Consider a smooth ,convex and bounded domain $K \subset \{ x_1 = 0 \} \subset R^n$ . Let $U \subset R^{n}_+ = \{ x = (x_1,..,x_n)\in R^n ; x_1 > 0\} $ with $K \subset \partial U$ and supoose that $U$ is smooth. Consider the trace operator $T : W^{1,2}(U) \rightarrow L^{2}(\partial U)$ . Exists $u \in W^{1,2}(U)$ such that $Tu = g$ where $ g= 1 $ on $K$ and zero otherwise ?
i believe that is tru , but i dont know how to prove this..
someone can give me a help ?
thanks in advance
Consider the set $K' = \{ (x_1,\dots,x_n) \in \mathbb{R^n} | x_1 \in (-1,1) \mbox{ and } (0,x_2,\dots,x_n) \in K \} $. Now consider the function :
$\varphi(x) = \begin{cases} e^{-1/(1-|x_1|^2)}& \text{ if } |x_1| < 1\\ 0& \text{ if } |x_1|\geq 1 \end{cases}$
$\varphi(x)$ occurs in reference to mollifiers, see wiki : http://en.wikipedia.org/wiki/Mollifier
$ u = \chi_{K'} \cdot e^2 \cdot \varphi $ should do the trick !
*Essentially I took a 1 width band on both sides of $K$, and produced a $C_0^{\infty}$-function in transversal manner with the required properties and stitched them together (possible) because $\partial U$ is smooth and measure of this band is finite because $K$ is of finite measure imposed by the fact that $g$ is in $L^2(\partial U)$.
Please correct me if I am wrong. I am just attempting an answer here.
