Uniformly convex Banach space

Let $X$ be a real or complex Banach space with closed unit ball $B_X$. The given norm, and hence $X$, is uniformly convex if for every $\epsilon \in (0,2]$ there is a $\delta >0$ such that

$x,y\in B_X\text{ and }\mathrm{\parallel }x-y\mathrm{\parallel }\ge \epsilon ⟹\parallel \frac{x+y}{2}\parallel \le 1-\delta \mathrm{.}$

The endpoint $2$ is included. Values $\epsilon >2$ need not be tested, because the triangle inequality gives $\mathrm{\parallel }x-y\mathrm{\parallel }\le 2$ on $B_X$. The zero space is uniformly convex vacuously: for each positive $\epsilon$ the antecedent has no witnesses.

Uniform convexity implies strict convexity of the unit ball. Indeed, for distinct unit vectors $x,y$, take $\epsilon =\mathrm{\parallel }x-y\mathrm{\parallel }>0$; the displayed condition makes the midpoint norm strictly less than one. The converse is not part of the definition.

The property belongs to the specified norm, not merely to the underlying topological vector space. For example, on ${\mathbb{R}}^2$ the parallelogram identity gives $\mathrm{\parallel }(x+y)\mathrm{/}2{\mathrm{\parallel }}_2^2=(\mathrm{\parallel }x{\mathrm{\parallel }}_2^2+\mathrm{\parallel }y{\mathrm{\parallel }}_2^2)\mathrm{/}2-\mathrm{\parallel }x-y{\mathrm{\parallel }}_2^2\mathrm{/}4\le 1-{\epsilon }^2\mathrm{/}4$, so the Euclidean norm is uniformly convex with $\delta =1-\sqrt{1-{\epsilon }^2\mathrm{/}4}>0$. The supremum norm is equivalent because $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }z{\mathrm{\parallel }}_2\le \sqrt{2}\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$, but it is not even strictly convex: $(1,1)$ and $(1,-1)$ are distinct unit vectors whose midpoint $(1,0)$ also has supremum norm one. Thus one may not transfer uniform convexity across an arbitrary equivalent renorming.