Weak convergence in a nonempty, closed, convex subset of Hilbert space

Question

Let $ (H, (·, ·)) $ be a (separable) Hilbert space and let $ K \subset H$ be a nonempty, closed, convex set. Let $x \in H$ and denote by $x_{k} \in K$ the unique element such that $ d(x,K) = \parallel x − x_{k} \parallel $.

(i) Prove that \begin{alignat*}{6} (x − x_{k}, y − x_{k}) \leq 0 \ \ \ \forall y \in K \end{alignat*}

(ii) Let $ \{ x_{n} \} \subset K$ be such that $ x_{n} \rightharpoonup x $. Prove that $ x \in K$ (hint: use (i))

I'm having some difficulty with the second point

Answer 1

To make the proof clearer, I will denote by $z_n \in K$ the sequence $z_{n} \rightharpoonup x$

Then, by $(i)$ you have $$ (x − x_{k}, z_n − x_{k}) \leq 0 $$

Now, making $n \to \infty$ you get $\| x-x_k\|=0$.

Answer 2

Proof of ii) without using 1). The fact that $x_n \to x$ weakly implies that $x$ is in the weak closure of $K$. For a convex set weak closure is same as norm closure. Since $K$ is given to be norm closed it follows that $x \in K$.