A Banach space $X$ is said to be uniformly convex if the following is satisfied: For $\epsilon>0, \exists \delta>0$ such that $x,y\in X, \|x\|, \|y\|\leq 1$, $\|x-y\|\geq \epsilon \Rightarrow \|x+y\|\leq 2(1-\delta)$.
I would like to show that this is equivalent to the following:
For $\epsilon>0, \exists \delta>0$ such that $\|x+\zeta y\|\leq1\forall \zeta\in [-1,1]$ and $\|y\|\geq \epsilon\Rightarrow \|x\|\leq 1-\delta$.
The first definition clearly implies the second, for if for $\epsilon>0$, $\|y\|\geq \epsilon$ and if $\|x+y\|\leq 1$ and $\|x-y\|\leq 1$, by the first definition of uniform convexity, $\exists \delta>0$ such that $\|(x+y)+(x-y)\|\leq 2(1-\delta)\Rightarrow \|x\|\leq 1-\delta$.
To prove that the second definition implies the first, here is what I have tried:
Suppose $\epsilon>0$ and $\|x\|\leq 1, \|y\|\leq 1$ with $\|x-y\|\geq \epsilon$. Let $\tilde{x}=\frac{x+y}{4}$ and $\tilde{y}=\frac{x-y}{4}$.
Then for $\zeta\in [-1,1], \|\tilde{x}+\zeta \tilde{y}\|=\frac{1}{4}\|(1+\zeta)x+(1-\zeta)y\|\leq \frac{1}{4}(1+|\zeta|)\|x+y\|\leq 1$, and $\|\tilde{y}\|\geq \frac{\epsilon}{4}$.
By the second definition, $\exists \delta>0$ such that $\|\tilde{x}\|\leq1-\delta$, i.e. $\|x+y\|\leq 4(1-\delta).$
What I require, however, is $\|x+y\|\leq 2(1-\delta).$
I would appreciate a hint on how to proceed!