Is there a notion of *"smooth bundle of Hilbert spaces"* (the base is a smooth finite dimensional manifold, and the fibers are Hilbert spaces) such that:



> 
> **1•** A smooth bundle of Hilbert spaces over a point is the same thing as a Hilbert space.
> 
> 
> **2•** If $E\to M$ is a smooth fiber bundle of orientable manifolds (say with compact fibers) equipped with a vertical volume form, then taking fiberwise $L^2$-functions produces a smooth bundle of Hilbert spaces over $M$.
> 
> 
> **3•** If the Hilbert space is finite dimensional, then this specializes to the usual notion of smooth vector bundle (with fiberwise inner product).
> 
> 
> 


I suspect that the answer is "no", because I couldn't figure out how it might work...  

If the answer is indeed no, then what is/are the best notion/s of smooth bundle of Hilbert spaces?



Does it make sense to talk about smooth bundles of Hilbert spaces?
**This is not an answer** but rather a comment to Peter Michor's answer. Anyway, I post it as an answer to get more flexibility in text formulation and to get more visibility. Namely, I think there is a crucial error which completely breaks down the argument so that generally *it is not possible* to perform the construction (**2•** of OP) of associating a fiberwise $L^2$ bundle over a given fiber bundle. The error lies in the following passage:



>  That it is smooth $U\times L^2(F, vol(g)) \to U\times L^2(F,vol(g))$ is seen as follows:
> It suffices to show that $(x,f)\mapsto \langle f\circ \rho\_x, \lambda\rangle\_{L^2}$ is smooth for all $\lambda$ in a subset $\subset L^2$ of linear functionals which together recognize bounded sets.
> We may take $C^\infty(F)\subset L^2(F,vol(g))$ as this set. By one of the two smooth uniform boundedness theorems from the book below it suffices to show that for each fixed $f\in L^2$ the function $F\to \mathbb R$ given by
> $$x\mapsto \langle f\circ \rho\_x, \lambda\rangle\_{L^2} = \int\_F f(\rho\_x(u))\lambda(u)\,vol(g)(u)= \int f(v) \lambda(\rho\_x^{-1}(v) ((\rho\_x^{-1})^\*vol(g))(v)$$ 
> is smooth.
> But this now obvious since $\lambda$ and $vol(g)$ are smooth. 
>  


Specifically, it is *not* sufficient to check "scalar smoothness" against the set of smooth (not even continuous) functions since it (generally) *does not* "recognize" bounded sets in $L^2$ . To give an explicit counterexample, understanding $\mathbb S^1$ as $\mathbb R$ mod $1$ , consider the map $f:\mathbb R\times L^2(\mathbb S^1)\to L^2(\mathbb S^1)$ defined by $(t,[\,x\,])\mapsto[\,\langle\,x(t+s):s\in\mathbb R\,\rangle\,]$ . If the argument in Peter Michor's answer were correct, then for any fixed $x$ in $L^2$ the map $c:t\mapsto f(t,[\,x\,])$ should be smooth $\mathbb R\to L^2(\mathbb S^1)$ . However, it is easily seen that it is not even once differentiable if for example one takes $x$ defined by $x(s)=1$ for $|\,s-n\,|\le\frac 14$ and $n\in\mathbb Z$ , and $x(s)=0$ otherwise, since then $\lim\_{\,t\to 0\,}\|\,t^{-1}(c(t)-c(0))\,\|\_{L^2}=+\infty$ .


