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Originally asked on MSE.

Let $T$ be a linear map from a normed space $E$ into a Banach space $F$.

Let $D\subset \overline{B}_{F^{\ast}}$ be norming, i.e., there is $r>0$ such that $\sup\limits_{v\in D}|\left<f,v\right>|\ge r\|f\|$, for every $f\in F$.

It is easy to see that a linear map $T:E\to F$ is bounded iff $T^{*}D$ is bounded.

If $T$ is (weakly) compact, then $T^{*}$ is a (weakly) compact, and then $T^{*}D$ is relatively (weakly) compact.

I am wondering about the converse of the last statement.

I can show that compactness of $T$ follows from relative compactness of $T^{*}D$ under the assumption that $E^{*}$ is separable, but no progress whatsoever about the weak compactness.

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If I understand you correctly, the following can be considered as a proof of the converse.

(1) We can assume that $D$ is closed, convex, and symmetric about zero (using Krein-Smulian theorem, see Dunford-Schwartz, volume 1, page 434).

(2) Under this assumption, by the result of Dixmier (Duke Math. J., 1948), the condition that $D$ is norming implies that the weak$^*$ closure of $D$ contains $r\bar B_{F^*}$.

(3) Since $T^*$ is weak$^*$-continuous, the image $T^*(r\bar B_{F^*})$ is contained in the weak$^*$ closure of $T^*D$.

(4) Since $T^*D$ is relatively weakly compact, the weak$^*$ closure of it is weakly compact. Hence $T^*(r\bar B_{F^*})$ is weakly compact.

(5) Using the V. Gantmacher theorem (Dunford-Schwartz, Volume 1, page 485) we get that $T$ is weakly compact.

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  • $\begingroup$ Thank you! But why is a weak* closure of a relatively weakly compact set weakly compact? $\endgroup$ – erz Jun 24 '18 at 6:25
  • $\begingroup$ Let $M$ be the weak closure. It is weakly compact by the assumption. Thus it is compact in any weaker topology, so is weak$^*$ compact. Therefore $M$ coincides with its weak$^*$ closure, and we are done. $\endgroup$ – Mikhail Ostrovskii Jun 24 '18 at 14:57
  • $\begingroup$ Thank you! And exactly the same argument works for usual compactness, right? $\endgroup$ – erz Jun 25 '18 at 4:20
  • $\begingroup$ Yes, similar argument works for strong topology. $\endgroup$ – Mikhail Ostrovskii Jun 25 '18 at 4:28

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