Fix $f \in H$ and let $u_n$ be the projection of $f$ onto $K_n$. Prove that $(u_n)$ converges in $H$

Question

Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $(K_n)$ be a non-increasing sequence of closed convex subsets of $H$ such that $K := \bigcap_n K_n \neq \emptyset$. Fix $f \in H$ and let $u_n := \pi_{K_n} (f)$ be the projection of $f$ onto $K_n$.

Prove that $(u_n)$ converges in $H$.

1. Could you have a check on my attempt?
2. Is there another proof that avoids the use of weak topology?

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Let $u := \pi_K (f)$. Let's prove that $u_n \to u$. We have $|u-u_n| \le |u-f|$ for all $n$. So $(u_n)$ is bounded. Let $\varphi$ be a sub-sequence of $\mathbb N$ and $x_n := u_{\varphi (n)}$. It suffices to prove that $(x_n)$ has a sub-sequence that converges to $u$.

Let $\sigma (H, H^*)$ be the weak topology of $H$. Because $(x_n)$ is bounded, there is a sub-sequence $\psi$ of $\mathbb N$ and $x\in H$ such that $x_{\psi (n)} \to x$ in $\sigma (H, H^*)$. It suffices to prove $x=u$. Because $x_{\psi (n)} \to x$ in $\sigma (H, H^*)$, we get $$ \begin{align} \langle x, f+v \rangle &= \lim_n \langle x_{\psi (n)} , f+v \rangle\\ |x|^2 &\le \liminf_n |x_{\psi (n)} |^2. \end{align} $$

Then $$ \begin{align} & \liminf_n \langle f - x_{\psi (n)} , v-x_{\psi (n)} \rangle \\ = & \liminf_n \big (\langle f, v \rangle - \langle x_{\psi (n)} , f+v \rangle + |x_{\psi (n)} |^2 \big ) \\ = & \langle f, v \rangle - \langle x , f+v \rangle + \liminf_n |x_{\psi (n)} |^2 \\ \ge & \langle f, v \rangle - \langle x , f+v \rangle + |x|^2 \\ = & \langle f - x, v-x \rangle. \end{align} $$

We have $\langle f - x_{\psi (n)} , v-x_{\psi (n)} \rangle \le 0$ for all $v \in K$, so $\langle f - x, v-x \rangle \le 0$ for all $v \in K$. This completes the proof.

Answer 1

As @Hidde pointed out in a comment, your proof is not complete. You need to also show that $(u_n)$ is convergent.

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We have $\langle f - u_n, v-u_n \rangle \le 0$ for all $v \in K_n$ implies $|v-u_n| \le |v-f|$ for all $v \in K_n$. Because $K \subset K_n$ for all $n$, we get $|u-u_n| \le |u-f|$ for all $n$. So $(u_n)$ is bounded. Let $d_n := |f-u_n|$. Because $K_{n+1} \subset K_n$, we get $d_n \le d_{n+1}$. Then $(d_n)$ is increasing and bounded. So $(d_n)$ is convergent. Fix $n > m$. By parallelogram law, $$ \begin{align} 2(d_n^2 + d_m^2) = 4|f - (u_n+u_m)/2|^2 + |u_n-u_m|^2. \end{align} $$

Notice that $(u_n+u_m)/2 \in K_m$, so $|f - (u_n+u_m)/2| \ge d_m$. Then $$ 2(d_n^2 + d_m^2) \ge 4 d_m^2 + |u_n-u_m|^2. $$

Then $$ 2(d_n^2 - d_m^2) \ge |u_n-u_m|^2. $$

Notice that $(d_n)$ is a Cauchy sequence, so is $(u_n)$.