I understand the meaning of separable Hilbert space. In wikipedia is quite clear. However, I read this definition which makes me confused:

In the separable Hilbert space $\mathcal{H}=L^2(T)$, there exist an orthonormal basis $f_j$, that is $\left\langle f_{j}, f_{k}\right\rangle=\delta_{j k}$, such that

$\mathcal{H}=\left\{x: x=\sum_{j=1}^{\infty} x_{j} f_{j}\right\} \supset \mathcal{H}_{k}=\left\{x: x=\sum_{j=1}^{K} x_{j} f_{j}\right\}$

I was wondering if $x_j$ is an infinite column vector of coefficients or not (first definition). So, could be also notated as e.g, $x=\sum_{j=1}^{\infty}a_{ij}f_{j}$? where $a_{ij}$ forms an infinte matrix $\left\{ a_{ij}\right\} _{ij=1}^{\infty}$.

On the other hand, $\mathcal{H}_{k}$ is a subspace of $\mathcal{H}$ which contains a finite countable basis, this is possible as $\mathcal{H}$ is infinite countable?

I'm quite new to functional analysis, but I would like to solve that type of doubts as I cannot discuss with a professor because I study on my own. Thank you,

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    $\begingroup$ For context: where did you read this definition? (I don't find it particularly clear — there must be better sources for this.) $\endgroup$ Commented Jul 1, 2020 at 8:48
  • $\begingroup$ It is from a presentation in Functional Data Analysis. Maybe it would be right to preserve anonymity, but I can send you the link. $\endgroup$
    – fina
    Commented Jul 1, 2020 at 8:56

1 Answer 1


I will try to answer in a "how to keep things in mind way".

First of all, the most common and most flexible definition of separability is the first: A topological space $X$ is called separable when there exists a countable subset $\{x_n\}_{n=1}^\infty$ that is dense in the space. Now we don't care about topology in general, we care about Hilbert spaces, but the definition is the same.

Using Zorn's lemma, one proves that every Hilbert space admits an orthonormal basis. An orthonormal basis of a Hilbert space $H$ is a set of vectors $(e_i)_{i\in I}$ such that $\|e_i\|=1$ for all $i\in I$, $\langle e_i,e_j\rangle=0$ when $i,j\in I$ and $i\neq j$ and it is true that for each $x\in H$ it is $$x=\sum_{i\in I}\langle x,e_i\rangle e_i$$ in the sense that for each $\varepsilon>0$ we may find a finite set $F_0\subset I$ such that for any finite set $F\subset I$ with $F_0\subset F$ it is $$\bigg{\|}x-\sum_{i\in F}\langle x,e_i\rangle e_i\bigg{\|}<\varepsilon.$$

Don't let this scare you! Once you learn about nets this will be very easy to digest. Just keep in mind that in the case that our index-set $I$ is countable, the sum above is simply a series (in the way that even your cat understands what a series is).

An equivalent definition is this: $(e_i)_{i\in I}$ is an orthonormal basis for $H$ if and only if $\|e_i\|=1$ for all $i$, $\langle e_i,e_j\rangle=0$ for all $i\neq j$ and the following implication is true for any $x\in H$: $$\bigg(\text{ for all i}\in I: \langle x,e_i\rangle=0\bigg)\implies x=0 $$

Anyway. It can be proved that if $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ are both orthonormal bases for a Hilbert space $H$, then $I$ and $J$ have the same cardinality. We say that this common cardinality is the Hilbert dimension of the Hilbert space.

One can easily convince themselves that a Hilbert space has finite Hilbert dimension if and only if it is finite dimensional in the usual sense.

Now one can also prove the following: a Hilbert space is separable if and only if its Hilbert dimension is (finite or) countably infinite.

Hint on how to prove the last thing: Obviously finite dimensional spaces are separable. If a space is separable, start from a dense sequence, pass to a dense linearly independent subsequence and apply Gram-Schmidt. Conversely, if a space has countably infinite Hilbert dimension, then the $\mathbb{Q}$-span of a countable orthonormal basis is a dense, countable set.

A good reference for all this is Conway's first Chapter. The exercises are difficult though, be warned about that!

  • $\begingroup$ But then $\langle x,e_{i}\rangle=x_{i}$ are scalar variables, which can be notated either as an infinite matrix, or an infinite vector column, am I right? On the other hand, what is the meaning of "Hilbert dimension"? Do you refer to the dimension of the Hilbert space? And then, I understand that the "nature" of the dimension is given from an index set. However, I don't have that capacity to understand all! But, anyway, fantastic answer! $\endgroup$
    – fina
    Commented Jul 1, 2020 at 9:30
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    $\begingroup$ @fina Yes, you can understand these vectors as infinite column matrices and these are indeed scalars. Hilbert dimension is defined in my answer: it is the number of elements that an orthonormal basis has. I hope this helps! $\endgroup$ Commented Jul 1, 2020 at 9:32
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    $\begingroup$ @fina I am confused with these notations, I can't understand what are $X_i(t)$ and what is $L_p$ here? $\endgroup$ Commented Jul 1, 2020 at 9:44
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    $\begingroup$ @fina I'm afraid I don't have what it takes to discuss these concepts. Sorry! $\endgroup$ Commented Jul 1, 2020 at 9:56
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    $\begingroup$ $\mathbb{Q}$ are the rational numbers. Instead of taking the linear span with real coefficients, take the linear span with rational coefficients. this is dense and $\mathbb{Q}$ is countable, which will make the $Q$-span of a countable set countable as well. $\endgroup$ Commented Jul 1, 2020 at 11:35

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