Showing the set $A+B$ is closed. 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. 
 A: Since $A$ and $B$ are closed subspaces of the Banach space $X$, they are themselves Banach spaces, and hence so is $A\times B$, endowed with the norm $\lVert (a,b)\rVert = \lVert a\rVert_X + \lVert b\rVert_X$. Now consider the map
$$T \colon A\times B \to X; \quad T(a,b) = a+b.$$
We have $\lVert T(a,b)\rVert_X \leqslant \lVert (a,b)\rVert$, so $T$ is continuous.
The condition
$$\delta := \inf \{ \lVert x-y\rVert : x\in A, y\in B, \lVert x\rVert_X = \lVert y\rVert_X = 1\} > 0$$
ensures first that $T$ is injective (equivalently $A\cap B = \{0\}$), and then that $T$ is an embedding, namely
$$\inf \{ \lVert T(a,b)\rVert_X : \lVert (a,b)\rVert = 1\} \geqslant \eta := \frac{\min \{1,\delta\}}{4}\tag{1}$$
is easy to show: Suppose in the following always $\lVert (a,b)\rVert = 1$.
If $\bigl\lvert\lVert a \rVert_X - \lVert b\rVert_X\bigr\rvert \geqslant \eta$, the triangle inequality yields $\lVert T(a,b)\rVert_X = \lVert a+b\rVert_X \geqslant \bigl\lvert \lVert a\rVert_X - \lVert b\rVert_X\bigr\rvert \geqslant\eta$ immediately.
If $\bigl\lvert\lVert a \rVert_X - \lVert b\rVert_X\bigr\rvert < \eta$, then in particular $a \neq 0 \neq b$, and
$$\begin{align}
\lVert T(a,b)\rVert_X &= \lVert a+b\rVert_X\\
&= \left\lVert \left(a - \frac{a}{2\lVert a\rVert_X}\right) + \left(\frac{a}{2\lVert a\rVert_X} + \frac{b}{2\lVert b\rVert_X}\right) + \left(b - \frac{b}{2\lVert b\rVert_X}\right)\right\rVert_X\\
&\geqslant \left\lVert\frac{a}{2\lVert a\rVert_X} + \frac{b}{2\lVert b\rVert_X}\right\rVert_X - \left\lvert 1 - \frac{1}{2\lVert a\rVert_X} \right\rvert\lVert a\rVert_X  - \left\lvert 1 - \frac{1}{2\lVert b\rVert_X} \right\rvert\lVert b\rVert_X\\
&\geqslant \frac{\delta}{2} - \left\lvert\lVert a\rVert_X - \frac{1}{2} \right\rvert - \left\lvert\lVert b\rVert_X - \frac{1}{2} \right\rvert\\
&= \frac{\delta}{2} - \bigl\lvert \lVert a\rVert_X - \lVert b\rVert_X\bigr\rvert\\
&> \frac{\delta}{2} - \eta\\
&\geqslant \frac{\delta}{4}\\
&\geqslant \eta,
\end{align}$$
where the equality
$$\left\lvert\lVert a\rVert_X - \frac{1}{2} \right\rvert + \left\lvert\lVert b\rVert_X - \frac{1}{2} \right\rvert = \bigl\lvert \lVert a\rVert_X - \lVert b\rVert_X\bigr\rvert$$
follows from $\lVert a\rVert_X + \lVert b\rVert_X = 1$, whence $\lVert a\rVert_X - \frac{1}{2}$ and $\lVert b\rVert_X - \frac{1}{2}$ have the same magnitude and opposite sign.
Having established that $T$ is an embedding, it follows that $A + B = \mathcal{R}(T)$ is complete, and hence closed.

Proving the inequality $(1)$, or a similar inequality that bounds $\lVert a\rVert_X$ (and $\lVert b\rVert_X$) in terms of $\lVert a+b\rVert_X$, is the crucial step also in other approaches to the proof.
A: 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 $z\in A+B$.
