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I am trying to prove an equivalence between two norms in the Sobolev space $H^1(\Omega)$ over a bounded Lipschitz domain $\Omega$, namely the standard norm $$||u||_{H^1(\Omega)}^2=\int_{\Omega} u^2 \,dx + \int_{\Omega} |\nabla u|^2\,dx$$ and the norm $$||u||_{\partial}^2= \int_{\partial \Omega} u_{|\partial \Omega}^2 \,d\sigma + \int_{\Omega} |\nabla u|^2\,dx$$

The second sumands are the same in both cases, so since in the first one we integrate the positve function $u^2$ over a larger domain, it is clear that $||u||_{\partial} \leq ||u||_{H^1(\Omega)}$, so it now suffices to find a positive constant $C$ such that $C||u||_{H^1(\Omega)}\leq ||u||_{\partial}$.

Now we can use the following projection theorem: given a Hilbert space $H$ and a closed subspace $V\subset H$, for every $X\in H$ there exists a unique $P_Vx\in V$ such that $$||x-P_vx||=\inf_{v\in V} \{||v-x||\}$$ and besides $x-P_Vx\in V^{\perp}$.

In our case, this yields a decomposition $$u=u_0+E(u_{|\partial \Omega})$$ where $E(u_{|\partial \Omega})$ is an extension to $\Omega$ of the restriction $u_{|\partial \Omega}$ and $u_0\in H_0^1(\Omega)$ and therefore $$||u||_{H^1(\Omega)}^2=||u_0||_{H^1(\Omega)}^2+||E(u_{|\partial \Omega})||_{H^1(\Omega)}^2$$

Note that by the projection theorem we have that $$||E(u_{|\partial \Omega})||_{H^1(\Omega)}^2=\inf\{||v||_{H^1(\Omega)}: v\in H^1(\Omega), v_{|\partial \Omega}=u_{|\partial \Omega}\} \stackrel{def}{=} ||u_{|\partial \Omega}||_{H^{1/2}(\partial \Omega)}$$ Does this lead towards our purpose?

Thanks in advance for any insight.

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First, the argument you used to get $\| u\|_\partial \leq \| u\|_{H^1}$ is wrong: $\partial\Omega$ is of (n-dimensional) Lebesgue measure zero so you're implying that $\int_{\partial \Omega} u^2 =0$ which is clearly not (necessarily) true. What you want to say here is that the trace map $u\mapsto u|_{\partial \Omega} = :g(u)$ is a bounded linear operator.

Now to prove the reverse inequality, assume it's not true, then we can find a sequence $\| u_k\|_{H^1} =1$ and $ \| u_k\|_{\partial} <\frac{1}{k}$. This implies that $\| \nabla u_k\|_{L^2} \to 0$, moreover, by Rellich we have an $u$ such that (you'll have a subsequence here) $u_k\to u$ in $L^2(\Omega)$ then $u_k \to u$ in $H^1$, $u\in H^1$ and is locally constant. Then $\| g(u)\|_{L^2(\partial \Omega)} = \lim \| g(u_k)\|_{L^2(\partial \Omega)}=0$, and so $u=0$ a contradiction.

Notice that actually, if $\Omega$ is connected, you can take any $\Gamma \subset \partial \Omega$ of positive measure in the definition of $\| \cdot \|_{\partial}$ and still get an equivalent norm.

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