Closed convex hull of weak convergent sequence: How to show that $x$ is the only element in $\bigcap K_n$?

Question

Le $H$ be a Hilbert space and Suppose $x_n$ converges weakly to $x$ in $H$. Let $K_n$ be the closed convex hull $\bar{co}\{x_k:k\geq n\}$. I would like to show $\bigcap K_n=\{x\}$.

What I know so far is that for convex set the weak closure is the same as norm closure in $H$. $x$ is clearly in $K_n$, since it is in the weak closure of the tail $\{x_k:k\geq n\}$ for every $n$. But how can I show that $x$ is the only element in $\bigcap K_n$?

Answer 1

Since $x_n$ converges weakly to $x$, one has for every $a \in H$. $$\lim_{n \to \infty} \langle a, x_n \rangle = \langle a, x \rangle.$$ Consequently, one has $$\lim_{n \to \infty} \sup_{z \in K_n} |\langle z, a\rangle - \langle x, a \rangle| = 0.$$ This implies for $y \in \bigcap K_n$, one has for every $a \in H$ $$\langle y, a \rangle = \langle x, a \rangle,$$ which yields $y = x$.

Answer 2

Suppose $\langle \phi, y \rangle \le \alpha$ for all $y \in K_n$, then $\langle \phi, x \rangle \le \alpha$. In particular, $x \in K_n$ for all $x$ and so $x \in \cap_n K_n$.

Suppose $z \neq x$, then there is some $\phi$ such that $\langle \phi, z \rangle > \langle \phi, x \rangle $, and since $\langle \phi, x_n \rangle \to \langle \phi, x \rangle$, we see that $z \notin K_n$ for $n$ sufficiently large and hence $\cap_n K_n = \{x\}$.