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

  • $\begingroup$ I'm not sure you always can. In $Y=\ell_1$, a sequence is weakly null if and only if it is norm null. $\endgroup$ Aug 20, 2018 at 13:09
  • $\begingroup$ Thank you David. This assersion was used in a paper wrote by William B. Johnson : eudml.org/doc/266092 (just under the first theorem in the first page). I ask my question here because I don't understand the reasons why this can be true. $\endgroup$
    – Kébir J
    Aug 20, 2018 at 13:50
  • $\begingroup$ There, it's assumed that $T$ is not compact. Can you deduce your result with that in mind? $\endgroup$ Aug 20, 2018 at 13:59
  • $\begingroup$ I suppose I can extract a subsequence for which $\|(T(x_n))_n$ is decreasing. Then I extract a sequence weakly-convergent. $\endgroup$
    – Kébir J
    Aug 20, 2018 at 14:23

1 Answer 1


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.

  • $\begingroup$ Thank you. I was very very far from this. $\endgroup$
    – Kébir J
    Aug 20, 2018 at 16:48

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