Prove that a function belongs (or does not belong) to $H^{\frac12}_{00}$

functional-analysistrace

Given a domain $\Omega$, the space $H^{1/2}(\partial \Omega)$ can be defined as the image of the trace operator $\gamma: H^{1}(\Omega) \rightarrow H^{1/2}(\partial \Omega)$, which (roughly speaking) is the linear operator that associates to each $\varphi \in D(\overline{\Omega})$ its restriction over $\partial \Omega$ i.e. $\gamma
\varphi = \varphi|_{\partial \Omega}$
. Note that extending the definition of $\gamma$ from $D(\overline{\Omega})$ to $H^1({\Omega})$ is legit because the latter is dense in the former.

In my understanding, given a portion of the boundary $\Gamma \subset \partial \Omega$, the space $H^{1/2}_{00}(\Gamma)$ is defined as the space of functions with support on $\Gamma$ whose trivial extension by zero outside of $\Gamma$ belongs to $H^{1/2}(\partial \Omega)$.

My question is: given a generic function on $\Gamma$, how can one be sure that it belong to $H^{1/2}_{00}(\Gamma)$?

I believe that this question can be linked to the well-posedness of the following problem. Let us consider the domain $\Omega = (0,1)^2$ and the problem $-\Delta u = f$ in $\Omega$, $u = g$ on $\Gamma_D = \{1\} \times (0,1)$ and $u = 0$ on $\partial \Omega \setminus \Gamma_D$. Does this equation admit a solution $u \in H^1(\Omega)$ for a general function $g$? If not, under which conditions does the solution live in such space?

Best Answer

Let $\Omega$ be of class $C^{1, 1}$ and $\Gamma_D$ open in $\Gamma : = \partial \Omega$ and $\Sigma := \text{Int} (\Gamma \backslash \Gamma_D)$. Indeed the space $H^{1/2}_{00}(\Gamma)$ (Lion-Magenes space) contains functions decaying fast enough at the border $\partial \Sigma$ for the extension by zero to be possible: let $\rho (x) = \text{dist} (x, \partial \Sigma)$ on $\Sigma$, then define

$$ H_{00}^{1 / 2} (\Sigma) : = \{ v \in H^{1 / 2} (\Sigma) : \rho^{- 1 / 2} v \in L^2 (\Sigma) \} $$

with the scalar product and corresponding (equivalent) norm

$$ (u, v)_{H_{00}^{1 / 2} (\Sigma)} : = (u, v)_{H^{1 / 2} (\Sigma)} + (\rho^{- 1 / 2} u, \rho^{- 1 / 2} v)_{L^2 (\Sigma)} $$ $$ \| u \|^2_{H_{00}^{1 / 2} (\Sigma)} : = \| u \|^2_{H^{1 / 2} (\Sigma)} + \| \rho^{- 1 / 2} u \|^2_{L^2 (\Sigma)} . $$

This is a Hilbert space properly and continuously embedded in $H^{1 / 2} (\Sigma)$, and the definition is actually independent of $\rho$, as long as we take some positive function decaying to zero as the distance to the boundary.

With this definition and $V : = \{ v \in H^1 (\Omega) : \gamma (v)_{| \Gamma_D} = 0 \}$, the trace operator

$$ \gamma^0 : V \rightarrow H^{1 / 2} (\Gamma) $$

is surjective onto $H_{00}^{1 / 2}$, which was the point of the definition.

For properties of this and other fractional spaces (and proof of the statement above), see "Pierre Grisvard. Elliptic Problems in Nonsmooth Domains. Classics in Applied Mathematics. Society for Industrial and Applied Mathematics, 2011." Chapter 1

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