Equivalence of the definition of rotund space

Question

I want to show that the following two statements are equivalent for a normed vector space $X$:

1. If $x,y\in X$ such that $x\neq y$ and $\|x\|=\|y\|=1$, then $\|\frac{x+y}{2}\|<1$.

2. If $x,y\in X$ such that $x\neq y$ and $\|x\|=\|y\|=1$, then $\|\alpha x+(1-\alpha)y\|<1$ for all $\alpha\in (0,1)$.

$2\implies 1$ is obvious. But how to show $1\implies 2$? Any suggestion will be appreciated.

Answer 1

Hint: Suppose $\alpha \in (0,{1 \over 2}]$. Write $\alpha x + (1-\alpha) y$ in terms of $y, {x+y\over 2}$ and use convexity.

Details:

Let $\beta = 2 \alpha$, we can write $\alpha x + (1-\alpha) y= \beta ({x+y\over 2}) + (1- \beta) y $, and so $\| \alpha x + (1-\alpha)y\| \le \beta \| {x+y\over 2} \| + (1 - \beta) \|y\| < 1$.