Let V be a Banach Space. Define $$\delta_V (t) = \inf \left\{ 1 - \frac{\|x+y\|}{2}: \|x\| = \|y\| = 1, \|x-y\| = t \right\}, \ \ \ \ t \in [0,2].$$ Consider $$ A_V = \inf \left \{ \frac{4 - \|x+y\|^2}{\|x-y\|^2} : \|x\| = \|y\| = 1, x \neq y \right\} $$ and $$ B_V = \inf \left\{ 4 \frac{1 - (1-\delta_V(t))^2}{t^2} : 0 < t \leq 2 \right\}.$$ Then $A_V = B_V$.
Proof. Let $x,y \in V$ s.t. $\|x\| = \|y\| = 1$ and $\|x-y\| = t$, as, by definition, $\delta_V(t) \leq 1 - \frac{\|x+y\|}{2} $, we have $$ (1 - \delta_V(t))^2 \geq \frac{\|x+y\|^2}{4} \Rightarrow 4 (1 - (1 - \delta_V(t))^2 ) \leq 4 - \|x+y\|^2$$ As $t^2 = \|x-y\|^2$, we have $$ \frac{4(1-(1 - \delta_V(t))^2)}{t^2} \leq \frac{4 - \|x+y\|^2}{\|x-y\|^2} \Rightarrow B_V \leq A_V $$ By the other side, given $0 < t \leq 2$ and $\epsilon > 0$, we can find $x,y \in V$ st $\|x\| = \|y\| = 1$ and $\|x-y\| = t$ satisfying: $$ 1 - \frac{\|x+y\|}{2} \leq \delta_V(t) + \epsilon t^2 \Rightarrow 2 - \|x+y\| \leq 2 \delta_V(t) + 2 \epsilon t^2 $$ The next step in the proof would be showing the following inequality: $$ 4 - \|x+y\|^2 \leq 4(\delta_V(t) + \epsilon t^2 (2 - \delta_V(t))) $$ However, I can't see why it's true. I would appreciate a hint.
Thank you.
I cannot follow why you want that last inequality. But here is something that works.
By definition of $\delta_V(t)$, you have $$ \delta_V(t)\leq\frac{2-\|x+y\|}2, $$ that is $$\|x+y\|\leq 2-2\delta_V(t).$$ Also, $$2+\|x+y\|\leq2+\|x\|+\|y\|=4.$$
Then \begin{align} 4-\|x+y\|^2&=(2+\|x+y\|)(2-\|x+y\|) \leq(2+\|x+y\|)(2\delta_V(t)+2\epsilon t^2)\\ \ \\ &\leq(2+\|x+y\|)2\delta_V(t)+8\epsilon t^2\\ \ \\ &\leq(2+2-2\delta_V(t))2\delta_V(t)+8\epsilon t^2\\ \ \\ &=8\delta_V(t)-4\delta_V(t)^2+8\epsilon t^2\\ \ \\ &=4(1-(1-\delta_V(t))^2)+8\epsilon t^2, \end{align} which implies $$ \frac{4-\|x+y\|^2}{\|x-y\|^2}\leq\frac{4(1-(1-\delta_V(t))^2)}{t^2}+8\epsilon. $$