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We knew that for every closed space $M$ of Hilbert space $H$,$H=M\oplus M^{\bot}$ i.e. $M$ is complemented.

And here is an example of inner product space that exist a closed subspace that is not complemented.

So I am questioning that is it true that if an inner product space satisfy that every closed subspace is complemented, then it be a Hilbert space?

Or any example that incomplete inner product space satisfy that every closed subspace is complemented is helpful enough for me.

Thanks if you can give me any advice of any form.

anyon
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1 Answers1

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Assume $V$ is a separable inner product space. Then $V$ admits an orthonormal basis, by the Gram-Schmidt procedure applied to a dense countable subset of $V.$ Let $\mathcal{B}$ be an orthonormal basis of $V.$ Denote by $\mathcal{H}$ the completion of $V.$ By assumption $V\subsetneq \mathcal{H}.$ Let $x_0\in \mathcal{H}\setminus V.$ Then $x_0$ cannot be represented by a finite linear combination of the elements of $\mathcal{B}.$ Hence there is a set $S_0:=\{e_k\}_{k=1}^\infty\subset \mathcal{B}$ such that $$x_0=\sum_{k=1}^\infty a_ke_{k},\qquad a_{k}\neq 0$$ Let $$M=\{v\in V\,:\, \langle v,x_0\rangle =0\}$$ Then $M$ is a closed subspace of $V.$ We have $e\in M$ for $e\in \mathcal{B}\setminus S_0.$ Moreover $ \overline{a_{k+1}}e_{k}-\overline{a_k}e_{k+1}\in M.$ Assume $0\neq v\in M^\perp.$ Then $v\perp e$ for $e\in \mathcal{B}\setminus S_0.$ Therefore $$v=\sum_{k=1}^\infty b_ke_{k}$$ Furthermore $$0=\langle v,\overline{a_{k+1}}e_k-\overline{a_k}e_{k+1}\rangle = b_ka_{k+1}-b_{k+1}a_k$$ Hence $b_{k+1}={a_{k+1}\over a_k}b_k$ for all $k,$ which implies $b_k=\lambda a_k$ for a nonzero constant $\lambda.$ Thus $v=\lambda x_0,$ i.e. $v\notin V,$ a contradiction.

Remarks

The separability assumption is crucial for the answer above due to this post

Ryszard Szwarc
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    $v = a_{{k+1}}e_{n_k}-a_{k}e_{n_{k+1}}$ does not satisfy the condition $\forall_{n\in S_0}\langle v, e_n\rangle=0$. I think this condition is not needed. – Chad K Jun 12 '23 at 06:49
  • Also, the conclusion is that $M=V$ which implies that $x_0\perp V$ and since $V$ is dense $x_0 =0$ - a contradiction. – Chad K Jun 12 '23 at 06:55
  • @Ryszard Szwarc: Thanks for first, maybe my asking is not clear enough, is it true that for an inner product space $H$ satisfy that $H=M\oplus N$ for closed subspace $M,N$ then $N=M^{\perp}$ ? – anyon Jun 12 '23 at 08:18
  • @anyon: No, that's not true and it's not what you asked. $\Bbb{R}^2$ is the direct sum of two non-orthogonal one-dimensional spaces. What you asked is whether if for all closed subspace $M$ of $H$, $H=M\oplus M^{\perp}$ does it follow that $H$ is complete. The answer is yes and that's what is proved. – Chad K Jun 12 '23 at 08:38
  • @JohnDoe There was a misprints:should be for all $n\notin S_0.$ Thanks – Ryszard Szwarc Jun 12 '23 at 08:49
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    @RyszardSzwarc: But now $e_n\not\in M$ for $n\not\in S_0$. I think the condition should be dropped and just take $M={x_0}^\perp\cap V$. Then everything following is correct. – Chad K Jun 12 '23 at 09:49
  • @JohnDoe: Thanks, I didn't realize that, somehow my brain isn't functioning today. – anyon Jun 12 '23 at 09:52
  • @JohnDoe You are right. Corrected – Ryszard Szwarc Jun 12 '23 at 11:43
  • @RyszardSzwarc: Sorry to bother you, but I still confusing in how to ensure the existence of the orthonormal basis of $V$? – anyon Jun 13 '23 at 02:54
  • My memory is bad. I will have to restrict to separable space $V.$ Otherwise it may occur that $V$ does not contain an orthonormal basis see – Ryszard Szwarc Jun 13 '23 at 04:18
  • @RyszardSzwarc: Thanks a lot again, I think I know how to deal with the nonseparable one: choose a Cauchy sequence ${ x_{n} } \subset H$ that have not limit point in $H$, then take $V=\overline{span{x_{n}}}$, it can be prove that $V$ satisfy that every closed subspace is complemented like $H$ and also incomplete, but $V$ have countable orthonormal basis. – anyon Jun 13 '23 at 07:34
  • I am unable to show the statement for non separable incomplete inner product space $V.$ – Ryszard Szwarc Jun 13 '23 at 13:42