Lemma 2.9.100821A7B52A3B

Proof. The multiplier is real and bounded by one in absolute value, so Plancherel's theorem shows that $T_{ab}$ is self-adjoint with norm at most one. For $v\in C_ c^\infty $, the first rule in (2.9.3) (lemma 2.9.2), applied twice, gives $\widehat{-\Delta v}(\xi )=4\pi ^2|\xi |^2\widehat v(\xi )$ and $\widehat{\partial _ a\partial _ bv}(\xi )=-4\pi ^2\xi _ a\xi _ b\widehat v(\xi )$. Multiplying the first identity by the multiplier of $T_{ab}$ gives the second, which proves (3).

To construct the kernel we use the heat kernel $H_ t(z)=(4\pi t)^{-d/2}\mathrm e^{-|z|^2/(4t)}$, $t>0$. By lemma 2.9.2, $\widehat H_ t(\xi )=\mathrm e^{-4\pi ^2t|\xi |^2}$, and hence $\widehat{\partial _ a\partial _ bH_ t}(\xi )=-4\pi ^2\xi _ a\xi _ b\mathrm e^{-4\pi ^2t|\xi |^2}$. For $0<\varepsilon <R<\infty $ put

Since $\partial _ a\partial _ bH_ t(z)=\bigl (z_ az_ b/(4t^2)-\delta _{ab}/(2t)\bigr )H_ t(z)$, where $\delta _{ab}$ is the Kronecker symbol, we have

Hence $\sup _ z|\partial _ a\partial _ bH_ t(z)|\leq C_ dt^{-d/2-1}$, and the substitution $z=\sqrt t\, w$ gives $\int |\partial _ a\partial _ bH_ t(z)|\, \mathrm d z\leq C_ dt^{-1}$. Therefore the kernel $k_{\varepsilon ,R}$ is bounded and integrable, and Fubini's theorem, applied with the integrable majorant $|\partial _ a\partial _ bH_ t(z)|$ on $\mathbb R^ d\times [\varepsilon ,R]$, gives its Fourier transform:

Let $f\in L^1\cap L^2$. Then $k_{\varepsilon ,R}*f$ is bounded and integrable, hence in $L^2$, and by Fubini's theorem and the last assertion of theorem 2.9.4 its Fourier transform is $\widehat{k_{\varepsilon ,R}}\widehat f$. As $\varepsilon \downarrow 0$ and $R\uparrow \infty $, these multipliers are bounded by one and converge to $-\xi _ a\xi _ b/|\xi |^2$ for $\xi \neq 0$. By dominated convergence on the Fourier side and Plancherel's theorem, $k_{\varepsilon ,R}*f\to T_{ab}f$ in $L^2$.

For $z\neq 0$ define

Since $H_ t(z)=t^{-d/2}H_1(z/\sqrt t)$, every derivative satisfies

and induction on $|\gamma |$ shows that $\partial ^\gamma H_1(w)=P_\gamma (w)\mathrm e^{-|w|^2/4}$ for a polynomial $P_\gamma $, of degree $m_\gamma $ say. Fix $0<c\leq c'$, and let $c\leq |z|\leq c'$ and $|\gamma |\geq 2$. Since $|P_\gamma (w)|\leq C_\gamma (1+|w|)^{m_\gamma }$ and $\mathrm e^{-|z|^2/(4t)}\leq \mathrm e^{-c^2/(4t)}$, we obtain $|\partial ^\gamma H_ t(z)|\leq g_\gamma (t)$, where

This function is integrable on $(0,\infty )$ and does not depend on $z$: near $t=0$ the factor $\mathrm e^{-c^2/(4t)}$ dominates every power of $t^{-1}$, and for $t\geq 1$ the exponent satisfies $(d+|\gamma |)/2\geq 3/2>1$. By the mean value theorem and dominated convergence, we may therefore differentiate under the integral sign any number of times. Hence $K_{ab}$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, and its derivatives are the integrals of the corresponding derivatives of $\partial _ a\partial _ bH_ t$. Moreover, the integrals $\int _0^\varepsilon $ and $\int _ R^\infty $ of $|\partial _ a\partial _ bH_ t(z)|$ tend to zero as $\varepsilon \downarrow 0$ and $R\uparrow \infty $, uniformly for $c\leq |z|\leq c'$.

Differentiating the identity $H_{r^2s}(rz)=r^{-d}H_ s(z)$ twice in $z$ and substituting $t=r^2s$, we find that $K_{ab}(rz)=r^{-d}K_{ab}(z)$ for $r>0$. Thus $K_{ab}$ is homogeneous of degree $-d$ and $\nabla K_{ab}$ is homogeneous of degree $-d-1$, and the bounds in (1) follow from the boundedness of the continuous functions $K_{ab}$ and $\nabla K_{ab}$ on the unit sphere. The same scaling shows that the function $z\mapsto \int _0^\infty |\partial _ a\partial _ bH_ t(z)|\, \mathrm d t$ is homogeneous of degree $-d$; it is bounded on the unit sphere by $\int _0^\infty g_\gamma (t)\, \mathrm d t$, where $\partial ^\gamma =\partial _ a\partial _ b$ and $c=c'=1$, so it is at most $C_ d|z|^{-d}$.

To prove (2), let $f$ and $F$ be as there. By the Cauchy–Schwarz inequality on $F$, the function $f$ is integrable, so $f\in L^1\cap L^2$. Let $L$ be a compact set disjoint from $F$. For $x\in L$,

where the outer integral may be taken over $F$. The differences $x-y$, with $x\in L$ and $y\in F$, form a compact set not containing the origin, on which $c\leq |x-y|\leq c'$ for some $0<c\leq c'$. By the uniform estimates just proved, the inner integral converges to $K_{ab}(x-y)$ uniformly on this set, and it is bounded by $\int _0^\infty |\partial _ a\partial _ bH_ t(x-y)|\, \mathrm d t\leq C_ dc^{-d}$. Hence, since $f$ is integrable, $k_{\varepsilon ,R}*f$ converges uniformly on $L$ to $\int K_{ab}(x-y)f(y)\, \mathrm d y$. Since it also converges to $T_{ab}f$ in $L^2$, a subsequence converges almost everywhere to $T_{ab}f$, and the two limits agree almost everywhere on $L$. Covering the open set $\mathbb R^ d\setminus F$ by countably many such compact sets proves (2). For $d=1$ the kernel vanishes, in accordance with $T_{11}=-1$: a direct differentiation gives $\partial _ z^2H_ t=\bigl (z^2/(4t^2)-1/(2t)\bigr )H_ t=\partial _ tH_ t$, so $K_{11}(z)=\lim _{t\uparrow \infty }H_ t(z)-\lim _{t\downarrow 0}H_ t(z)=0$ for $z\neq 0$. $\square$