---
abstract: 'This paper presents a closed-form expression for the integral kernels associated with the derivatives of the Ornstein-Uhlenbeck semigroup ${\mathrm{e}}^{tL}$. Our approach is to expand the Mehler kernel into Hermite polynomials and applying the powers $L^N$ of the Ornstein-Uhlenbeck operator to it, where we exploit the fact that the Hermite polynomials are eigenfunctions for $L$. As an application we give an alternative proof of the kernel estimates by [@Portal2014], making all relevant quantities explicit.'
address: |
    Delft Institute of Applied Mathematics, Delft University of Technology\
    P.O. Box 5031\
    2600 GA Delft\
    The Netherlands\
    j.j.b.teuwen@tudelft.nl
author:
- Jonas Teuwen
bibliography:
- 'library.bib'
title: 'On the integral kernels of derivatives of the Ornstein-Uhlenbeck semigroup'
---

Introduction
============

Much effort [@Harboure2000; @Kemppainen2015; @Maas2010b; @MaasNeervenPortal2011; @MauceriMeda2007; @MauceriMeda2012; @Muckenhaupt; @Pineda2008; @Portal2014; @Sjogren1997; @Teuwen2015] has gone into developing the harmonic analysis of the [*Ornstein-Uhlenbeck operator*]{} $$\label{eq:OU-operator}
 L := \frac12 \Delta - \langle x, \nabla \rangle.$$ On the space $L^2({\mathbf{R}}^d, {\mathrm{d}}\gamma)$, where $\gamma$ is the Gaussian measure $$\label{eq:Gaussian-measure}
 {\mathrm{d}}\gamma(x) := \pi^{-d/2} {\mathrm{e}}^{-|x|^2} {\mathrm{d}}{x},$$ this operator can be viewed as the Gaussian counterpart of the Laplace operator $\Delta$. Indeed, one has $L = -\nabla^*\nabla$, where $\nabla$ is the the usual gradient and the $\nabla^*$ is its adjoint in $L^2({\mathbf{R}}^d, {\mathrm{d}}\gamma)$. It is a classical fact that the semigroup operators $e^{tL}$, $t>0$, are integral operators of the form $${\mathrm{e}}^{tL} u(\cdot) = \int_{{\mathbf{R}}^d} M_t(\cdot, y) u(y) {\,\mathrm{d}}\gamma(y),$$ where $M_t$ is the so-called [*Mehler kernel*]{} [@Sjogren1997] $$\label{eq:Mehler-kernel-oneformula}
  \begin{aligned}
  M_t(x, y)
  & = \frac{\exp\Bigl(\displaystyle-\frac{|{\mathrm{e}}^{-t}x - y|^2}{1 - {\mathrm{e}}^{-2t}}
    \Bigr)}{(1 - {\mathrm{e}}^{-2t})^{d/2}} {\mathrm{e}}^{|y|^2}
  \end{aligned}$$ (see [@Teuwen2015] for a representation of $M_t$ which makes the symmetry in $x$ and $y$ explicit).

Developing a Hardy space theory for $L$ is the subject of active current research [@Portal2014; @MauceriMeda2007]. In this theory the derivatives $(d^k/dt^k)e^{t L} = L^N e^{tL}$ play an important role. The aim of the present paper is to derive closed form expressions for the integral kernels of these derivatives, that is, to determine explicitly the kernels $M_t^N$ such that we have the identity $$\label{eq:Mehler}
  L^N{\mathrm{e}}^{tL} u(\cdot)  = \int_{{\mathbf{R}}^d} M_t^N(\cdot, y) u(y) {\,\mathrm{d}}\gamma(y).$$ Direct application of the derivatives $d^N/dt^N$ to the Mehler kernel yields expressions which become intractible even for small values of $N$. Our approach will be to expand the Mehler kernel in terms of the $L^2$-normalised Hermite polynomials and then to apply $L^N$ to it, thus exploiting the fact that the Hermite polynomials are eigenfunctions for $L$.

As an application of our main result, which is proved in section \[sec:mainresult\] after developing some preliminary material in the sections \[sec:prelim\]-\[sec:Weyl\], we shall give a direct proof for the kernel bounds of [@Portal2014] in section \[sec:application\].

Preliminaries {#sec:prelim}
=============

In this preliminary section we collect some standard properties of Hermite polynomials and their relationship with the Ornstein-Uhlenbeck operator. Most of this material is classical and can be found in [@Sjogren1997; @MR1215939].

Hermite polynomials
-------------------

The [*Hermite polynomials*]{} $H_n$, $n\ge 0$, are defined by Rodrigues’s formula $$\label{eq:Hermite-Rodrigues}
  H_n(x) := (-1)^n {\mathrm{e}}^{x^2} \partial_x^n {\mathrm{e}}^{-x^2}.$$ Their $L^2$-normalizations, $$h_n := \frac{H_n}{\sqrt{2^{n} n!}}$$ form an orthonormal basis for $L^2({\mathbf{R}},{\mathrm{d}}\gamma)$. We shall use the fact that the generating function for the Hermite polynomials is given by $$\label{eq:Generating-function-identity} \sum_{n = 0}^\infty \frac{H_n(x)}{n!} t^n ={\mathrm{e}}^{2 tx - t^2}.$$ The relationship with the Ornstein-Uhlenbeck operator is encoded in the eigenvalue identity $L H_n = -n H_n$, from which it follows that for all $t \ge 0$ we have ${\mathrm{e}}^{tL} H_n =
{\mathrm{e}}^{-t n} H_n$. From this one quickly deduces that the Mehler kernel is given by $$\label{eq:Mehler_kernel_non_comp}
  M_t(x, y) := \sum_{n = 0}^\infty {\mathrm{e}}^{-t n} h_n(x)h_n(y).$$ We will need two further identities for the Hermite polynomials which can be found, e.g., in [@NIST:DLMF Chapter 18]: the integral representation $$\label{eq:Hermite-integral}
  H_n(x) = \frac{(-2i)^n {\mathrm{e}}^{x^2}}{\sqrt\pi} \int_{-\infty}^\infty
  \xi^n {\mathrm{e}}^{2ix \xi} {\mathrm{e}}^{-\xi^2} {\,\mathrm{d}}\xi$$ and the “binomial” identity $$\label{eq:Hermite-binomial-type}
  H_n(x + y) = \sum_{k = 0}^n \binom{n}{k} (2y)^{n - k} H_k(x) .$$

Hermite polynomials in several variables
----------------------------------------

The Hermite polynomials on ${\mathbf{R}}^d$ are defined, for multiindices $\alpha = (\alpha_1, \dots, \alpha_d)\in{\mathbf{N}}^d$ by the tensor extensions $$\label{eq:Hermite-Rodrigues-Rd}
  H_\alpha(x) := \prod_{n = 1}^{d} h_{\alpha_i}(x_i),$$ for $x=(x_1,\dots,x_d)$ in ${\mathbf{R}}^d$. The normalized Hermite polynomials $$\label{eq:Hermite-normalized}
  h_\alpha := \frac{H_\alpha}{\sqrt{2^{|\alpha|} \alpha!}}, \  \text{  where } \alpha! = \alpha_1! \cdot \dots \cdot \alpha_d!$$ form an orthonormal basis in ${{L^2({\mathbf{R}}^d, {\mathrm{d}}\gamma)}}$, and we have the eigenvalue identity $$\label{eq:OU-Hermite-action}
  LH_{\alpha} = -|\alpha| H_\alpha, \ \text{  where } |\alpha| = \alpha_1 + \dots + \alpha_d.$$ If we consider the action of $L^N {\mathrm{e}}^{tL}$ on a Hermite polynomial $h_\alpha$, through the multinomial theorem applied to $|\alpha|^k$ we get (writing $L_d$ for the operator $L$ in dimension $d$ and $L_1$ for the operator $L$ in dimension $1$) $$\label{eq:reduction-to-d1}
\begin{aligned}  L_d^N {\mathrm{e}}^{tL} & h_\alpha(x) = |\alpha|^N {\mathrm{e}}^{-t|\alpha|}
                            h_{\alpha_1}(x_1) \cdot \hdots \cdot
                            h_{\alpha_d}(x_d)\\
&= \sum_{|n| = N} \binom{N}{n_1, n_2, \hdots, n_d} \alpha_1^{n_1}
    \cdot \hdots \cdot \alpha_d^{n_d} {\mathrm{e}}^{-t\alpha_1}\cdot\hdots\cdot {\mathrm{e}}^{\alpha_d}
                            h_{\alpha_1}(x_1) \cdot \hdots\cdot
                            h_{\alpha_d}(x_d)
\\ & = \sum_{|n| = N} \binom{N}{n_1, n_2, \hdots, n_d} L_1^{n_1}
      {\mathrm{e}}^{tL_1} h_{\alpha_1}(x_1) \cdot \hdots \cdot L_1^{n_d}
      {\mathrm{e}}^{tL_1} h_{\alpha_d}(x_d).
\end{aligned}$$ This implies that we can reduce the question of computing the $d$-dimensional version of the integral kernel to the one-dimension one.

A combinatorial lemma {#sec:setup}
=====================

From now on we concentrate on the Ornstein-Uhlenbeck operator $L$ in one dimension, i.e., in $L^2({\mathbf{R}},{\mathrm{d}}\gamma)$. We are going to follow the approach of [@Sjogren1997]. Recalling the identity $L h_n = -n h_n$, we will apply $L^N$ to the generating function of the Hermite polynomials . A problem which immediately occurs is that $\Delta$ and $\langle x, \nabla \rangle$ do not commute, and because of this we cannot use a standard binomial formula to evaluate $L^N$. Instead, we note that $$L g = -t \partial_t g.$$ In particular this implies that $$L^N g = (-1)^k D_N g,$$ where $$\label{eq:Differential-operator-generated}
  D_N := \underbrace{t \partial_t \circ t \partial_t \circ \dots \circ t \partial_t}_{\text{$k$ times}} = (t\partial_t)^k.$$ The following lemma will be very useful.

\[lem:Lk-powers-expanded\] We have $$\label{eq:Expanded-Differentiatial-operator-generated}
    D_N = \sum_{n = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{n}} t^n \partial_t^n, $$ where ${\genfrac{\{}{\}}{0pt}{}{N}{n}}$ are the Stirling numbers of the second kind.

The Stirling numbers of the second kind are quite well-known combinatorial objects. For the sake of completeness we will state their definition and recall some relevant properties below. For more information we refer the reader to [@LINT]. The related Stirling numbers of the first kind will not be needed here.

We begin by recalling the definition of [*falling factorial*]{} $$\label{eq:falling-factorial}
  (j)_n := j(j - 1)\dots (j - n + 1) = \frac{j!}{(j - n)!},$$ for non-negative integers $k \geq n$.

\[def:stirling-numbers-second-kind\] For non-negative integers $N \ge n$, the number [*Stirling number of the second kind*]{} ${\genfrac{\{}{\}}{0pt}{}{N}{n}}$ is defined as the number of partitions of an $N$-set into $n$ non-empty subsets.

These numbers satisfy the recursion identity $$\label{eq:Stirling-numbers-second-kind-recursion}
    {\genfrac{\{}{\}}{0pt}{}{N}{n}} = N {\genfrac{\{}{\}}{0pt}{}{N - 1}{n}} + {\genfrac{\{}{\}}{0pt}{}{N - 1}{n - 1}}.$$ For all non-negative integers $j$ and $k$ one has the generating function identity $$\label{eq:Stirling-numbers-second-kind-generating-function-1}
    j^N = \sum_{n = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{n}} (j)_n.$$

Weyl Polynomials {#sec:Weyl}
================

Before turning to the proof of lemma \[lem:Lk-powers-expanded\], let us already mention that it only depends on the commutator identity $[t,\partial_t] = -1$. This brings us to the observation that *Weyl polynomials* provide the natural habitat for our expressions. Rougly speaking, a Weyl polynomial is a polynomial in two non-commuting variables $x$ and $y$ which satisfy the commutator identity $[x, y] = -1$. This is made more precise in the following definition.

The [*Weyl algebra*]{} over a field ${\mathbf{F}}$ of characteristic zero is the ring ${\mathbf{F}}\langle x,y\rangle$ of all polynomials of the form $p(x,y) = \sum_{m=0}^M \sum_{n=0}^N c_{mn} x^m y^n$ with coefficients $c_{mn}\in{\mathbf{F}}$ in two noncommuting variables $x$ and $y$ which satisfy the commutator identity $$[x,y]:=xy-yx = -1.$$

We now have the following abstract version of lemma \[lem:Lk-powers-expanded\]:

\[lem:Weyl-expansion\] In the Weyl algebra ${\mathbf{F}}\langle x, y\rangle$ we have the identity $$\label{eq:Weyl-base-thm}
    (x y)^m = \sum_{i = 1}^m {\genfrac{\{}{\}}{0pt}{}{m}{i}} x^i y^i,$$ where ${\genfrac{\{}{\}}{0pt}{}{m}{i}}$ are the Stirling numbers of the second kind.

As a preparation for the proof of lemma \[lem:Weyl-expansion\] we make a couple of easy computations. If we set $D := xy$, then $$\begin{aligned}
  D x^m &= x^m D + m x^m\label{eq:Weyl-push-1},\\
  D y^m &= y^m D - m y^m\label{eq:Weyl-push-2}.\end{aligned}$$ This can be shown by induction on $m$. For instance, note that $$D x^m = x(D + 1)x^{m - 1} = x D x^{m - 1} + x^m.$$ If we take this a bit further and have $p \in {\mathbf{F}}[D]$, then $$\begin{aligned}
  \label{eq:Weyl-poly-1}
  p(D) x^m &= x^m p(D + m),\\
  \label{eq:Weyl-poly-2}
  p(D) y^m &= y^m p(D - m).\end{aligned}$$ The $m$-th powers, $m\ge 1$, of $x$ and $y$ satisfy $$\begin{aligned}
  \label{eq:Weyl-xmym}
  x^m y^m &= \prod_{i = 0}^{m - 1} (D - i),\\
  \label{eq:Weyl-ymxm}
  y^m x^m &= \prod_{i = 1}^m (D + i).\end{aligned}$$ This can be seen using induction: $$x^{m + 1} y^{m + 1} = x^m D y^m \overset{\eqref{eq:Weyl-poly-2}}{=} x^m y^m (D - m)$$ and $$y^{m + 1} x^{m + 1} = y^m (D + 1) x^m = y^m D x^m + y^m x^m \overset{\eqref{eq:Weyl-poly-1}}{=} y^m x^m (D + (m + 1)).$$

The [*weighted degree*]{} of a monomial $x^m y^n\in {\mathbf{F}}\langle x,y\rangle$ is the integer $m-n$. A polynomial in ${\mathbf{F}}\langle x,y \rangle$ is said to be [*homogeneous of weighted degree $j$*]{} if all its constituting monomials have weighted degree $j$.

Left multiplication by $xy$ is [*homogeneity preserving*]{}, i.e., for all $j\in{\mathbf{Z}}$ it maps the set of homogeneous monomials of weighted degree $j$ into itself. To prove this, first consider a monomial $x^m y^n$ of dweighted egree $j = m-n$. Then, $$(xy) x^m y^n \mathrel{\overset{\eqref{eq:Weyl-push-1}}{=}} (x^m (xy) + m x^m)y^n = x^m(xy)y^n +mx^m y^n = x^{m + 1} y^{n + 1} + m x^m y^n,$$ and we see that weighted degree of homogeneity is indeed preserved. The general case follows immediately. Through we conclude that left multiplication $x^k y^k$ is homogeneity preserving as well, for all non-negative integers $k$. We claim that left multiplication by $x^i y^j$ is homogeneity preserving only if $i = j$. To see this note that $$y x^i y^j = x^i y^{i + j} + i
x^{i - 1} y^j$$ from which we can deduce that $$x^m y^M x^n y^N = x^{n + m} y^{N + M} + \text{ lower order terms}.$$ From which the claim follows.

Finally, a polynomial is homogeneity preserving if and only if all of its constituting monomials have this property. If this were not to be the case we could look at the highest-order non-homogeneous term and note from above $x^m y^M x^n y^N$ would give terms of a lower order in the polynomial expansion which cannot cancel as they have different powers of $x$ or $y$.

It follows from these observations that $$\label{eq:Weyl-G0}
  F_0 := \biggl\{\sum_{n=0}^N c_{n} x^n y^n \ \bigg| \ N\in{\mathbf{N}}, \, c_1,\dots,c_N\in{\mathbf{F}}\biggr\}$$ is precisely the set of *homogeneity preserving polynomials* in ${\mathbf{F}}\langle
x, y\rangle$.

Now everything is in place to give the proof of lemma \[lem:Weyl-expansion\].

As $(xy)^k$ is homogeneity preserving, we infer that there are coefficients $a_i^k$ in ${\mathbf{F}}$ such that $$\label{eq:Weyl-lemma-intermediate-step-wanted-form}
  (xy)^k = \sum_{i = 0}^k a_i^k x^i y^i.$$ We will apply $x^j$ to the right on both sides of and derive an expression for the $a_i^k$. First note that gives $$(xy)^k x^j = x^j (xy + j)^k,$$ and together with gives $$x^i y^i x^j \overset{\eqref{eq:Weyl-xmym}}{=}
\prod_{\ell = 0}^{i - 1} (xy - \ell) x^j \overset{\eqref{eq:Weyl-poly-1}}{=} x^j \prod_{\ell = 0}^{i - 1} (xy - \ell + j).$$ Hence, to find the coefficients $a_i^k$ it is sufficient to consider $$(xy + j)^k = \sum_{i = 0}^k a_n^k \prod_{\ell = 0}^{i - 1}(xy - \ell + j).$$ Comparing the constant terms on both the left-hand side and right-hand side, we find $$\label{eq:Weyl-result-generating-function}
  j^k = \sum_{i = 0}^k a_i^k \prod_{\ell = 0}^{i - 1} (j - \ell) = \sum_{i = 0}^k a_i^k (j)_i,$$ where $(j)_i$ is the falling factorial as in . Comparing with the generating function of the Stirling numbers of the second kind ${\genfrac{\{}{\}}{0pt}{}{k}{i}}$ as given in , we see that $a_i^k = {\genfrac{\{}{\}}{0pt}{}{k}{i}}$. This concludes the proof of lemma \[lem:Weyl-expansion\].

The integral kernel of $L^N {\mathrm{e}}^{t L}$ {#sec:mainresult}
===============================================

As mentioned before, as a first step we would like to apply $D_N$ to the generating function $g(x,t) := {\mathrm{e}}^{-2tx + t^2 } = {\mathrm{e}}^{-(x - t)^2 + x^2}$ for the Hermite polynomials . We first compute the action of $\partial_t^N$ on the generating function.

We have $$\label{eq:derivatives-generating-function-Hermite}
    \partial_t^N {\mathrm{e}}^{-(x - t)^2 + x^2}  = {\mathrm{e}}^{-(x - t)^2 + x^2} H_N(x - t).$$

We first note that, $$\begin{aligned}
  \partial_t {\mathrm{e}}^{-(x - t)^2} &= 2(x - t) {\mathrm{e}}^{-(x - t)^2} = - \partial_x {\mathrm{e}}^{-(x - t)^2}.\end{aligned}$$ Using this we get $$\begin{aligned}
  \partial_t^N {\mathrm{e}}^{-(x - t)^2 + x^2} &= {\mathrm{e}}^{x^2} \partial_t^N {\mathrm{e}}^{-(x -
  t)^2}\\
  &= {\mathrm{e}}^{x^2} \partial_t^{N - 1} \partial_t {\mathrm{e}}^{-(x - t)^2}\\
  &= -{\mathrm{e}}^{x^2} \partial_t^{N - 1} \partial_x {\mathrm{e}}^{-(x - t)^2}\\
  &= (-)^2{\mathrm{e}}^{x^2} \partial_t^{N - 2} \partial_x^2 {\mathrm{e}}^{-(x - t)^2}\\
  &= \dots\\
  &= (-1)^N {\mathrm{e}}^{x^2} \partial_x^N {\mathrm{e}}^{-(x - t)^2}.\end{aligned}$$ By a change of variables, $$\partial_t^N {\mathrm{e}}^{-(x - t)^2 + x^2} = (-1)^N {\mathrm{e}}^{-(x - t)^2 + x^2} {\mathrm{e}}^{(x - t)^2} \partial_y^N {\mathrm{e}}^{-y^2} \Bigr|_{y = x - t}.$$ Hence, by , $$\partial_t^N {\mathrm{e}}^{-(x - t)^2 + x^2} = {\mathrm{e}}^{-(x - t)^2 + x^2} H_N(x - t).$$

\[lem:power-Ornstein-Uhlenbeck-generating-Hermite\] For all $x\in{\mathbf{R}}$ and $t>0$ we have $$\label{eq:power-Ornstein-Uhlenbeck-generating-Hermite}
    L^N {\mathrm{e}}^{-(x - t)^2 + x^2} = (-1)^N {\mathrm{e}}^{-(x - t)^2 + x^2} \sum_{n = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{n}} t^n H_n(x - t).$$

This is now easy to prove. Recalling that $L = -t \partial_t$ and using , we get $$\begin{aligned}
  L^N {\mathrm{e}}^{-(x - t)^2 + x^2} &= D_N {\mathrm{e}}^{-(x - t)^2 + x^2}\\
  &= (-1)^N \sum_{n = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{n}} t^n \partial_t^n {\mathrm{e}}^{-(x - t)^2 + x^2}\\
  &= (-1)^N {\mathrm{e}}^{-(x - t)^2 + x^2} \sum_{n = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{n}} t^n H_n(x - t).\end{aligned}$$

Our next theorem is the main result of this paper and provides an explicit expression for the integral kernel of $L^N e^{t L}$.

\[th:integral-kernel\] Let $L$ be the Ornstein-Uhlenbeck operator on $L^2({\mathbf{R}}^d,{\mathrm{d}}\gamma)$, let $t>0$, and let $N\ge 0$ be an integer. The integral kernel $M_t^N$ of $L^N {\mathrm{e}}^{tL}$ is given by $$\label{eq:Mehler-kernel-of-powers}
  \begin{split}
  M_t^N(x, y) &= M_t(x, y) \sum_{|n| = N} \binom{N}{n_1, \dots, n_d} \prod_{i = 1}^d
  \sum_{m_i = 0}^{n_i} \sum_{\ell_i = 0}^{m_i}  2^{-m_i}
  {\genfrac{\{}{\}}{0pt}{}{n_i}{m_i}} \binom{m_i}{\ell_i}\\
  &\quad \times \biggl(-\frac{{\mathrm{e}}^{-t}}{\sqrt{1 - {\mathrm{e}}^{-2t}}} \biggr)^{2m_i - \ell_i}
H_{\ell_i}(x_i) H_{2m_i - \ell_i}\biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1 - {\mathrm{e}}^{-2t}}}\biggr).
  \end{split}$$

We first prove the result for $d = 1$. Pulling $L^N$ through the integral expression for $e^{t L}$ involving the Mehler kernel, we must find a suitable expression for the kernel $M_t^N(\cdot, y) = L^N
M_t(\cdot, y)$. Using and the normalization of $H_m$ in we get $$\begin{aligned}
  M_t^N(x, y) &= L^N \sum_{m = 0}^\infty \frac{{\mathrm{e}}^{-t m}}{m!}\frac1{2^m} H_m(x) H_m(y)\\
  &\overset{\eqref{eq:Hermite-integral}}{=} L^N \sum_{m = 0}^\infty \frac{{\mathrm{e}}^{-t m}}{m!}\frac1{2^m} H_m(x) \frac{(-2i)^m}{\sqrt{\pi}} {\mathrm{e}}^{y^2} \int_{-\infty}^\infty {\mathrm{e}}^{-\xi^2} \xi^m {\mathrm{e}}^{2 i y \xi} {\,\mathrm{d}}\xi\\
  &= L^N \frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}} \int_{-\infty}^\infty  {\mathrm{e}}^{-\xi^2} {\mathrm{e}}^{2 i y \xi} \sum_{m = 0}^\infty \frac1{m!} H_m(x) (-i \xi {\mathrm{e}}^{-t})^m{\,\mathrm{d}}\xi\\
  &\overset{\eqref{eq:Generating-function-identity}}{=} L^N
  \frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}} \int_{-\infty}^\infty {\mathrm{e}}^{-\xi^2} {\mathrm{e}}^{2
    i y \xi} {\mathrm{e}}^{-(x + i \xi {\mathrm{e}}^{-t})^2 + x^2} {\,\mathrm{d}}\xi.\end{aligned}$$ The operator $L^N$ is applied with respect to $x$ here, so by lemma \[lem:power-Ornstein-Uhlenbeck-generating-Hermite\] we get $$\begin{aligned}
  M_t^N(x, y) &= \frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}}
  \int_{-\infty}^\infty {\mathrm{e}}^{-\xi^2} {\mathrm{e}}^{2 i y \xi} L^N {\mathrm{e}}^{-(x + i \xi {\mathrm{e}}^{-t})^2 + x^2} {\,\mathrm{d}}\xi\\
  &\overset{\eqref{eq:power-Ornstein-Uhlenbeck-generating-Hermite}}{=}(-1)^N
  \frac{{\mathrm{e}}^{x^2 + y^2}}{\sqrt{\pi}} \sum_{m = 0}^N {\genfrac{\{}{\}}{0pt}{}{N}{m}}
  \int_{-\infty}^\infty {\mathrm{e}}^{2 i y \xi} {\mathrm{e}}^{-(x + i \xi {\mathrm{e}}^{-t})^2} (i
  \xi {\mathrm{e}}^{-t})^m H_m(y + i \xi {\mathrm{e}}^{-t}) {\mathrm{e}}^{-\xi^2} {\,\mathrm{d}}\xi,\end{aligned}$$ where in last line we have used the analytic continuation of the algebraic identity . Similarly we can expand $H_m(y + i \xi {\mathrm{e}}^{-t})$ using . This gives $$H_m(y + i \xi {\mathrm{e}}^{-t}) = \sum_{\ell = 0}^m \binom{m}{\ell} H_\ell(y)
  (2 i \xi {\mathrm{e}}^{-t})^{m - \ell},$$ so that $M_t^N$ can be written as $$\label{eq:M_k-integral}
(-1)^N  \frac{{\mathrm{e}}^{x^2 + y^2}}{\sqrt{\pi}} \sum_{m = 0}^N \sum_{\ell = 0}^m {\genfrac{\{}{\}}{0pt}{}{N}{m}} \binom{m}{\ell} H_\ell(y) 2^{m - \ell} \int_{-\infty}^\infty {\mathrm{e}}^{2 i y \xi - \xi^2} {\mathrm{e}}^{-(x + i \xi {\mathrm{e}}^{-t})^2}  (i \xi {\mathrm{e}}^{-t})^{2m - \ell} {\,\mathrm{d}}\xi.$$ If we set $M = 2m - \ell$, this reduces our task to computing the integral $$\begin{aligned}
  \notag
  \frac{{\mathrm{e}}^{x^2 + y^2}}{\sqrt{\pi}} \int_{-\infty}^\infty &{\mathrm{e}}^{2 i y \xi - \xi^2} {\mathrm{e}}^{-(x + i \xi {\mathrm{e}}^{-t})^2}  (i \xi {\mathrm{e}}^{-t})^M {\,\mathrm{d}}{\xi}\\
  \label{eq:Hermite-integral-derivation-1}
  &= \frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}} \int_{-\infty}^\infty {\mathrm{e}}^{2 i (x {\mathrm{e}}^{-t} - y) \xi} {\mathrm{e}}^{-(1 - {\mathrm{e}}^{-2 t}) \xi^2} (i \xi {\mathrm{e}}^{-t})^M
  {\,\mathrm{d}}{\xi}.\end{aligned}$$ To make the computation less convolved, let us set $$\alpha_t := \sqrt{1 - {\mathrm{e}}^{-2t}}, \text{ and, } \beta_t(x, y) := \frac{x {\mathrm{e}}^{-t} - y}{\sqrt{1 - {\mathrm{e}}^{-2t}}}.$$ This allows us to write the exponential in the integral as $${\mathrm{e}}^{2 i (x{\mathrm{e}}^{-t} - y) \xi} {\mathrm{e}}^{-(1 - {\mathrm{e}}^{-2 t}) \xi^2} = {\mathrm{e}}^{2 i \alpha_t \beta_t(x,y) \xi} {\mathrm{e}}^{-\alpha_t^2 \xi^2}.$$ This reduces the problem, after the substitution $\alpha_t \xi \to \xi$, to computing the integral $$\frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}} \frac{i^M {\mathrm{e}}^{-M t}}{\alpha_t^{M + 1}}
  \int_{-\infty}^\infty {\mathrm{e}}^{2 i \beta_t(x, y) \xi} {\mathrm{e}}^{-\xi^2} \xi^M {\,\mathrm{d}}{\xi}.$$ The final integral is an integral expression for the Hermite polynomials , so $$\begin{aligned}
\ &   \frac{{\mathrm{e}}^{y^2}}{\sqrt{\pi}} \frac{i^M {\mathrm{e}}^{-M t}}{\alpha_t^{M + 1}}
  \int_{-\infty}^\infty {\mathrm{e}}^{2 i \beta_t(x,y) \xi} {\mathrm{e}}^{-\xi^2} \xi^M {\,\mathrm{d}}{\xi}
\\ & \qquad  \overset{\eqref{eq:Hermite-integral}}{=} {\mathrm{e}}^{y^2-\beta_t(x,y)^2}
  \frac1{\alpha_t^{M + 1}} \frac{(-1)^M {\mathrm{e}}^{-M t}}{2^M} H_M(\beta_t(x,y)).\end{aligned}$$ Next we note that $\exp(y^2 - \beta_t(x,y)^2) \alpha_t^{-1} = M_t$, the Mehler kernel from . Hence, $$\label{eq:Mehler-kernel-Lk-intermediate-1}
  \begin{split}
  M_t^N(x, y) &= M_t(x, y) \sum_{m = 0}^N \sum_{\ell = 0}^m
  \binom{m}{\ell} {\genfrac{\{}{\}}{0pt}{}{N}{m}}
 \biggl(-\frac{{\mathrm{e}}^{-t}}{\sqrt{1 - {\mathrm{e}}^{-2t}}} \biggr)^{2m - \ell}
  2^{-m}\\
  &\quad \times H_\ell(x) H_{2m - \ell}\biggl(\frac{x {\mathrm{e}}^{-t} - y}{\sqrt{1 - {\mathrm{e}}^{-2t}}}\biggr).
  \end{split}$$ Applying we get the result in $d$ dimensions.

An application {#sec:application}
==============

As an application of our main result, in this section we give an alternative proof of the bounds on the kernels $K$ and $\tilde K$ of [@Portal2014] (see the definition below), making the dependence on the parameters more explicit. These kernels play a central role in the study of the Hardy space $H^1({\mathbf{R}}^d,{\mathrm{d}}\gamma)$ in [@Portal2014], where the standard Calderón reproducing formula is replaced by $$u = C \int_0^\infty (t^2 L)^{N + 1} {\mathrm{e}}^{\frac{t^2}{\alpha} L} u \frac{{\,\mathrm{d}}{t}}{t} + \int_{{\mathbf{R}}^d} u {\,\mathrm{d}}\gamma,$$ where $C$ is a suitable constant only depending on $N$ and $\alpha$ (this can be seen by letting $u$ be a Hermite polynomial). The kernels $K$ and $\tilde K$ then occur in several decompositions, and the estimates below allow them to be related to classical results about the Mehler kernel.

We define the kernels $K$ and $\tilde K$ by $$\begin{aligned}
    \int_{{\mathbf{R}}^d} K_{t^2, N, \alpha}(x, y) u(y) {\,\mathrm{d}}\gamma(y) &= (t^2 L)^N
                                                            {\mathrm{e}}^{\frac{t^2}{\alpha}
                                                            L} u(x),\\
    \int_{{\mathbf{R}}^d} \tilde{K}_{t^2, N, \alpha, j}(x, y) u(y) {\,\mathrm{d}}\gamma(y) &= (t^2 L)^N
                                                            {\mathrm{e}}^{\frac{t^2}{\alpha}
                                                            L} t \partial_{x_j}^* u(x).
  \end{aligned}$$

Note that the operators on the right-hand sides are indeed given by integral kernels: the first is a scaled version of the operator we have already been studying, and a duality argument implies that the second is given by the integral kernel $$\tilde{K}_{t^2, N, \alpha,j}(x, y) =
t \partial_{x_j} K_{t^2, N, \alpha}(x, y).$$ Thus, both kernels are given as appropriate derivatives of the Mehler kernel.

We begin with a technical lemma which is a rephrased version of [@Portal2014 Lemma 3.4]. One should take note that we define the kernels with respect to the Gaussian measure whereas, [@Portal2014] defines these with respect to the Lebesgue measure.

\[lem:Mehler-alpha-efficient1\] For all $\alpha > 1$ and all $t>0$ and $x, y$ in ${\mathbf{R}}^d$ we have $$\label{eq:Mehler-alpha-efficient2}
   \frac{|{\mathrm{e}}^{-\frac{t}{\alpha}}x - y|^2}{1 - e^{-2\frac{t}{\alpha}}}
   \geq\frac{\alpha}2 {\mathrm{e}}^{-2t}  \frac{|{\mathrm{e}}^{-t}x - y|^2}{1 -
            {\mathrm{e}}^{-2t}} - \frac{t^2 \min{(|x|^2, |y|^2)}}{1 - {\mathrm{e}}^{-2\frac{t}{\alpha}}}.$$ Additionally, we have $$\label{eq:Mehler-alpha-efficient3}
    \alpha {\mathrm{e}}^{-2t} \leq \frac{1 - {\mathrm{e}}^{-2t}}{1 - {\mathrm{e}}^{-2\frac{t}{\alpha}}} \leq \alpha.$$

Let $N$ be a positive integer, $0 < t < T$. The for all large enough $\alpha>1$ we have

1.  If $t |x| \leq C$, then $$\displaystyle |K_{t^2, \alpha, N}(x, y)| \lesssim_{T, N}
      \alpha \exp(\frac{\alpha}2C^2)
      M_{t^2}(x, y)  \exp\biggl(-\frac{\alpha}{8{\mathrm{e}}^{2T}}  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
                {\mathrm{e}}^{-2t^2}} \biggr).$$

2.  If $t |x| \leq C$, then $$\displaystyle |\tilde{K}_{t^2, \alpha, N,j}(x, y)|
      \lesssim_{T, N} \alpha \exp(\frac{\alpha}2C^2) M_{t^2}(x, y) \exp\biggl(-\frac{\alpha}{8 {\mathrm{e}}^{2T}}  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
                {\mathrm{e}}^{-2t^2}} \biggr).$$

For $K_{t^2, \alpha, N}$, we use Theorem \[th:integral-kernel\] to obtain, after taking absolute values, $$\begin{aligned}
  |K_{t^2, \alpha, N}(x, y)| &\leq  M_{\frac{t^2}{\alpha}}(x, y) \sum_{|k| = N} \binom{N}{n_1, \dots, n_d}
  \prod_{i = 1}^d  t^{2k_i }\sum_{\ell_i = 0}^{n_i} \sum_{m_i = 0}^{m_i} 2^{-m_i}
  \binom{m_i}{\ell_i}  {\genfrac{\{}{\}}{0pt}{}{n_i}{m_i}}\\
  &\quad \times \biggl(\frac{{\mathrm{e}}^{-\frac{t^2}{\alpha}}}{\sqrt{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}} \biggr)^{2m_i - \ell_i}
| H_{\ell_i}(x_i)| \biggl| H_{2m_i - \ell_i}\biggl(\frac{x_i {\mathrm{e}}^{-\frac{t^2}{\alpha}} - y_i}{\sqrt{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\biggr) \biggr |.\end{aligned}$$ Recalling that $\ell_1 + \dots + \ell_d \leq N$, using the assumptions $t \leq T$ and $t |x| \leq
C$ we can bound $t^{2k_i} |H_{\ell_i}(x)|$ by considering the highest order term to obtain $$t^{2k_i} |H_{\ell_i}(x)| \lesssim_{C,N,T} 1.$$ Using we proceed by looking at $$\begin{aligned}
  M_{\frac{t^2}{\alpha}}(x, y) &= M_{t^2}(x, y) \bigg(\frac{{1 - {\mathrm{e}}^{-2t^2}}}{{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\bigg)^{1/2}\exp\biggl(\frac{|{\mathrm{e}}^{-t^2} x - y|^2}{1 - {\mathrm{e}}^{-2t^2}}
  \biggr) \exp\biggl(-\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}} x - y|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}
  \biggr)\\
&\le \alpha M_{t^2}(x, y) \exp\biggl(\frac{|{\mathrm{e}}^{-t^2} x - y|^2}{1 - {\mathrm{e}}^{-2t^2}}
  \biggr) \Bigg[\exp\biggl(-\frac12\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}} x - y|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}
  \biggr) \bigg]^2.
$$ We can now bound the final Hermite polynomial in the expression of the kernel. Setting $M_i = 2m_i - \ell_i$ we get $$\begin{aligned}
 \biggl(\frac{{\mathrm{e}}^{-\frac{t^2}{\alpha}}}{\sqrt{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}} \biggr)^{M_i}
 \biggl|H_{M_i}\biggl(\frac{{\mathrm{e}}^{-\frac{t^2}{\alpha}}x_i - y_i}{\sqrt{1 -
  {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\biggr)\biggr| &\lesssim_N
                                    \biggl(\frac{{\mathrm{e}}^{-\frac{t^2}{\alpha}}}{\sqrt{1
                                    - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}} \biggr)^{M_i}
                                    \biggl(\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}}x_i - y_i|}{\sqrt{1 -
  {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\biggr)^{M_i}\\
&\le \biggl(\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}}x_i - y_i|}{{1 -
  {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\biggr)^{M_i}.\end{aligned}$$ Also, $$\begin{aligned}
 \biggl(\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}}x_i - y_i|}{{1 -
  {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}}\biggr)^{M_i} \exp\biggl(-\frac1{2}\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}} x_i - y_i|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}
  \biggr) \lesssim 1.\end{aligned}$$ Putting these estimates together, using Lemma \[lem:Mehler-alpha-efficient1\], and taking $\alpha>1$ so large that $$1-\frac{\alpha}{4 {\mathrm{e}}^{2T}} \le -\frac{\alpha}{8 {\mathrm{e}}^{2T}} \ \hbox{ and } \
1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}} \ge \frac{t^2}{\alpha},$$ we obtain $$\begin{aligned}
  |K_{t^2, \alpha, N}(x, y)| & \lesssim_{T, N}\alpha M_{t^2}(x, y) \exp\biggl(\frac{|{\mathrm{e}}^{-t^2} x - y|^2}{1 - {\mathrm{e}}^{-2t^2}}
  \biggr) \exp\biggl(-\frac12\frac{|{\mathrm{e}}^{-\frac{t^2}{\alpha}} x - y|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}}
  \biggr)\\
&\leq \alpha M_{t^2}(x, y) \exp\biggl(\frac{|{\mathrm{e}}^{-t^2} x - y|^2}{1 - {\mathrm{e}}^{-2t^2}}
  \biggr) \exp\biggl(-\frac{\alpha}{4 {\mathrm{e}}^{2T}}  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
            {\mathrm{e}}^{-2t^2}} \biggr) \exp\biggl(\frac12 \frac{t^4 |x|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}} \biggr)\\
&= \alpha M_{t^2}(x, y) \exp\biggl(\Big(1-\frac{\alpha}{4 {\mathrm{e}}^{2T}}\Big)  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
            {\mathrm{e}}^{-2t^2}} \biggr) \exp\biggl(\frac12 \frac{t^4 |x|^2}{1 - {\mathrm{e}}^{-2\frac{t^2}{\alpha}}} \biggr)\\
            &\leq \alpha M_{t^2}(x, y)  \exp\biggl(-\frac{\alpha}{8 {\mathrm{e}}^{2T}}  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
            {\mathrm{e}}^{-2t^2}} \biggr) \exp\biggl(\frac\alpha2 t^2 |x|^2 \biggr)
            \\  &\leq \alpha \exp\biggl(\frac\alpha2 C^2 \biggr) M_{t^2}(x, y)
            \exp\biggl(-\frac{\alpha}{8 {\mathrm{e}}^{2T}}  \frac{|{\mathrm{e}}^{-t^2}x - y|^2}{1 -
            {\mathrm{e}}^{-2t^2}} \biggr),\end{aligned}$$ using $t|x| \leq C$ in the last step.

For the bound on $\tilde{K}$ we consider $$\begin{aligned}
  t \partial_{x_i} \biggl[H_{\ell_i}(x_i) H_{m_i} \biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1
  - {\mathrm{e}}^{-2t}}} \biggr) \biggr] &=  t H_{m_i} \biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1
  - {\mathrm{e}}^{-2t}}} \biggr) \partial_{x_i} H_\ell(x_i)\\
  &\quad + H_{\ell_i}(x_i) t \partial_{x_i}  H_{m_i} \biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1
  - {\mathrm{e}}^{-2t}}} \biggr).\end{aligned}$$ So, as the first term on the right-hand side just decreases in degree we look at $$\begin{aligned}
  t \partial_{x_i}  \biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1 - {\mathrm{e}}^{-2t}}}
  \biggr)^{m_i} =  m_i \biggl(\frac{x_i {\mathrm{e}}^{-t} - y_i}{\sqrt{1 - {\mathrm{e}}^{-2t}}}
  \biggr)^{m_i - 1} t \frac{{\mathrm{e}}^{-t}}{\sqrt{1 - {\mathrm{e}}^{-2t}}}\end{aligned}$$ The last term is bounded as $t \downarrow 0$, and the rest of the proof is as before.

Acknowledgments {#acknowledgments .unnumbered}
---------------

This work was partially supported by NWO-VICI grant 639.033.604 of the Netherlands Organisation for Scientific Research (NWO).

The author wishes to thank Alex Amenta and Mikko Kemppainen for inspiring discussions.
