Let $X,Y$ be two Banach spaces. Assume that $x_n \rightharpoonup x$ weakly in $X$. Let $T:X\rightarrow Y$ be a bounded linear map.
Are the two following statements equivalent:
1) The sequence $\{Tx_n\}_n$ is relatively compact in $Y$.
2) $Tx_n \rightarrow Tx$ strongly in $Y$ ?
I am reading an article where the author uses 1) as an assumption but if 1) and 2) are equivalent, I find this formulation a bit cumbersome…
My attempt:
2) $\Longrightarrow$ 1) is obvious. For 1) $\Longrightarrow$ 2), we will prove the following points:
(a) any subsequence of $\{Tx_n\}_n$ has a converging subsequence.
(b) the limit of any such subsequence is necessarly $Tx$.
It will follow that the whole sequence in fact converges to $Tx$ strongly in $Y$ (not only up to a subsequence).
Proof of (a): note that $\{x_n\}_n$ is bounded since $x_n \rightharpoonup x$.
Thus, $\{Tx_n\}_n$ is also bounded. Thus, any subsequence of $\{Tx_n\}_n$ is bounded and, by assumption 1), it has a converging subsequence, say of limit $y \in Y$.
Proof of (b): note that $T$ is weakly continuous since it is a strongly continuous linear map. Thus, $Tx_n \rightharpoonup Tx$ weakly. By uniqueness of the weak limit, we necessarily have $y=Tx$.
Best Answer
Your proof is OK but you don't have to show that $(Tx_n)$ is bounded. Every sequence in a relatively compact set has a convergent subsequence.