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This is an optinal problem in my functional analysis homework:

$T \in B(X,l_1)$, $X$ is a reflexive Banach space, then $T$ is compact.

MY IDEA: show for any bounded $\left\{x_n\right\}$, $\left\{T(x_n)\right\}$ has a convergent subsequence.

Pick $f \in l_1'$, $f(Tx_n)$ is uniformly bounded in $\mathbb{F}$, then $\exists \left\{x_{n_k}\right\}$ such that $f(Tx_{n_k})$ is convergent. Denote the limit by $c_{f,T}$.

$S:= \left\{fT: f\in l_1'\right\}$ is a subspace of $X'$, define $A: S \to \mathbb{F}$ by $$A(fT) := c_{f,T}.$$ It can be shown that $A$ is linear and bounded. Then, by Hahn-Banach, extend $A$ to $\hat A \in X''$.

By reflexivity, $\exists x$ such that $\hat A = \Phi(x)$, where $\Phi$ is the canonical mapping. Then, $$fTx_{n_k} \to c_{f,T} = A(fT) = \Phi(x)(fT) = fT(x), \forall f \in l_1'.$$ By Schur property of $l_1$, $Tx_{n_k} \to Tx$, then we are done.

Question: It seems that X is complete is not used in the proof. Is there any mistake in my proof? Any help will be appreciated.

Siamese
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    I think your proof is correct. As for the completeness of $X$, note that every reflexive space is complete, since it is isometrically isomorphic to it's double dual - a complete space. A different more standard approach to showing that $T$ is compact is the following. If $x_n$ is bounded in $X$ then by the Eberlein Smulian theorem $x_n$ admits a weakly convergent subsequence, say $u_n$. Since $T$ is bounded, it is weak to weak continuous, so that $Tu_n$ converges weakly in $l_1$. Now apply Schur' s Lemma. – Evangelopoulos Foivos Apr 23 '22 at 05:43
  • @EvangelopoulosF. Thanks. Your comment is quite helpful. – Siamese Apr 23 '22 at 06:51
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    I doubt that $A$ is linear because the subsequence depends on $f$. For $g=2f$ nothing prevents you from getting a very different subsequence. – Jochen Apr 23 '22 at 09:25
  • Concerning completeness: Reflexive normed spaces are always complete (because the bidual is Banach). – Jochen Apr 23 '22 at 09:26
  • @Jochen I do not know if it is okay to define $A$ like this: for each $f \in X'$, we know that there is such a subsequence. For $\alpha f$, define $A(\alpha fT) := \lim fT(\alpha x_{n_k})$. For $f + g$, define $A( (f+g)T ):=\lim fT(z_{n_k} ) + gT(z_{n_k})$, where $\left{z_{n_k}\right}$ is the union of $f$'s and $g$''s subseqs. – Siamese Apr 23 '22 at 10:36
  • Maybe you can do something like this if you fix a Hamel basis of $X'$. – Jochen Apr 23 '22 at 13:06
  • In $\ell^1$, every weakly convergent sequence is convergent, see here: https://en.m.wikipedia.org/wiki/Schur%27s_property. This easily yields the claim. See also here: https://math.stackexchange.com/questions/42609/strong-and-weak-convergence-in-ell1 – PhoemueX Apr 23 '22 at 19:17

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