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?
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asked Jun 18, 2019 at 12:30
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3$\begingroup$ 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. $\endgroup$– Bill JohnsonJun 18, 2019 at 13:09
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2$\begingroup$ 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? $\endgroup$– Dirk WernerJun 18, 2019 at 19:38
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1$\begingroup$ 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. $\endgroup$– Dirk WernerJun 18, 2019 at 19:44
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1$\begingroup$ Correction: Not $S$ is weakly compact, but its unit ball... $\endgroup$– Dirk WernerJun 18, 2019 at 20:39
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2 Answers
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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.
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2$\begingroup$ As an aside, the converse fails: the whole space is weak * separable does not imply that the ball is weak * separable. math.stackexchange.com/q/408624/442 $\endgroup$ Jun 20, 2019 at 9:53
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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.
answered Jun 22, 2019 at 6:28