# The heat kernel on $\mathbb{R}^{n}$

Jordan Bell
March 28, 2014

## 1 Notation

For $f\in L^{1}(\mathbb{R}^{n})$, we define $\hat{f}:\mathbb{R}^{n}\to\mathbb{C}$ by

 $\hat{f}(\xi)=(\mathscr{F}f)(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-2\pi i\xi x}dx,% \qquad\xi\in\mathbb{R}^{n}.$

The statement of the Riemann-Lebesgue lemma is that $\hat{f}\in C_{0}(\mathbb{R}^{n})$.

We denote by $\mathscr{S}_{n}$ the Fréchet space of Schwartz functions $\mathbb{R}^{n}\to\mathbb{C}$.

If $\alpha$ is a multi-index, we define

 $D^{\alpha}=D_{1}^{\alpha_{1}}\cdots D_{n}^{\alpha_{n}},$
 $D_{\alpha}=i^{-|\alpha|}D^{\alpha}=\left(\frac{1}{i}D_{1}\right)^{\alpha_{1}}% \cdots\left(\frac{1}{i}D_{n}\right)^{\alpha_{n}},$

and

 $\Delta=D_{1}^{2}+\cdots+D_{n}^{2}.$

## 2 The heat equation

Fix $n$, and for $t>0$, $x\in\mathbb{R}^{n}$, define

 $k_{t}(x)=k(t,x)=(4\pi t)^{-n/2}\exp\left(-\frac{|x|^{2}}{4t}\right).$

We call $k$ the heat kernel. It is straightforward to check for any $t>0$ that $k_{t}\in\mathscr{S}_{n}$. The heat kernel satisfies

 $k_{t}(x)=(t^{-1/2})^{n}k_{1}(t^{-1/2}x),\qquad t>0,x\in\mathbb{R}^{n}.$

For $a>0$ and $f(x)=e^{-\pi a|x|^{2}}$, it is a fact that $\hat{f}(\xi)=a^{-n/2}e^{-\pi|\xi|^{2}/a}$. Using this, for any $t>0$ we get

 $\hat{k}_{t}(\xi)=e^{-4\pi^{2}|\xi|^{2}t},\qquad\xi\in\mathbb{R}^{n}.$

Thus for any $t>0$,

 $\int_{\mathbb{R}^{n}}k_{t}(x)dx=\hat{k}_{t}(0)=1.$

Then the heat kernel is an approximate identity: if $f\in L^{p}(\mathbb{R}^{n})$, $1\leq p<\infty$, then $\|f*k_{t}-f\|_{p}\to 0$ as $t\to 0$, and if $f$ is a function on $\mathbb{R}^{n}$ that is bounded and continuous, then for every $x\in\mathbb{R}^{n}$, $f*k_{t}(x)\to f(x)$ as $t\to 0$.11 1 $k_{1}$, and any $k_{t}$, belong merely to $\mathscr{S}_{n}$ and not to $\mathscr{D}(\mathbb{R}^{n})$, which is demanded in the definition of an approximate identity in Rudin’s Functional Analysis, second ed. For each $t>0$, because $k_{t}\in\mathscr{S}_{n}$ we have $f*k_{t}\in C^{\infty}(\mathbb{R}^{n})$, and $D^{\alpha}(f*k_{t})=f*D^{\alpha}k_{t}$ for any multi-index $\alpha$.22 2 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 11, Theorem 0.14.

The heat operator is $D_{t}-\Delta$ and the heat equation is $(D_{t}-\Delta)u=0$. It is straightforward to check that

 $(D_{t}-\Delta)k(t,x)=0,\qquad t>0,x\in\mathbb{R}^{n},$

that is, the heat kernel is a solution of the heat equation.

To get some practice proving things about solutions of the heat equation, we work out the following theorem from Folland.33 3 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 144, Theorem 4.4. In Folland’s proof it is not apparent how the hypotheses on $u$ and $D_{x}$ are used, and we make this explicit.

###### Theorem 1.

Suppose that $u:[0,\infty)\times\mathbb{R}^{n}\to\mathbb{C}$ is continuous, that $u$ is $C^{2}$ on $(0,\infty)\times\mathbb{R}^{n}$, that

 $(D_{t}-\Delta)u(t,x)=0,\qquad t>0,x\in\mathbb{R}^{n},$

and that $u(0,x)=0$ for $x\in\mathbb{R}^{n}$. If for every $\epsilon>0$ there is some $C$ such that

 $|u(t,x)|\leq Ce^{\epsilon|x|^{2}},\qquad|D_{x}u(t,x)|\leq Ce^{\epsilon|x|^{2}}% ,\qquad t>0,x\in\mathbb{R}^{n},$

then $u=0$.

###### Proof.

If $f$ and $g$ are $C^{2}$ functions on some open set in $\mathbb{R}\times\mathbb{R}^{n}$, such as $(0,\infty)\times\mathbb{R}^{n}$, then

 $\displaystyle g(\partial_{t}f-\Delta f)+f(\partial_{t}g+\Delta g)$ $\displaystyle=$ $\displaystyle\partial_{t}(fg)-g\sum_{j=1}^{n}\partial_{j}^{2}f+f\sum_{j=1}^{n}% \partial_{j}^{2}g$ $\displaystyle=$ $\displaystyle\partial_{t}(fg)+\sum_{j=1}^{n}\partial_{j}(f\partial_{j}g-g% \partial_{j}f)$ $\displaystyle=$ $\displaystyle\mathrm{div}_{t,x}F,$

where

 $F=(fg,f\partial_{1}g-g\partial_{1}f,\ldots,f\partial_{n}g-g\partial_{n}f).$

Take $t_{0}>0$, $x_{0}\in\mathbb{R}^{n}$, and let $f(t,x)=u(t,x)$ and $g(t,x)=k(t_{0}-t,x-x_{0})$ for $t>0$, $x\in\mathbb{R}^{n}$. Let $0 and $r>0$, and define

 $\Omega=\{(t,x):|x|

In $\Omega$ we check that $(\partial_{t}-\Delta)f=0$ and $(\partial_{t}+\Delta)g=0$, so by the divergence theorem,

 $\int_{\partial\Omega}F\cdot\nu=\int_{\Omega}\mathrm{div}_{t,x}F=\int_{\Omega}g% (\partial_{t}f-\Delta f)+f(\partial_{t}g+\Delta g)=\int_{\Omega}g\cdot 0+f% \cdot 0=0.$

On the other hand, as

 $\partial\Omega=\{(b,x):|x|\leq r\}\cup\{(a,x):|x|\leq r\}\cup\{(t,x):a

we have

 $\displaystyle\int_{\partial\Omega}F\cdot\nu$ $\displaystyle=$ $\displaystyle\int_{|x|\leq r}F(b,x)\cdot(1,0,\ldots,0)dx+\int_{|x|\leq r}F(a,x% )\cdot(-1,0,\ldots,0)dx$ $\displaystyle+\int_{a}^{b}\int_{|x|=r}F(t,x)\cdot\frac{x}{r}d\sigma(x)t^{n-1}dt$ $\displaystyle=$ $\displaystyle\int_{|x|\leq r}f(b,x)g(b,x)dx-\int_{|x|\leq r}f(a,x)g(a,x)dx$ $\displaystyle+\int_{a}^{b}\int_{|x|=r}\sum_{j=1}^{n}(f\partial_{j}g-g\partial_% {j}f)(t,x)\frac{x_{j}}{r}d\sigma(x)t^{n-1}dt$ $\displaystyle=$ $\displaystyle\int_{|x|\leq r}u(b,x)k(t_{0}-b,x-x_{0})dx-\int_{|x|\leq r}u(a,x)% k(t_{0}-a,x-x_{0})dx$ $\displaystyle+\int_{a}^{b}\int_{|x|=r}\sum_{j=1}^{n}\Big{(}u(t,x)\partial_{j}k% (t_{0}-t,x-x_{0})$ $\displaystyle-k(t_{0}-t,x-x_{0})\partial_{j}u(t,x)\Big{)}\frac{x_{j}}{r}d% \sigma(x)t^{n-1}dt,$

where $\sigma$ is surface measure on $\{|x|=r\}=rS^{n-1}$. As $r\to\infty$, the first two terms tend to

 $\int_{\mathbb{R}^{n}}u(b,x)k_{t_{0}-b}(x-x_{0})dx=\int_{\mathbb{R}^{n}}u(b,x)k% _{t_{0}-b}(x_{0}-x)dx=u(b,\cdot)*k_{t_{0}-b}(x_{0})$

and

 $\int_{\mathbb{R}^{n}}u(a,x)k_{t_{0}-a}(x-x_{0})dx=\int_{\mathbb{R}^{n}}u(a,x)k% _{t_{0}-a}(x_{0}-x)dx=u(a,\cdot)*k_{t_{0}-a}(x_{0})$

respectively. Let $\epsilon<\frac{1}{4(t_{0}-a)}$, and let $C$ be as given in the statement of the theorem. Using $\partial_{j}k(t,x)=-\frac{x_{j}}{2t}k(t,x)$, for any $r>0$ the third term is bounded by

 $n\int_{a}^{b}\int_{|x|=r}\Big{(}Ce^{\epsilon r^{2}}\frac{|x-x_{0}|}{2t}k(t_{0}% -t,x-x_{0})+k(t_{0}-t,x-x_{0})Ce^{\epsilon r^{2}}\Big{)}d\sigma(x)t^{n-1}dt,$

which is bounded by

 $n\int_{a}^{b}\int_{|x|=r}Ce^{\epsilon r^{2}}\left(\frac{|x_{0}|+r}{2a}+1\right% )(4\pi(t_{0}-b))^{-n/2}\exp\left(-\frac{r^{2}}{4(t_{0}-a)}\right)d\sigma(x)t^{% n-1}dt,$

and writing $\eta=\frac{1}{4(t_{0}-a)}-\epsilon$ and $\omega_{n}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$, the surface area of the sphere of radius $1$ in $\mathbb{R}^{n}$, this is equal to

 $(b-a)^{n}r^{n-1}\omega_{n}Ce^{-\eta r^{2}}\left(\frac{|x_{0}|+r}{2a}+1\right)(% 4\pi(t_{0}-b))^{-n/2},$

which tends to $0$ as $r\to\infty$. Therefore,

 $u(b,\cdot)*k_{t_{0}-b}(x_{0})=u(a,\cdot)*k_{t_{0}-a}(x_{0}).$

One checks that as $b\to t_{0}$, the left-hand side tends to $u(t_{0},x_{0})$, and that as $a\to 0$, the right-hand side tends to $u(0,x_{0})=0$. Therefore,

 $u(t_{0},x_{0})=0.$

This is true for any $t_{0}>0$, $x_{0}\in\mathbb{R}^{n}$, and as $u:[0,\infty)\times\mathbb{R}^{n}\to\mathbb{C}$ is continuous, it follows that $u$ is identically $0$. ∎

## 3 Fundamental solutions

We extend $k$ to $\mathbb{R}\times\mathbb{R}^{n}$ as

 $k(t,x)=\begin{cases}(4\pi t)^{-n/2}\exp\left(-\frac{|x|^{2}}{4t}\right)&t>0,x% \in\mathbb{R}^{n}\\ 0&t\leq 0,x\in\mathbb{R}^{n}.\end{cases}$

This function is locally integrable in $\mathbb{R}\times\mathbb{R}^{n}$, so it makes sense to define $\Lambda_{k}\in\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})$ by

 $\Lambda_{k}\phi=\int_{\mathbb{R}}\int_{\mathbb{R}^{n}}\phi(t,x)k(t,x)dxdt,% \qquad\phi\in\mathscr{D}(\mathbb{R}\times\mathbb{R}^{n}).$

Suppose that $P$ is a polynomial in $n$ variables:

 $P(\xi)=\sum c_{\alpha}\xi^{\alpha}=\sum c_{\alpha}\xi_{1}^{\alpha_{1}}\cdots% \xi_{n}^{\alpha_{n}}.$

We say that $E\in\mathscr{D}^{\prime}(\mathbb{R}^{n})$ is a fundamental solution of the differential operator

 $P(D)=\sum c_{\alpha}D_{\alpha}=\sum c_{\alpha}i^{-|\alpha|}D^{\alpha}$

if $P(D)E=\delta$. If $E=\Lambda_{f}$ for some locally integrable $f$, $\Lambda_{f}\phi=\int_{\mathbb{R}^{n}}\phi(x)f(x)dx$, we also say that the function $f$ is a fundamental solution of the differential operator $P(D)$. We now prove that the heat kernel extended to $\mathbb{R}\times\mathbb{R}^{n}$ in the above way is a fundamental solution of the heat operator.44 4 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 146, Theorem 4.6.

###### Theorem 2.

$\Lambda_{k}$ is a fundamental solution of $D_{t}-\Delta$.

###### Proof.

For $\epsilon>0$, define $K_{\epsilon}(t,x)=k(t,x)$ if $t>\epsilon$ and $K_{\epsilon}(t,x)=0$ otherwise. For any $\phi\in\mathscr{D}(\mathbb{R}\times\mathbb{R}^{n})$,

 $\displaystyle\left|\int_{\mathbb{R}}\int_{\mathbb{R}^{n}}(k(t,x)-K_{\epsilon}(% t,x))\phi(t,x)dxdt\right|$ $\displaystyle=$ $\displaystyle\left|\int_{0}^{\epsilon}\int_{\mathbb{R}^{n}}k(t,x)\phi(t,x)dxdt\right|$ $\displaystyle\leq$ $\displaystyle\|\phi\|_{\infty}\int_{0}^{\epsilon}\int_{\mathbb{R}^{n}}k(t,x)dxdt$ $\displaystyle=$ $\displaystyle\|\phi\|_{\infty}\int_{0}^{\epsilon}dt$ $\displaystyle=$ $\displaystyle\|\phi\|_{\infty}\epsilon.$

This shows that $\Lambda_{K_{\epsilon}}\to\Lambda_{k}$ in $\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})$, with the weak-* topology. It is a fact that for any multi-index, $E\mapsto D^{\alpha}E$ is continuous $\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})\to\mathscr{D}^{\prime}(% \mathbb{R}\times\mathbb{R}^{n})$, and hence $(D_{t}-\Delta)\Lambda_{K_{\epsilon}}\to(D_{t}-\Delta)\Lambda_{k}$ in $\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})$. Therefore, to prove the theorem it suffices to prove that $(D_{t}-\Delta)\Lambda_{K_{\epsilon}}\to\delta$ (because $\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})$ with the weak-* topology is Hausdorff).

Let $\phi\in\mathscr{D}(\mathbb{R}\times\mathbb{R}^{n})$. Doing integration by parts,

 $\displaystyle(D_{t}-\Delta)\Lambda_{K_{\epsilon}}(\phi)$ $\displaystyle=$ $\displaystyle\Lambda_{K_{\epsilon}}\left((D_{t}-\Delta)\phi\right)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}}\int_{\mathbb{R}^{n}}K_{\epsilon}(t,x)(D_{t}\phi% (t,x)-\Delta\phi(t,x))dxtx$ $\displaystyle=$ $\displaystyle\int_{\epsilon}^{\infty}\int_{\mathbb{R}^{n}}k(t,x)D_{t}\phi(t,x)% -k(t,x)\Delta\phi(t,x)dxtx$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}\left(k(\epsilon,x)\phi(\epsilon,x)-\int_{% \epsilon}^{\infty}\phi(t,x)D_{t}k(t,x)dt\right)dx$ $\displaystyle+\int_{\epsilon}^{\infty}\int_{\mathbb{R}^{n}}\phi(t,x)\Delta k(t% ,x)dxdt$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}k(\epsilon,x)\phi(\epsilon,x)dx$ $\displaystyle-\int_{\epsilon}^{\infty}\int_{\mathbb{R}^{n}}\phi(t,x)(D_{t}-% \Delta)k(t,x)dtdx$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}k(\epsilon,x)\phi(\epsilon,x)dx.$

So, using $k_{t}(x)=k_{t}(-x)$ and writing $\phi_{t}(x)=\phi(t,x)$,

 $\displaystyle(D_{t}-\Delta)\Lambda_{K_{\epsilon}}(\phi)$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}^{n}}k_{\epsilon}(-x)\phi_{\epsilon}(x)dx$ $\displaystyle=$ $\displaystyle k_{\epsilon}*\phi_{\epsilon}(0)$ $\displaystyle=$ $\displaystyle k_{\epsilon}*\phi_{0}(0)+k_{\epsilon}*(\phi_{\epsilon}-\phi_{0})% (0).$

Using the definition of convolution, the second term is bounded by

 $\sup_{x\in\mathbb{R}^{n}}|\phi_{\epsilon}(x)-\phi_{0}(x)|\|k_{\epsilon}\|_{1}=% \sup_{x\in\mathbb{R}^{n}}|\phi_{\epsilon}(x)-\phi_{0}(x)|,$

which tends to $0$ as $\epsilon\to 0$. Because $k$ is an approximate identity, $k_{\epsilon}*\phi_{0}(0)\to\phi_{0}(0)$ as $\epsilon\to 0$. That is,

 $(D_{t}-\Delta)\Lambda_{K_{\epsilon}}(\phi)\to\phi_{0}(0)=\delta(\phi)$

as $\epsilon\to 0$, showing that $(D_{t}-\Delta)\Lambda_{K_{\epsilon}}\to\delta$ in $\mathscr{D}^{\prime}(\mathbb{R}\times\mathbb{R}^{n})$ and completing the proof. ∎

## 4 Functions of the Laplacian

This section is my working through of material in Folland.55 5 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., pp. 149–152, §4B. For $f\in\mathscr{S}_{n}$ and for any nonnegative integer $k$, doing integration by parts we get

 $\mathscr{F}((-\Delta)^{k}f)(\xi)=\int_{\mathbb{R}^{n}}((-\Delta)^{k}f)(x)e^{-2% \pi i\xi x}dx=(4\pi^{2}|\xi|^{2})^{k}(\mathscr{F}f)(\xi),\qquad\xi\in\mathbb{R% }^{n}.$

Suppose that $P$ is a polynomial in one variable: $P(x)=\sum c_{k}x^{k}$. Then, writing $P(-\Delta)=\sum c_{k}(-\Delta)^{k}$, we have

 $\displaystyle\mathscr{F}(P(-\Delta)f)(\xi)$ $\displaystyle=$ $\displaystyle\sum c_{k}\mathscr{F}((-\Delta)^{k}f)(\xi)$ $\displaystyle=$ $\displaystyle\sum c_{k}(4\pi^{2}|\xi|^{2})^{k}(\mathscr{F}f)(\xi)$ $\displaystyle=$ $\displaystyle(\mathscr{F}f)(\xi)P(4\pi^{2}|\xi|^{2}).$

We remind ourselves that tempered distributions are elements of $\mathscr{S}_{n}^{\prime}$, i.e. continuous linear maps $\mathscr{S}_{n}\to\mathbb{C}$. The Fourier transform of a tempered distribution $\Lambda$ is defined by $\widehat{\Lambda}f=(\mathscr{F}\Lambda)f=\Lambda\hat{f}$, $f\in\mathscr{S}_{n}$. It is a fact that the Fourier transform is an isomorphism of locally convex spaces $\mathscr{S}_{n}^{\prime}\to\mathscr{S}_{n}^{\prime}$.66 6 Walter Rudin, Functional Analysis, second ed., p. 192, Theorem 7.15.

Suppose that $\psi:(0,\infty)\to\mathbb{C}$ is a function such that

 $\Lambda f=\int_{\mathbb{R}^{n}}f(\xi)\psi(4\pi^{2}|\xi|^{2})d\xi,\qquad f\in% \mathscr{S}_{n},$

is a tempered distribution. We define $\psi(-\Delta):\mathscr{S}_{n}\to\mathscr{S}_{n}^{\prime}$ by

 $\psi(-\Delta)f=\mathscr{F}^{-1}(\hat{f}\Lambda),\qquad f\in\mathscr{S}_{n}.$

Define $\check{f}(x)=f(-x)$; this is not the inverse Fourier transform of $f$, which we denote by $\mathscr{F}^{-1}$. As well, write $\tau_{x}f(y)=f(y-x)$. For $u\in\mathscr{S}_{n}^{\prime}$ and $\phi\in\mathscr{S}_{n}$, we define the convolution $u*\phi:\mathbb{R}^{n}\to\mathbb{C}$ by

 $(u*\phi)(x)=u(\tau_{x}\check{\phi}),\qquad x\in\mathbb{R}^{n}.$

One proves that $u*\phi\in C^{\infty}(\mathbb{R}^{n})$, that

 $D^{\alpha}(u*\phi)=(D^{\alpha}u)*\phi=u*(D^{\alpha}\phi)$

for any multi-index, that $u*\phi$ is a tempered distribution, that $\mathscr{F}(u*\phi)=\hat{\phi}\hat{u}$, and that $\hat{u}*\hat{\phi}=\mathscr{F}(\phi u)$.77 7 Walter Rudin, Functional Analysis, second ed., p. 195, Theorem 7.19.

We can also write $\psi(-\Delta)$ in the following way. There is a unique $\kappa_{\psi}\in\mathscr{S}_{n}^{\prime}$ such that

 $\mathscr{F}\kappa_{\psi}=\Lambda.$

For $f\in\mathscr{S}_{n}$, we have $\mathscr{F}(\kappa_{\psi}*f)=\hat{f}\hat{\kappa}_{\psi}=\hat{f}\Lambda$, but, using the definition of $\psi(-\Delta)$ we also have $\mathscr{F}(\psi(-\Delta)f)=\mathscr{F}\mathscr{F}^{-1}(\hat{f}\Lambda)=\hat{f}\Lambda$, so

 $\kappa_{\psi}*f=\psi(-\Delta)f.$

Moreover, $\kappa_{\psi}*f\in C^{\infty}(\mathbb{R}^{n})$; this shows that $\psi(-\Delta)f$ can be interpreted as a tempered distribution or as a function. We call $\kappa_{\psi}$ the convolution kernel of $\psi(-\Delta)$.

For a fixed $t>0$, define $\psi(s)=e^{-ts}$. Then $\Lambda:\mathscr{S}_{n}\to\mathbb{C}$ defined by

 $\Lambda f=\int_{\mathbb{R}^{n}}f(\xi)\psi(4\pi^{2}|\xi|^{2})d\xi=\int_{\mathbb% {R}^{n}}f(\xi)\exp\left(-4\pi^{2}|\xi|^{2}t\right)d\xi=\int_{\mathbb{R}^{n}}f(% \xi)\hat{k}_{t}(\xi)d\xi$

is a tempered distribution. Using the Plancherel theorem, we have

 $\Lambda f=\int_{\mathbb{R}^{n}}\hat{f}(\xi)k_{t}(\xi)d\xi.$

With $\kappa_{\psi}\in\mathscr{S}_{n}^{\prime}$ such that $\mathscr{F}\kappa_{\psi}=\Lambda$, we have

 $\Lambda f=(\mathscr{F}\kappa_{\psi})(f)=\kappa_{\psi}(\hat{f}).$

Because $f\mapsto\hat{f}$ is a bijection $\mathscr{S}_{n}\to\mathscr{S}_{n}$, this shows that for any $f\in\mathscr{S}_{n}$ we have

 $\kappa_{\psi}(f)=\int_{\mathbb{R}^{n}}f(\xi)k_{t}(\xi)d\xi.$

Hence,

 $e^{t\Delta}f=\kappa_{\psi}*f=k_{t}*f,\qquad t>0,f\in\mathscr{S}_{n}.$ (1)

Suppose that $\phi:(0,\infty)\to\mathbb{C}$ and $\omega:(0,\infty)\to(0,\infty)$ are functions and that

 $\psi(s)=\int_{0}^{\infty}\phi(\tau)e^{-s\omega(\tau)}d\tau,\qquad s>0.$

Manipulating symbols suggests that it may be true that

 $\psi(-\Delta)=\int_{0}^{\infty}\phi(\tau)e^{\omega(\tau)\Delta}d\tau,$

and then, for $f\in\mathscr{S}_{n}$,

 $\psi(-\Delta)f=\int_{0}^{\infty}\phi(\tau)e^{\omega(\tau)\Delta}fd\tau=\int_{0% }^{\infty}\phi(\tau)(k_{\omega(\tau)}*f)d\tau,$

and hence

 $\kappa_{\psi}(x)=\int_{0}^{\infty}\phi(\tau)k_{\omega(\tau)}(x)d\tau,\qquad x% \in\mathbb{R}^{n}.$ (2)

Take $\psi(s)=s^{-\beta}$ with $0<\mathrm{Re}\,\beta<\frac{n}{2}$. Because $\mathrm{Re}\,\beta<\frac{n}{2}$, one checks that

 $\Lambda f=\int_{\mathbb{R}^{n}}f(\xi)(4\pi^{2}|\xi|^{2})^{-\beta}d\xi$

is a tempered distribution. As $\mathrm{Re}\,\beta>0$, we have

 $s^{-\beta}=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{-s\tau}d\tau$

and writing $\phi(\tau)=\frac{\tau^{\beta-1}}{\Gamma(\beta)}$ and $\omega(\tau)=\tau$, we suspect from (2) that the convolution kernel of $(-\Delta)^{-\beta}$ is

 $\kappa_{\psi}(x)=\int_{0}^{\infty}\frac{\tau^{\beta-1}}{\Gamma(\beta)}k_{\tau}% (x)d\tau,$

which one calculates is equal to

 $\frac{\Gamma\left(\frac{n}{2}-\beta\right)}{\Gamma(\beta)4^{\beta}\pi^{n/2}|x|% ^{n-2\beta}}.$ (3)

What we have written so far does not prove that this is the convolution kernel of $(-\Delta)^{-\beta}$ because it used (2), but it is straightforward to calculate that indeed the convolution kernel of $(-\Delta)^{-\beta}$ is (3). This calculation is explained in an exercise in Folland.88 8 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 154, Exercise 1.

Taking $\alpha=2\beta$ and defining

 $R_{\alpha}(x)=\frac{\Gamma\left(\frac{n-\alpha}{2}\right)}{\Gamma\left(\frac{% \alpha}{2}\right)2^{\alpha}\pi^{n/2}|x|^{n-\alpha}},\qquad 0<\mathrm{Re}\,% \alpha

we call $R_{\alpha}$ the Riesz potential of order $\alpha$. Taking as granted that (3) is the convolution kernel of $(-\Delta)^{-\beta}$, we have

 $(-\Delta)^{-\alpha/2}f=R_{\alpha}*f,\qquad f\in\mathscr{S}_{n}.$

Then, if $n>2$ and $\alpha=2$ satisfies $0<\mathrm{Re}\,\alpha, we work out that

 $R_{2}(x)=\frac{1}{(n-2)\omega_{n}|x|^{n-2}},$

where $\omega_{n}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$, and hence

 $(-\Delta)^{-1}f=R_{2}*f,\qquad f\in\mathscr{S}_{n},$

and applying $-\Delta$ we obtain

 $f=-\Delta(R_{2}*f)=(-\Delta R_{2})*f,$

hence $-\Delta R_{2}=\delta$. That is, $R_{2}$ is the fundamental solution for $-\Delta$.

Suppose that $\mathrm{Re}\,\beta>0$. Then, using the definition of $\Gamma(\beta)$ as an integral, with $\psi(s)=(1+s)^{-\beta}$, we have

 $\psi(s)=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{-(1+s)\tau}d% \tau,\qquad s>0.$

Manipulating symbols suggests that

 $\psi(-\Delta)=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{-\tau}e% ^{\tau\Delta}d\tau,$

and using (1), assuming the above is true we would have for all $f\in\mathscr{S}_{n}$,

 $\psi(-\Delta)f=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{-\tau}% e^{\tau\Delta}fd\tau=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{% -\tau}(k_{\tau}*f)d\tau,$

whose convolution kernel is

 $\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}\tau^{\beta-1}e^{-\tau}k_{\tau}d\tau.$

We write $\alpha=2\beta$ and define, for $\mathrm{Re}\,\alpha>0$,

 $B_{\alpha}(x)=\frac{1}{\Gamma\left(\frac{\alpha}{2}\right)(4\pi)^{n/2}}\int_{0% }^{\infty}\tau^{\frac{\alpha-n}{2}-1}e^{-\tau-\frac{|x|^{2}}{4\tau}}d\tau,% \qquad x\neq 0.$

We call $B_{\alpha}$ the Bessel potential of order $\alpha$. It is straightforward to show, and shown in Folland, that $\|B_{\alpha}\|_{1}<\infty$, so $B_{\alpha}\in L^{1}(\mathbb{R}^{n})$. Therefore we can take the Fourier transform of $B_{\alpha}$, and one calculates that it is

 $\widehat{B}_{\alpha}(\xi)=(1+4\pi^{2}|\xi|^{2})^{-\alpha/2},\qquad\xi\in% \mathbb{R}^{n},$

and then

 $\psi(-\Delta)=(1-\Delta)^{-\alpha/2}f=B_{\alpha}*f,\qquad f\in\mathscr{S}_{n}.$

## 5 Gaussian measure

If $\mu$ is a measure on $\mathbb{R}^{n}$ and $f:\mathbb{R}^{n}\to\mathbb{C}$ is a function such that for every $x\in\mathbb{R}^{n}$ the integral $\int_{\mathbb{R}^{n}}f(x-y)d\mu(y)$ converges, we define the convolution $\mu*f:\mathbb{R}^{n}\to\mathbb{C}$ by

 $(\mu*f)(x)=\mu(\tau_{x}\check{f})=\int_{\mathbb{R}^{n}}(\tau_{x}\check{f})(y)d% \mu(y)=\int_{\mathbb{R}^{n}}\check{f}(y-x)d\mu(y)=\int_{\mathbb{R}^{n}}f(x-y)d% \mu(y).$

Let $\nu_{t}$ be the measure on $\mathbb{R}^{n}$ with density $k_{t}$. We call $\nu_{t}$ Gaussian measure. It satisfies

 $\nu_{t}*f(x)=\int_{\mathbb{R}^{n}}f(x-y)d\nu_{t}(y)=\int_{\mathbb{R}^{n}}f(x-y% )k_{t}(y)dy=f*k_{t}(x),\qquad x\in\mathbb{R}^{n}.$