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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.