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Let $X$ be an inner product space.

An orthonormal subset $B \subseteq X$ is called:

(i) a maximal orthonormal set if there is no other orthonormal subset of $X$ that contains $B$

(ii) an orthonormal basis if $\text{span}(B)$ is dense in $X$

Consider the following statements about $X$.

(a) $X$ is complete

(b) every maximal orthonormal subset of $X$ is an orthonormal basis for $X$

Note 1. The converse of (b) (i.e., every orthonormal basis for $X$ is a maximal orthonormal subset of $X$) is always true and easy to prove.

Note 2. The proof that (a) implies (b) is in essentially every textbook that covers Hilbert spaces.

Note 3. If $X$ is separable, then (b) implies (a). See:

Question. If X is not separable, does (b) imply (a)?

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Yes, (b) implies (a) also in the case of non-separable inner product spaces. Assuming $X$ is not complete we construct below a maximal orthonormal set, the span of which is not dense in $X$. You need the Axiom of Choice to get the maximal set but with a little twist added in order to show maximality. Two useful lemmas:

Lemma 1: If $(u_n)_n$ is a Cauchy sequence in $X$ then $\ell(x)=\lim_n \langle u_n, x\rangle$ defines a continuous linear functional of $x\in X$. If $\ker\ell = X$ then $u_n \to 0$ (i.e. a Cauchy sequence that goes weakly to zero, converges to zero). Proof-hint: $(\|u_n\|)_n$ is a Cauchy sequence in ${\Bbb R}_+$.

Lemme 2: Let $(u_n)_n$ be a Cauchy sequence with $\|u_n\|=1$ but which does not converge in $X$ and let $\ell$ be the associated linear functional. Then $M=\ker\ell$ is a proper closed subspace of $X$ and $M^\perp = \{0\}$. Proof-hint: If $M^\perp$ contains a non-trivial element $z$ use Lemma 1 to show that $u_n\to c z$ for some constant $c$, contradicting the hypotheses.

Now, consider the collection $\Lambda$ of orthonormal subsets of $M$ with a partial order given by inclusion. Then $\Lambda$ is inductively ordered and by Zorn, $\Lambda$ has a maximal element which we denote ${\cal B}$. Let $N\subset M$ be the closure of the span of ${\cal B}$. Several cases must be considered.

  1. If $N=M$ the proof is over by Lemma 2. ${\cal B}$ is maximal but does not generate $X$.

  2. if $N^\perp=\{0\}$ the proof is also over (for the same reason).

  3. If $N^\perp\cap M\neq \{0\}$ that would contradict maximality of ${\cal B}$ so we may assume that this intersection is trivial.

  4. If $N^\perp$ contains at least two independent elements $p_1,p_2$ then $q=p_1/\ell(p_1)-p_2/\ell(p_2) \in N^\perp\cap M$, again contradicting maximality of ${\cal B}$.

  5. There remains just one case to consider, i.e. that $N^\perp$ is one-dimensional, whence generated by one unit vector $p\in X\setminus M$ and that $M\setminus N$ contains some non-zero vector $q$. Let ${\cal B}'={\cal B} \cup \{p\}$ and let $H$ be the closed span of ${\cal B}'$. By orthonormality, it is the direct sum of $N$ and the span of $p$. First, we must have $H^\perp=\{0\}$ so ${\cal B}'$ is indeed a maximal orthonormal set in $X$. Second, suppose that $q\in H$. Then $q=cp + n$ with $c$ a constant and $n\in N$. But as $\ell(p)\neq 0$ and $0=\ell(q)=c \ell(p) + \ell(n) = c \ell(p)$ we wee that $c=0$ so $q\in N$ contrary to hypothesis. Thus, the orthonormal set ${\cal B}'$ must be maximal but not generating.

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  • $\begingroup$ I am not sure I can follow your proof entirely. First question: What's the relation between V and M? $\endgroup$ Commented Apr 10, 2022 at 20:12
  • $\begingroup$ I realize that I wrote $V$ instead of $X$ for the incomplete inner product space. (I think simply because the first link does that - I should perhaps change that...). I never use a completion of $V$. I define the linear function $\ell$ from a non-converging Cauchy sequence and then define $M$ as the kernel of $\ell$ (in the incomplete space $V$). $\endgroup$ Commented Apr 11, 2022 at 16:45
  • $\begingroup$ I have two questions: (a) In Lemma 1, why is $\ell$ continuous? (b) In Lemma 2, why is $M^{\perp} = \{0\}$? (Your hint wasn't enough for me.) I understand the rest of the proof perfectly $\endgroup$ Commented Apr 18, 2022 at 20:43
  • $\begingroup$ For Lemma 1, linearity should be clear? and $(u_n)$ is Cauchy whence bounded so the same is true for $\ell$ as a linear functional. For Lemma 2 suppose there is $z\in M^\perp$ of norm one and let $c=\ell(z)$. Then for any $x\in X$ the vector $x-z<z,x>$ is in $M$, i.e. in the kernel of $\ell$. Thus, $0=\lim_n <u_n,x-z<z,x>>= \lim_n<u_n-cz,x>$. By lemma 1 we must have $u_n\to cz$, so $u_n$ did in fact converge in $X$. $\endgroup$ Commented Apr 18, 2022 at 21:12
  • $\begingroup$ Ah, I see about Lemma 2 now, you are using that $M^{\perp}$ cannot have dimension greater than 1. And I see now that the continuity in Lemma 1 is consequence of the continuity of the norm, Cauchy-Schwarz, and the boundedness of the Cauchy sequences (as you said). Thank you. Great answer. $\endgroup$ Commented Apr 18, 2022 at 23:03

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