Let $E$ and $F$ be two Banach spaces, and let $T \in \mathcal{L}(E, F)$. Consider the following property (P).

For every weakly convergent sequence $(u_n)$ in $E$, $u_n \rightharpoonup u$, then $Tu_n \to Tu$ strongly in $F$.

Assume that either $E = \ell^1$ or $F = \ell^1$. Does every operator $T \in \mathcal{L}(E, F)$ satisfy (P)?

Edit. Here, we denote by $\mathcal{L}(E, F)$ the space of continuous, i.e. bounded, linear operators from $E$ into $F$ equipped with the norm$$\|T\|_{\mathcal{L}(E, F)} = \sup_{x \in E,\,\|x\| \le 1} \|Tx\|.$$

  • $\begingroup$ well if $\mathcal{L}(E,F)$ denotes the space of linear, not necessarily bounded operators then certainly not. $\endgroup$ – user159517 Feb 14 '16 at 21:52
  • $\begingroup$ @user159517, I've clarified via edit. $\endgroup$ – Student Feb 14 '16 at 21:54

Operators with the property you are interested in are called completely continuous. The answer in both cases is yes.

Every weakly convervent sequence in $\ell_1$ converges strongly, so the answer is trivially yes if $E=\ell_1$. Suppose now that $F=\ell_1$ and let $(x_n)_{n=1}^\infty$ be a weakly convergent sequence in $E$. As $T$, being bounded, is weak-to-weak continuous, $(Tx_n)_{n=1}^\infty$ is weakly convergent, hence also strongly convergent.


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