Below is the proof of Poincaré's inequality for open, convex sets. It is taken from "An Introduction to the Regularity Theory for Elliptic Systems, Harmonic Maps and Minimal Graphs" by Giaquinta and Martinazzi. I believe the way the authors use the Fundamental Theorem of Calculus after they write "Noticing that..." cannot be justified on an arbitrary convex set. What they do amounts to connecting any two points by a sequence of $n$ lines, with the $k$-th line parallel to the $x_k$-axis. I am pretty sure this is only possible in an $n$-dimensional cube. Am I wrong about this and can their proof still be made to work for an arbitrary convex set?

Incidentally, I've been looking for a straightforward proof of this this proposition and this is the best I could find, although the proof does not look solid. Could anyone give me a reference with a working proof? It is important that the exponent of the diameter should be exactly $p$ (or $1$ if you take $p$-th roots) and the constant should only depend on $p$ and $n$ (and not on $\Omega$).

Poincaré on convex sets

  • $\begingroup$ You are right, this proof does not work. The is a proof in Evans & Gariepy. That proof is slightly more technical. $\endgroup$
    – daw
    Commented Jan 23, 2023 at 19:24
  • $\begingroup$ @daw I have seen that proof but it is only formulated for balls. When I try to generalize it to convex sets, I can't get the exponent of the diameter to be p. $\endgroup$ Commented Jan 24, 2023 at 6:56
  • $\begingroup$ If you go from $x$ to $y$ with a straight line, which you can do by convexity, then you can salvage the proof. If I am not mistaken you get $c(n,p)=\frac{2^n-2}{n-1}$ for $n\ge 2$ (which will not be optimal but does only depend on $n$) $\endgroup$
    – Del
    Commented Nov 15, 2023 at 19:48

1 Answer 1


I'll post an answer since I thought this'd be an easy problem, but I couldn't get anything better than what Gilbarg-Trudinger get (see equation 7.45 on pg. 164 on the third printing and set $S=\Omega$ there), and for them the constant $c$ blows up if the measure of $\Omega$ goes to 0 (so for example it's bad for rectangles with sides $a, 1$, with $0<a<<1$).

I'll use $c(n,p)$ to denote the best constant in the inequality $$ \lVert u-u_\Omega\rVert_p^p \leq c(n,p)d^p\lVert \nabla u\rVert_p^p, $$ where $d:=\text{diam}\Omega$, and norms are taken over $\Omega$.

Turns out, it's a much more intricate problem. As far as I know it was only semi-recently answered:

  1. The case $p=2$ is due to Payne and Weinberger, and their technique of reducing things to a weighted, one-dimensional Poincaré inequality is used in all subsequent works (there's also a lot of nice geometry arguments going on in all the papers discussed). They prove that the optimal constant is the same as that for the 1-d problem on the interval $[0,d]$, $d=\text{diam}\Omega$, i.e. $c(n,2)=(1/\pi)^2$. In particular it's dimension free!

  2. The case $p=1$ is due to Acosta and Durán. The constant here is $c(n,1)=1/2$.

  3. The case of general $p>2$ is due to Esposito, Nitsch, and Trombetti. They show that $C(n,p)=(1/\pi_p)^p$, where $\pi_p$ is some explicit constant ($\pi_2=\pi$). Here $C(n,p)$ denotes the best constant for $$ \inf_{t\in \mathbb{R}}\lVert u-t\rVert_p^p \leq C(n,p)d^p \lVert\nabla u \rVert_p^p. $$ The reason is that this is the formulation that corresponds to a Rayleigh quotient, as in the case $p=2$, and it's this quotient that gets related to a weighted, 1-d estimate. This is not a problem as far as getting an upper bound for $c(n,p)$ since $$ \inf_{t\in \mathbb{R}}\lVert u-t\rVert_p \leq \lVert u-u_\Omega\rVert_p \leq 2\inf_{t\in \mathbb{R}}\lVert u-t\rVert_p, $$ and so $c(n,p)\leq 2^p (1/\pi_p)^p$. I didn't look too much to see if they find the optimal value of $c(n,p)$.

  4. The case $p>1$ was done by Valtorta and Valtorta, Naber in a more general context, by estimating the eigenvalues of the p-Laplacian on manifolds with certain conditions (here I must admit I know little about differential geometry, and so some of the terminology in these papers is unknown to me).

Short of it then, you can actually get away with dimension-free constants, but you have to work for them. I'm not aware of a simpler (perhaps dimension-dependent) proof of a bound involving only the diameter.


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