# Weaky compact subset of Banach space with separable predual

Let $$X$$ be a Banach space and $$S\subseteq X$$ be a subspace such that the unit sphere of $$S$$ is weakly compact. If $$Y^*=X$$ for some separable Banach space $$Y,$$ is it true that $$S$$ is separable?

• Exercise: If $X$ is an infinite dimensional normed space, then its unit sphere is weakly dense in its unit ball. Hint: First prove that zero is in the weak closure of the unit sphere. – Bill Johnson Jun 18 '19 at 13:09
• Maybe the OP wants to consider the unit ball rather than the unit sphere, which in light of Bill's comments is more reasonable to consider? – Dirk Werner Jun 18 '19 at 19:38
• Now, every $Y$ as above is a quotient of $L_1[0,1]$, and hence every $X$ as above is a subspace of $L_\infty$. Thus, $S$ is a weakly compact subset of $L_\infty$ and therefore norm separable. – Dirk Werner Jun 18 '19 at 19:44
• Correction: Not $S$ is weakly compact, but its unit ball... – Dirk Werner Jun 18 '19 at 20:39

Assume that Samya Kumar Ray meant that the unit ball of $$S$$ is weakly compact. If $$S$$ is really a sphere, the fact that the answer is "Yes" follows from the argument outlined by Bill Johnson (and the additional assumptions are not needed).
The answer is "yes" anyway. Possibly a bit more straightforward argument than the one of Dirk Werner: the weak$$^*$$ topology on the unit ball of $$S$$ is weaker than the weak topology, so by the well-known result on compacts this implies that on the unit ball of $$S$$ weak$$^*$$ topology is the same as the weak topology. It is well-known and easy to show that bounded sets in the dual to a separable Banach space are weak$$^*$$ separable. Thus the unit ball of $$S$$ is weak separable. It remains to prove the easy statement that Banach spaces whose balls are weakly separable are separable.
I found a proof using a theorem in "Topcis In Banach Space Theory" by Albiac and Kalton. Theorem: Let $$X$$ be a Banach space. The Dual $$X^*$$ contains a sequence $$(x_n^*)$$ having the property $$X_n^*(x)=0$$ for all $$n\geq 1$$ implies $$x=0.$$ Then any weakly compact subset of $$X$$ is metrizable in weak topology.
Now to the original question take a sequence $$(y_n)\subseteq Y\subseteq Y^{**}=X^*$$ such that $$(y_n)$$ is dense in $$Y$$. Clearly $$(y_n)$$ is total. The result follows applying the above theorem.