I'm doing Ex 3.29.1 in Brezis's book of Functional Analysis
Let $(E, |\cdot|)$ be a uniformly convex Banach space. Then $\forall M>0, \forall \varepsilon>0, \exists \delta>0$ such that the inequality
$$
\frac{|x|^2}{2} + \frac{|y|^2}{2} -\left | \frac{x + y}{2} \right |^2 \ge \delta
$$
holds for all $x,y\in E$ with $|x|\le M,|y|\le M, |x-y|>\varepsilon$.
My strategy is to assume the contrary and then obtain $x,y\in E$ such that
$$
\frac{|x|^2}{2} + \frac{|y|^2}{2} \le \left | \frac{x + y}{2} \right |^2.
$$
Below is my failed attempt. Could you leave me some hints (not full solution) to finish the proof?
By normalizing, we can assume $M=1$. Assume the contrary that there exist $\varepsilon>0$ and a sequence $(x_n,y_n)$ with $|x_n|\le 1,|y_n|\le 1, |x_n-y_n|>\varepsilon$ such that
$$
0\le \frac{|x_n|^2}{2} + \frac{|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 < \frac{1}{n}, \quad \forall n.
$$
Then
$$
\lim_n \left [\frac{|x_n|^2}{2} + \frac{|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 \right ] =0.
$$
Notice that $({|x_n|^2}/{2}), ({|y_n|^2}/{2})$ are bounded. WLOG, we can assume that they are convergent. Then
$$
\lim_n \left [\frac{|x_n|^2}{2} + \frac{|y_n|^2}{2} \right ] = \lim_n \left | \frac{x_n+y_n}{2} \right |^2.
$$
Because $E$ is uniformly convex, it's reflexive. WLOG, we assume that there are $x,y\in E$ such that $x_n \rightharpoonup x$ and $y_n \rightharpoonup y$ in weak topology. Notice that the norm function is lower semi-continuous in weak topology, so
$$
\frac{|x|^2}{2} + \frac{|y|^2}{2} \le \lim_n \left [\frac{|x_n|^2}{2} + \frac{|y_n|^2}{2} \right ] = \lim_n \left | \frac{x_n+y_n}{2} \right |^2 \ge \left | \frac{x + y}{2} \right |^2.
$$
Best Answer
Luckily, I have found the proof which I post it below.
By normalizing, we can assume $M=1$. Assume the contrary that there exist $\varepsilon>0$ and a sequence $(x_n,y_n)$ with $|x_n|\le 1,|y_n|\le 1, |x_n-y_n|>\varepsilon$ such that $$ 0\le \frac{|x_n|^2+|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 < \frac{1}{n}, \quad \forall n. $$
Then $$ \lim_n \left [\frac{|x_n|^2+|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 \right ] =0. $$
Notice that $({|x_n|}), ({|y_n|})$ are bounded. WLOG, we assume they are convergent, i.e., $|x_n| \to a$ and $|y_n| \to b$ with $a,b \ge 0$. Then $$ \frac{a^2+b^2}{2} = \lim_n \left |\frac{x_n+y_n}{2} \right |^2 \le \lim_n \left [ \frac{|x_n|+|y_n|}{2} \right ]^2 = \left ( \frac{a+b}{2} \right )^2. $$
This implies $a=b$ and thus $$ \lim_n |x_n| = \lim_n |y_n| = \lim_n \left |\frac{x_n+y_n}{2} \right |=a. $$
Let $\delta>0$ be the modulus of convexity of $\varepsilon$, i.e., $$ \left |\frac{x_n+y_n}{2} \right | \le 1-\delta, \quad \forall n. $$
It follows that $a \le 1-\delta<1$. Next we pick $\eta,\lambda >0$ such that $\lambda(a+\eta) \le 1$ and $1-\delta < \lambda a <1$. WLOG, we assume $|x_n|,|y_n| \le a+\eta$ for all $n$. We define $x_n' := \lambda x_n$ and $y_n':=\lambda y_n$. It follows that $|x'_n|, |y'_n|\le 1$ and $|x_n'-y_n'| \ge \varepsilon$. We also get $$ \lim_n \left |\frac{x'_n+y'_n}{2} \right | = \lambda a>1-\delta, $$ which is a contradiction. This completes the proof.