Let \( D \subset \mathbb{R}^{d-1} \) be a domain, and \( \eta \in C_c^\infty(\mathbb{R}^d_+) \), with \( 0 \le \eta(x) \le 1 \) and \( \eta = 1 \) on \( D \). Then, for \( u \in C^\infty(\mathbb{R}^d_+) \cap H^1(\mathbb{R}^n_+) \),
\[ \|u\|_{H^{1/2}(D)}^2 \le C \bigl(1+\|\nabla\eta\|_\infty^2\bigr)\, \|u\|_{H^1(\operatorname{supp}\eta)}^2 , \]where \(C>0\) is a constant, depending on \(d\).
Remarks
- It is sufficient if \(u \in C^\infty (\supp \eta)\).
Proof
By the trace theorem on \( \mathbb{R}^d_+ \) 1 , we obtain
\begin{equation}\label{eq:1} \|u\|_{H^{1/2}(D)}^2 \le \|u\eta\|_{H^{1/2}(\partial \mathbb{R}^n_+)}^2 \le C \, \|u\eta\|_{H^1(\mathbb{R}^n_+)}^2. \end{equation}Expanding the right-hand side gives
\begin{equation}\label{eq:2} \begin{aligned} \int_{\mathbb{R}^n_+}\!\bigl(|u\eta|^2+|\nabla(u\eta)|^2\bigr) &\le 2 \int_{\operatorname{supp}\eta}\!\Bigl(|u|^2+\bigl(|\nabla u|^2+|\nabla\eta|^2|u|^2\bigr)\Bigr) \\ &\le 2\bigl(1+\|\nabla\eta\|_\infty^2\bigr)\, \|u\|_{H^1(\operatorname{supp}\eta)}^2 . \end{aligned} \end{equation}Combining \(\eqref{eq:1}\) and \(\eqref{eq:2}\) yields the claim.
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not fully proven yet. ↩︎