# Is every Hilbert space separable?

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?

• 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. – user275377 Sep 27 '17 at 22:20
• @Alex $L^2$ is separable since $L^2([0,1])$ is separable, see also the Hermite functions. Now it is not a RKHS – reuns Sep 27 '17 at 22:21
• @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. – Rhys Steele Sep 27 '17 at 22:31
• Thanks everyone, I appreciate your help. So can we say at least all RKHS are separable? – user1609656 Sep 27 '17 at 22:42
• @nobody You take a non-locally finite measure to obtain non-separability ? – reuns Sep 27 '17 at 22:45

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.

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 uncountable.

• Thanks a lot. I can connect to this example much better and now I can see how to frame it for my own problem. Also, your answer should be correct too, but I think I can only pick one. – user1609656 Sep 28 '17 at 18:51
• 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. – Heimdall Jan 24 '19 at 9:10
• 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. – Heimdall Jan 24 '19 at 9:22
• @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). – MSIS Sep 13 '20 at 19:46
• Separability of $\mathbb R$ does not matter here/does not help. The uncountability of $\mathbb R$ is the property that gives non-separability of $l^2(\mathbb R)$. – daw Sep 14 '20 at 8:15

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).