Schwartz kernel theorem for topological spaces

Question

Is there some regularizing version of Schwartz kernel theorem for topological spaces, i.e., in the form of

Every continuous linear map $A\colon C\prime(X_2) \to C(X_1)$ is given by a kernel $k \in C(X_1 \times X_2)$?

Here $C(\cdot)$ is the space of continuous functions, equipped with a suitable topology, and $C\prime(\cdot)$ its dual space. The topological spaces $X_1$ and $X_2$ may be as nice as you want (e.g., locally compact Hausdorff).

Note that I need the kernel to be continuous (this is the reason why the operator $A$ is per assumption regularizing).

Or maybe I have to use other spaces like $C_c(\cdot)$ (compactly supported continuous functions) in the formulation of the theorem? Because I'm not even sure about the right formulation of such a theorem.

Thanks in advance, Alex

Answer 1

As has been pointed out by Peter Michor above, this is false, even for compact spaces. However, it is perhaps of interest that one can characterise in a natural way those operators from $C(K_1)'$ to $C(K_2)$ which are represented by kernels of the above type. They are the linear operators whose restrictions to the unit ball of $C(K_1)'$ are continuous for the weak star topology. This can be proved directly with standard methods but the following background to the result might shed more light. There is a natural locally convex topology on the dual of $C(K_1)$ (indeed on the dual of any Banach space) which is complete and compatible with the duality with $C(K)$. It has many descriptions, perhaps the simplest being as the topology of uniform convergence on the compact subsets of $C(K_1)$. Then $C(K_1)'$ with this structure contains $K_1$ as a topological subspace (in the natural manner---identifying a point in $K_1$ with the corresponding $\delta$ measure). This embedding of $K_1$ has the universal property that each continuous mapping from $K_1$ into a Banach space $F$ can be lifted in a unique way to a linear mapping on $C(K_1)'$ which is continuous in the above sense. If we take $F$ to be $C(K_2)$ then the required result pops out---use the fact that there is a natural identification between $C(K_1\times K_2)$ and the continuous functions from $K_1$ into the Banach space $C(K_2)$.

There is a natural extension to a corresponding result for bounded continuous functions on completely regular spaces but one needs some more elaborate functional analytic structures to give a precise version.

Answer 2

If $X_i$ are both compact, then $C(X_1\times X_2)= C(X_1)\hat{\hat\otimes} C(X_2)$ for the completed inductive tensor product. So you only get the closure of the space of continuous finite rank operators in the space of all continuous operators in this way. If $X_i$ are locally compact, then the same works for $C_0(X_i)=$ space of continuous functions vanishing at infinity. For $C_c(X_i)=$ space of continuous functions with compact support this is already wrong, since the inductive tensor product does not commute with direct limits.

The kernel theorem needs: The space are all nuclear, so (1) all tensor products between the inductive and the projective coincide, and (2) you can approximate each bounded operator by finite rank ones.