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Let $X$, $Y$ be Banach spaces. A bounded linear operator $T:X\to Y$ is called $\textbf{completely continuous}$ if it maps a weak convergence sequence to a norm convergence sequence. We know that Banach adjoint $T^*:Y^*\to X^*$ of a completely continuous operator $T$ also a completely continuous operator. Also, the adjoint of a compact operator sends a weak* convergence sequence to norm convergence sequence. In this proof it has been used that a compact operator sends a bounded set to a relatively compact subset (or, precompact subset).

Now, my question is does the adjoint of a completely continuous operator sends weak* convergence sequence to norm convergence sequence?

Any help is appreciated. Thank you in advance.

Tuh
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  • Thank you for your comment. I am trying. – Tuh Jan 09 '25 at 05:34
  • @geetha290krm The adjoint of a completely continuous operator is again a completely continuous operator. It follows that $T^$ sends weak convergence sequence, say, ${y_\alpha^}$ to norm convergence sequence. As, weak* topology weaker than weak topology, the weak convergence of ${y_\alpha^}$ becomes weak convergence. Is this argument right? – Tuh Jan 09 '25 at 05:44
  • @geetha290krm Thank you for the hint. But in my previous comment, I mentioned weak convergence implies weak* convergence. Isn't that right? – Tuh Jan 09 '25 at 05:55
  • My mistake. I got your point. – Tuh Jan 09 '25 at 05:58
  • @geetha290krm is it true that for every weak convergence sequence there is a weak* convergence subsequence? – Tuh Jan 09 '25 at 06:02
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    weak convergence in $X^$ implies weak-star convergence in $X^$ since $X \subset X^{**}$. – daw Jan 09 '25 at 07:41
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    There is a lot of confusion about definitions here. On p. 274 Yoisda (Functional Nalysis) the author says that an operator between Banach spaces is compact or completely continuous if the image of the unit sphere is relatively compact. Perhaps, you should include your definitions in the question. – Kavi Rama Murthy Jan 09 '25 at 08:50
  • @geetha290krm I have now added the definition in the question for avoiding the confusion. Sorry for the late edit. – Tuh Jan 10 '25 at 01:07

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The operator $id: l^1 \to l^1$ is completely continuous, its adjoint $id^*$ is the identity on $l^\infty=(l^1)^*$. The unit vectors $e_n \in l^\infty$ converge weak-star to $0$, but not strongly. Hence $id^*$ does not send weak-star to strongly converging sequences.

daw
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  • Thank you very much for the answer. – Tuh Jan 10 '25 at 01:01
  • According to the definition of completely continuous operator, identity operator on an infinite dimensional normed space is not completely continuous. The definition is given in the question. What can you say about this? – Tuh Jan 10 '25 at 11:24
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    No: the identity on $l^1$ is completely continuous, see https://en.wikipedia.org/wiki/Schur%27s_property – daw Jan 10 '25 at 11:28
  • Ok sir. Convinced. – Tuh Jan 10 '25 at 11:36