Let $X$ be a Banach space. For $\epsilon\in(0,2]$, define: $$ \delta_X(\epsilon) = \inf_{x,y \in X}\bigg\{1 - \|\frac{1}{2}(x + y)\| : \|x\| = \|y\| = 1, \|x-y\| \ge \epsilon\bigg\}. $$ 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 $p$ if there exists some $C>0$ such taht $\delta_X(\epsilon) \ge C\epsilon^p$ for any $\epsilon \in (0,2]$.
We know that any Hilbert space is convex of power type $2$. But it seems that Hilbert spaces (or at least $\mathbb{R}^n$) are convex of power type $p$ for any $p\geq 2$. Is it true?
This follows directly from the inequality in the definition. If $q>p$ and $X$ is convex of power $p$ then $$ \delta_X(\epsilon)\ge C \epsilon^p = C \epsilon^q \epsilon^{p-q} \ge C \epsilon^q 2^{p-q} $$ for all $\epsilon\in (0,2]$. So the smallest power $p$ for which this inequality is fulfilled is interesting.