In this question [Operators on Hilbert $C^\*$-module and families of Fredholm operators](https://mathoverflow.net/questions/239978/operators-on-hilbert-c-module-and-families-of-fredholm-operators) I asked about the relation between being a family of compact operators $F:X \to K(H)$ on Hilbert space $H=\ell^2$ and being $C(X)$-compact operator on Hilbert module $H\_{C(X)}$ (here $X$ is a compact space). If I take a family $F$ like above then I can view it as an operator on $H\_{C(X)}$ by the formula $F(\xi)(x)=F\_x(\xi(x))$ where $\xi \in H\_{C(X)}$ and therefore $\xi(X) \in H=\ell^2$ so $F\_x(\xi(x))$ does makes sense. I wonder whether the opposite is true, namely if we have a $C(X)$-compact operator on $H\_{C(X)}$ may we somehow interpret it as a continuous family of compact operators? I suspect that this no longer will be true for *all* operators, so in the case if the answer is affirmative for compact operators, and negative for general operators, I would also like to understand where the obstruction lies. 



$C(X)$-compact operators and families of compact operators
The Hilbert module $H\_{C(X)}$ is the space of all continuous functions $X \to H$, and hence for each $s \in H\_{C(X)}$, we can talk about $s(x) \in H$, for each $x \in X$. 
Assume that $F:H\_{C(X)} \to H\_{C(X)}$ is $C(X)$-linear. We get individual operators $F\_x: H \to H$, namely
$$F\_x (v):= (F\tilde{v})(x), $$
where $v \in H$ yields the constant function $\tilde{v}: X \to H$. It is then not hard to show that for each $s \in H\_{C(X)}$ and $x \in X$, the relation 
$$(Fs)(x)= F\_x (s(x))$$
holds.
Claim: if $F$ is $C(X)$-compact, then $F\_x$ is compact for each $x \in X$ and $x \mapsto F\_x$ is (norm-)continuous. Proof: let $\epsilon>0$. Then, by the definition of a $C(X)$-compact operator, there exists finitely many $s\_i, t\_i \in H\_{C(X)}$, such that the operator 
$$G=\sum\_i t\_i \langle s\_i, \\_ \rangle$$
on $H\_{C(X)}$ satisfies $\| G-F\| \leq \epsilon$. It is clear that $G\_x$ is compact for each $ x\in X$ and $x\mapsto G\_x$ is norm-continuous. Note that $\| F\_x - G\_x\| \leq \epsilon$. Since $\epsilon$ was arbitrary, the proof is complete. 


WARNING: if $F$ is a $C(X)$-Fredholm operator on $H\_{C(X)}$, then you also get Fredholm operators $F\_x$ on $H$, but the map $x\mapsto F\_x$ is not continuous, at least not if the target has the norm topology.


