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