Parametrised Hilbert spaces; can we put a norm on the following space of Hilbert spaces? For each $s \in [0,\infty)$, let $H(s)$ be a Hilbert space. Let us suppose for simplicity that $H(s) = L^2(\Omega_s)$, where $\Omega_s$ is some nice domain that depends on $s$ in a nice way.
Define $H = \{H(s) : s \in [0,\infty)\}$ the set containing all Hilbert spaces $H(s)$.
What kind of space is $H$? Can we put a norm on it or a vector space structure? What is known about such spaces of Hilbert spaces? Assume whatever is necessary for $\Omega_s$ (eg. $\Omega_0 \subset \Omega_s$ for all $s > 0$).
I ask this question because I wish to think of convergence of (not in) Hilbert spaces.
 A: I think that the concept you are looking for is that of a measurable field of Hilbert spaces.  These are studied in detail in Dixmier's classic "von Neumann algebras" which is available on Google Books.  In particular he shows how to define
a Hilbert space from such a field, the latter being exactly as in your query a family of Hilbert spaces depending in a suitable manner on a parameter.
A: If you want to put a norm on $H$, you should first think about the vector space structure that you want to impose on $H$ (I don't see a canonical one). If you are satisfied with a metric on $H$ (sounds like that, if you are interested in convergence of Hilbert spaces) have a look at:


*

*The Gromov-Hausdorff distance, a distance between two metric spaces, or

*the concept of Banach-Mazur distance, a distance between two isomorphic Banach spaces (the English Wikipedia states that it is only used for finite dimensional spaces, but I've heard that it is also useful in infinite dimensions). However, as Jochen Wengenroth noted in the comments, this notion is not helpful for isomorphic Hilbert spaces as they are also isometric in this case, and hence their Banach-Mazur distance is always zero. 


I don't know of a specialized metric between Hilbert spaces but there may be some…
A: The question is a bit vague, and the answer below may  not be what you want. I assume that by the union of those spaces you mean disjoint union.  Here is  one  quite general answer when  one can  equip such an union with a structure of   vector bundle. Assume that $\Omega$ is a domain in a fixed Euclidean space $\newcommand{\bR}{\mathbb{R}}$  $\bR^n$. Suppose  $(\Psi_s)_{s\geq 0}$ is a smooth family  of diffeomorphisms $\Psi_s:\bR^n\to\bR^n$ and $\Omega_s=\Psi_s(\Omega)$.  For each $s\geq 0$  denote by  $J_s: \Omega_s\to (0,\infty)$ the  Jacobian of $\Psi_s$, i.e., the absolute value of the determinant of  the differential of $\Psi_s$. For any $s$ we have a natural Hilbert space isomorphism,
$$T_s: H_s:=L^2(\Omega_s)\to H:=L^2(\Omega),\;\; L^2(\Omega_s)\ni f \mapsto  J_s^{\frac{1}{2}} \Psi_s^* (f)\in L^2(\Omega). $$
Indeed 
$$ \int_\Omega \bigl( \; J_s^{\frac{1}{2}} \Psi_s^* (f)\bigr)^2= \int_\Omega J_s \Psi_s(f^2) dV= \int_{\Omega_s} f^2 dV, $$
where $dV$ denote the Euclidean volume element in $\bR^n$.      We now have a bijection $\newcommand{\T}{\mathscr{T}}$ $\newcommand{\H}{\mathscr{H}}$
$$  \T: \H:=\bigsqcup_{s\geq 0} H_s =\bigcup_{s\geq 0} \{s\}\times H_s \to [0,\infty)\times  H, $$
$$ \{s\} \times H_s\ni (s, f) \mapsto (s, T_s f)\in [0,\infty)\times  H. $$
There exists a unique topology on $\H$  which makes $\T$ a homeomorphism onto the space $[0,\infty)\times H$ equipped with the product topology.
