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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.

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

Daniel Montealegre
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    Just as a note, $A + B$ will not be closed for general closed sets $A$ and $B$, one of these must be compact. – user40276 Apr 10 '14 at 02:32
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    I would first try to show $\inf\lbrace |a-b|: a\in A, b\in B, |a|\ge 1,|b|\ge 1\rbrace >0$ (I think this is $\ge c/2$) and using this I would try to show that for the unique decomposition $z=a+b$ one has $|a|\le k |z|$ and $|b|\le k|z|$ for some constant $k$. Following the idea in your question, one could conclude that $a_n$ is Cauchy and hence convergent. – Jochen Apr 10 '14 at 11:35

2 Answers2

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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.

Daniel Fischer
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  • Not quite. We need more than $c_1 \leqslant \lVert T\rVert \leqslant c_2$, we need that $$\bigl(\forall a,b\bigr)\bigl(c_1(\lVert a\rVert + \lVert b\rVert) \leqslant \lVert T(a,b)\rVert \leqslant c_2(\lVert a\rVert + \lVert b\rVert).$$ $c_2$ comes simply from the triangle inequality, and $c_1$ can be chosen as $\eta$. These inequalities show that $T$ is an embedding, and hence $A\times B$ is homeomorphic to the image $A+B$ of $T$. In general, not isometrically isomorphic. But, for topological vector spaces, the topology determines the uniform structure, and hence uniform properties like ... – Daniel Fischer Apr 16 '14 at 15:17
  • ... completeness are determined by the topology alone for topological vector spaces. Since $A\times B$ is complete, it follows that $A+B$ is also complete. – Daniel Fischer Apr 16 '14 at 15:18
  • ok thanks! that was really cool. I didnt know much about embeddings. – Daniel Montealegre Apr 16 '14 at 16:02
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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 $z\in A+B$.

Alex Becker
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