Sketch: Consider the domain $$U_\epsilon=\{x\in U : \mbox{dist}{(x,\partial U)}>\epsilon\}$$. For $x\in U_\epsilon$ consider the function $$u_\epsilon(x)=\int_U \eta_\epsilon(x-y)u(y)\,dy$$ is a standard mollifier. Then $u_\epsilon\in C^\infty(U_\epsilon)$ and $$Du_\epsilon(x)=\int_U\eta_\epsilon(x-y)Du(y)\,dy$$ by the definition of the weak derivative. Since $Du=0$ a.e., we have that $Du_\epsilon(x)=0$ for all $x\in U_\epsilon$ and hence $u_\epsilon$ is constant in $U_\epsilon$. Since $\|u_\epsilon−u\|_{L^p(U_\epsilon)}\to 0$ as $\epsilon\to 0$, we have that $u$ is constant a.e.
Observations: This demostration uses some results and notation from the text Lawrence C. Evans - Partial Differential Equations_ Second Edition.
Definitions: (i) Define $\eta\in C^\infty(\mathbb{R}^n)$ by $$\eta(x):=\left\{\begin{array}{ll}C\exp\left(\frac{1}{|x|^2-1}\right), & \mbox{if }|x|<1\\0, & \mbox{if } |x|\geq 1\end{array}\right.$$ the constant $C>0$ selected so that $\displaystyle{\int_{\mathbb{R}^n}}\eta\,dx=1$.
(ii) For each $\epsilon>0$, set $$\eta_\epsilon(x):=\frac{1}{\epsilon^n}\eta\left(\frac{x}{\epsilon}\right).$$
We call $\eta$ standard mollifier. The functions $\eta_\epsilon$ are $C^\infty$ and satisfy $$\int_{\mathbb{R}^n}\eta_\epsilon\,dx=1, \mbox{spt}(\eta_\epsilon)\subset B(0,\epsilon).$$ spt denotes support.
It should be noted that this exercise has already been posted several times, but I am particularly focused on the demonstration whose outline is shown.
My questions are the following:
1. Is $U_\epsilon$ really a domain?
2. By definition $Du=0$ a.e, implies that the gradient vector of u is equal to zero, except in a set of measure zero, then in the next line, it is stated that
Since $Du=0$ a.e., we have that $Du_\epsilon(x)=0$ for all $x\in U_\epsilon$ and hence $u_\epsilon$ is constant in $U_\epsilon$.
therefore, for this to be true, $U_\epsilon$ must have a nonzero measure. How could I prove such a thing?
- I do not understand why "Since $\|u_\epsilon−u\|_{L^p(U_\epsilon)}\to 0$ as $\epsilon\to 0$, we have that $u$ is constant a.e."