I would like to understand how the Schwartz kernel theorem works for some more difficult cases and therefore would like to discuss an example from scratch:

Let the Dirichlet Laplacian on the half-line $-\Delta:H_0^1((0,\infty))\cap H^2((0,\infty)) \rightarrow L^2((0,\infty))$ be given. Then $f(-\Delta)$, for $f \in C_c^{\infty}(\mathbb{R})$ maps by the functional calculus, as $(-\Delta-i)^{n}f(-\Delta)$ is still bounded on $L^2$ clearly $L^2(0,\infty)$ into any Sobolev space $H^n(0,\infty).$

By the Schwartz kernel theorem $f(-\Delta)(\psi)(\phi) = K(\psi \otimes \phi)$ for $\psi,\phi \in C_c^{\infty}(\mathbb{R})$ and some Distributions $K \in D'((0,\infty) \times (0,\infty)).$

What I would like to know is whether it follows already that $K$ is actually an integral kernel operator, where the kernel is given by a measurable function? So why does it already follow that $$f(-\Delta) \psi(x)= \int_{(0,\infty)}K(x,y)\psi(y)dy$$ (In fact I know that even $C^{\infty}$ kernel is true, but I am more interested in the techniques at this point)

And can we somehow show that $f(-\Delta)$ also maps $H^{-n}(0,\infty)$ into $H^n(0,\infty)?$


1 Answer 1


By the Helffer-Sjöstrand formula, $$ f(-\Delta_D) = \frac1\pi \int_{\mathbb C} (\partial_{\bar z} \tilde f)(z) (-\Delta_D -z)^{-1}\,L(dz), $$ where $L(dz)$ is the Lebesgue measure and $\tilde f$ is an almost-analytic extension of $f$. Now the resolvent $(-\Delta_D-z)^{-1}$ for $z\in\mathbb C\setminus[0,\infty)$ has kernel $$ K_z(x,y) = \begin{cases} \frac1{2\mu}\left(e^{-\mu(x-y)}-e^{-\mu(x+y)}\right), & 0< y< x, \\ \frac1{2\mu} \left(e^{\mu(x-y)}-e^{-\mu(x+y)}\right), & x< y< \infty, \end{cases} $$
where $\mu=\sqrt{-z}$ with $\Re\mu>0$. After a short computation this reveals \begin{align*} K_f(x,y) &= \frac1\pi \,\int_{\mathbb C} (\partial_{\bar z}\tilde f)(z) K_z(x,y)\,L(dz) \\ &= \frac1\pi \int_0^\infty f(\lambda)\,\frac{\sin(\sqrt{\lambda} x)\sin(\sqrt{\lambda} y)}{\sqrt{\lambda}}\,d\lambda \end{align*} for the kernel of $f(-\Delta_D)$. (An alternative approach uses the spectral resolution of $-\Delta_D$, see e.g. on page 261 here.)

If you want $f(-\Delta_D)\colon H^{-n}(\mathbb R_+) \to L^2(\mathbb R_+)$ for $n\in \mathbb N$, then by duality you need $\bar f(-\Delta_D)\colon L^2(\mathbb R_+) \to H_0^n(\mathbb R_+)$. However, $$ (\partial_x K_{\bar f})(0,y) = \frac1\pi \int_0^\infty \bar f(\lambda)\sin(\sqrt{\lambda} y)\,d\lambda, $$ which is not identically zero (except when $f|_{\,\mathbb R_+}\equiv0$). So, you already fail to map $L^2(\mathbb R_+)$ into $H_0^2(\mathbb R_+)$.


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