# Poincaré’s inequality for a bounded open set in Brezis' book

I am trying to understand the Corollary 9.19 (Poincaré’s inequality) in Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis.

Suppose that $$1 \le p < \infty$$ and $$\Omega$$ is a bounded open set. Then there exists a constant $$C$$ (depending on $$\Omega$$ and $$p$$) such that $$\left\Vert u \right\Vert_{L^p(\Omega)} \le C \left\Vert \nabla u \right\Vert_{L^p(\Omega)}$$ for all $$u \in W^{1,p}_0(\Omega)$$.

So far, I know that the zero extension $$\bar u$$ of $$u\in W^{1,p}_0(\Omega)$$ is an element in $$W^{1,p}(\mathbb{R}^N)$$ (which is a consequence of the preceding Proposition 9.18).

Case 1: $$1\le p

With the help of Theorem 9.9 (Sobolev, Gagliardo, Nirenberg),

$$\left\Vert \bar u \right\Vert_{L^{p^*}(\mathbb{R}^N)} \le C \left\Vert \nabla \bar u \right\Vert_{L^p(\mathbb{R}^N)}$$, where $$\frac{1}{p*} = \frac{1}{p} - \frac{1}{N}$$,

I can show the Poincaré’s inequality due to compact support and $$p.

Case 2: $$p>N$$

Theorem 9.12 (Morrey) asserts $$|\bar u(x) - \bar u(y)|\le C |x-y|^\alpha \left\Vert \nabla \bar u \right\Vert_{L^p(\mathbb{R}^N)}$$ a.e.

Take the continuous representative and still denote $$\bar u$$. By translation, I may take $$y = 0$$ and $$\bar u(y) = 0$$. Then taking p-norm (or $$\infty$$-norm) yields the result.

My question is how to conclude the Poincaré’s inequality when $$p=N$$?

For any $$1\le p_1 we have the embedding $$W^{1,N}_0 (\Omega) \to W^{1,p_1}_0 (\Omega)$$ since $$\Omega$$ is bounded. Now choose $$p_1 so that

$$p_1^* = \frac{p_1N}{N-p_1}> N.$$

Then there is a compact embedding

$$W^{1,N}_0 (\Omega) \to W^{1,p_1} (\Omega) \to L^N (\Omega).$$

(the second $$\to$$ is compact). Then for all $$u\in W^{1,N}_0(\Omega)$$,

$$\| u\|_{L^N} \le C \|u\|_{L^{p_1^*}} \le C \| \nabla u\|_{L^{p_1}} \le C \| \nabla u\|_{L^N}.$$

• I picked $1 = \left\Vert u_k \right\Vert_{L^N} > k \left\Vert \nabla u_k \right\Vert_{L^N}$ with $\int u_k = 0$. Using compactness, I saw $u_k \to u$ (use the same notation for the subsequence) in $L^N$ with $\left\Vert u \right\Vert_{L^N} = 1$ and $\int u_k = 0$. I could argue $\nabla u = 0$ a.e. by considering $\int u \partial_{x_i} \phi$. But then I got stuck. $u$ may not have a continuous representative. Even if it has, the domain is not assumed to be connected. Dec 18, 2021 at 18:36
• @JustinLien: You don't need a contradiction argument, the Sobolev inequality is enough: $$\|u\|_N \lesssim \| u\|_{p_1^*}\lesssim \|\nabla u\|_{p_1}\lesssim\|\nabla u\|_N,$$ where the first and last inequalities are Hölder's (so there's some dependence on $|\Omega|$). Dec 19, 2021 at 6:59
• @JustinLien Jose's commment is correct. If you consider only elements in $W^{1,p}_0$ instead of $W^{1,p}$, you can restrict to smooth function with compact support, then one has $\|u\|_{p_1^*} \le C\|\nabla u\|_{p_1}$. The fact that $\Omega$ is not connected is fine, since again the functions are in $W^{1,p}_0$. The fact that you added the condition $\int u_k =0$ indicates that you might want to something else. Can you state that? Dec 19, 2021 at 9:27
• Thanks @Jose27 (I have made an edit). Dec 19, 2021 at 9:30
• Yes, but then the statement would be different. I suppose you are trying to prove that there is $C$ so that if $u \in W^{1,p}$ and $\int \nabla u =0$ then $\| u\|_{L^p} \le C\|\nabla u\|_{L^p}$. This is one version. Another version is what I state here: that there is $C$ such that $\| u\|_{L^p} \le C\|\nabla u\|_{L^p}$ for all $u\in W^{1,p}_0$. @JustinLien Dec 19, 2021 at 12:22