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In Brezis' Functional Analysis, exercise 6.2, item 1 is the following

Let $E$ and $F$ two Banach spaces, and $T: E \rightarrow F$ a continuous linear operator.

Assume $E$ is reflexive. Prove that $T(B_E)$ is closed (strongly).

In pg 62 ones read

suppose that T is continuous from $E$ weak into $F$ weak. Then $G(T)$ is closed in $E \times F$ equipped with the product topology $\sigma(E, E^*) \times \sigma(F, F^*)$ (...). It fallows that $G(T)$ is strongly closed.

Theorem 3.10 gives the above continuity for $T$ in weak topology and with Kakutani's Theorem ensures that $B_E$ is compact in the weak topology. The above comment is enough to finish the proof for the question?

Thanks in advance

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What you've stated definitely helps. As you've noted, $T$ is weak-to-weak continuous, and since $E$ is reflexive, $B_E$ is weakly compact. Thus $T(B_E)$ is weakly compact, hence weakly closed, and therefore norm closed.

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  • $\begingroup$ Thanks very much! I'm a little confused for the operators here. Can I assert that if T is linear continuous and A is (weak) closed then T(A) is closed (weak) strong? $\endgroup$ – gdlm Jun 20 '17 at 21:21
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    $\begingroup$ You don't need convexity in the last step. A weakly closed set is always norm closed. You need convexity for the converse. $\endgroup$ – Nate Eldredge Jun 20 '17 at 21:28
  • $\begingroup$ @NateEldredge, yes but my question is whether can I say that T(A) is closed or not $\endgroup$ – gdlm Jun 20 '17 at 21:36
  • $\begingroup$ @NateEldredge Duh I don't know what I was thinking. $\endgroup$ – Aweygan Jun 20 '17 at 22:01
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    $\begingroup$ @gdlm In general the assertion made in your comment is not true; $T$ need not map closed sets to closed sets, regardless of the topologies used. $\endgroup$ – Aweygan Jun 20 '17 at 22:04

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