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Let $X,Y \subset \mathbb{R}^n$ be open subsets. Denote by $C^\infty(X)$ the smooth functions on $X$, let $\mathcal{E}'(Y)$ be its dual space considered as a space of distributions. Let $L(C^\infty(X), \mathcal{E}'(Y))$ be the continuous linear operators. This space itself is equipped with the topology of bounded convergence (see for example Operator topology).

An instance of the Schwartz kernel theorem states that

$L(C^\infty(X), \mathcal{E}'(Y)) \simeq \mathcal{E}'(X \times Y)$,

which morally says that I can write every continuous linear operator $T \colon C^\infty(X) \to \mathcal{E}'(Y)$ as an integral operator with a distributional kernel $k(y,x) \in \mathcal{E}'(X \times Y)$ as

$ (Tu)(y) = \int k(y,x)u(x) dx $

(where the integral denotes the pairing of functions with distributions).

Now let $E$ and $F$ be Hilbert $A$-modules for a $C^*$-algebra $A$. I can still consider smooth functions with values in the Banach space $E$. Also $F$-valued distributions still make sense. They can be defined by

$\mathcal{E'}(Y,F) = L(Y,F) = \mathcal{E}'(Y) \hat{\otimes} F$,

where the tensor product is the completion of the projective tensor product (well, since $\mathcal{E}'(Y)$ is nuclear this doesn't really matter).

I was wondering if in this case there is an analogue of the Schwartz kernel theorem with $A$-linearity and adjointability build into it, thus stating something like

$L_A^*(C^\infty(X,E), > \mathcal{E}'(Y,F)) \simeq > \mathcal{E}'(X \times Y, > \text{Hom}_A^*(E,F))$

Here the left hand side should denote adjointable continuous $A$-linear operators (part of the question is to find the right notion of adjointability, $A$-linearity should be clear) and $\text{Hom}_A^*(E,F)$ are the adjointable $A$-linear operators from $E$ to $F$.

I know that there exists a Schwartz kernel theorem for vector-valued distributions (see e.g. theorem 1.8.9 in This looks like

$\mathcal{E}'(X \times Y,Z') \simeq L(C^\infty(X), \mathcal{E}'(Y,Z'))$

for a Banach space $Z$ and I heard that Schwartz has proven similar theorems in much greater generality, but without considering Hilbert modules of course. The above theorem is close to what I want, if $\text{Hom}_A^*(E,F)$ has a predual, then it implies

$\mathcal{E}'(X \times Y,\text{Hom}_A^*(E,F)) \simeq L(C^\infty(X), \mathcal{E}'(Y,\text{Hom}_A^*(E,F)))$.

share|cite|improve this question
Sorry; you probably know this. In general $\hom^*_A(E,F)$ does not have a predual-- for example, if $E=F=A$ then $\hom^*_A(A,A)$ is just the multiplier algebra of $A$. If $E=F$ is a "self-dual" module then $\hom^*_A(E,E)$ is a von Neumann algebra, and so does have a predual. This is in Paschke's original paper. – Matthew Daws Dec 2 '11 at 10:24

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