When reading "Chebyshev centers and uniform convexity" by Dan Amir I encountered the following result which is apparently "known and easy to prove". I'm sure it is, but I can't find a proof and am failing to prove it myself.


The result (slightly simplified) is


If $X$ is a uniformly convex space (i.e. if $||x|| = ||y|| = 1$ with $||x - y|| \geq \epsilon$ then there exists $\delta(\epsilon) > 0$ such that $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$) then for any $x, y$ with $||x|| \leq 1$ and $||y|| \leq 1$, and $||x - y|| \geq \epsilon$, $||\frac{x + y}{2}|| \leq 1 - \delta(\epsilon)$.


Part of the problem is that I think this isn't true without making some additional restrictions to reduce the value of $\delta(\epsilon)$. e.g. by considering $||x|| = 1$ and $y = (1 - \epsilon) x$ you can see that this requires that $\delta(\epsilon) \leq \frac{1}{2} \epsilon$. So I think the true result is probably just that you can choose $\delta$ so that this is true.


I'm sure this should be easy and I'm just missing an obvious trick, but oh well.



Inequalities for uniformly convex normed spaces
If the second $\delta(\varepsilon)$ is allowed to differ from the first one, then there is a simple implicit argument: Suppose the contrary, then there is a sequence $X\_n$ of 2-dimensional normed spaces satisfying the definition with the same function $\delta(\varepsilon)$ and points $x\_n,y\_n\in X\_n$ with $\|x\_n\|\le 1$, $\|y\_n\le 1\|$, $\|x\_n-y\_n\|\ge\varepsilon$ but $\|(x\_n+y\_n)/2\|\ge 1-\delta\_n$ where $\delta\_n\to 0$. Since the Banach--Mazur compactum is compact, there is a converging subsequence, and the limit space satisfies the definition for the same $\delta(\varepsilon)$ but contains two points $x,y$ with $\|x\|\le 1$, $\|y\|\le 1$, $\|x-y\|\ge\varepsilon$ and $\|(x+y)/2\|\ge 1$, a contradiction.


In fact, you can always choose the same $\delta(\varepsilon)$ in the second case provided that $\dim X\ge 2$. Suppose the contrary, then there are points $x,y\in X$ such that $\|x\|\le 1$, $\|y\|\le 1$, $\|x-y\|\ge\varepsilon$ but $\|(x+y)/2\|=1-\delta\_1$ where $\delta\_1<\delta=\delta(\varepsilon)$. We may assume that $X$ is 2-dimensional (otherwise restrict to a 2-dimensional subspace containing $x$ and $y$). Fix $\delta\_1$ and from all such pairs $x,y$ choose one that minimize $\big|\|x\|-\|y\|\big|$. I claim that this minimizing pair satisfies $\|x\|=\|y\|$.


Suppose the contrary: let $\|x\|>\|y\|$. Denote $z=(x+y)/2$, $v=(x-y)/2$. If $v$ is proportional to $x$, choose any $v'$ with $\|v'\|=\|v\|$ such that $\|z+v\|\ne \|z\|\pm\|v\|$. Then the points $x'=z+v'$ and $y'=z-v'$ show that $x$ and $y$ did not minimize $\big|\|x\|-\|y\|\big|$. If $v$ is not proportional to $x$, choose a vector $w$ parallel to a supporting line to the unit sphere of $\|\cdot\|$ at the point $v/\|v\|$. Note that $w$ cannot be parallel to a supporting line at $x/\|x\|$, so either $\|x+tw\|<\|x\|$ or $\|x-tw\|<\|x\|$ for a sufficiently small $t>0$. Hence the points $x'=x+tw$ and $y'=y-tw$ or $x'=x-tw$ and $y'=y+tw$ provide a counter-example with $\big|\|x'\|-\|y'\|\big|<\big|\|x\|-\|y\|\big|$.


Thus the minimizing pair satisfies $\|x\|=\|y\|$. Multiplying by $\|x\|^{-1}$ we get a counter-example with $\|x\|=\|y\|=1$.


