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.


1 Answer 1


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.


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

  • 1
    $\begingroup$ $v = a_{{k+1}}e_{n_k}-a_{k}e_{n_{k+1}}$ does not satisfy the condition $\forall_{n\in S_0}\langle v, e_n\rangle=0$. I think this condition is not needed. $\endgroup$
    – Chad K
    Jun 12, 2023 at 6:49
  • $\begingroup$ Also, the conclusion is that $M=V$ which implies that $x_0\perp V$ and since $V$ is dense $x_0 =0$ - a contradiction. $\endgroup$
    – Chad K
    Jun 12, 2023 at 6:55
  • $\begingroup$ @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 $N=M^{\perp}$ ? $\endgroup$
    – anyon
    Jun 12, 2023 at 8:18
  • $\begingroup$ @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. $\endgroup$
    – Chad K
    Jun 12, 2023 at 8:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Chad K
    Jun 12, 2023 at 9:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .