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David Milman was a mathematician who helped develop the geometry of Banach spaces and the use of extreme points in functional analysis. He studied with Mark Krein in Odessa and later worked at Tel Aviv University. Their Krein-Milman theorem describes a compact convex set through the closed convex hull of its extreme points.^[ref] He also proved that a uniformly convex Banach space is a reflexive Banach space. This connection makes a geometric property of the norm control a duality property of the entire space.

Memoir: www.math.tau.ac.il/images/stories/documents/david_milman.pdf

Let $x,y\in {\ell }^2$ satisfy $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$. The parallelogram law givesIf $\Vert x-y{\Vert }_2\ge \epsilon$, thenand thereforeThis depends only on $\epsilon$, so the l2 sequence space is a uniformly convex Banach space, as asserted by the uniform convexity of l2.

For $p\ge 2$, integrate the stated scalar inequality to obtain the Clarkson inequalityIf $f,g$ belong to the unit ball and $\Vert f-g{\Vert }_p\ge \epsilon$, thenThus $L^p[0,1]$ is uniformly convex.

Now let $X$ be uniformly convex. It is enough to show that every $\Phi \in S_{X^{\ast \ast }}$ lies in the canonical image of $X$. Given $\epsilon >0$, choose the corresponding uniform-convexity constant $\delta$, and choose $f\in B_{X^{\ast }}$ with $\Phi (f)>1-\delta$. If $x,y\in B_X$ both satisfy $f(x),f(y)>1-\delta$, thenso $\Vert x-y\Vert <\epsilon$. By Goldstine theorem, every weak-star neighbourhood of $\Phi$ contains some $Jx$ with $x\in B_X$. Directing these neighbourhoods produces a norm-Cauchy net $(x_{\alpha })$; completeness gives $x_{\alpha }\to x\in B_X$, and weak-star convergence then gives $Jx=\Phi$. Scaling handles the whole bidual ball, so $X$ is reflexive.

For $p\ge 2$, uniform convexity therefore makes $L^p$ reflexive. If $1<p<2$, its conjugate exponent $q$ is greater than two, so $L^q$ is reflexive. Since $(L^p{)}^{\ast }=L^q$ and a Banach space whose dual is reflexive is itself reflexive, $L^p$ is reflexive for every $1<p<\infty$.

The l2 sequence space is a uniformly convex Banach space. If $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$, the parallelogram law givesThus $\Vert x-y{\Vert }_2\ge \epsilon$ permits the modulus $\delta (\epsilon )=1-\sqrt{1-{\epsilon }^2/4}$.

A normed vector space is uniformly convex if, for each $\epsilon >0$, there is $\delta >0$ such that unit-ball vectors $x,y$ with $\Vert x-y\Vert \ge \epsilon$ satisfy $\Vert (x+y)/2\Vert \le 1-\delta$. Completeness is not part of this definition; a complete example is a uniformly convex Banach space.