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The following screenshot is taken from the book 'Topics in Banach Space Theory'.

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I don't understand how the author obtains the second sentence, which states that

'Therefore in order to prove that $T$ is compact it suffices to show that $T|_{B_X}$ is weak-to-norm continuous'. Here, $B_X$ refers to the closed unit ball of $X$.

Can anyone enlighten me?

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To show that T is a compact operator it suffices to show that the closure of the image of the closed unit ball $B_X$ is compact. Now the continuous image of a compact space is compact.And $B_X$ with the weak topology is a compact space. So if $f$ is any function $f:B_X\to Y$ which is continuous with respect to the weak topology on $B_X$,then the image $f(B_X)$ is a compact space. In this Q, with $f=T$ ,the target space is $Y=l_p$ with the $l_p$-norm topology. If $T$ is continuous with respect to these topologies then $T(B_X)$ is compact and also closed in $l_p$ (because it's compact in the metric space $l_p$.)

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  • $\begingroup$ Since $T(B_X)$ is norm-compact, it is also norm-closed in $l_p$. Hence, $\overline{T(B_X)} = T(B_X)$ is norm-compact. Am I right? $\endgroup$ – Idonknow Oct 24 '15 at 5:53
  • $\begingroup$ Yes.Exactly. I didn't read the whole proof. I just tried to explain that one sentence. $\endgroup$ – DanielWainfleet Oct 24 '15 at 6:28
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Let us suppose that $T|_{B_X}$ is weak-to-norm continuous. To show that $T$ is compact, we must show that $T(B_X)$ is precompact in $Y$. But $B_X$ is compact with respect to the weak topology on $X$ and $T|_{B_X}$ is continuous with respect to that topology. Hence $T(B_X)$ is compact, being the continuous image of a compact set (using the weak topology on $B_X$). In particular, this means $T(B_X)$ is precompact, so $T$ is compact.

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