5

I proved that if $X,Y$ are Banach spaces and $T: X \to Y$ is compact and $x_n \to x$ weakly then $Tx_n \to Tx$ strongly. I am now wondering if there is a shorter proof?

Here is my proof:

Let $x_n \to x$ weakly. By this result here $T$ is norm-norm continuous if and only if it is weak-weak continuous. Hence $Tx_n \to Tx$ weakly.

Next note that because $x_n\to x$ weakly the sequence $x_n$ is bounded. Hence $S=\{x_n\}\cup \{x\}$ is bounded. Since $T$ is compact it follows that $Tx_n$ has a (norm) convergent subsequence. Let's call it $Tx_{n_k}$. Since strong convergence implies weak convergence, $Tx_{n_k}$ converges weakly and since the weak topology is Hausdorff we use uniqueness of the limit to get that $Tx_{n_k}$ converges to $Tx$ weakly. Finally, since $Tx_n$ has the same limit in the weak topology and in the strong topology, $Tx_{n_k} \to Tx$ strongly.

The proof seems to long in general but I am particularly unhappy with using the result I link to. Any ideas how to make this proof neater?

Community
  • 1
  • 1
    But, you need to show (the whole sequence) $(Tx_n)$ converges to $Tx$. – David Mitra Aug 06 '14 at 11:54
  • @DavidMitra Thank you for the comment. Good point. It's not clear to me at the moment if I can fix this mistake or if I will have to rewrite the proof. –  Aug 06 '14 at 12:30
  • 1
    Use your argument to show that every subsequence of $(Tx_n)$ has a further subsequence that converges to $Tx$. This will suffice. – David Mitra Aug 06 '14 at 12:31
  • Your linked result isn't overpowered here. If you wish, you could just incorporate its proof in your argument. (Given $f\in X^$, $f\circ T$ is in $X^$, so $(f\circ T) x_n\rightarrow (f\circ T)(x)$. This implies $(Tx_n)$ converges weakly to $Tx$.) – David Mitra Aug 06 '14 at 12:40
  • @DavidMitra Thanks for the hint! –  Aug 08 '14 at 08:27

0 Answers0