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Prove the following:

Every non trivial separable Hilbert space $H$ has an orthonormal basis, i.e., an orthonormal set whose linear span is dense in $H$

My attempt:

Let $V = (v_n)_{n\geq1}$ be a dense countable set of vectors in $H$. Remove all vectors that are $0$, and then, skim the sequence from the left to the right and remove every vector that is a linear combination of the previous ones. Call the obtained sequence, which is possibly finite, $(y_n)_{n\geq1}$.

By construction, this sequence of vectors is linearly independent, so we can apply the Gram-Schmidt (GS) algorithm to find a sequence $(\phi_n)_n$ of vectors such that $O := \{\phi_n\mid n\}$ is an orthonormal set of vectors.

It now suffices to show that the linear span of $O$ is dense in $H$, in order to conclude that $O$ is an orthonormal basis. For this, it suffices to show that $O^\perp = \{0\}$.

So, let $v \in O^\perp$. Then $\langle v,\phi\rangle = 0$ for every $\phi \in O$, and by linearity of the inner product, it follows that $v \in \operatorname{span}(O)^\perp$. By continuity of the inner product, it follows that $v \in \left(\overline{\operatorname{span}(O)}\right)^\perp$.

Now, $\operatorname{span} O = \operatorname{span}\{\phi_n \mid n\} =^{GS} \operatorname{span}\{y_n\mid n\} = \operatorname{span} V$

so that $\left(\overline{\operatorname{span}(O)}\right)^\perp = \left(\overline{\operatorname{span(V)}}\right)^\perp \subseteq \operatorname{span}(\overline{V})^\perp = \operatorname{span}(H)^\perp = H^\perp = \{0\}$, where we used that $X \subseteq Y \implies Y^\perp \subseteq X^\perp$ and hence $v = 0$

Is this correct?

G Tony Jacobs
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    For the record, a trivial ($={0}$) Hilbert space has an orthonormal basis: a trivial ($=\emptyset$) one. And non-separable Hilbert spaces have orthonormal basis (of uncountable cardinality). –  Mar 26 '18 at 21:35
  • You are right. Useful comment thanks. –  Mar 26 '18 at 21:37
  • I think you've added a few unnecessary steps: As you've shown, $\text{span}(O) = \text{span}(V)$, and your assumption at the beginning was that $V$ is dense, which implies $\text{span}(O)$ is dense, without using the inner product. – Alex Zorn Mar 26 '18 at 21:43

1 Answers1

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Your proof is correct.

However, you can make it shorter. You're trying to prove that $\overline{\operatorname{span}\{\phi_n | n \in \mathbb{N}\}} = X$.

As you noticed, by the span-preserving property of Gram-Schmidt, we have

$$\operatorname{span}\{\phi_n | n \in \mathbb{N}\} = \operatorname{span}\{y_n | n \in \mathbb{N}\} = \operatorname{span}\{v_n | n \in \mathbb{N}\}$$

and hence $$\overline{\operatorname{span}\{\phi_n | n \in \mathbb{N}\}} = \overline{\operatorname{span}\{v_n | n \in \mathbb{N}\}} = X$$

since $\operatorname{span}\{v_n | n \in \mathbb{N}\}$ is assumed to be dense in $X$.

Your proof, as is, relies on the fact that $M^{\perp\perp} = M$ for all closed subspaces $M$ of $X$, a fact which only holds in Hilbert spaces. Indeed, in a non-Hilbert space it can hold $M^\perp = \{0\}$ but $M$ doesn't have to be dense.

So if you use the shorter proof above, you can in fact conclude:

Every separable inner product space has an orthonormal basis.

mechanodroid
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  • I intentionally didn't index $O$ with natural numbers because it might be finite (think about a finite dimensional space) –  Mar 27 '18 at 07:07
  • The essence of the answer was of course most helpful. I will look at it later in detail, and accept it if everything is clear. –  Mar 27 '18 at 07:09
  • Why wouldn't this work for separable Banach space in general without the orthogonalization so we can construct a Schauder basis? – Krum Kutsarov Aug 30 '24 at 22:30
  • @KrumKutsarov No, there are separable Banach spaces without a Schauder basis. You just get a linearly independent sequence with a dense span, it doesn't follow that it is a Schauder basis. – mechanodroid Aug 31 '24 at 07:17