Putting a Lower Bound On The Power Type Of Modulus Of Uniform Convexity

Question

Take a Banach space $(X,\|\cdot\|_X)$ and define, on the interval $[0,2]$, the modulus of uniform convexity to be $\delta_X(\epsilon) = \inf\{1-\frac{\|x+y\|}{2}:\|x\|=\|y\|=1, \|x-y\|=\epsilon\}$.

$X$ has modulus of convexity of power type $q$ if, for some constant $c\in(0,\infty)$, $\delta_X(\epsilon)\geq c\epsilon^q$.

I'm trying to show that $q$ is necessarily greater than $2$. I'm a little confused about how I should do this. After trying a couple things, I think that a Hilbert space has power type 2, and so somehow this should be our best case scenario. How can I show that this is the case?

Answer 1

Pick any two-dimensional subspace of $X$, thinking of it as a plane $\mathbb{R}^2$ with another norm $\|\cdot \|$. Let $B=\{(x,y)\in \mathbb{R}^2 : \|(x,y)\|\le 1\}$; this is a bounded closed convex set. Let $P$ be the point of $B$ that is the furthest from $(0,0)$ in the Euclidean norm. We can choose coordinates so that this point is $(1,0)$. In particular, $B$ is contained in the Euclidean closed unit disk $D$.

Consider the intersection of $B$ with the line $x=t$ for a fixed $t\in (0,1)$. Since $B\subset D$, this intersection is a line segment of length at most $2\sqrt{1-t^2}$. Let $A,B$ be its endpoints. The segment $AB$ contains the point $(t,0)$ which has norm $t$. By using the triangle inequality involving $(t,0)$ and one of the points $A,B$, we arrive at $$\|(A+B)/2\| \le \frac12 (1+t)$$

So, for some $\epsilon \le 2\sqrt{1-t^2}\le \sqrt{8(1-t)}$ we have $$\delta_X(\epsilon)\ge (1-t)/2 \ge \epsilon^2/16$$