= Solution
First suppose the <convex set> $K$ is closed in the <norm topology>, and $x_n\in K$ satisfies $x_n\rightharpoonup x$. Apply the <weak Banach–Saks theorem in a Hilbert space> to the <weakly null sequence> $x_n-x$. Along the resulting subsequence,
$$
z_m=\frac1m\sum_{k=1}^m x_{n_k}\longrightarrow x
$$
in <norm>. Each $z_m$ is a <convex combination> of points in $K$, so $z_m\in K$. Since $K$ is a <closed set> in the <norm topology>, $x\in K$.
Conversely, suppose $K$ is a <sequentially weakly closed set>. If $x\in\overline K$ in the <norm topology>, choose $x_n\in K$ with $\|x_n-x\|<1/n$. The <Cauchy-Schwarz inequality> gives $|\langle x_n-x,z\rangle|\leq\|x_n-x\|\|z\|\to0$ for every $z$, hence <weak convergence> to $x$. The assumed closure property gives $x\in K$. The empty <set> satisfies both properties too. Thus
$$
\boxed{K\text{ convex}:\quad K\text{ norm-closed}\ \Longleftrightarrow\ K\text{ sequentially weakly closed}.}
$$
Back to article page
