Uniformly convex Banach space

A Banach space $X$ is uniformly convex when, for every $\epsilon >0$, some $\delta >0$ satisfies

This is a complete uniformly convex space.

Every nonempty closed convex subset of a uniformly convex Banach space has a unique nearest point to each vector. A minimizing sequence is Cauchy: otherwise uniform convexity would put some midpoint strictly closer than the infimum. Completeness and closedness give its limiting point. In ${\ell }^p$, $p\ge 2$, the Clarkson inequality makes this quantitative: the distance between two minimizing vectors tends to zero after subtracting the midpoint's lower distance bound. Nonemptiness is required.

For $2\le p<\infty$, the Clarkson inequality isFor $1<p\le 2$ and $q=p/(p-1)$, the other Clarkson inequality isTogether these inequalities prove uniform convexity of $L^p$ for $1<p<\infty$.

For $p\ge 2$ and $p^{\prime }=p/(p-1)$, interpolate the two-coordinate map $(x,y)\mapsto (x+y,x-y)$ between ${\ell }^1\to {\ell }^{\infty }$ norm one and ${\ell }^2\to {\ell }^2$ norm $\sqrt{2}$. This proves the scalar inequality with factor $2^{1/p}$. Integrating that estimate and applying Minkowski inequality to $|f{|}^{p^{\prime }}+|g{|}^{p^{\prime }}$ in $L^{p/p^{\prime }}$ gives the displayed function inequality.