Projection is uniformly continuous on bounded sets of a uniformly convex Banach space I'm doing questions 2., 3., and 4. of Ex 3.32 in Brezis's book of Functional Analysis. Let $(E, |\cdot|)$ be a Banach space and $C \subset E$ a nonempty closed convex set.

Assume that $E$ is uniformly convex.

*

*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 $E$.


He gives a hint "Use Exercise 3.29" to solve question  4. Exercise 3.29 is about another characterization of uniform convexity, i.e.,

If $E$ is uniformly convex, then $\forall M>0, \forall \varepsilon>0, \exists \delta>0$ such that the inequality
$$
\frac{|x|^2}{2} + \frac{|y|^2}{2} -\left | \frac{x + y}{2} \right |^2 \ge \delta, \quad \forall x,y\in E \, \text{ such that } \, |x|\le M,|y|\le M, |x-y|>\varepsilon.
$$

Below is my attempt.

*

*To solve 2., I use  the fact that "in a topological space, a sequence converges to a point if and only if each of its subsequences has a further subsequence converging to that point". I would like to ask for a proof that does not use this characterization.


*Assume that the author did not give us the hint. Can we solve 4. without applying Ex 3.29?

My attempt:
2.
Assume $(y_n)$ is a sequence in $C$ such that $|y_n-x| \to |Px-x|$ as $n \to \infty$. Because $E$ is uniformly convex and thus reflexive, it suffices to show that $y_n \rightharpoonup Px$ in weak topology. We will prove that every subsequence $(z_{n})$ of $(y_n)$ has a further subsequence that converges weakly to $Px$.
Clearly, $(y_n)$ and thus $(z_{n})$ are bounded. Also, $E$ is reflexive, so there exist $z\in C$ and a subsequence $(t_n)$ of $(z_n)$ such that $t_n \rightharpoonup z$ in weak topology. Notice that norm is l.s.c. in weak topology, so $|z-x| \le \liminf_n |t_n-x| = |Px-x|$. Then $z=Px$.
3.
Let $(x_n)$ be a sequence in $E$ that converges in norm topology to $x\in E$. Notice that
$$
\begin{aligned}
|Px-x| &\le \lim_n |Px_n-x| \\
& \le \lim_n |Px_n-x_n| + \lim_n|x_n-x| \\
&= \lim_n |Px_n-x_n| \\
&\le \lim_n |Px-x_n| \\
&= |Px-x|.
\end{aligned}
$$
It follows that $(Px_n)$ is a minimizing sequence in $C$, so $Px_n \to  Px$ in norm topology.
4.
Let $B \subseteq E$ and $M >0$ such that $|x| \le M$ for all $x\in B$. Assume the contrary that $P$ is not uniformly continuous on $B$. Then there exist $\varepsilon>0$ and a sequence $(x_n, y_n)$ in $B^2$ such that $|x_n-y_n| < 1/n$ and $|Px_n -Py_n| > \varepsilon$.
We have $|Px_n -x_n| \le |Py_n-x_n| \le |Py_n-y_n| + |y_n-x_n|$. So $\lim_n |Px_n -x_n| \le \lim_n |Py_n -y_n|$. By symmetry, we get
$$
\lim_n |Px_n -x_n| = \lim_n |Py_n -y_n|.
$$
WLOG, we assume $|x_n-y_n| \le \varepsilon/2$ for all $n$. Then
$$
\begin{aligned}
|(Px_n -x_n)-(Py_n -y_n)| &= |(Px_n-Py_n)-(x_n-y_n)| \\
& \ge |Px_n-Py_n|-|x_n-y_n| \\
& \ge \varepsilon/2.
\end{aligned}
$$
Because the sequences $(x_n), (y_n)$ are bounded, so are $(Px_n-x_n)$ and $(Py_n-y_n)$. Due to the uniform convexity of $E$, there is $\delta>0$ such that
$$
\frac{|Px_n-x_n|^2}{2} + \frac{|Py_n-y_n|^2}{2}  \ge \delta + \left | \frac{(Px_n-x_n) + (Py_n-y_n)}{2} \right |^2, \quad \forall n.
$$
It follows that
$$
\frac{|Px_n-x_n|^2}{2} + \frac{|Py_n-y_n|^2}{2}  \ge \delta + \left [ \frac{|Px_n-x_n| + |Py_n-y_n|}{2} \right ]^2, \quad \forall n.
$$
By taking the limit, we get a contradiction.
 A: @Evangelopoulos pointed out a logical mistake of above attempt. Below is my fix.

Consider $f:x \mapsto Px -x$. For all $x,y \in E$, we have
$$
\frac{|f(x) + f(y)|}{2} = \frac{|(Px + Py) -(x+y)|}{2}  \ge \left | P \frac{x+y}{2} - \frac{x+y}{2}  \right | = \left | f \left ( \frac{x+y}{2} \right ) \right |
$$
Let $B \subseteq E$ and $M >0$ such that $|x| \le M$ for all $x\in B$. Assume the contrary that $P$ is not uniformly continuous on $B$. Then there exist $\varepsilon>0$ and a sequence $(x_n, y_n)$ in $B^2$ such that $|x_n-y_n| < 1/n$ and $|Px_n -Py_n| > \varepsilon$. We have
$$
|f(x_n)| = |Px_n -x_n| \le |Py_n-x_n| \le |Py_n-y_n| + |y_n-x_n| =|f(y_n)| + |y_n-x_n|.
$$
So $\lim_n |f(x_n)| \le \lim_n |f(y_n)|$. By symmetry, we get $\lim_n |f(x_n)| = \lim_n |f(y_n)|$.
WLOG, we assume $|x_n-y_n| \le \varepsilon/2$ for all $n$. Then
$$
|f(x_n)-f(y_n)| = |(Px_n-Py_n)-(x_n-y_n)|  \ge |Px_n-Py_n|-|x_n-y_n| \ge \varepsilon/2
$$
Because the sequences $(x_n), (y_n)$ are bounded, so are $(f(x_n))$ and $(f(y_n))$. Due to the uniform convexity of $E$, there is $\delta>0$ such that
$$
\frac{|f(x_n)|^2}{2} + \frac{|f(y_n)|^2}{2}  \ge \delta + \left | \frac{f(x_n) + f(y_n)}{2} \right |^2, \quad \forall n.
$$
It follows that
$$
\frac{|f(x_n)|^2}{2} + \frac{|f(y_n)|^2}{2}  \ge \delta + \left | f \left ( \frac{x_n+y_n}{2} \right ) \right |^2, \quad \forall n.
$$
It follows that
$$
\lim_n |f(x_n)| > \lim_n \left | f \left ( \frac{x_n+y_n}{2} \right ) \right |.
$$
Let $z_n := \frac{x_n+y_n}{2}$. Then $x_n -z_n \to 0$. Just as the case of $(x_n,y_n)$, we get $\lim_n |f(x_n)| = \lim_n |f(z_n)|$. It leads to a contradiction. This completes the proof.
