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: $1\le p<N$

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 $p<p^*$.

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 $p=N$?


1 Answer 1


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

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

  • $\begingroup$ 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. $\endgroup$ Dec 18, 2021 at 18:36
  • $\begingroup$ @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 $|\Omega|$). $\endgroup$
    – Jose27
    Dec 19, 2021 at 6:59
  • $\begingroup$ @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? $\endgroup$ Dec 19, 2021 at 9:27
  • $\begingroup$ Thanks @Jose27 (I have made an edit). $\endgroup$ Dec 19, 2021 at 9:30
  • 1
    $\begingroup$ 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 $\endgroup$ Dec 19, 2021 at 12:22

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