If $u \in H^1(U)$, then $Du = 0$ almost everywhere on the set $\{u = 0\}$, auxiliary result

Question

Let $\phi$ be a smooth, bounded and nondecreasing function, such that $\phi'$ is bounded and $\phi(z) = z$ if $|z| \le 1$. Set$$u^\epsilon(x) := \epsilon \phi(u/\epsilon).$$Do we necessarily have that$$\int_U Du^\epsilon \cdot Du\,dx = \int_U \phi'(u/\epsilon)|Du|^2\,dx \to 0?$$

Answer 1

We can assume $U$ is bounded.

On one hand, $\|u^{\epsilon}\|_{L^2}\rightarrow0$ as $\epsilon\rightarrow0$. And $\|Du^{\epsilon}\|_{L^2}$ is uniformly bounded. Thus we infer $u^\epsilon$ converges weakly to $0$ in $H^1(U)$, which implies that $$\int_UDu^\epsilon\cdot Dudx=(u^\epsilon,u)_{H^1}-\int_Uu^\epsilon udx\rightarrow0,\ as\ \epsilon\rightarrow0.$$

On the other hand, $$\int_U\phi'(u/\epsilon)|Du|^2dx\geq\int_{\{u=0\}}|Du|^2dx.$$ We finally conclude $$\int_{\{u=0\}}|Du|^2dx=0.$$ Thus we get the desired result.