# If $X$ is reflexive, $T\in B(X,Y)$ and $\| Tx_n - Tx \| \rightarrow 0$ whenever $x_n \rightarrow x$ weakly then $T$ is compact.

I want to prove:

If $X$ is reflexive, $T\in B(X,Y)$ and $\| Tx_n - Tx \| \rightarrow 0$ whenever $x_n \rightarrow x$ weakly then $T$ is compact.

Here is my attempt at the proof: Assuming all the conditions hold, we'll want to show $T(ball X )$ has compact closure, or equivalently, any sequence in $T(ball X)$ has convergent subsequence, where $ballX$ is the unit ball of $X$.

So, let $(x_n)$ be a net in $ballX$ which converges weakly to $x$ then $Tx_n$ converges (strongly) to $Tx$

(?) then $(x_n)$ has a convergent subsequence, then $T(ballX)$ has compact closure.

This doesn't sound right... and I don't know where to use the fact that $X$ is reflexive.

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A bounded sequence in $X$ has a weakly convergent subsequence (this is where you use reflexivity). So, by hypothesis, every bounded sequence in $X$ has a subsequence whose image under $T$ is norm convergent. –  David Mitra Jun 5 '12 at 0:31