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Consider the following Sobolev spaces

${\bf H}^1(\Omega)= \{{\bf u } \in {\bf L}^2(\Omega) : \nabla {\bf u}\in {\bf L}^2(\Omega) \}$, with inner product $(\bf u,\bf v)_{{\bf H}^1} = \int_{\Omega} (\bf u\bf v + \nabla \bf u \nabla \bf v) dx$ and norm $\|\bf u\|_{{\bf H}^1} = (\|\bf u\|^2 + \|\nabla \bf u \|^2)^{1/2}$.

The spaces

${\bf H}^1_0(\Omega)= \{{\bf u}\in {\bf H}^1(\Omega)\,:\,{\bf u}=0\,\mbox{ on }\Gamma\}$ where $\Gamma$ is boundary of $\Omega$. With inner product $({\bf u},{\bf v})_{{\bf H}^1_0} = ( \nabla {\bf u}, \nabla {\bf v})$ and norm $\|{\bf u}\|_{{\bf H}^1_0} = \| \nabla {\bf u}\|$.

${\bf V}= \{{\bf u}\in{\bf H}^1_0(\Omega)\,:\,{\rm div}{\bf u}=0\mbox{ en }\Omega\} \ \mbox{ with norm } \|{\bf u}\|_{\bf V}=\|\nabla{\bf u}\|$

Further, the Poincaré inequality

Let $p$, so that $1 ≤ p < \infty$ and $\Omega$ a subset with at least one bound. Then there exists a constant $C$, depending only on $\Omega$ and $p$, so that, for every function $u$ of the $W^{1,p}(\Omega)$ Sobolev space,

$${\displaystyle \|u\|_{L^{p}(\Omega )}\leq C\|\nabla u\|_{L^{p}(\Omega )}}$$

Now, how is the demonstration of the equivalence of norms $\|{\bf u}\|_{{\bf H}^1_0} \equiv \| \nabla {\bf u}\|$ and $\|{\bf u}\|_{\bf V} \equiv \|\nabla{\bf u}\|$ using Poincaré inequality?

  • $||u||_{H_0^1}=||\nabla u||=||u||_V$ are identical as halfnorms on $V \cap H_0^1$. I don´t understand the question. – Peter Melech Jun 01 '17 at 11:24
  • I tried the following $|{\bf u}|{{\bf H}^1} = (|{\bf u}|^2 + |\nabla {\bf u}|^2)^{1/2}$ for definition. Now, bound first term the right-hand side for Poincaré, i get $|{\bf u}|{{\bf H}^1} \leq C|\nabla {\bf u}|$. I do not know how to associate the norm with space ${\bf H}^1_0$ – NoNewtoniano Jun 01 '17 at 15:34
  • @PeterMelech i know that identical as halfnorms on ${\bf V} \subset {\bf H}^1_0$. Now, why are they? Just for contention? – NoNewtoniano Jun 01 '17 at 15:38
  • @NoNewtoniano by definition. But this has nothing to do with Poincaré´s inequality. as i said i don´t understand the question. – Peter Melech Jun 02 '17 at 07:54
  • @PeterMelech You're right, sorry. Thank you. – NoNewtoniano Jun 09 '17 at 05:50

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