Showing that $A(\overline{B_1^X}) \subseteq Y$ is closed when $A \in \mathcal{L}(X,Y)$.

Exercise :

Let $$X,Y$$ be Banach spaces and $$X$$ be reflexive, $$A \in \mathcal{L}(X,Y)$$. If $$\overline{B_1^X} = \{ u \in X : \|u\|_X \leq 1\}$$, show that $$A(\overline{B_1^X}) \subseteq Y$$ is closed.

Intuition :

The closed unit ball $$\overline{B_1^X}$$ is closed, convex and bounded in $$X$$. Let $$b_n \in \overline{B_1^X}$$ and $$Ab_n \xrightarrow{w} y \in Y$$. Since $$X$$ is reflexive, then $$\overline{B_1^X}$$ is weakly compact and of course $$b_n$$ lies in a weakly compact set. Hence we have a subsequence $$b'_n \xrightarrow{w} x \in X$$. But since $$\overline{B_1^X}$$ is also weakly closed, it would be $$x \in \overline{B_1^X}$$. Now, it is $$A \in \mathcal{L}(X,Y) \Leftrightarrow A \in \mathcal{L}(X_w,Y_w)$$. Hence $$Ab'_n \xrightarrow{w} Ax$$ which must coincide with $$y$$ as we showed a weak convergence earlier, thus $$y = Ax \in \overline{B_1^X}$$ and the desired closedness is proven.

Question (1) : Is my approach correct and rigorous enough ?

Question (2): I feel like this is a bit too much working around with all the sequences. I know sometimes that with some statement-theorems one may tackle it faster. Any alternatives would be much appreciated !

• "rigorous enough?" This is subjective, in the end. – SK19 May 9 at 20:45
• @SK19 Sure. But at least, is it correct ? – Rebellos May 9 at 20:45
• It is correct. I would be inclined to write $A b_n \stackrel{w}{\to}y$ to emphasise the convergence type. – copper.hat May 9 at 20:51
• @copper.hat Thanks for the heads up ! – Rebellos May 9 at 21:08
• You can avoid sequences: $X$ is reflexive, so $\overline{B_1}$ is weakly compact. Since $A$ is weak-weak continuous $A(\overline{B_1})$ is weakly compact, thus weakly closed, thus norm closed. – David Mitra May 10 at 3:26