I'm doing Ex 3.32.(5. and 6.) in Brezis's book of Functional Analysis. Could you have a check on my attempt?
Let $(E, |\cdot|)$ be a uniformly convex Banach space and $C \subset E$ a nonempty.
- Prove that for every $x \in E$,
$$
\inf _{y \in C}|x-y|
$$
is achieved by some unique point in $C$, denoted by $P x$.- Prove that every minimizing sequence $\left(y_{n}\right)$ in $C$ converges strongly to $P x$.
- Prove that the map $x \mapsto P x$ is continuous from $E$ strong into $E$ strong.
- More precisely, prove that $P$ is uniformly continuous on bounded subsets of $E$.
Let $\varphi: E \rightarrow(-\infty,+\infty]$ be a convex l.s.c. function, $\varphi \not \equiv+\infty$.
- Prove that for every $x \in E$ and every integer $n \geq 1$,
$$
\inf _{y \in E}\left\{n|x-y|^{2}+\varphi(y)\right\}
$$
is achieved at some unique point, denoted by $y_{n}$.- Prove that $y_{n} \underset{n \rightarrow \infty}{\longrightarrow} P x$, where $C=\overline{D(\varphi)}$.
I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.
Best Answer
Fix $x\in E$ and let $(y_m)$ be a sequence that minimizes $n|x-y|^{2}+\varphi(y)$. Because $\varphi$ is proper l.s.c. convex, it is bounded from below by a continuous affine function. It follows that $(y_m)$ is bounded. On the other hand, $n|x-y|^{2}+\varphi(y)$ is convex and l.s.c. in norm topology, so it is in weak topology. Also, bounded convex set is weakly compact in reflexive space. Because $E$ is uniformly convex, it is reflexive. It follows that the minimizer indeed exists. Uniformly convex space is strictly convex, i.e., $y \mapsto |y|^2$ is strictly convex, so the minimizer is unique.
6.
We fix $x\in E$ and consider $f_n (y) : = n|x-y|^2 +\varphi(y)$. Then $f_n(y_n) \le f_n(y)$ for all $y\in E$ and $n \ge 1$. In particular, $f_{n+1}(y_{n+1}) \le f_{n+1}(y_n)$ and $f_{n}(y_{n}) \le f_{n}(y_{n+1})$, i.e., $$ \begin{cases} (n+1)|x-y_{n+1}|^2+\varphi (y_{n+1}) \le (n+1)|x-y_{n}|^2+\varphi (y_{n}) \\ n|x-y_{n}|^2+\varphi (y_{n}) \le n|x-y_{n+1}|^2+\varphi (y_{n+1}). \end{cases} $$ This implies $\varphi (y_{n}) \le \varphi (y_{n+1})$. We have $$ n|x-y_n|^2+\varphi(y_n) \le n|x-y|^2+\varphi(y), \quad \forall y \in E. $$
It follows that $\varphi(y_n) < +\infty$ and thus $y_n \in D(\varphi)$ for all $n$. Also, $$ |x-y_n|^2 - |x-y|^2 \le \frac{\varphi(y)-\varphi(y_n)}{n} \le \frac{\varphi(y)-\varphi(y_1)}{n}, \quad \forall n \ge 1, \forall y \in E. $$
This implies $$ \lim_n |x-y_n|^2 - |x-y|^2 \le 0, \quad \forall y \in D(\varphi). $$
Hence $$ \lim_n |x-y_n|^2 \le |x-y|^2, \quad \forall y \in \overline{D(\varphi)}. $$
This means $(y_n)$ is a minimizing sequence. By 2., we get $y_n \to Px$.