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Let $U$ be bounded connected, open subset of $\mathbb R^n$ with $C^1$ boundary. Let $q=\frac{pn}{n-p}$. Prove the following Poincare inequality for $u \in W^{1,p}(U)$:

$$\|u- (u)_U\|_{L^q(U)}\leq C \|Du\|_{L^p(U)}.$$

My idea: I read the Poincare inequality proof in Evan's book, that said $\|u- (u)_U\|_{L^p(U)}\leq C \|Du\|_{L^p(U)}$ for each function $u \in W^{1,p}(U)$.

I used the general embedding theorem, according to this embedding if $u \in W^{1,p}(U)$, then $u \in L^q(U)$ under some conditions.

Now I am little confuse to connect these two result to make conclusion.

Can anyone suggest, how do I connect both result to solve the given inequality?

Michael Hardy
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User124356
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  • If the typographical difference between $||u||$ and $|u|$ is not conspicuous to you, look at this difference: $$ \begin{align} & |u||v| \ & ||u|| ||v|| \end{align}$$ The former is coded as \|u\|\|v\| and the latter as ||u|| ||v||. The former usage is standard. (Also, I changed one line of your question from inline to display.) $\qquad$ – Michael Hardy Sep 26 '20 at 16:31
  • Thanks for letting me know. – User124356 Sep 26 '20 at 16:33

1 Answers1

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You know $W^{1,p} \hookrightarrow L^q$, i.e., $\|u\|_{L^q} \leq \tilde C \|u\|_{W^{1,p}}=\tilde C (\|u\|_{L^p}^p+\|Du\|_{L^p}^p)^{1/p}$. Hence we conclude

$$\begin{aligned}\|u-(u)_U\|_{L^q} &\leq \tilde C(\|u-(u)_U\|_{L^p}^p+\|Du\|_{L^p}^p)^{1/p} \\ &\leq \tilde C(C^p\|Du\|_{L^p}^p+\|Du\|_{L^p}^p)^{1/p} \\ &=\tilde C (C^p+1)^{1/p} \|Du\|_{L^p}. \end{aligned}$$

Cahn
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