Let $X$ be a Banach space. For $\epsilon \in (0,2]$, define:

$$\delta_X(\epsilon) = \inf_{x,y \in X}\{1 - \|\frac{1}{2}(x + y)\| : \|x\| = \|y\| = 1, \|x-y\| \ge \epsilon\}.$$

Then we say that $X$ is uniformly convex if $\delta_X(\epsilon) > 0$ for any$\epsilon \in (0,2]$. Furthermore, we say that $X$ is convex of power type $p$ (or $p$-convex) if there exists some $C>0$ so that $\delta_X(\epsilon) \ge C\epsilon^p$ for any $\epsilon \in (0,2]$.

I would like the show that, if $(\mathcal{H}, (\cdot,\cdot))$ is a Hilbert space, then $\mathcal{H}$ is of power type $2$.

I do know that every Hilbert space is uniformly convex, and that this proof uses the parallelogram law. For if we have $x, y \in \mathcal{H}$ and $\epsilon \in (0, 2]$ with $\|x\| = \|y\| = 1$ and $\|x-y\| \ge \epsilon$, we get:

$$\|x+y\|^2 = \|x+y\|^2 +\|x-y\|^2 - \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2 - \|x - y\|^2 \le 4- \epsilon^2.$$

We can then set $\eta = 1 - \frac{1}{2} \sqrt{4 - \epsilon^2}>0$ and perform a few manipulations to get $\eta \le 1 - \|\frac{1}{2}(x+y)\| \implies \delta_{\mathcal{H}}(\epsilon) \ge \eta.$

I'm wondering if the parallelogram law is also enough to prove $2$-convexity, or if I need some more powerful machinery?

Hints or solutions are greatly appreciated.


No, you don't need another tool. Parallelogram law is good enough:

One has $$4=\|x+y\|^2+\|x-y\|^2\ge \|x+y\|^2+\epsilon^2.$$ So $$\epsilon^2\le 4-\|x+y\|^2=(2-\|x+y\|)(2+\|x+y\|).$$ Can you find the constant $C$ so that $1-\|\frac12(x+y)\|\ge C\epsilon^2$ from here?

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    $\begingroup$ To finish, we use the fact that $\|x\| = \|y\| = 1$. So this gives $(2- \|x+y\|)(2 + \|x + y\|) \le 2(1- \frac{1}{2}\|x+y\|)(2 + \|x\| + \|y\|) \le 8(1 - \frac{1}{2} \|x + y\|)$. So we can use $C = \frac{1}{8}$. $\endgroup$ – JZS Aug 13 '14 at 23:27

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