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 $H^n(0,\infty)?$



Schwartz kernel theorem
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](http://link.springer.com/book/10.1007%2FBFb0077960).)


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 $H\_0^2(\mathbb R\_+)$.


