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 $C$ be a non-empty closed convex subset of $H$. Let $T:C \to C$ be $1$-Lipschitz, i.e., $$ |Tu-T v| \le |u-v| \quad \forall u,v \in C. $$
- Let $(u_n) \subset C$ and $u,f \in H$ such that $u_n \to u$ in the weak topology $\sigma(H, H^*)$ and $(u_n-Tu_n) \to f$ in norm topology. Then $u-Tu = f$. [Hint: Start with the case $C=H$ and use the inequality $\langle (u-Tu) - (v-Tv), u-v \ge 0$ for all $u,v \in H$]
My below proof does not follow the hint by the author. It seems more direct and simpler than the one following the hint.
Could you confirm if my attempt contains some logical mistakes?
We define $F:C \to H$ by $Fv:= v-Tv$. We have $C$ is closed in $\sigma(H, H^*)$, so $u \in C$. We need to prove that $Fu=f$. Because $T$ is $1$-Lipschitz, we get $$ \begin{align} |u-u_n|^2 &\ge|Tu-Tu_n|^2 \\ &= |(Fu-Fu_n)-(u-u_n)|^2 \\ &= |Fu-Fu_n|^2 - 2 \langle Fu-Fu_n, u-u_n \rangle + |u-u_n|^2. \end{align} $$
Then $2 \langle Fu-Fu_n, u-u_n \rangle \ge |Fu-Fu_n|^2$. We have $(Fu-Fu_n)_n$ converges in norm, while $(u-u_n)_n$ converges to $0$ in $\sigma(H, H^*)$. This implies $\lim_n \langle Fu-Fu_n, u-u_n \rangle=0$. The claim then follows from $0 \ge \lim_n |Fu-Fu_n|^2 = |Fu-f|^2$.