# Orthonormal basis in every Hilbert space?

Is it true that in every Hilbert space (not separable) there exists an orthonormal basis (complete and countable orthogonal system)? So equivalently asked can we have a Hilbert space in which it is not sufficient a countable set for which the closure of it's span to be the entirely Hilbert space?

• It is true that every Hilbert space has an orthonormal basis (assuming the axiom of choice), but it is not true that every Hilbert space has a countable orthonormal basis. – Lorenzo Quarisa Aug 29 '18 at 10:26
• A Hilbert space has a countable orthonormal basis if and only if it is separable. – gerw Aug 29 '18 at 10:59

## 3 Answers

If $H$ is the space of all functions from $[0,1] \to \mathbb R$ such that $\sum_{0\leq t \leq 1} |f(t)|^{2} <\infty$ with the inner product $\langle f, g \rangle =\sum_{0\leq t \leq 1} f(t)g(t)$ the there is no countable orthonormal basis. A Hilbert space has a countable orthonormal basis iff it is separable. [In my example $\sum_{0\leq t \leq 1} |f(t)|^{2}$ is defined as the supremum of all finite sums. When this supremum is finite all but countable number of values $\{f(t),0\leq t \leq 1\}$ are zero which makes the definition of inner product meaningful. What I have done is to give explicit construction of a non -separable Hilbert space.

Let $H$ be a Hilbert space (and $H \ne \{0\}$).

Let $u \in H$ with $||u||=1$ and put $S_0 = \{u\}$. Then $S_0$ is an orthonormal systen in $H$.

Let $M$ be the set of all orthonormal systems $S$ with $S_0 \subseteq S$. On $M$ we can define an order relation by inclusion. By Zorn's Lemma, $M$ contains a maximal element, which is an orthonormal basis of $H$.

• Why the downvote ??????????????????????????? – Fred Aug 29 '18 at 10:33
• I think you relied on the title instead of reading the question fully. OP is asking for a countable ONB. – Kavi Rama Murthy Aug 29 '18 at 10:34
• @Kavi: Hi, Sheriff. The OP asked: "Is it true that in every Hilbert space (not separable) there exists an orthonormal basis". In my answer I said: yes it is true (and I gave a short proof). So why you are beefing about my answer ???? – Fred Aug 29 '18 at 10:42
• To clarify matters I did not downvote you. I am just trying to guess why you were downvoted. I still think you have not interpreted the question correctly. TH eOP does not know that countability of an ONB is tied to separability and he is asking if non-separable spaces also have countable ONB's. I am sorry if my interpretation is wrong. If I find a mistake in an answer I make a comment. I don't downvote. – Kavi Rama Murthy Aug 29 '18 at 11:48

Let $\mathcal{H}$ consist of all functions $x : S\rightarrow\mathbb{C}$ such that $x(s)=0$ except for at most countably many values of $x$, and such that $\|x\|^2=\sum_{s\in S}|x(s)|^2 < \infty$. Then $\mathcal{H}$ is a Hilbert space under $\langle x,y\rangle = \sum_{s\in S}x(s)\overline{y(s)}$. Let $e_s$ be the function that is $0$ except at $s$, where it is $1$. Then $\{ e_s \}_{s\in S}$ is an orthonormal basis of $\mathcal{H}$ with the same cardinality as $S$. $\mathcal{H}$ is separable iff $S$ is countable or finite.