TITLE: Showing the set $A+B$ is closed.
QUESTION [5 upvotes]: Let $X$ be a banach space, and let $A$, and $B$ be closed linear subspaces. Assume that $$\inf\{\|x-y\|\mid x\in A, y\in B, \|x\|=\|y\|=1\}>0$$
I want to show that $A+B$ is closed. 
I was thinking of doing something like, let $z$ be a limit point of $A+B$, then there exist $z_n\in A+B$ such that $z_n\rightarrow z$, each $z_n$ can be written as $a_n+b_n$. Then I wanted to do something along the lines of determining whether $a_n$ and $b_n$ have a limit (if they do, then call them $a$ and $b$ and then $a\in A$, $b\in B$, and $z=a+b\in A+B$), but I can't seem to be able to use the condition given. 
I am studying for a qual, so you can go ahead and either give a solution or sketch it.

REPLY [3 votes]: First we want to show that 
$$c'=\inf\{\|a-b\|:a\in A,b\in B, \|a\|,\|b\|\ge 1\}>0$$
as suggested by Jochen. Suppose $a_n-b_n\to 0$ but $\|a_n\|,\|b_n\|\ge 1$. Then 
$$0\le|\|a_n\|-\|b_n\||\le \|a_n-b_n\|\to 0$$
so $\|a_n\|-\|b_n\|\to 0$. Let $c=\inf\{\|a-b\|:a\in A,b\in B, \|a\|=\|b\|= 1\}$. Then
$$\begin{align}
\|a_n-b_n\| &\ge \left\|\frac{\|a_n\|+\|b_n\|}{2\|a_n\|}a_n-\frac{\|a_n\|+\|b_n\|}{2\|b_n\|}b_n\right\| - \left\|a_n-\frac{\|a_n\|+\|b_n\|}{2\|a_n\|}a_n\right\| -\left\|b_n-\frac{\|a_n\|+\|b_n\|}{2\|b_n\|}b_n\right\|\\
&\ge \frac{\|a_n\|+\|b_n\|}{2}c - \frac{|\|a_n\|-\|b_n\||}{\|2a_n\|} - \frac{|\|a_n\|-\|b_n\||}{\|2b_n\|}\\
&\ge c - |\|a_n\|-\|b_n\||\to c
\end{align}$$
contradicting $a_n-b_n\to 0$. 
Define $\phi:A+B\to A$ by $\phi(a+b)=a$, which is well-defined since $A\cap B=\{0\}$ and is bounded since if $\|a+b\|< c'$ then either $\|a\|<1$ or $\|b\|<1$, and if $\|b\|<1$ then $\|a\|\le \|a+b\|+\|b\|=1+c'$. Thus it is uniformly continuous, so we can extend it to a function $\psi:\overline{A+B}\to A$. If $z_n\to z$ then $a_n=\psi(z_n)\to \psi(z)$ and $b_n\to z-\psi(z)$, which are in $A$ and $B$ respectively by closure, so $z\in A+B$.