Orthonormal basis in every Hilbert space?

Question

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?

Answer 1

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.

Answer 2

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 $H$.

Answer 3

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.