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.
Best Answer
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