Let $X,Y$ be Banach spaces and $T : X \rightarrow Y$ a bounded operator that is weakly compact.
How to prove that there exists $\epsilon > 0$ and a sequence $(x_n)_n \subset X$ such that $\|x_n\| = 1, \|T(x_n)\| \geq \epsilon$ forall $n \in \mathbb N$, and the sequence $(T(x_n))_n$ is congerging weakly to $0$ ?
All what I can say is that I can extract a subsequence from $(Tx_n)_n$ which converges to a $y \in Y$.
Thanks for any help
Best Answer
As in the paper you referenced in the comments, we'll assume $T$ is not compact.
Then, we may (and do) choose a sequence $(x_n)_n$ in $S_X$ such that no subsequence of $(Tx_n)_n$ is norm convergent. Now choose a subsequence $(x_{n_k})_k$ of $(x_n)_n$ so that $(T x_{n_k})_k$ is weakly convergent.
Now, $(T x_{n_k})_k$ is not norm-Cauchy. So we find $\epsilon>0$ and subsequences $(p_k)$ and $(q_k)$ of $(n_k)$ with
$$\Vert T( x_{p_k}- x_{q_k})\Vert>\epsilon,\quad k=1,2,\ldots.\tag{1}$$
Set $y_k= x_{p_k}- x_{q_k}$.
Then $(Ty_k)$ is weakly null (since $(T(x_{n_k}))_k$ is weakly convergent).
Since $T$ is bounded and since (1) holds, it follows that $(\Vert y_k\Vert)_k$ is bounded away from $0$. From this, it follows that $\bigl(T( y_k/\Vert y_k\Vert)\bigr)_k$ is weakly null. Appealing to (1) again, $\bigl\Vert T(y_k/\Vert y_k \Vert)\bigr\Vert>\epsilon/2 $ for each $k$.
So, $(y_k/\Vert y_k\Vert)$ is the sought after sequence.