Suppose that $U$ is a bounded domain in $\mathbb{R}^n$. Poincaré's inequality states that (for $U$ sufficiently "nice") there exists a constant $C>0$ such that if $u\in H^1(U)$ satisfies $\int_U u = 0$ then $$ \int_U \lvert u\rvert^2 \leq C\int_U \lvert \nabla u \rvert^2 $$ I was wondering if it was possible to find an example of a bounded domain on which the inequality does not hold. The proof for this inequality in Evan's PDE requires $U$ to be an extension domain. So I though that maybe the inequality would be false on a domain as simple as $U = \{(x,y) : 0<x<1, 0<y<x^2\}$.

I tried to construct a sequence of functions $u_k\in H^1(U)$ satisfying $\int_U u_k = 0$ and such that $$ \frac{\int_U \lvert u_k\rvert^2}{\int_U \lvert \nabla u_k \rvert^2} \to\infty $$ but unforunately I was unable to obtain such a sequence.

Is the inequality always true on bounded domains? Or is it indeed possible to find a counter-example

  • $\begingroup$ What is your definition of "domain"? $\endgroup$ – gerw Nov 27 '18 at 7:05
  • $\begingroup$ I think Evans takes domain to be “open and connected”. @gerw $\endgroup$ – DaveNine Nov 27 '18 at 8:04

The classical counter-example is called "Rooms and passages" or "Rooms and corridors". You take a sequence of squares $R_n$ of side-length $1/n^p$ and in each of them, you assign $u$ to be a large constant, say $n^q$. Then you connect each square with a narrow corridor and in this corridor, you take $u$ to be affine. If the corridors are narrow enough the gradient will be in $L^2$ but the function will not be in $L^2$. The details are here Room and Passages


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