let $H$ be a Hilbert space and let M be a dense linear subspace of H. Can we find a complete orthonormal set $\{u_{\alpha}: \alpha \in A\}$ for H in M?

I think the answer is negative in general, but I cannot find a counterexample. Thank you very much in advance for your help.

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    $\begingroup$ most probably we can... but I dont know if that works without choice axiom if $H$ is not separable. $\endgroup$
    – Max
    Mar 3, 2014 at 18:50
  • $\begingroup$ Have you tried Gram-Schmidt in the separable case? $\endgroup$
    – gerw
    Mar 3, 2014 at 18:52
  • $\begingroup$ I'm inclined to agree with Max. $\endgroup$ Mar 3, 2014 at 18:53
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    $\begingroup$ Obviously, if H is separable, the Gram-Schmidt procedure gives an affirmative answer. $\endgroup$ Mar 3, 2014 at 19:14
  • $\begingroup$ This discussion came up in a previous question. math.stackexchange.com/questions/201119/… . One of the contributors gave a counterexample for the non-separable case. $\endgroup$ Mar 6, 2014 at 4:08

2 Answers 2


Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis?

It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2.

I tried to solve the exercise, but I'm not quite sure that I understood it.

Let $H_1=\ell_2(\mathbb N)$, and $H_2=\ell_2(I)$, where $I$ is a set with cardinality $\mathfrak c$ (the continuum). Choose a linearly independent family $(x_i)_{i\in I}\subset H_1$ and an orthonormal basis $(e_i)_{i\in I}$ of $H_2$. Now let $H:=H_1\oplus H_2$ and $M:={\rm span}\,\{ x_i\oplus e_i;\; i\in I\}\subset H$. Let us try to show that $M$ admits no orthonormal basis.

First, note that, identifying $H_2$ with $\{ 0\} \oplus H_2\subset H$, we have $M\cap H_2=\{ 0\}$. This is rather easy to check using that fact that the $x_i$ are linearly independent.

Now, let us show that any orthonormal family $\mathcal Z\subset M$ is countable. To see this, choose an orthonormal basis $(f_n)_{n\in\mathbb N}$ of $H_1=\ell_2(\mathbb N)\subset H$ and observe that for any $n\in\mathbb N$, we have $\langle f_n, z\rangle=0$ for all but countably many $z\in\mathcal Z$, because $\sum_{z\in\mathcal Z}\vert \langle f_n,z\rangle\vert^2<\infty$. Since by the previous remark, any $z\in\mathcal Z$ must satisfy $\langle f_n ,z\rangle\neq 0$ for at least one $n\in\mathbb N$ (otherwise we would have $z\in H_1^{\perp}=H_2$ and hence $z=0$), it follows that $\mathcal Z$ must be countable. Indeed, we have $\mathcal Z=\bigcup_{n\in\mathbb N} \mathcal Z_n$, where $\mathcal Z_n=\{ z;\; \langle f_n,z\rangle\neq 0\}$ is countable for each $n$.

It follows that if $M$ were to admit an orthonormal basis, then it would be separable. But this is not possible because $\pi_2(M)$ is dense in $H_2$ (since it contains all the vectors $e_i$) and $H_2$ is not separable.

  • $\begingroup$ This looks pretty good to me too, but then what do I know? Anyway, +1! $\endgroup$ Mar 4, 2014 at 0:19
  • $\begingroup$ Your proof looks fine to me too. Nice example! $\endgroup$ Mar 4, 2014 at 0:22
  • $\begingroup$ Thanks. One can find many nice things in Bourbaki's exercises... $\endgroup$
    – Etienne
    Mar 4, 2014 at 2:06
  • $\begingroup$ Thanks, very very much for your fantastic example, Etienne. I checked it in all the details carefully, and it is perfectly correct! As I supposed, the answer was negative in general. $\endgroup$ Mar 4, 2014 at 15:13
  • $\begingroup$ Yes, nonseparable spaces are sometimes strange objects... Glad that you found the answer useful! $\endgroup$
    – Etienne
    Mar 4, 2014 at 15:30

consider $X:=\left\{A\in\mathcal{P}\left(M\right): A \text{ is an ON-System }\right\}$ and $\subseteq$. Obviously each partially ordered chain $C$ has an upper bound in $X$: simply consider $\cup_{a\in C}a$. By Zorn we get that there exists at least one maximal element of $X$. Take the maximal element, by definition it's an orthonormal Hilbert-basis.

see comments. i will not delete this. (for documentation)

  • $\begingroup$ That looks right, +1! Close to my thoughts on the matter, in any event; that's probably why it looks right to me! ;-) Will get back if I have any further ideas. $\endgroup$ Mar 3, 2014 at 19:03
  • $\begingroup$ i had to edit, there is an important point missing. $\endgroup$
    – Max
    Mar 3, 2014 at 19:04
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    $\begingroup$ How is $M$ a subspace? $\endgroup$
    – copper.hat
    Mar 3, 2014 at 19:42
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    $\begingroup$ @Daniel: then it wouldnt be dense (if its not in the orthogonal but somewhere else project it to the orthogonal of the span of the vectors of the maximal elements. it' s in $M$ by definition) $\endgroup$
    – Max
    Mar 3, 2014 at 20:18
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    $\begingroup$ No. Consider $M\subset \ell^2$ the dense subspace of all sequences with only finitely many nonzero terms. Let $\xi = (\frac{1}{n+1})$, and $K = \langle\xi\rangle^\perp \cap M$. Take the system $(n+1)\cdot e_n-(n+2)\cdot e_{n+1}$ in $K$ and orthonormalise it. A maximal orthonormal subset of $M$ containing that system is contained in $\langle\xi\rangle^\perp$. $\endgroup$ Mar 3, 2014 at 20:28

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