Question: Let $E$ be a Banach space and let $A\subset E$ be a subset that is closed in the weak topology $\sigma(E, E')$. Let $B\subset E$ be a subset that is compact in the weak topology $\sigma(E, E')$. Prove that $A+B$ is closed in $\sigma(E, E')$.
Brezis' solution is:
For every $y\in B$ $\underbrace{\text{there exists a convex neighborhood}\; V(y)\; of\; 0\; \text{such that}\;(x+W)\cap(A+B)=\varnothing}_{(1)}$
(since $A+y$ is closed and $x\notin A+y$).
$\underbrace{\text{Clearly}}_{(2)}$
$$B\subset\displaystyle\bigcup_{y\in B}\left(y-\frac{1}{2} V(y)\right),$$
and since $B$ is compact, there is some finite set $I$ such that
$$B\subset\displaystyle\bigcup_{i\in I}\left(y_i-\frac{1}{2} V(y_i)\right)\; \text{with}\; y_i\in B.$$
Set
$$W=\frac{1}{2}\displaystyle\bigcap_{i\in I} V(y_i).$$
We claim that $(x+W)\cap (A+B)=\varnothing$. Indeed, suppose by contradiction that there exists some $\omega \in W$ such that $x+\omega \in (A+B)$. Hence there is some $i\in I$ such that $$x+\omega \in A+y_i-\frac{1}{2} V(y_i).$$
Since $V(y_i)$ is convex it follows that there exists some $\omega'\in V(y_i)$ such that $\underbrace{x+\omega'\in A+y_i}_{(3)}$. Consequently $(x+V(y_i))\cap (A+y_i)\neq \varnothing;$ absurd.
Doubt:
Would you give me a hint to solve it in another way? (Different from the solution in Brezis' Book?) OR Explain the solution in Brezi's Book more step by step? (Explaining why every conclusion?)
In case, of choose for explaining Brezis' solution, I would like to understand the following:
- Which results assure me that really exists a convex neighborhood of zero for every $y \in B$?
- For me was not clear that
$$B\subset\displaystyle\bigcup_{y\in B}\left(y-\frac{1}{2} V(y)\right)$$
and also that this is an open cover of opens set? (is this interpretation right, isn't?) - Why $x+\omega'\in A+y_i$ and not $x+\omega'\in A+y_i-\frac{1}{2}V(y_i)$, since $\omega'\in V(y_i)$?
Best Answer
The $\sigma(E,E')$ topology is a topology that makes $E$ a topological vector space, so, also a topological group with addition. Now, the proof of the fact that $A+B$ is closed if $A$ compact, and $B$ closed works almost like in the case of $E= \mathbb{R}$. It is easiest to work with (generalized) sequences. Let $a_n + b_n \to c$. Now, $a_n$ being from $A$, has a convergent subsequence $a_{n_k}\to a$. So $a_{n_k}\to a$, $a_{n_k} + b_{n_k} \to c$ implies $b_{n_k} \to c-a$. Since $B$ is closed, we have $c-a= b\in B$. Therefore $c = a+ (c-a) \in A+B$.
Now, in general, usual sequences might not work, but all of the above works as well for generalized sequences. And they feel like the usual ones.