1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ What is your definition of "domain"? $\endgroup$
    – gerw
    Commented Nov 27, 2018 at 7:05
  • $\begingroup$ I think Evans takes domain to be “open and connected”. @gerw $\endgroup$
    – DaveNine
    Commented Nov 27, 2018 at 8:04

1 Answer 1

2
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .