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 $T(x,y) \in H^{s}$.

  • $\begingroup$ I think the obstacle is that $H^{\infty}(\mathbb R^n)\to L^2(\mathbb R^n)$ is not nuclear... $\endgroup$ Mar 19 '18 at 18:45
  • $\begingroup$ @paulgarrett, could you elaborate? What is that an obstacle to? $\endgroup$ Mar 19 '18 at 22:01
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    $\begingroup$ Or, more bluntly, $V=H^\infty(\mathbb R^n)$ is not nuclear, in the sense that there is no categorical tensor product (e.g.) of it with itself. If there were a TVS $V\otimes V$, then we have a Cartan-Eilenberg-type adjunction ${\mathrm Hom}(V\otimes V,\mathbb C)\approx {\mathrm Hom}(V,{\mathrm Hom}(V,\mathbb C))$. That is $(V\otimes V)^*\approx {\mathrm Hom}(V,V^*)$. $\endgroup$ Mar 19 '18 at 23:52
  • $\begingroup$ This is a great comment. I was thinking about nuclear spaces but I did not know the subject very well. $\endgroup$ Mar 20 '18 at 3:11
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    $\begingroup$ Only barely relevant as I am not even attempting to look at the Schwartz angle, but Thm. 4.4.7 in Duistermaat's book on Fourier Integral Operators (or Thm. 5.4.1 in Duistermaat and Hörmander's FIO II paper) gives some information on the Sobolev regularity of the kernel of a FIO. And of course some FIOs map $H^{s_1}$ to $H^{s_2}$ for some $s_1, s_2$, see e.g. Cor. 4.4.5 in Duistermaat's book. $\endgroup$ Mar 28 '18 at 12:16

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.


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