# Is there a bounded domain on which Poincaré's inequality does not hold?

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.

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

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