Let $(E, |\cdot|)$ be a Banach space. The usual definition of uniformly convex space is
Definition: $E$ is uniformly convex if and only if $\forall \varepsilon>0, \exists \delta>0$ such that the inequality
$$
\left | \frac{x + y}{2} \right | \le 1- \delta
$$
holds for all $x,y\in E$ with $|x|\le 1,|y|\le 1, |x-y|>\varepsilon$.
I guess above definition is equivalent to the below one, i.e.,
$E$ is uniformly convex if and only if $\forall \varepsilon>0, \exists \delta>0$ such that the inequality
$$
\left |\frac{x + y}{2} \right | \le \frac{|x|}{2} + \frac{|y|}{2} -\delta
$$
holds for all $x,y\in E$ with $|x|\le 1,|y|\le 1, |x-y|>\varepsilon$.
I mimic another proof from this thread. My attempt does not succeed because I can not guarantee that the system
$$
\begin{cases}
\lambda(a+\eta) \le 1 \\
\lambda(b+\eta) \le 1 \\
1-\delta < \lambda c <1
\end{cases}
$$
has a solution $(\eta, \lambda)$. Could you confirm if they are indeed equivalent and suggest a way to fix my below proof?
My attempt: 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
$$
0\le \frac{|x_n| + |y_n|}{2} -\left | \frac{x_n+y_n}{2} \right | < \frac{1}{n}, \quad \forall n.
$$
Then
$$
\lim_n \left [\frac{|x_n|+|y_n|}{2} -\left | \frac{x_n+y_n}{2} \right | \right ] =0.
$$
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 $a,b \ge 0$. Let $c:= (a+b)/2$. Then
$$
\lim_n \left |\frac{x_n+y_n}{2} \right |= c.
$$
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 $c \le 1-\delta<1$. Next we pick $\color{blue}{\eta,\lambda >0}$ such that $\lambda(a+\eta) \le 1, \lambda(b+\eta) \le 1$, and $1-\delta < \lambda c <1$. WLOG, we assume $|x_n| \le a+\eta$ and $|y_n| \le b+\eta$ 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
$$
\lim_n \left |\frac{x'_n+y'_n}{2} \right | = \lambda c>1-\delta,
$$
which is a contradiction. This completes the proof.
Best Answer
No spaces satisfy your condition. For, fix $\epsilon$. If $\|x\|\in(\epsilon,1)$ and $y=0$, then your condition implies that there exists $\delta$ with $$\|x\|=2\cdot\frac{\|x+y\|}{2}\leq2\left(\frac{\|x\|}{2}+\frac{\|y\|}{2}-\delta\right)=\|x\|-2\delta$$
But this is impossible.