Let $E$ be a Banach space and $B(E)$ (resp.~$S(E)$) be the closed unit ball (resp.~the unit sphere) of the Banach space $E$. $E$ has strictly convex norm if for each pair of elements $x, y \in S(E)$ with $x\neq y$, it is true that $\| \frac12 (x+y)\|<1$.
If given norm is strictly convex, then each convex subset of $E$ contains at most one element of minimal norm.
Proof: Let $\|x\|=\|y\|=\inf_{z\in E}\|z\|=:d$ where $x\neq y$. Then $d\leqslant \left\|\frac{1}{2}(x+y)\right\|\leqslant \frac12\left(\|x\|+\|y\|\right)=d$ so $\|\frac{x+y}{2}\|=d$. However, this is a contradiction to strict convexity of norm. Therefore if a minimal element exists, it is unique.
I don't understand why the conclusion above damages the strict convexity? I don't see its relation with the condition $\| \frac12 (x+y)\|<1$.
I appreciate any help.
Thank you