On the closed convex hull of a weakly compact set

Question

Let $H$ be an infinite-dimensional real Hilbert space and let $B$ be the closed unit ball of $H$. Let $K\subset B$ be a weakly compact set whose closed convex hull agrees with $B$. Question: does $K$ necessarily contain $0$ ?

Answer 1

If $K$ does not contain $0$, then there is a weak neighbourhood of $0$ disjoint from it, i.e. there is a finite set $\{y_1, \ldots, y_n \} \subset H$ such that $\max_{i=1}^n |\langle y_i, x \rangle| \ge 1$ for all $x \in K$. Since $H$ is infinite-dimensional, there is $z \in H$ such that $\|z\| = 1$ and $\langle y_i, z \rangle = 0$ for all $i$, thus $z \notin K$. Since $K$ is weakly compact, the functional $f: x \to \langle z, x \rangle$ attains a maximum on $K$, and that maximum can't be $1$ because the only member $b$ of $B$ with $\langle z, b \rangle = 1$ is $z$. So for some $\epsilon > 0$, $K \subset \{z: \langle z, x \rangle < 1-\epsilon \}$, and its closed convex hull can't include $z$.

Answer 2

I don't want to think about the non-separable case, but if $H$ is separable so that $(B,\mathcal T_w)$ is metrizable, then the answer is yes. Suppose that $x_n\to x$ weakly and recall two basic facts: (1) $\|x\|\le\liminf \|x_n\|$; (2) if also $\|x_n\|\to\|x\|$, then $\|x_n-x\|\to 0$.

So the unit vectors $\|x\|=1$ are actually norm limits of vectors from the convex hull of $K$. But then such approximating convex combinations $\sum c_j y_j$ must contain vectors $y_j$ norm close to $x$, since $x$ is an extreme point (or make $x$ the first vector of an ONB and expand the $y_j$'s). Since this holds for every $x$, $\|x\|=1$, we obtain a sequence $y_n\in K$ which is almost orthogonal and hence $y_n\to 0$ weakly.