# Does $A+A^\perp=\scr H$ in a Hilbert space imply $A$ is closed?

This is just the converse to the Hilbert projection theorem, which says that if $A$ is a closed subspace of a Hilbert space $\scr H$ then $A+A^\perp=\scr H$. If $A$ is a linear subspace of $\scr H$ and $A+A^\perp=\scr H$, can we show $A$ is closed?

Context: Just a problem I thought up, and I don't have any work specifically tacking this problem to show. (If I did, I would post it as answer instead of cluttering the OP.)

Yes. Suppose $x_n \in A$ and $x_n \to x$. If $A + A^\perp = \mathcal{H}$ then we can write $x = y + z$ where $y \in A$ and $z \in A^\perp$. That is, $x-y \in A^\perp$. But then $$\|x-y\|^2 = \langle x-y, x-y \rangle = \lim_{n \to \infty} \langle x_n - y, x-y \rangle = 0$$ since $x-y \in A^\perp$ and $x_n - y \in A$ for all $n$. So $x=y$, meaning $x \in A$.
Let $v\in A^{\perp\perp}$. Then there is a decomposition $v=x+y$ where $x\in A$, $y\in A^\perp$. But since $A\subseteq A^{\perp\perp}$, $x+y$ is also a projection decomposition in $A^{\perp\perp}+A^\perp$, which also has the decomposition $v=v+0$ since $v\in A^{\perp\perp},0\in A^\perp$, so by uniqueness of projections $x=v,y=0$, and $v\in A$. Thus $A=A^{\perp\perp}$ is closed.