Given an arbitrary Hilbert space $\scr H$ and closed subspaces $A,B\subseteq\scr H$ with trivial intersection, is it true that $A+B=\{x+y:x\in A,y\in B\}$ is closed? So far, I have the following:
Let $P_A(x)$ be the unique $z\in A$ such that $x-z\in B$. This is a well defined quantity for all $x\in A+B$ because $A\cap B=0$, and $x=P_A(x)+P_B(x)$ is a decomposition of $x$ into $A$ and $B$. Given any $(x_n)_{n\in\Bbb N}\subseteq A+B$ converging to $x$, we wish to show that $x\in A+B$. Assuming that $(P_A(x_n))_{n\in\Bbb N}$ and $(P_B(x_n))_{n\in\Bbb N}$ converge, they must converge to points $p_A\in A,p_B\in B$ respectively, since $A,B$ are closed, and then $x=p_A+p_B$ because of the continuity of $+$.
It remains to show that the projected sequence is Cauchy, but I'm stuck at showing something along the lines of $||P_A(x_n)-P_A(x_m)||\le C||x_n-x_m||$, because there is no reason to believe (I think) that $P_A$ is bounded.
