A point of the closed unit ball is the midpoint of two points of the unit sphere

Question

Let $(E, \lVert \cdot \rVert)$ be a real normed space of dimension $n \ge 2$. Any point of the closed unit ball $B$ is the midpoint of two points of the unit sphere $S$. Note: $E$ is not supposed to be Euclidean, and therefore $\lVert \cdot \rVert$ is whatever norm.

My attempts

I investigated a proof by contradiction and take $x \in B$. I suppose that $x \neq 0$, as the result is trivial in that case: $0$ is the midpoint of any diameter of $S$. I consider the map

$$\begin{array}{l|rcl} f : & E \times E & \longrightarrow & \mathbb R \\ & (y_1,y_2) & \longmapsto & \left\lVert\frac{y_1 + y_2}{2} - x\right\rVert \end{array}$$

$f$ is continuous for whatever norm equips $E \times E$ and all those norms are equivalent as $E$ is supposed to have a finite dimension. $S \times S$ is a compact subset of $E \times E$, again because the dimension of $E$ is finite. Therefore $f$ attains its minimum in $S \times S$ at let say $(x_1,x_2)$. If the minimum value is equal to zero, we're done. So I suppose that $f(x_1,x_2) \gt 0$, and I have to seek a contradiction. For this, I try to find $(x_1 + \epsilon_1, x_2+\epsilon_2) \in S \times S$ "close to $(x_1,x_2)$" such that $f(x_1 + \epsilon_1, x_2+\epsilon_2) \lt f(x_1,x_2)$... without much result so far. Any good idea?

Subsidiary question: does the result also hold for infinite dimensional spaces $E$?

Answer 1

After various attempts, I was finally able to find an elementary and self contained proof.

Let $x \in B$ and consider the map:

$$\begin{array}{l|rcl} g : & E & \longrightarrow & \mathbb R \\ & y & \longmapsto & \lVert 2x-y \rVert \end{array}$$

$g$ is continuous. As the dimension of $E$ is supposed to be greater or equal to $2$, $S$ is connected (it is path-connected). Notice that

$$g\left(\frac{x}{\lVert x \rVert }\right) \le 1 \le g\left(-\frac{x}{\lVert x \rVert }\right)$$

The image of $S$ under $g$ is an interval that contains $1$. Therefore, it exists $a \in S$ such that $g(a) = 1$. $a$ and $2x-a$ belong to $S$ and their midpoint is $x$. This allows to conclude to the desired result.

Answer 2

I don't have an idea to complete your proof, but here is an alternative

First, as noted in the comments that it is enough to consider the case $n = 2$ (even without the assumption that $E$ is finite dimensional). Indeed, for $E$ of arbitrary dimension $n \ge 2$ and $x$ in the closed unit ball of $E$, we can consider the two dimensional subspace $E_{x,y} = \operatorname{span}\{x,y\}$ for any $y \in E$ which is linearly independent with $x$. If we equip $E_{x,y}$ with the norm induced by $E$ then $x$ is in the closed unit ball of $E_{x,y}$ and any two points on the unit sphere of $E_{x,y}$ are also points on the unit sphere of $E$.

Therefore I can assume without loss of generality that $E = \mathbb{R}^2$ (but not necessarily with the Euclidean norm). Let $x \in B$ and define $\lambda_\pm(\theta) = \inf\{\lambda: \|x \pm \lambda A_\theta e \| = 1\}$ where $A_\theta$ is the rotation matrix by angle $\theta$ and $e$ is an arbitrary norm $1$ vector. Then $x$ is the midpoint of the line connecting $x + \lambda_+(\theta)$ and $x - \lambda_-(\theta)$ if and only if $\lambda_+(\theta) = \lambda_-(\theta)$. This motivates us to consider $g(\theta) = \lambda_+(\theta) - \lambda_-(\theta)$. Note that $\lambda_\pm(\theta + \pi) = \lambda_\mp(\theta)$ so that $g(\theta + \pi) = - g(\theta)$ which implies that if $g$ takes non-zero values then it takes both positive and negative values. Therefore by the intermediate value theorem, it suffices to check that $g$ is continuous.

For this, it is enough to realise that $\lambda_\pm(\theta) = p_{B - x}(\pm A_\theta e)$ where $p_K$ denotes the Minkowski Functional of $K$. Both $A_\theta$ and the Minkowski functional are continuous (where for the latter claim we use that $x \in B$: see here for a proof) so that the result follows.