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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 !

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  • $\begingroup$ "rigorous enough?" This is subjective, in the end. $\endgroup$ – SK19 May 9 at 20:45
  • $\begingroup$ @SK19 Sure. But at least, is it correct ? $\endgroup$ – Rebellos May 9 at 20:45
  • $\begingroup$ It is correct. I would be inclined to write $A b_n \stackrel{w}{\to}y$ to emphasise the convergence type. $\endgroup$ – copper.hat May 9 at 20:51
  • $\begingroup$ @copper.hat Thanks for the heads up ! $\endgroup$ – Rebellos May 9 at 21:08
  • $\begingroup$ 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. $\endgroup$ – David Mitra May 10 at 3:26

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