The image of a relatively weakly compact is also rel. weakly compact

Question

Let $T: X \rightarrow Y$ be a bounded linear operator, where $X$, $Y$ are Banach spaces.

Claim: If $A$ is relatively weakly compact, then $T(A)$ is too.

Actually, I don't know whether this statement is true or false, but it what I need to prove my main goal:

Let $S \circ T: X \rightarrow Z \rightarrow Y$, where $Z$ is reflexive. Then $S \circ T$ is a weakly compact operator.

As it turns out that $T(B_X)$ is rel. $w$-compact, I only need the claim to end this.

So, is the claim true? If so, how to prove it?

Answer 1

Yes, this is true.

1. Choose any net $(x_i)$ in $A$.
2. By relative weak compactness, there is a subnet $(x'_j)$ of $(x_i)$ weakly convergent to some $x'$.
3. Since $T$ is bounded, it is also weak-weak continuous and $(T(x'_j))$ converges weakly to $T(x')$.
4. Since $(x_i)$ was arbitrary (and hence so was $(T(x_i))$) and $(T(x'_j))$ is a weakly convergent subnet of $(T(x_i))$, $T(A)$ is weakly compact.

(If you don't like nets, you can apply Eberlein–Šmulian and use sequences instead, but it's not necessary.)

More generally, the continuous image of a relatively compact set is relatively compact by essentially the same argument.

Answer 2

You can think of relative compactness being a topological invariant under continuous mappings between topological spaces.