0
$\begingroup$

Let $u \in H^1_0(B)$, where $B$ is the unit ball in $\mathbb{R}^N$. Given $\epsilon > 0$, I need to show there exists a function $\chi_\epsilon \in C^\infty_0(\mathbb{R}^N)$ such that $$ \| u - \chi_\epsilon u\|_{H^1_0(B)} < \epsilon $$ and $\text{supp} \chi_\epsilon \subset \subset B$.

My attemps I considered $f \in C^\infty_0(\mathbb{R}^N)$ such that $\text{supp} f \subset B_{1/4}(0)$ and defined $\chi_\epsilon = 1-\epsilon + \epsilon f.$ The bad part: The support of $\chi_\epsilon$ is all $\mathbb{R}^N$.

$\endgroup$

1 Answer 1

2
$\begingroup$

Do it first for the half-space $\{x_n >0\}=\Sigma$. If $u$ vanishes at the boundary then $u(x',x_n)^2=2\int_0^{x_n} uD_n u$ and so ($\Sigma_\delta=\{0 <x_n <\delta\}$) $$ \int_{\Sigma_\delta} |u|^2 \leq 2 \delta \int_{\Sigma_\delta} |u||D_n u|\leq 2\delta \left (\int_{\Sigma_\delta}|u|^2\right)^{1/2}\left (\int_{\Sigma_\delta}|\nabla u|^2\right)^{1/2} $$ which gives $\left (\int_{\Sigma_\delta} |u|^2 \right )^{1/2} \leq 2\delta \left (\int_{\Sigma_\delta} |\nabla u|^2 \right )^{1/2}$. This inequality for smooth functions extends to $H^1_0$ functions by density. With this at hand, consider cut-off functions $\eta_k$ which are equal to 1 for $x_n \geq 2/k$ and which vanish for $x_n \leq 1/k$. Then $\eta_k u \to u$ in $H_1$. In a smooth $\Omega$ one can do the same after flattening the boundary and in the ball one can do the same explicitely using the radial derivative.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .