A Schwartz kernel type theorem for Sobolev spaces The well-known Schwartz kernel theorem states that a continuous operator from smooth test functions to distributions, $T \colon C^\infty_c(\mathbb{R}^n) \to C^\infty_c(\mathbb{R}^n)'$ is continuous iff it is of the form $T[f](x) = \int_{\mathbb{R}^n} T(x,y) f(y) \, dy$, where the kernel is a distribution $K(x,y) \in C^\infty_c(\mathbb{R}^n \times \mathbb{R}^n)'$.
Q: Is there any analog of such a theorem where smooth test functions and distributions are replaced by Sobolev spaces $H^{s}$ or $W^s_p$?
For instance, $H^{s}$ with $s\ge 0$ could play the role of test functions, while those with $s\le 0$ could play the role of distributions. The simplest example of a multiplication operator $T\colon H^0 \to H^0$, where $T[f](x) = t(x)f(x)$ with compactly supported and sufficiently regular $t(x)$, and of course $H^0 = L^2(\mathbb{R}^n)$, shows that a simple transcription of the result is false, since $T(x,y) = t(x)\delta(x-y)$, and $t(x)\delta(x-y) \not\in L^2(\mathbb{R}^n \times \mathbb{R}^n)$. But $t(x)\delta(x-y) \in H^{-1}(\mathbb{R}^n \times \mathbb{R}^n)$. So perhaps for any continuous operator $T\colon H^{s_1} \to H^{s_2}$ there necessarily exists an $s$ made out of $s_1$, $s_2$ and some appropriate shift, such that $T(x,y) \in H^{s}$.
 A: This might be off topic since it has already been pointed out that in the absence of nuclearity you cannot expect such a result in your context. However, the following general remarks might be of interest to you.  If $A$ is an unbounded self-adjoint operator on a separable Hilbert space $H$, which we assume to be $\geq \text{Id}$, then we can define a so-called hilbert scale $H^t$ of spaces, where for positive $t$, $H^t$ is the  domain of definition of $A^t$ and $H^{-t}$ is its dual under the scalar product on $H$.  These can be regarded as abstract Sobolev spaces and if we choose the classical operators (Sturm-Liouville operators, Laplace operators, Schrödinger operators) of mathematical physics, we get a vast spectrum of generalised Sobolev spaces, including the one in the above posting. Using the spectral theorem, we can represent them as so-called weighted $L^2$-spaces.  Thus in general there is no kernel theorem since there is no realisation of bilinear forms on such spaces using measurable kernels.  However, if the operator has a discrete spectrum consisting of a sequence of eigenvalues, say $(\lambda_n)$, then the $L^2$'s are in fact $\ell^2$'s and things work fine. In particular, if the eigenvalues satisfy a condition of type $\sum \dfrac 1 {\lambda_n^\alpha}<\infty$ for some positive $\alpha$, we have nuclearity and so results of the type you are looking for.  It is also possible to get explicit results on the relationship between the indices as requested.
