For convex domains of diameter d, Poincaré inequality holds with constant d/pi 

You need to know: Euclidean space {\mathbb R}^n, norm ||x||=\sqrt{(x,x)} of x\in{\mathbb R}^n, convex domain \Omega\subset {\mathbb R}^n, diameter of \Omega, integral \int_\Omega u(x) dx of function u:\Omega \to {\mathbb R}, gradient \nabla u: \Omega \to {\mathbb R}^n of u.

Background: For convex domain \Omega\subset {\mathbb R}^n, let H^1(\Omega) be the set of functions u:\Omega \to {\mathbb R} for which \nabla u exists on \Omega and both ||u||_{L^2(\Omega)}=\sqrt{\int_\Omega u(x)^2 dx} and ||\nabla u||_{L^2(\Omega)}=\sqrt{\int_\Omega ||\nabla u(x)||^2 dx} are finite.

The Theorem: On 28th February 2003, Mario Bebendorf submitted to the Journal of Analysis and its Applications a paper in which he proved that, for any convex domain \Omega\subset {\mathbb R}^n with diameter d, inequality ||u||_{L^2(\Omega)} \leq \frac{d}{\pi}||\nabla u||_{L^2(\Omega)} holds for all u \in H^1(\Omega) such that \int_\Omega u(x) dx = 0.

Short context: The inequality ||u||_{L^2(\Omega)} \leq c_\Omega||\nabla u||_{L^2(\Omega)}  holds for any (non necessarily convex) subset \Omega\subset {\mathbb R}^n, bounded at least in one direction. It is known as the Poincaré inequality. However, in this generality, it is unclear how to explicitly express constant c_\Omega in terms of parameters of set \Omega. In 1960, Payne and Weinberger published a theorem stating that if \Omega\subset {\mathbb R}^n is convex with diameter d, then Poincaré inequality holds with c_\Omega=\frac{d}{\pi}. However, their proof is correct only for n=2. The Theorem proves this result for all n.

Links: The original paper is available at here.

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