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 $0<a<b<t_{0}$ and $r>0$ , and define

\Omega=\{(t,x):|x|<r,a<t<b\}.

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<t<b,|x|% =r\},

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<n,x\in\mathbb{R}^{n},

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 $0<\mathrm{Re}\,\alpha<n$ , 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}.

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

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

The heat kernel on ℝn

1 Notation

2 The heat equation

Theorem 1.

Proof.

3 Fundamental solutions

Theorem 2.

Proof.

4 Functions of the Laplacian

5 Gaussian measure

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