This is likely a silly question, but I was wondering if $T$ is some nice integral transform, e.g. a Hilbert-Schmidt integral operator, with an, say, $L^2(\mathbb{R}^n)$ kernel, what then guarantees that the adjoint of $T$ is again an integral operator with a kernel? If
$$Tf(x) = \int_{\mathbb{R}^n}f(y)k(x,y)dy$$
then we know that $$\|Tf\|\leq \|k\|\|f\|$$
in $L^2$ and the Ries representation theorem guarantess the existence of $T^*$. But how can we justify representing $T^*f$ as
$$T^*f(x) = \int_{\mathbb{R}^n}f(y)s(x,y)dy$$
for some $s(.,.)$? Or is it so that in general we only know that the adjoints of some suitable classes of operators, like those of the Hilbert-Schmidt ones, have such representations?