21
$\begingroup$

A Hilbert space is a complete inner product space; that is any Cauchy sequence is convergent using the metric induced by the inner product.

From Wikipedia: A Hilbert space is separable if and only if it has a countable orthonormal basis.

What are the examples of non-separable Hilbert spaces? From an applied point of view, are all interesting (finite or infinite) Hilbert spaces separable?

$\endgroup$
10
  • 3
    $\begingroup$ An example of a non-separable Hilbert space is $L^2$, the space of square integrable functions. This space is widely applicable in quantum mechanics and probability theory. $\endgroup$ Commented Sep 27, 2017 at 22:20
  • 7
    $\begingroup$ @Alex $L^2$ is separable since $L^2([0,1])$ is separable, see also the Hermite functions. Now it is not a RKHS $\endgroup$ Commented Sep 27, 2017 at 22:21
  • 3
    $\begingroup$ @reuns separability of $L^2(\Omega, \mathcal{F}, \mu)$ depends on the choice of measure space $(\Omega, \mathcal{F}, \mu)$. $L^2([0,1])$ (with the lebesgue measure and borel $\sigma$-algebra) is separable but that has nothing to do with separability of other $L^2$-spaces. $\endgroup$ Commented Sep 27, 2017 at 22:31
  • 1
    $\begingroup$ @reuns You're free to call whatever you want an $L^2$ space. Unfortunately here you disagree with common usage since $([0,1], P([0,1]), \mu)$ is a measure space against which we have a notion of the Lebesgue integral. $\endgroup$ Commented Sep 27, 2017 at 22:58
  • 1
    $\begingroup$ @nobody Locally finite means Integrating locally constant functions makes sense. Otherwise it becomes very different so at the very least you should mention that it is. $\endgroup$ Commented Sep 27, 2017 at 23:02

3 Answers 3

Reset to default
31
$\begingroup$

The space $l^2(\mathbb R)$ is another example of a non-separable Hilbert space: It consists of all functions $f:\mathbb R\to\mathbb R$ such that $f(x)\ne0$ only for countable many $x$, and $$ \sum_{x\in \mathbb R}f(x)^2 <\infty. $$ It is easy to see that this is a Hilbert space, the crucial argument is that the countable union of countable sets is countable.

The functions $f_y$ defined by $$ f_y(x) = \begin{cases} 1 &\text{ if } x=y\\ 0 & \text{ else}\end{cases} $$ are an uncountable set of elements with distance $\sqrt2$, hence $l^2(\mathbb R)$ is not separable.

$\endgroup$
5
  • $\begingroup$ Instead of $l^2(\mathbb R)$ I would say $l^2(S)$ where $S$ is an arbitrary uncountable set. Also, the functions $f_y$ form a base which means the dimension of a Hilbert space can be any cardinal number. $\endgroup$ Commented Jan 24, 2019 at 9:10
  • $\begingroup$ What I wonder, though, is: has there been any research on non-separable / uncountably-dimensional Hilbert spaces? In particular, we know that for any countable cardinal number $k$ (so finite ones and $\aleph_0$) there is (up to isomorphism) exactly one $k$-dimensional Hilbert space. Is it so for uncountable cardinals, too? The $l^2(S)$ (where $S$ is some set of cardinality $k$) contruction only tells there is at least one $k$-dimensional Hilbert space. $\endgroup$ Commented Jan 24, 2019 at 9:22
  • 1
    $\begingroup$ @daw: I'm not sure I get your argument for why $l^2(\mathbb R)$ is not separable.True that countable union of countable sets is countable but $\mathbb R$ is uncountable but separable ( by being 2nd countable). $\endgroup$ Commented Sep 13, 2020 at 19:46
  • 3
    $\begingroup$ Separability of $\mathbb R$ does not matter here/does not help. The uncountability of $\mathbb R$ is the property that gives non-separability of $l^2(\mathbb R)$. $\endgroup$ Commented Sep 14, 2020 at 8:15
  • $\begingroup$ In particular this is an example of a non-separable reproducing kernel Hilbert space. $\endgroup$ Commented Jul 29, 2023 at 8:56
20
$\begingroup$

The set of almost periodic functions with the inner product $$\langle f, g \rangle = \lim_{N \to \infty} \frac{1}{2N} \int_{-N}^N f(x) \overline{g(x)}dx$$ has an uncountable orthonormal family $\{e^{i \omega x}\}_{\omega \in \mathbb{R}}$. Its completion is a non-separable Hilbert space.

$\endgroup$
0
2
$\begingroup$

Yes, some researchers have worked on Fredholm theory using non separable Hilbert spaces. For instance you may go through Approximation by semi-Fredholm and semi-alpha-Fredholm Operators in Hilbert spaces of arbitrary dimension by Laura Burlando in Acta Sci.Math(Szeged) 65(1999).

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.