# 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.

• Maybe mathoverflow.net/questions/101526 can help. Dec 10, 2013 at 12:17
• In your problem, do you have an explicit dependence of $\Omega_s$ from $s$ which allows you to define a continuous map from $H(s)$ to $H(s')$ for every $s < s'$ or $s > s'$? In that case you might want to have a look at the inverse and direct limit construction of topological vector spaces. In this way you don't have a topology on $H$ but have nonetheless a limit space $\lim H(s)$. Dec 10, 2013 at 13:54
• @Tobias Yes that's (continuous maps between the $H(s)$) exactly what I have. I will have a look at the things you mentioned. Thanks. Dec 10, 2013 at 14:14

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.

• Thanks for this answer. The topic is very interesting. But I think it is not what I am seeking: it seems properties of vector fields are studied instead (so objects such that $f(t) \in H(t)$ for each $t$ which are square integrable form a Hilbert space). I am more interested in issues such as what it means to say $H(t) \to H(\infty)$ as $t \to \infty.$ Or maybe I missed something. Dec 10, 2013 at 18:08
• i did Independent misunderstand your query. but you van embed all your Hilbert spaces into a large tvs,say the measurable functions and then you have a situation which has been studied by L. Schwartz who considers the family of all hilbertien subspaces of a gives tvs (that is all Hilbert spaces which embed continuously into it) Dec 10, 2013 at 19:25
• as a space (with structure) in its own right. I don't recall him using a topology but he does use an Dec 10, 2013 at 19:26
• ordering which can be used to define notions of convergence---see his article in Jour. d'Analyse Math. Dec 10, 2013 at 19:30
• 13(1964). Sorry about the chopped up form of my answer. It's the software. Dec 10, 2013 at 19:31

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.

• Thank you. What i meant was the set $H= \{H(0), ..., H(1), ..., H(2), ...\}$ which contains as elements all the Hilbert spaces $H(s)$, not the union of these Hilbert spaces. Sorry for the confusion. I will think about your answer a bit more.. Dec 10, 2013 at 18:46

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…

• If two Hilbert spaces are isomorphic then orthonormal bases have the same cardinality. Hence they are isometric so that the Banach-Mazur distance is $0$. Dec 10, 2013 at 12:15
• @JochenWengenroth Right! Not particularly help notion here…
– Dirk
Dec 10, 2013 at 15:37
• Thank you for the answer. It would be nice to have a norm but let me see how I can progress with Gromov-Hausdorff. Dec 10, 2013 at 21:09