I'm trying to solve an exercise in Brezis' Functional Analysis
Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $K$ be a non-empty closed convex subset of $H$. Let $(u_n) \subset H$ such that for each $v \in K$, the sequence $(|u_n-v|)_n$ is non-increasing.
- Check that the sequence $(d(u_n, K))_n$ is non-increasing. Here $d(u_n, K) := \inf_{v\in K} |u_n-v|$.
- Prove that the sequence $(\pi_K u_n)_n$ converges to a limit $\ell \in K$. Here $\pi_K :H \to K$ is the orthogonal projection onto $K$.
- Assume $(u_n)$ has a property that whenever a subsequence $(u_{n_k})_k$ converges in the weak topology $\sigma(H, H^*)$ to some limit $u \in H$, then $u \in K$. Prove that $u_n \to \ell$ in $\sigma(H, H^*)$.
Could you check if my attempt contains some logical mistakes?
1.This is clear. It follows that $(d(u_n, K))_n$ is convergent.
Let $v_n := \pi_K u_n$. Then $|v-v_n|^2 \le |v-u_n|^2 - |v_n-u_n|^2$ for all $v \in K$. Then for all $n \ge m$, $$ \begin{align} |v_m-v_n|^2 &\le |v_m-u_n|^2 - |v_n-u_n|^2 \\ &\le |v_m-u_m|^2 - |v_n-u_n|^2 \\ &= d^2(u_m, K) - d^2(u_n, K). \end{align} $$ Because $(d(u_n, K))_n$ is a Cauchy sequence, so is $(v_n)$.
Let $\varphi$ be a subsequence of $\mathbb N$. Let $x_n := u_{\varphi (n)}$. It suffices to prove that $(x_n)$ has a subsequence that converges to $\ell$ in $\sigma(H, H^*)$. Clearly, $(x_n)$ is bounded. So there is a subsequence of $(x_n)$ and $x \in H$ such that this subsequence converges to $x$ in $\sigma(H, H^*)$. To simplify notations, we still denote this subsequence by $(x_n)$. Then $x \in K$. Let $y_n := v_{\varphi (n)}$. Then $y_n = \pi_K x_n$ and $y_n \to \ell$. We have $$ \begin{align} |x-\ell|^2 &= \lim_n \langle x-y_n, x-y_n \rangle \\ &= \lim_n \langle x-y_n, x_n-y_n \rangle. \end{align} $$
We have $x \in K$, so by characterization of $\pi_K x_n$ we have $\langle x_n-y_n, x - y_n \rangle \le 0$. Then $x=\ell$. This completes the proof.