Originally [asked](https://math.stackexchange.com/questions/2817700/compactness-of-operators-and-norming-sets) 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.



Compactness of operators and norming sets
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. 


