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The answer must be no, otherwise the identity would be a H.S. operator. However, whats wrong with the following:

We have by Schwartz kernel theorem

$k_I( \phi\otimes\psi)=(I\phi,\psi)_{L^2}$

Therefore, for $\psi,\phi \in D(x)$ we have

$|k_I( \phi\otimes\psi)|=|(I\phi,\psi)_{L^2}|\le ||\phi||_{L^2}||\psi||_{L^2}=||\phi\otimes\psi||_{L^2\otimes L^2}$

Since $D(X)\otimes D(X)$ is dense in $L^2(X)\otimes L^2(X)$ and $L^2(X)\otimes L^2(X)\cong L^2({X^2})$ the above lines seems to show that $k_I$ is in $(L^2({X^2}))^*=L^2({X^2})$.

yess
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1 Answers1

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I believe the problem is that the Schwartz Kernel Theorem simply does not hold in the $L^2$-setting: You will not be able to find a $k_I \in L^2(X^2)^* = L^2(X^2)$ with $k_I(\phi \otimes \psi) = \langle \phi, \psi \rangle_{L^2}$ for all test functions $\phi, \psi$. In fact, I think your reasoning constitutes a proof of this fact.

There are many intuitive reasons for this: An expression like $\phi \otimes \psi \mapsto \langle \phi, \psi \rangle_{L^2}$ is supported in the diagonal of $X^2$, which is a set of measure zero, so this can't be a well-defined functional on $L^2(X^2)$.

What you can find is a distribution $k_I \in \mathcal D'(X^2)$ fulfilling the property, simply the delta distribution $\delta(x-y)$ along the diagonal. But if you stay in $L^2$, this will not be possible. If you want to get into the functional-analytic details: The problem is that $L^2(X)$ is not a nuclear space.

Lukas Miaskiwskyi
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