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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

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2 Answers

up vote 4 down vote accepted

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.

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The answers of jbc and Peter Michor are more or less the same as both describe the $\varepsilon$-product $C(K_1) \varepsilon C(K_2)$ of Laurent Schwartz. –  Jochen Wengenroth Feb 1 '13 at 7:49
    
By "unit ball of $C(K_1)\prime$" you mean $C(K_1)\prime$ equipped with the operator norm and then the unit ball of it? And has the "natural locally convex topology on the dual of $C(K_1)$" a name, so that I can find more information about it? –  AlexE Feb 1 '13 at 11:25
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I thought that I had given the topology a name---that of uniform convergence on the compact subsets of $E$. Another description is as the finest locally convex topology which agrees with the weak* topology on the unit ball for tne norm on $C(K_1)'$ as the dual of $C(K_1)$, i.e., the bounded weak* topology . Indeed it is the finest toplogy (not necessarily locally convex or even linear ) which agrees with this topology on the balls. This all works in suitable form for the dual of any Banach space (even Fr\'echet space)---it is the essential content of the Banach-Dieudonn\'e theorem. –  jbc Feb 1 '13 at 14:57
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A suitable reference is the two volume treatise on topological vector spaces by G. Koethe. –  jbc Feb 1 '13 at 14:58
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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.

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Thanks for your answer! But I can't figure out what you mean by "since the inductive tensor product does not commute with direct limits". I'm now especially interested in the following case: If $X$ is a metric space, denote by $C_u(X)$ the $C^\ast$-algebra of all bounded, uniformly continuous functions on $X$ (with the sup-norm). Do we have $C_u(X_1 \times X_2) = C_u(X_1) \widehat{\otimes} C_u(X_2)$? –  AlexE Apr 26 '13 at 12:46
    
I asked the above question as a separate one: mathoverflow.net/questions/128901 –  AlexE Apr 27 '13 at 7:41
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