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Let $\Omega\subset\mathbb R^d$ a Lipschitz domain and $\Gamma:=\partial\Omega$. For $u\in C^{\infty}(\Gamma)$ we define $$||u||_{H^{1/2}(\Gamma)} = \inf_{\substack{v\in H^1(\Omega) \\ v|_\Gamma =u}} ||v||_{H^1(\Omega)}$$

I'm trying to show that this indeed defines a norm, but so far I've only been able to show that $||u||_{H^{1/2}} =0$ if and only if $u\equiv 0$: 1. If $u\equiv 0$ we have $||0||_{H^{1/2}} = \inf_{\substack{v\in H^1 \\ v|_\Gamma = 0}} ||v||_{H^1} = 0$ since $0\in H^1$ and $0|_\Gamma = 0$. 2. If $v=0$ we have $||v||_{H^1} =0\Leftrightarrow 0=\inf_{\substack{v\in H^1 \\ v|_\Gamma = u}} ||v||_{H^1} = ||u||_{H^{1/2}}$.

I've not been able to proof the other characteristics of a norm. How do I proof those?

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It's not clear how much of the theory of traces you developed. To begin with, how do you understand "$v_{|\Gamma}=u$"? Here $v$ is a function (or rather, equivalence class of functions) on $\Omega$ while $u$ is a function (actually, also an equivalence class) on $\partial \Omega$.

On an abstract level, the norm property of the infimum in your definition follows from the general construction of a quotient norm. Given a closed subspace $Y$ of a Hilbert space $X$, we can define a norm on $X/Y$ as $\|x+Y\|=\inf \{\|z\|_X: z\in x+Y\}$. The axioms of a norm are easy to verify in this generality. Your definition corresponds to the special case $X=H^1(\Omega)$ and $Y=H^1_0(\Omega)$.

So, if $v$ is understood abstractly as an element of the quotient space $H^1(\Omega)/H^1_0(\Omega)$, there is not much to do. The difficult part is to relate the quotient space to actual functions defined on the boundary of $\partial \Omega$.

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  • $\begingroup$ Thank you for your answer! I know the trace theorem, which says there exists a continuous linear functional $\gamma: H^1(\Omega)\to L^2(\partial\Omega)$ such that $(\gamma u)(x) = u(x)$ for all $u\in C^1(\overline\Omega)$ and $x\in\partial\Omega$. I'm still having trouble with the quotient space: If I have $u\in C^\infty(\partial\Omega)$ I can write it as the sum of a $H^1$ function and a $H^1_0$ function and define the norm as you did it. Can I say that for each function in $H^1$ we find by the trace theorem a function that fulfills our boundary condition $v|_\Gamma=u$ hence we have a norm? $\endgroup$ – dinosaur May 11 '13 at 9:04

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