I'm studying periodic pseudo-differential operators on torus and I have a question concearning the Schwartz kernel theorem:

If $A:C^\infty(\mathbb T^n)\rightarrow \mathcal{D}^{'}(\mathbb T^n)$ is a continuous linear functional then by the Schwartz kernel theorem there is an unique $K_A\in \mathcal{D}^{'}(\mathbb T^{2n})$ such that, $$\langle A\varphi, \psi\rangle=\langle K_A, \psi\otimes \varphi\rangle.$$ Here $D^{'}(\mathbb T^n)$ is the set of all distributions on $\mathbb T^n$ (linear and continuous functionals on $\mathbb T^n$).

In the case the distribution $Af$ is induced by $Af$ (as a mapping) in the standard way, $$\psi\mapsto \int_{\mathbb T^n}Af(x)\psi(x)\ dx,$$ then the distribution $K_A$ is also induced by a function?


1 Answer 1


The short answer is "no", in effect because not every continuous linear map of a Hilbert space to itself is Hilbert-Schmidt (nor even compact).

That is, suppose $A$ were the identity on test functions on a product of circles. We imagine the kernel to be something like $\sum_i f_i\otimes \overline{f}_i$ for an orthonormal basis $f_i$, except this cannot be quite right because the space of test functions is not a Hilbert space. Suspending concern about that important fact for a moment, such a sum, in effect the reproducing kernel for a Hilbert space $L^2(\mathbb T^n)$ or similar, is never in $L^2(\mathbb T^n\times \mathbb T^n)$. The expressions $\sum_{ij} A_{ij} f_i\otimes \overline{f}_j$ in $L^2$ of the product are exactly the kernels for Hilbert-Schmidt operators, possibly by definition, depending how one defines these things. In any case, all such are {\it compact} operators, and the identity map is rarely compact.

It is our good fortune that, by Sobolev imbedding, the test functions on products of circles are the projective limits of the Levi-Sobolev spaces there, and the strong topologized distributions are the inductive limit of the negatively-indexed such. It is an interesting exercise to prove that any map from a projective limit of Hilbert spaces to a normed space factors through a limitand. Thus, a map from test functions to a negatively-indexed Sobolev space (limitand of distributions) factors through some Sobolev space. Thus, all such maps arise from continuous $H^s(\mathbb T^n)\to H^{-t}(\mathbb T^n)$.

The injection $H^{s-{n\over 2}-1}\to H^s$ is Hilbert-Schmidt, so the composition is Hilbert-Schmidt from $H^{s-{n\over 2}-1}\to H^{-t}$, and is given by a "function" in a Levi-Sobolev space on $\mathbb T^n\times \mathbb T^n$.

For example, the identity map $A:f\to f$ is not given by a kernel in $L^2(\mathbb T^n\times \mathbb T^n)$, but by a kernel from $H^{-n/2-1}(\mathbb T^n\times \mathbb T^n)$.

  • $\begingroup$ You mean $H^{-n/2-1}$ in your last sentence, don't you? The kernel of the identity is a "delta distribution on the diagonal". $\endgroup$
    – Kofi
    Jun 5, 2014 at 8:24
  • $\begingroup$ @Kofi, oops, yes, thanks! $\endgroup$ Jun 5, 2014 at 12:28

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