and $0 < \delta \le 1$. For sufficiently small $\delta ' > 0$ we have $\Vert \pi _1 u \Vert _1 \lesssim \Vert u \Vert _{\mathfrak {t}^1(\gamma )}$. Moreover, the function $u(\cdot , t) = 1_D (\cdot , t) t^2L e^{-\delta t^2L}f$ satisfies $\Vert u \Vert _{\mathfrak {t}^1(\gamma )} \lesssim \Vert Sf \Vert _1$.

Proof

By the atomic decomposition, it suffices to show that $\Vert \pi _1 a \Vert _1 \lesssim 1$ for any atom a associated with a ball $B\in \widetilde{\mathscr {B}}_5$. Let us consider the annuli $C_k(B) = 2^{k+1}B \setminus 2^k B$ for $k\ge 1$, and $C_0(B) = 2B$. By Hölder’s inequality we have

$$\begin{aligned} \Vert \pi _1 a \Vert _1&= \left\| \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2L e^{-\delta ' t^2L} a(\cdot , t) \, \frac{dt}{t} \right\| _1 \nonumber \\&\le \sum _{k=0}^\infty \gamma (2^{k+1}B)^{1/2} \left\| 1_{C_k(B)} \int _0^{r_B\wedge 2^{-k-1}/\kappa } \widetilde{\Phi } (t^2) t^2L e^{-\delta ' t^2L} a(\cdot , t) \, \frac{dt}{t} \right\| _2 . \end{aligned}$$

(2)

We estimate the norms on the right hand side of (2) by pairing with a $g\in L^2(\gamma )$ and relying on the assumption that $\Phi$ is bounded:

$$\begin{aligned} \begin{array}{ll} &{}\left| \int _{\mathbb {R}^n} \int _0^{r_B\wedge 2^{-k-1}/\kappa } \widetilde{\Phi } (t^2) t^2Le^{-\delta ' t^2L} a(\cdot , t) \, \frac{dt}{t} \, g \, d\gamma \right| \\ &{}= \left| \int _0^{r_B\wedge 2^{-k-1}/\kappa } \int _B a(\cdot , t) \widetilde{\Phi } (t^2) t^2Le^{-\delta ' t^2L} g \, d\gamma \, \frac{dt}{t} \right| \\ &{}\lesssim \left( \int _0^{r_B} \Vert a(\cdot , t) \Vert _2^2 \, \frac{dt}{t} \right) ^{1/2} \left( \int _0^{r_B} \Vert 1_B t^2L e^{-\delta ' t^2L} g \Vert _2^2 \, \frac{dt}{t} \right) ^{1/2} . \end{array} \end{aligned}$$

Now, for g supported in $C_0(B) = 2B$ we have

$$\begin{aligned} \begin{array}{ll} \left( \int _0^{r_B} \Vert 1_B t^2L e^{-\delta ' t^2L}g \Vert _2^2 \, \frac{dt}{t} \right) ^{1/2} &{}= \left( \sum \limits _{\beta \in \mathbb {N}^n} |\langle g , h_\beta \rangle |^2 \int _0^{r_B} (t^2|\beta |)^2 e^{-2\delta ' t^2|\beta |} \, \frac{dt}{t} \right) ^{1/2} \\ &{}\lesssim \Vert g \Vert _2 . \end{array} \end{aligned}$$

When $k\ge 1$ we have $d(C_k(B),B) \ge (2^k - 1)r_B \ge 2^{k-1} r_B$ and so, by Lemma 1, it follows that for $0 < t \le r_B$,

$$\begin{aligned} \Vert 1_B t^2L e^{-\delta ' t^2L} 1_{C_k(B)} \Vert _{2\rightarrow 2} \lesssim \exp \left( -\frac{4^{k-1}r_B^2}{c_{od}\delta ' t^2} \right) \lesssim \exp \left( -\frac{4^{k-2}}{c_{od}\delta '} \right) \left( \frac{t}{r_B}\right) ^{1/2}. \end{aligned}$$

Hence, for g supported in $C_k(B)$, $k\ge 1$, we have

$$\begin{aligned} \left( \int _0^{r_B} \Vert 1_B t^2L e^{-\delta ' t^2L} g \Vert _2^2 \, \frac{dt}{t}\right) ^{1/2} \lesssim \exp \left( -\frac{4^{k-2}}{c_{od}\delta '} \right) \Vert g \Vert _2 . \end{aligned}$$

We have therefore shown that, for $k\ge 0$,

$$\begin{aligned} \left\| 1_{C_k(B)} \int _0^{r_B} \widetilde{\Phi } (t^2) t^2L e^{-\delta ' t^2L} a(\cdot , t) \, \frac{dt}{t} \right\| _2 \lesssim \exp \left( -\frac{4^{k-2}}{c_{od}\delta '}\right) \gamma (B)^{-1/2} . \end{aligned}$$

According to the doubling inequality (1), we have $\gamma (2^{k+1}B)^{1/2} \lesssim e^{2\cdot 4^{k+1} \cdot 25} \gamma (B)^{1/2}$ and therefore

$$\begin{aligned}&\Vert \pi _1 a \Vert _1 \lesssim \sum _{k=0}^\infty \gamma (2^{k+1}B)^{1/2} \exp \left( -\frac{4^{k-2}}{c_{od}\delta '} \right) \gamma (B)^{-1/2}\\&\lesssim \sum _{k=0}^\infty \exp \left( 50\cdot 4^{k+1} - \frac{4^{k-2}}{c_{od}\delta '} \right) \lesssim 1 \end{aligned}$$

as soon as $\delta ' < 1/(3200c_{od})$. This proves the first claim.

For the second claim, let $u(\cdot , t) = 1_D (\cdot , t) t^2Le^{-\delta t^2L} f$. We perform a change of variable, $\delta t^2 = s^2$, i.e. $t = s / \sqrt{\delta }$ so that

$$\begin{aligned} (y,t)\in \Gamma (x)&\Leftrightarrow |y-x| < t < \widetilde{m}(y) \\&\Leftrightarrow |y-x| < s / \sqrt{\delta } < \widetilde{m}(y) \\&\Leftrightarrow (y,s) \in \Gamma '_{1/\sqrt{\delta }}(x) := \{ (y,s)\in D' : |y-x| < s/\sqrt{\delta } \} , \end{aligned}$$

where $D' := \{ (y,s) \in \mathbb {R}^n \times (0,\infty ) : s < \sqrt{\delta } \widetilde{m}(y) \}$. Now, change of aperture in the Gaussian tent space on $D'$ (see [1, Corollary 3.5]) guarantees that

$$\begin{aligned} \Vert u \Vert _{\mathfrak {t}^1(\gamma )}&= \int _{\mathbb {R}^n} \left( \iint _{\Gamma (x)} |t^2L e^{-\delta t^2 L}f(y)|^2 \,\frac{d\gamma (y)\, dt}{t\gamma (B(y,t))} \right) ^{1/2} d\gamma (x) \\&= \int _{\mathbb {R}^n} \left( \iint _{\Gamma '_{1/\sqrt{\delta }} (x)} |\delta ^{-1} s^2 L e^{-s^2L}f(y)|^2 \,\frac{d\gamma (y)\, ds}{s \gamma (B(y,s/\sqrt{\delta }))} \right) ^{1/2} d\gamma (x) \\&\lesssim \int _{\mathbb {R}^n} \left( \iint _{\Gamma '(x)} |s^2L e^{-s^2L}f(y)|^2 \,\frac{d\gamma (y)\, ds}{s \gamma (B(y,s))} \right) ^{1/2} d\gamma (x). \end{aligned}$$

We then observe (see [9, Lemma 2.3]) that for any $x,y\in \mathbb {R}^n$, $|y-x| < s < m(y)$ implies $s < 2m(x)$, and therefore

$$\begin{aligned} \Gamma '(x) \subset \Gamma (x) \subset \bigcup _{0<s<2m(x)} B(x,s) \times \{ s \} . \end{aligned}$$

Moreover, $\gamma (B(y,s)) \eqsim \gamma (B(x,s))$ when $|y-x| < s < \delta \widetilde{m}(y)$, and hence

$$\begin{aligned}&\iint _{\Gamma '(x)} |s^2L e^{-s^2L}f(y)|^2 \,\frac{d\gamma (y)\, ds}{s \gamma (B(y,s))}\\&\quad \lesssim \,\int _0^{2m(x)} \frac{1}{\gamma (B(x,s))}\int _{B(x,s)} |s^2L e^{-s^2L}f(y)|^2 \, d\gamma (y) \, \frac{ds}{s} \end{aligned}$$

for every $x\in \mathbb {R}^n$, which shows that $\Vert u \Vert _{\mathfrak {t}^1(\gamma )} \lesssim \Vert Sf \Vert _1$ as required.$\square$

For $\pi _2$ and $\pi _3$ (more precisely, for Proposition 5 and Lemma 6) we need the following two lemmas concerning pointwise kernel estimates.

Lemma 3

Let $j=0,1$. For all $x,y\in \mathbb {R}^n$ we have the pointwise kernel estimate

$$\begin{aligned} |t^j \partial _t^j M_t(x,y)| \lesssim t^{-n/2} \exp \left( -\frac{|x-y|^2}{8t} \right) \exp \left( \frac{|x|^2 + |y|^2}{2} \right) , \quad 0 < t \le 1 . \end{aligned}$$

As a consequence, for all $0 < t \le 1$ we have

$$\begin{aligned} \Vert 1_{E'} (tL)^j e^{-tL} 1_E \Vert _{1\rightarrow \infty } \lesssim t^{-n/2} \exp \left( -\frac{d(E,E')^2}{8t} \right) \sup _{\begin{array}{c} x\in E \\ y\in E' \end{array}} \exp \left( \frac{|x|^2 + |y|^2}{2} \right) , \end{aligned}$$

whenever $E,E'\subset \mathbb {R}^n$.

Proof

For $0 < t \le 1$ we have the elementary estimates

$$\begin{aligned} \frac{1}{1-e^{-2t}} \eqsim \frac{1}{t}, \quad \frac{1}{4t} \le \frac{e^{-t}}{1-e^{-2t}} \le \frac{1}{2t}, \quad \frac{1}{8} \le \frac{e^{-t}}{1 + e^{-t}} \le \frac{1}{2} \end{aligned}$$

and the case $j=0$ follows immediately:

$$\begin{aligned} \begin{array}{ll} M_t(x,y) &{}= \frac{1}{(1-e^{-2t})^{n/2}} \exp \left( -\frac{e^{-t}}{1-e^{-2t}} |x-y|^2 \right) \exp \left( \frac{e^{-t}}{1+e^{-t}} (|x|^2 + |y|^2) \right) \\ &{}\lesssim t^{-n/2} \exp \left( -\frac{|x-y|^2}{4t} \right) \exp \left( \frac{|x|^2 + |y|^2}{2} \right) . \end{array} \end{aligned}$$

For $j=1$ we calculate:

$$\begin{aligned}&\partial _t M_t(x,y) = \left( -\frac{ne^{-2t}}{1-e^{-2t}} + |x-y|^2 \frac{e^{-t}(1+e^{-2t})}{(1-e^{-2t})^2}\right. \\&\left. - (|x|^2 + |y|^2) \frac{e^{-t}}{(1+e^{-t})^2}\right) M_t(x,y) . \end{aligned}$$

Using the previous case $j=0$ we then see that

$$\begin{aligned} \begin{array}{ll} |t\partial _t M_t(x,y)| &{}\lesssim \left( 1 + \frac{|x-y|^2}{t} + |x|^2 + |y|^2 \right) M_t(x,y) \\ &{}\lesssim t^{-n/2} \exp \left( -\frac{|x-y|^2}{8t} \right) \exp \left( \frac{|x|^2 + |y|^2}{2} \right) . \end{array} \end{aligned}$$

The consequence is also immediate: for any $x\in E'$ we have

$$\begin{aligned} \begin{array}{ll} |(tL)^je^{-tL}f(x)| &{}\lesssim t^{-n/2} \int _E \exp \left( -\frac{|x-y|^2}{8t} \right) \exp \left( \frac{|x|^2 + |y|^2}{2}\right) |f(y)| \, d\gamma (y) \\ &{}\lesssim t^{-n/2} \exp \left( -\frac{d(E,E')^2}{8t} \right) \sup _{y\in E} \exp \left( \frac{|x|^2 + |y|^2}{2}\right) \int _E |f(y)| \, d\gamma (y) . \end{array} \end{aligned}$$

$\square$

Lemma 4

For $\alpha$ large enough there exists a constant $c>0$ such that for all $x,y\in \mathbb {R}^n$ and all $0 < t \le 1$ we have

$$\begin{aligned} M_{t/\alpha }(x,y) \lesssim \exp \left( -\frac{|x-y|^2}{ct} \right) \exp \left( \alpha t \min (|x|^2,|y|^2) \right) M_t(x,y) , \end{aligned}$$

and, consequently,

$$\begin{aligned} |t\partial _t M_{t/\alpha }(x,y)| \lesssim \exp \left( \alpha t \min (|x|^2,|y|^2) \right) M_t(x,y) . \end{aligned}$$

Proof

An alternative way to express the Mehler kernel is (see [12])

$$\begin{aligned} M_t(x,y) = \frac{1}{(1-e^{-2t})^{n/2}} \exp \left( -\frac{|e^{-t}x - y|^2}{1-e^{-2t}} \right) e^{|y|^2} . \end{aligned}$$

By [11, Lemma 3.4] for $\alpha$ large enough we have for all $x,y\in \mathbb {R}^n$ and all $0 < t \le 1$ that

$$\begin{aligned} \exp \left( -\frac{|e^{-t/\alpha }x - y|^2}{1-e^{-2t/\alpha }} \right) \le \exp \left( -2\frac{|e^{-t}x - y|^2}{1-e^{-2t}} \right) \exp \left( \frac{t^2\min (|x|^2,|y|^2)}{1-e^{-2t/\alpha }} \right) . \end{aligned}$$

Therefore

$$\begin{aligned} M_{t/\alpha }(x,y) \lesssim \exp \left( -\frac{|e^{-t}x - y|^2}{1-e^{-2t}} \right) \exp \left( \frac{t^2\min (|x|^2,|y|^2)}{1-e^{-2t/\alpha }} \right) M_t(x,y), \end{aligned}$$

where, by symmetry, the first exponential factor can be replaced by

$$\begin{aligned} \exp \left( -\frac{\max (|e^{-t}x - y|^2, |x-e^{-t}y|^2)}{1-e^{-2t}} \right) . \end{aligned}$$

The first claim now follows because for all $x,y\in \mathbb {R}^n$ and all $0 < t \le 1$ we have

$$\begin{aligned} |x-y|^2 \lesssim \max (|e^{-t}x - y|^2, |x-e^{-t}y|^2) . \end{aligned}$$

In order to see this, let us assume, with no loss of generality, that $|x|\le |y|$, and show that $|x-y|^2 \lesssim |e^{-t}x - y|^2$. Then

$$\begin{aligned} \begin{array}{ll} |x-y|^2 &{}\le e ( e^{-t}|x|^2 - 2e^{-t} x\cdot y + e^{-t}|y|^2) \\ &{}=e(e^{-t}|x|^2 - (1-e^{-t})|y|^2 - 2e^{-t}x\cdot y + |y|^2) , \end{array} \end{aligned}$$

where

$$\begin{aligned} e^{-t}|x|^2 - (1-e^{-t})|y|^2 \le e^{-2t}|x|^2 , \end{aligned}$$

because $|x| \le |y|$. Indeed,

$$\begin{aligned} e^{-t}|x|^2 - (1-e^{-t})|x|^2 - e^{-2t}|x|^2 =(2e^{-t} - 1 - e^{-2t})|x|^2 , \end{aligned}$$

where $2e^{-t} - 1 - e^{-2t} \le 0$ for all $t>0$.

The second claim now follows from the first one since

$$\begin{aligned} \begin{array}{ll} |t\partial _t M_{t/\alpha }(x,y)| &{}\lesssim \left( 1 + \frac{|x-y|^2}{t} + |x|^2 + |y|^2 \right) M_{t/\alpha }(x,y) \\ &{}\lesssim \exp \left( \alpha t \min (|x|^2,|y|^2) \right) M_t(x,y) . \end{array} \end{aligned}$$

Here the first inequality is obtained as in the proof of Lemma 3 (case $j=1$).$\square$

Let us then consider

$$\begin{aligned} \pi _2 f = \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2L e^{-\delta ' t^2L} (1_{D^c}(\cdot , t) t^2L e^{-\delta t^2L}f) \, \frac{dt}{t} . \end{aligned}$$

Proposition 5

Let $\kappa \ge 4$. For sufficiently small $\delta > \delta ' > 0$ we have $\Vert \pi _2f \Vert _1 \lesssim \Vert f \Vert _1$.

Proof

We begin by observing that if $t\le \widetilde{m}(x)/4$ and $t > 2^{-k-1}$ for some $k\ge 2$, then $|x| < 2^{k-2}$. Moreover, if $t\ge \widetilde{m}(y)$ and $t\le 2^{-k}$, then $|y| \ge 2^{k-1}$.

We then decompose $\pi _2f$ (using boundedness of $\Phi$) as follows:

$$\begin{aligned} \begin{array}{ll} \Vert \pi _2 f \Vert _1 &{}= \left\| \int _0^{\widetilde{m}(\cdot )/\kappa } \widetilde{\Phi }(t^2) t^2Le^{-\delta ' t^2L} (1_{D^c}(\cdot , t) t^2Le^{-\delta t^2L}f ) \, \frac{dt}{t} \right\| _1 \\ &{}\lesssim \sum \limits _{k=2}^\infty \int _{2^{-k-1}}^{2^{-k}} \Vert 1_{B(0,2^{k-2})} t^2L e^{-\delta ' t^2L} (1_{\mathbb {R}^n\setminus B(0,2^{k-1})} t^2Le^{-\delta t^2L}f ) \Vert _1 \, \frac{dt}{t}\\ &{}\le \sum \limits _{k=2}^\infty \sum \limits _{l=1}^\infty \int _{2^{-k-1}}^{2^{-k}} \Vert 1_{B(0,2^{k-2})} t^2L e^{-\delta ' t^2L} (1_{C_{k+l-1}} t^2Le^{-\delta t^2L}f) \Vert _1 \, \frac{dt}{t} , \end{array} \end{aligned}$$

(3)

where $C_{k+l-1} := B(0,2^{k+l-1})\setminus B(0,2^{k+l-2})$.

First, by Lemma 4, we choose a $\delta > 0$ such that for all $0 < t \le 1$ we have

$$\begin{aligned} |t^2L e^{-\delta t^2L}f(x)| \lesssim \exp \left( \frac{t^2|x|^2}{\delta } \right) |e^{-tL}f(x)| , \quad x\in \mathbb {R}^n . \end{aligned}$$

Hence, for $2^{-k-1} < t \le 2^{-k}$ we have

$$\begin{aligned} \Vert 1_{C_{k+l-1}} t^2L e^{-\delta t^2L}f \Vert _1 \lesssim \exp \left( \frac{4^{-k} \cdot 4^{k+l-1}}{\delta } \right) \Vert e^{-tL} f \Vert _1 \lesssim \exp \left( \frac{4^{l-1}}{\delta } \right) \Vert f \Vert _1 . \end{aligned}$$

Then, since the distance between $B(0,2^{k-2})$ and $C_{k+l-1}$ is at least $2^{k+l-3}$, we have, by Lemma 3, for $2^{-k-1} < t \le 2^{-k}$ that

$$\begin{aligned} \begin{array}{ll} \Vert 1_{B(0,2^{k-2})} t^2L e^{-\delta ' t^2L} 1_{C_{k+l-1}} \Vert _{1\rightarrow 1} &{}\lesssim t^{-n} \exp \left( -\frac{4^{k+l-3}}{8\delta ' t^2} \right) \exp \left( \frac{4^{k-2} + 4^{k+l-1}}{2} \right) \\ &{}\lesssim 2^{kn} \exp \left( -\frac{4^{2k+l-5}}{\delta '} + 4^{k+l-1} \right) . \end{array} \end{aligned}$$

Combining the two estimates we see that for $2^{-k-1} < t \le 2^{-k}$ we have

$$\begin{aligned} \begin{array}{ll} &{}\Vert 1_{B(0,2^{k-2})} t^2L e^{-\delta ' t^2L} (1_{C_{k+l-1}} t^2L e^{-\delta t^2L}f) \Vert _1 \\ &{}\lesssim 2^{kn} \exp \left( -\frac{4^{2k+l-5}}{\delta '} + 4^{k+l-1} + \frac{4^{l-1}}{\delta } \right) \Vert f \Vert _1 \\ &{}= 2^{kn} \exp \left( -4^{k+l+1} \left( \frac{4^{k-6}}{\delta '} - 4^{-2} - \frac{4^{-2}}{\delta } \right) \right) \Vert f \Vert _1 \\ &{}\lesssim \exp (-4^{k+l}) \Vert f \Vert _1 , \end{array} \end{aligned}$$

where in the last step we chose $\delta ' < \delta$ small enough.

The right-hand side of (3) is therefore dominated by

$$\begin{aligned} \sum _{k=2}^\infty \sum _{l=1}^\infty \exp (-4^{k+l}) \Vert f \Vert _1 \int _{2^{-k-1}}^{2^{-k}} \frac{dt}{t} \lesssim \Vert f \Vert _1 . \end{aligned}$$

$\square$

Lemma 6

For any $\alpha > 0$ we have

$$\begin{aligned} \Vert (e^{-tL}f)|_{t={\widetilde{m}(\cdot )^2/\alpha }} \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

Moreover, for $\alpha$ large enough we have

$$\begin{aligned} \Vert (tLe^{-tL}f)|_{t={\widetilde{m}(\cdot )^2/\alpha }} \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

Proof

Write $C_0 = B(0,1)$ and $C_k = B(0,2^k)\setminus B(0,2^{k-1})$ for $k\ge 1$. Moreover, let $C_0^* = B(0,2)$, $C_1^* = B(0,4)$, and $C_k^* = B(0,2^{k+1})\setminus B(0,2^{k-2})$ for $k\ge 2$.

We first show that for any $\alpha > 0$,

$$\begin{aligned} \Vert (e^{-tL}f)|_{t={\widetilde{m}(\cdot )^2}/\alpha } \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

(4)

Denote $\varepsilon = 1/\alpha$ for notational convenience. For $x\in C_k$ we have $\widetilde{m}(x)^2 = 4^{-k}$ and hence

$$\begin{aligned} \Vert (e^{-tL}f)|_{t=\varepsilon \widetilde{m}(\cdot )^2} \Vert _1 = \sum _{k=0}^\infty \Vert 1_{C_k} e^{-\varepsilon 4^{-k} L} f \Vert _1 . \end{aligned}$$

We split f into $1_{C_k^*}f$ and $1_{\mathbb {R}^n\setminus C_k^*} f$, and first estimate

$$\begin{aligned} \sum _{k=0}^\infty \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} (1_{C_k^*}f) \Vert _1 \le \sum _{k=0}^\infty \Vert 1_{C_k^*} f \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

Fixing an integer N for which $8\varepsilon \le 4^N$, we use the trivial estimate for $k=0,1,\ldots , N+3$:

$$\begin{aligned} \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} (1_{\mathbb {R}^n\setminus C_k^*} f) \Vert _1 \le \Vert f \Vert _1 . \end{aligned}$$

For $k\ge N+4$ we have the decomposition

$$\begin{aligned} \mathbb {R}^n \setminus C_k^* = B(0,2^{k-2}) \cup \bigcup _{l=2}^\infty C_{k+l} . \end{aligned}$$

Observing that $d(C_k , B(0,2^{k-2})) = 2^{k-2}$ we obtain, by Lemma 3,

$$\begin{aligned} \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} 1_{B(0,2^{k-2})} \Vert _{1\rightarrow 1}\lesssim & {} 2^{kn} \exp \left( - \frac{4^{k-2}}{8\varepsilon 4^{-k}} \right) \exp \left( \frac{4^k + 4^{k-2}}{2} \right) \\\le & {} 2^{kn} \exp ( - 4^{2k-2-N} + 4^k ) \\\lesssim & {} \exp (-4^k) . \end{aligned}$$

Furthermore, since $d(C_k , C_{k+l}) = 2^{k+l-2}$, Lemma 3 implies that

$$\begin{aligned} \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} 1_{C_{k+l}} \Vert _{1\rightarrow 1}\lesssim & {} 2^{kn} \exp \left( -\frac{4^{k+l-2}}{8\varepsilon 4^{-k}}\right) \exp \left( \frac{4^k + 4^{k+l}}{2} \right) \\\le & {} 2^{kn} \exp ( -4^{2k+l-2-N} + 4^{k+l}) \\\lesssim & {} \exp (-4^{k+l}) . \end{aligned}$$

Therefore,

$$\begin{aligned} \sum \limits _{k=N+4}^\infty \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} (1_{\mathbb {R}^n\setminus C_k^*} f) \Vert _1= & {} \sum \limits _{k=N+4}^\infty \left( \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} (1_{B(0,2^{k-2})} f) \Vert _1 \right. \\&\left. + \sum \limits _{l=2}^\infty \Vert 1_{C_k} e^{-\varepsilon 4^{-k}L} (1_{C_{k+l}} f) \Vert _1 \right) \\\lesssim & {} \sum \limits _{k=N+4}^\infty \left( \exp (-4^k) \Vert f \Vert _1 + \sum \limits _{l=2}^\infty \exp (-4^{k+l})\Vert f \Vert _1 \right) \\\lesssim & {} \Vert f \Vert _1. \end{aligned}$$

We have now proven (4) which includes the first case from the statement of the lemma.

The second case follows by using Lemma 4, which guarantees that there exists an $\alpha > 0$ such that for all $x,y\in \mathbb {R}^n$

$$\begin{aligned} \left| (t\partial _t M_t(x,y))|_{t=\widetilde{m}(x)^2/\alpha } \right| \lesssim \exp \left( \alpha \frac{\widetilde{m}(x)^2}{\alpha } |x|^2 \right) M_{\widetilde{m}(x)^2}(x,y) \lesssim M_{\widetilde{m}(x)^2}(x,y) . \end{aligned}$$

Then

$$\begin{aligned} \Vert (tL e^{-tL}f)|_{t=\widetilde{m}(\cdot )^2 / \alpha } \Vert _1 \lesssim \Vert (e^{-tL}f)|_{t=\widetilde{m}(\cdot )^2} \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

$\square$

Finally, we turn to

$$\begin{aligned} \pi _3f = \int _{\widetilde{m}(\cdot )/\kappa }^\infty \widetilde{\Phi }(t^2) (t^2L)^2 e^{-(\delta '+\delta )t^2L}f \, \frac{dt}{t} . \end{aligned}$$

Proposition 7

Let $0 < \delta ,\delta ' \le 1/2$. For $\kappa$ large enough we have $\Vert \pi _3f \Vert _1 \lesssim \Vert f \Vert _1 + \Vert (1+\log _+ |\cdot |) \, Mf \Vert _1$, where $Mf(x) = \sup _{\varepsilon m(x)^2<t\le 1} |e^{-tL}f(x)|$ and $\varepsilon > 0$ does not depend on f.

Proof

Integrating by parts we obtain

$$\begin{aligned}&\int _{\widetilde{m}(\cdot )/\kappa }^\infty \widetilde{\Phi } (t^2) (t^2L)^2 e^{-(\delta '+\delta )t^2L}f \, \frac{dt}{t} \\&\quad = c \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^\infty \widetilde{\Phi } (t) t\partial _t^2 e^{-(\delta '+\delta )tL}f \, dt \\&\quad = c \left[ \widetilde{\Phi } (t) t\partial _t e^{-(\delta '+\delta )tL}f \right] _{t=\widetilde{m}(\cdot )^2/\kappa ^2}^\infty \\&\qquad +\, c' \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^\infty (\widetilde{\Phi } (t) + t\widetilde{\Phi }'(t)) \partial _t e^{-(\delta '+\delta )tL}f \, dt. \end{aligned}$$

Repeating for the last term we get

$$\begin{aligned}&\int _{\widetilde{m}(\cdot )^2/\kappa ^2}^\infty (\widetilde{\Phi } (t) + t\widetilde{\Phi }'(t)) \partial _t e^{-(\delta '+\delta )tL}f \, dt \\&\quad = c \left[ (\widetilde{\Phi } (t) + t\widetilde{\Phi }'(t)) e^{-(\delta '+\delta )tL}f \right] _{t=\widetilde{m}(\cdot )^2/\kappa ^2}^\infty \\&\qquad + \, c' \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^\infty (2\widetilde{\Phi }'(t) + t\widetilde{\Phi }''(t)) e^{-(\delta '+\delta )tL}f \, dt . \end{aligned}$$

Now, having assumed that $\sup _{0 < t < \infty } (|\Phi (t)| + t|\Phi '(t)|) < \infty$, we may use Lemma 6 to choose $\kappa$ large enough so that

$$\begin{aligned} \left\| \left[ \widetilde{\Phi } (t) t\partial _t e^{-(\delta '+\delta )tL}f \right] _{t=\widetilde{m}(\cdot )^2/\kappa ^2}^\infty \right\| _1 \lesssim \Vert (tLe^{-(\delta '+\delta )tL}f)|_{t={\widetilde{m}(\cdot )^2/\kappa ^2}} \Vert _1 \lesssim \Vert f \Vert _1 \end{aligned}$$

and

$$\begin{aligned} \left\| \left[ (\widetilde{\Phi } (t) + t\widetilde{\Phi }'(t)) e^{-(\delta '+\delta )tL}f \right] _{t=\widetilde{m}(\cdot )^2/\kappa ^2}^\infty \right\| _1 \lesssim \Vert (e^{-(\delta '+\delta )tL}f)|_{t={\widetilde{m}(\cdot )^2/\kappa ^2}} \Vert _1 \lesssim \Vert f \Vert _1 . \end{aligned}$$

Moreover,

$$\begin{aligned}&\left\| \int _1^\infty (2\widetilde{\Phi }'(t) + t\widetilde{\Phi }''(t)) e^{-(\delta ' +\delta )tL}f \, dt \right\| _1\\&\quad \lesssim \int _1^\infty (|\widetilde{\Phi }'(t)| + t|\widetilde{\Phi }''(t)|) \Vert e^{-(\delta ' +\delta )tL}f \Vert _1 \, dt \\&\quad \lesssim \Vert f \Vert _1 . \end{aligned}$$

Finally, having assumed that $\sup _{0 < t < \infty } (t|\Phi '(t)| + t^2|\Phi ''(t)|) < \infty$, we get

$$\begin{aligned}&\left| \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^1 (2\widetilde{\Phi }'(t) + t\widetilde{\Phi }''(t)) e^{-(\delta '+\delta )tL}f \, dt \right| \\&\quad \lesssim \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^1 (|\widetilde{\Phi }'(t)| + t|\widetilde{\Phi }''(t)|) \, |e^{-(\delta '+\delta )tL}f| \, dt \\&\quad \lesssim \sup \limits _{\varepsilon m(\cdot )^2<t\le 1} |e^{-tL}f| \int _{\widetilde{m}(\cdot )^2/\kappa ^2}^1 \frac{dt}{t} \\&\quad \lesssim (1 + \log _+ |\cdot |) \, Mf , \end{aligned}$$

where $\varepsilon > 0$ is chosen small enough depending on $\delta$, $\delta '$ and $\kappa$.$\square$