Uniqueness of midpoint in strictly convex space

Question

Inspired by this post, I want to know if for a strictly convex normed vector space (that is, $\lVert \lambda x+(1-\lambda)y\rVert<1$ for any $x,y\in B_X$ with $x\not=y$ and $\lambda \in(0,1)$) it holds that if $$\lVert a-v \rVert=\lVert b-v \rVert=\lVert a-b \rVert/2$$ then $v=\frac{a+b}{2}$

Answer 1

Definitely.

First consider the case where $\|a - v\| = \|b - v\| = \|a - b\|/2 = 0$. In this case, $a = b = v$, and $v = \frac{a + b}{2}$ follows immediately.

Otherwise, suppose the norms are non-zero. Let \begin{align*} x &:= \frac{a - v}{\|a - v\|} \\ y &:= \frac{v - b}{\|b - v\|}. \end{align*} Note that $\|x\| = \|y\| = 1$, so $x, y \in B_X$. Then,

\begin{align*} \left\|\frac{1}{2}x + \frac{1}{2}y\right\| &= \frac{1}{2}\left\|\frac{a - v}{\|a - v\|} + \frac{v - b}{\|b - v\|}\right\| \\ &= \frac{1}{2} \cdot \frac{2}{\|a - b\|}\|(a - v) + (v - b)\| = 1. \end{align*}

According to strict convexity we must therefore have $x = y$, which implies that $v - b = a - v$. Thus, $v = \frac{a + b}{2}$.