Using the closest point property of Hilbert spaces we can prove that for any closed subspace $A \subset H$ we can decompose H as
$$H = A \oplus A^{\bot}.$$
Is this property characterizing Hilbert spaces? Namely, if we have an inner product space H such that for every closed subspace $A \subset H$ we can write
$$H = A \oplus A^{\bot}$$
then space is Hilbert? I've seen some specific cases, where the space was not complete and $H \neq A \oplus A^{\bot}$ for some closed $A$; for example for $H = c_{00}$ (space of sequences with finite support) with inner product inherited from $\ell^2$ and $A = \{ x \in c_{00}: \sum_{n=1}^{\infty} \frac{x_n}{n}=0\}$ so I wonder if we can generalize example like this and always find a closed subspace $A$ of an incomplete inner product space $H$, such that $H \neq A \oplus A^{\bot}$? Or if there is a different way of approaching this question perhaps.
Best Answer
In fact, $H$ is complete iff for every closed subspace $K\subseteq H$, one has that $H=K\oplus K^\perp$. Actually, we will prove something a bit stronger:
Theorem. Let $H$ be a separable, infinite dimensional inner-product space such that $K^\perp\neq \{0\}$, for every proper closed subspace $K\subseteq H$. Then $H$ is complete.
Proof. Find an orthonormal basis $\{e_n\}_n$ for $H$ (the existence of bases relies only on Graham-Schmidt and hence works on any inner-product space).
As usual we may then use $\{e_n\}_n$ to view $H$ as a subspace of $\ell ^2$ containing the compactly supported sequences.
If $H$ is not complete, then $H\subsetneq \ell ^2$, so we may pick a vector $\xi \in \ell ^2\setminus H$. It is then obvious that $$ K:= H\cap \{\xi \}^\perp $$ is a proper closed subspace of $H$, and I claim that the orthogonal complement of $K$ within $H$ coincides with $\{0\}$. In other words, we claim that $$ K^\perp\cap H=\{0\}. $$
To see this, suppose by contradiction that $\eta $ is a nonzero vector in $K^\perp\cap H$.
Fix any $m$ such that $\langle e_m, \xi \rangle \not=0$, and for every $n$ consider the vector $$ u_n = e_n - \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }e_m. $$ Observing that $\langle u_n, \xi \rangle =0$, we see that $u_n\in K$, so $$ 0 = \langle u_n, \eta \rangle = \langle e_n, \eta \rangle - \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }\langle e_m, \eta \rangle , $$ hence $$ \langle e_n, \eta \rangle = \frac{\langle e_n, \xi \rangle }{\langle e_m, \xi \rangle }\langle e_m, \eta \rangle = \langle e_n, \xi \rangle \frac{\langle e_m, \eta \rangle }{\langle e_m, \xi \rangle }. $$ Since $m$ is fixed we see that the coefficients of $\eta $ and $\xi $ are proportional, meaning that also $\eta $ and $\xi $ themselves are proportional, but this is a contradiction because $\eta $ lies in $H$ and $\xi $ does not.