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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.

Maurizio Barbato
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2 Answers2

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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.

Etienne
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  • This looks pretty good to me too, but then what do I know? Anyway, +1! – Robert Lewis Mar 04 '14 at 00:19
  • Your proof looks fine to me too. Nice example! – Nate Eldredge Mar 04 '14 at 00:22
  • Thanks. One can find many nice things in Bourbaki's exercises... – Etienne Mar 04 '14 at 02:06
  • 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. – Maurizio Barbato Mar 04 '14 at 15:13
  • Yes, nonseparable spaces are sometimes strange objects... Glad that you found the answer useful! – Etienne Mar 04 '14 at 15:30
  • I understand I can find a linearly independent family in $H_1$, indexed by $I$ with cardinality $\mathfrak c$. But may you explain more about how do we define $l^2(I)$, and why can we find an orthonormal basis $(e_i)$ of $H_2$ indexed by $I$? – John Oct 17 '15 at 09:38
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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)

Max
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  • 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. – Robert Lewis Mar 03 '14 at 19:03
  • i had to edit, there is an important point missing. – Max Mar 03 '14 at 19:04
  • OK, but I still think you're on the right track. As of now, of course. Keep me posted! – Robert Lewis Mar 03 '14 at 19:14
  • bad news... i'm sorry – Max Mar 03 '14 at 19:27
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    How is $M$ a subspace? – copper.hat Mar 03 '14 at 19:42
  • what do you need that statement for? can you maybe drop orthogonality? – Max Mar 03 '14 at 19:43
  • ah, space not set. sorry, my mistake. give me a moment :-) – Max Mar 03 '14 at 19:44
  • @Max: so what is the bad news, exactly? – Robert Lewis Mar 03 '14 at 19:55
  • @robert-lewis i edited. now i edit again and i think it's good news. please check it. – Max Mar 03 '14 at 19:57
  • @Max: I'll try and check it but I'm really busy right now so this may take awhile. Best, RKL – Robert Lewis Mar 03 '14 at 19:59
  • What do you mean by ${ \cal P}(M)$? – copper.hat Mar 03 '14 at 20:02
  • the set of all subsets – Max Mar 03 '14 at 20:05
  • That every maximal orthonormal subset of $M$ must be a Hilbert basis of $H$ is - if it is even true - nontrivial. How do you know its closed span is not a proper subspace whose orthogonal complement happens to have trivial intersection with $M$? – Daniel Fischer Mar 03 '14 at 20:10
  • edited again, but this was my last edit. if there is another problem i will move the wrong/incomplete answer to the comments of the original question. – Max Mar 03 '14 at 20:10
  • But the vector orthogonal to the ON system need not lie in $M$. You only know there is one in $H$. – Daniel Fischer Mar 03 '14 at 20:14
  • How do you conclude that the maximal element is a basis for $H$? – copper.hat Mar 03 '14 at 20:14
  • if it wouldnt be a basis i could find a normed vector in it's orthogonal and add it to my maximal element – Max Mar 03 '14 at 20:17
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    @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) – Max Mar 03 '14 at 20:18
  • I love playing around with Zorn :-) using it way to seldomly. – Max Mar 03 '14 at 20:26
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    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 $\langle\xi\rangle^\perp$. – Daniel Fischer Mar 03 '14 at 20:28
  • you didn't prove $M$ does not have any on-subsystem, you only proved that $\xi$ is not perpendicular to $span\left(2e_{1}-3e_{2},2e_{2}-4e_{3},...\right)$ – Max Mar 03 '14 at 21:01
  • bah, sorry. you proved that $K\perp \xi$. still i dont see why this should show that there is no complete on-subsequence of $M$. and btw: $K$ should be normalized. (so should $\xi$ and that's possible since $\sum_{n}\frac{1}{n^{2}}$ exists) – Max Mar 03 '14 at 21:07
  • @DanielFischer: Nice counterexample. It looks like the straightforward "take a maximal orthonormal subset" construction won't work. max: It's a counterexample to your construction, though not to the original conjecture, which we still haven't solved. $M$ definitely does have a complete orthonormal subset, since it contains the standard orthonormal basis of $\ell^2$. – Nate Eldredge Mar 03 '14 at 21:09
  • the standard basis of $l^{2}$ would indeed be one example for a complete ons for $l^{2}$. when you state it's in $M$ that means............. ? ($M$ does have a complete ons for $l^{2}$) – Max Mar 03 '14 at 21:14
  • please prove in detail why this should be a counterexample, i really don't see it. (and i think it isn't.) – Max Mar 03 '14 at 21:20
  • have you read my comment "normalize $K$ and $\xi$"? – Max Mar 03 '14 at 21:26
  • gentlemen behavour looks differently. you come here, claim something without proof and then you don't give a proper proof when asked for. i'll keep waiting though... – Max Mar 03 '14 at 21:39
  • @Max If you want people to react to your messages, address them with an @, so they get notified and don't need to happen across the message by chance. $K$ is a subspace, what would normalising it mean? Normalising $\xi$ isn't necessary, since we're only interested in its orthogonal complement. The example shows that not every maximal orthonormal subset of $M$ needs to be a Hilbert basis of $H$. Note in fact that if you apply Gram-Schmidt to $(n+1)e_n - (n+2)e_{n+1}$, you obtain an orthonormal basis of $\langle\xi\rangle^\perp$, so that is already maximal (as an ON subset of $M$), – Daniel Fischer Mar 03 '14 at 21:53
  • since $\xi \in H\setminus M$. I tentatively expect that a dense subspace of a Hilbert space contains a Hilbert basis, but proving that is not an easy application of Zorn's lemma. – Daniel Fischer Mar 03 '14 at 21:56
  • $K$ is no subspace. $K$ is a subsystem. and it's neither orthogonal nor is it normal. just to give some examples: $\left<\left(n+1\right)e_{n}-\left(n+2\right)e_{n+1},\left(n+2\right)e_{n+1}-\left(n+3\right)e_{n+2}\right>=n^{2}+n$ and $\left<\left(n+1\right)e_{n}+\left(n+2\right)e{n+1},\left(n+1\right)e_{n}+\left(n+2\right)e{n+1}\right>=\left(n+1\right)^{2}+n^{2}$ so i ask you again: how is this a counter example? what exactly do you get by applying gram-schmidt? – Max Mar 03 '14 at 22:01
  • i couldn't debug this comment, if i copy pasted it to the answer window everything looked fine in the preview. – Max Mar 03 '14 at 22:08
  • Max: You claimed that the a maximal element is a basis for $H$, Daniel's example shows a maximal element of $M$ that is not a basis for $H$. You need to show why a maximal element constructed as above is a basis for $H$. – copper.hat Mar 03 '14 at 22:11
  • no he didnt since he didnt take any on system – Max Mar 03 '14 at 22:13
  • Regardless, you need to show why a maximal element constructed as above is a basis for $H$. – copper.hat Mar 03 '14 at 22:15
  • if the maximal element (let's call it $m$) wouldn't be a basis there would be an element $e$ in it's complement and so would $f:=pr_{m^{\perp}}\left(e\right)$ (w.l.o.g. normed) because $M$ is a subspace thus $\left{f\right}\cup m$ would be a ons in $X$ and biger than $m$ which is not possible. ($pr_{m^{\perp}}$ is the projection on the orthogonal of the span of $m$) – Max Mar 03 '14 at 22:17
  • anyway i'm done with this discussion, unless i see a rigurous counter example (remember: we need a maximal orthonormal set in $M$ which does have an orthogonal complement in $H$. is it possible that you still dont see your problem? hint: $M$ is dense in $H$) – Max Mar 03 '14 at 22:26
  • I'm still not seeing it. How do you know the element in the complement is in (or has a representative in) $M$? $M$ is not closed. – copper.hat Mar 03 '14 at 22:54
  • @Max $K$ is the intersection of the orthogonal complement of the span of $\xi$ - a subspace - with $M$ (also a subspace). Thus it is a subspace. "what exactly do you get by applying gram-schmidt?" By applying the Gram-Schmidt algorithm to ${ (n+1)e_n - (n+2)e_{n+1} : n \in \mathbb{N}}$ we get an orthonormal system in $M$. In fact, an orthonormal basis of $\langle\xi\rangle^\perp$, and therefore a maximal orthonormal system in $M$. – Daniel Fischer Mar 03 '14 at 23:42
  • ye, the closedness it is, thanks! daniel: on the first glance i didn't see that $\left(1,0,...\right)$ is not in the span of $K$. – Max Mar 04 '14 at 00:14