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Let $V$ be a vector space. Let $H_{1}$ and $H_{2}$ be two Hilbert spaces contained in $V$. Suppose some orthonormal basis of $H_{1}$ is also an orthonormal basis of $H_{2}$. Is $H_{1}=H_{2}$?

My approach:

Suppose $f$ is in $H_{1}$. Then there exists $\{f_{n}\}$, in the linear span of the orthonormal basis, such that $f_{n}\to f$. Now $\{f_{n}\}$ is also Cauchy in $H_{2}$. But $f_{n}$ may converge to $g$ (wrt the norm on $H_{2}$) that need not be equal to $f$. How can I show that $H_{1}\subseteq H_{2}$ and vice versa?

I have figured out that if both $H_{1}$ and $H_{2}$ are an RKHS then $H_{1}=H_{2}$. That is by norm convergence imply pointwise convergence in an RKHS. Proof: Suppose $f_{n}$ converges to f in $H_{1}$ and $f_{n}$ converges to g in $H_{2}$. Then norm convergence imply pointwise convergence. this implies $f_{n}(x)\to f(x)$ for all $x\in H_{1}$ and similiarly $f_{n}(x)\to g(x)$ for all $x\in H_{2}$. Then f=g.

I am not able to think of any counterexample so far. Kindly help me in this.

Anne Bauval
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Santosh Negi
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  • @Rmal What common orthonormal basis have these two Hilbert spaces? – Anne Bauval Jan 25 '25 at 05:15
  • @SantoshNegi I removed "(inner product need not be equal in both $H_1$ and $H_2$)" which made no sense if these sets are not the same. I also made explicit that your sequence ${f_n}$ is in the linear span of the common orthonormal basis. Still, your paragraph about reproducing kernel Hilbert spaces is completely unclear to me. Please edit your post to include your proof of $H_{1}=H_{2}$ in this case. – Anne Bauval Jan 25 '25 at 05:25
  • @Rmal but the orthonormal basis of both Hilbert space are not same. – Santosh Negi Jan 25 '25 at 11:35
  • @AnneBauval yeah make sense. – Santosh Negi Jan 25 '25 at 11:37
  • You are welcome. Now, as I suspected, your RKHS case doesn't make sense to me, because in your problem, $V$ is not supposed to be a space of functions. – Anne Bauval Jan 25 '25 at 12:55
  • Your new explicit requirement that all orthonormal basis of $H_1$ and $H_2$ are the same (instead of just one) makes your question vacuous because it compels all unit vectors (and therefore all vectors) of $H_1$ to belong to $H_2$, and conversely. – Anne Bauval Jan 25 '25 at 13:13
  • But the norm on H_1 and H_2 may be different. I want to phase my question one last time to you. You have a vector space. Fix a set A. Can set A become orthonormal basis for 2 different Hilbert space contained in that same vector space? Please give me precise answer that take H_1 and H_2 ,A and vector space like this so it make sense to me. – Santosh Negi Jan 25 '25 at 16:53
  • Understood, so your edit changed the meaning of your post: "[the] set of orthonormal basis of H1 and [the] set of orthonormal basis of H2 are |the] same" means: each orthonormal basis of H1 is an orthonormal basis of H2, and conversely. Since this is not what you intended to mean, I shall revert that edit and restore "Suppose some orthonormal basis of $H_{1}$ is also an orthonormal basis of $H_{2}$." – Anne Bauval Jan 25 '25 at 17:59
  • Thanks but then this question become trivial after your modification. Can you figure it out for the question that I raised in my last comment? – Santosh Negi Jan 26 '25 at 03:16
  • The question as I re-edited it is precisely the one in your last comment, which I had already answered below (I elaborated it some more, to adress your complain; can you get it now?). The question which was trivial was the one before my modification, as explained above 17 hours ago: if "[the] set of orthonormal basis of H1 and [the] set of orthonormal basis of H2 are |the] same", then all unit vectors (and therefore all vectors) of H1 to belong to H2 , and conversely. – Anne Bauval Jan 26 '25 at 07:08
  • Sir I am not able to understand that if orthonormal basis of H1 and H2 are same, then why H1 will contained in H2. Can you give proof of it. It may be trivial but I am not able to see it. – Santosh Negi Jan 27 '25 at 04:10
  • Convergence of same cauchy sequence in H1 and H2 may be different – Santosh Negi Jan 27 '25 at 04:12
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    If (as you wrote) "[the] set of orthonormal basis of H1 and [the] set of orthonormal basis of H2 are [the] same" (i.e. all of them, instead of just one) then, for any $x\ne0$ in H1, $y:=x/|x|$ belongs to some orthonormal basis of H1, which is also an orthonormal basis of H2, so $y\in$H2, and therefore $x=|x|y\in$H2. – Anne Bauval Jan 27 '25 at 05:10
  • I think I had confused you. How to think for for just one set of orthonormal basis for both H1 and H2? I also want to thank you for your patience while dealing with my question. – Santosh Negi Jan 27 '25 at 05:37
  • You are welcome and I was never confused. Your confusion seems to be a problem of language. If you (rightly) want to talk about just one common orthonormal basis (as I immediately understood and answered negatively), don't insist on adding the words "set of", which, as repeated, would mean all, and would make the question trivial (this is detailed in my previous comment). – Anne Bauval Jan 27 '25 at 14:28
  • Thanks a lot sir. – Santosh Negi Jan 28 '25 at 04:43

1 Answers1

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The answer is no.

Consider an orthonormal basis $(e_i)_{i\in I}$ of a Hilbert space $H_1$, and a Hamel basis $(e_i)_{i\in I}\cup(f_j)_{j\in J}$ of it, and a Hamel basis $(e_i)_{i\in I}\cup(g_j)_{j\in J}$ of another Hilbert space $H_2$ with the same orthonormal basis $(e_i)_{i\in I}$. Both are subspaces of a vector space $V$ with Hamel basis $(e_i)_{i\in I}\cup(f_j)_{j\in J}\cup(g_j)_{j\in J}$.

For instance, let $(e_n)_{n\in\Bbb N}$ be an orthonormal basis of $\ell^2$ and $(h_t)_{t\in\Bbb R}$ be such that $(e_n)_{n\in\Bbb N}\cup(h_t)_{t\in\Bbb R}$ is a Hamel basis of $\ell^2$. Consider the free vector space $V$ over the set $\Bbb N\sqcup\Bbb R\sqcup\Bbb R$ and denote its canonical Hamel basis by $(e_n)_{n\in\Bbb N}\cup(f_t)_{t\in\Bbb R}\cup(g_t)_{t\in\Bbb R}$. Endow $H_1:=\operatorname{span}\left((e_n)_{n\in\Bbb N}\cup(f_t)_{t\in\Bbb R}\right)$ and $H_2:=\operatorname{span}\left((e_n)_{n\in\Bbb N}\cup(g_t)_{t\in\Bbb R}\right)$ with the inner product making it naturally isomorphic to $\ell^2=\operatorname{span}\left((e_n)_{n\in\Bbb N}\cup(h_t)_{t\in\Bbb R}\right)$, and you get a counterexample.

Anne Bauval
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  • My construction is explicit. You can take two (isomorphic) copies of any Hilbert space, for instance $\ell^2$, and identify an orthonormal basis of the 1st one with an orthonormal basis of the 2nd one. – Anne Bauval Jan 25 '25 at 13:09
  • Sir, I have one more question. As you have taken vector space $V_{1}$,and $V_{2}$ different and their completion is same Hilbert space. So you have done that both Hilbert space are different as it corresponds to two different vector space. Is it possible that their completion are not equal ? – Santosh Negi Jan 30 '25 at 12:03
  • ? I just gave a counterexample. I cannot see what you are missing. My two Hilbert (hence complete) spaces $H_1,H_2$, though distinct as subsets of $V$, have a common orthonormal basis. – Anne Bauval Jan 30 '25 at 13:23
  • So, what i understood from your example is. You have $H_{1}$ that is an vector space and you induce some inner product on it and its completion is isomorphic to $\ell^{2}$. Correct? – Santosh Negi Jan 31 '25 at 04:36
  • No, no completion! Again: $H_1$ and $H_2$ are already complete since (as written in my answer) they are already isomorphic to $\ell^2$. – Anne Bauval Jan 31 '25 at 06:39
  • isn't $H_{1}$=span$(e_{n}){n\in \mathbb{N}}?$. I am new to understand free vector space $V$. is $f{t}$ is an sequence in $\ell^{2}$? If yes, can $f_{t}$ be written as span of $e_{n}$ if we consider $e_{n}$ as standard orthonormal basis of $\ell^{2}$? – Santosh Negi Jan 31 '25 at 09:52
  • No $H_1$ is not $\operatorname{span}\left((e_n){n\in\Bbb N}\right)$. I wrote in my answer $H_1:=\operatorname{span}\left((e_n){n\in\Bbb N}\cup(f_t){t\in\Bbb R}\right)$ and $H_2:=\operatorname{span}\left((e_n){n\in\Bbb N}\cup(g_t){t\in\Bbb R}\right)$. 2) No $f_t$ and $g_t$ are not sequences in $\ell^2$. It is $h_t$ (for each $t\in\Bbb R$) which belongs to $\ell^2$, but it is not in the span of the "orthonormal basis" $(e_n){n\in\Bbb N}$ (which is not a Hamel basis): $h_t=\sum_{n\in\Bbb N}\langle e_n, h_t\rangle e_n$ is not a finite linear combination of the $e_n$'s.
    • – Anne Bauval Jan 31 '25 at 15:12
  • As for the notion of free vector space, follow the link given in my answer, or type the three words free vector space as a query to the search engine on this site https://math.stackexchange.com/search?q=free+vector+space or more generally on your favorite web browser. – Anne Bauval Jan 31 '25 at 15:18
  • Okay thankyou so much sir for your help and suggestions, i couldn't reply earlier because of fever. – Santosh Negi Feb 01 '25 at 16:03