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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.

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I am studying for a qual, so you can go ahead and either give a solution or sketch it.

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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.

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 $z\in A+B$.

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