# Converse to standard Sobolev Proposition

I want to prove that if $u \in W^{1,p}(U)$, then $$Du = 0 \ \ \text{a.e. on the set} \ \{ u =0 \}.$$

Proposed proof: Suppose $Du = v$ for some nonzero $v$ and let $U = \{ u =0 \}$. Then for all $\phi \in C_c^{\infty}(U)$, \begin{eqnarray*} \int_U uD\phi dx &=& - \int_U v\phi dx. \end{eqnarray*} But $$\int_U u D\phi dx = 0,$$ and therefore $$\int_U v \phi dx =0, \ \ \ \forall \phi \in C_c^{\infty}(U).$$ This yields that $v \equiv 0$.

I don't feel this is correct however.

This doesn't work, because the fundamental lemma of calculus of variations $$\forall_{\varphi \in C_c^\infty(U)} \int_U v \varphi = 0 \quad \Rightarrow \quad v \equiv 0$$ works only for open $U$.

I propose the following proof. Choose a $C^1$ function $\psi \colon \mathbb R \to \mathbb R$ such that $$|\psi(t)| \le |t| \text{ for } t \in \mathbb R, \quad \psi(t) = t \text{ for } |t| \ge 1, \quad \psi'(0) = 0;$$ then $|\psi'|$ is bounded by some constant $M>1$. Consider the composition $u_\varepsilon(x) = \varepsilon \cdot \psi(u(x)/\varepsilon)$ for $\varepsilon > 0$. By formal computation, its derivative is $$\nabla u_\varepsilon(x) = \psi'(u(x)/\varepsilon) \cdot \nabla u(x),$$ hence $|\nabla u_\varepsilon| \le M|\nabla u|$ pointwise. Approximating $u$ by smooth functions, one can show that $u_\varepsilon \in W^{1,p}$ and the above calculations hold in the weak sense (in general, this is true for compositions with suitable $C^1$ functions).

Now take the limit $\varepsilon \to 0$. It is easy to see that $u_\varepsilon \to u$ pointwise and this sequence is dominated by $|u|$ (in fact, $u_\varepsilon \to u$ in $L^p$). The derivative is also pointwise convergent: $$\nabla u_\varepsilon(x) \to V(x) := \begin{cases} \nabla u(x) & \text{if } u(x) \neq 0 \\ 0 & \text{if } u(x) = 0 \end{cases}$$ and this sequence is dominated by $M|\nabla u|$. For any test function $\varphi$, Lebesgue dominated convergence theorem shows that $$\int V \varphi \leftarrow \int \nabla u_\varepsilon \varphi = - \int u_\varepsilon \operatorname{div} \varphi \rightarrow - \int u \operatorname{div} \varphi = \int \nabla u \varphi.$$ Hence $\nabla u = V$ a.e. and $\nabla u = 0$ a.e. in $\{ u=0 \}$.

• How would I alter the proof?
– user419593
Apr 27, 2017 at 21:34
• I thought about it, but failed to adapt the proof. Apr 27, 2017 at 22:07
• Perhaps, approximating $u$ by mollifier functions?
– user419593
Apr 27, 2017 at 22:15
• @J.Doetry If you look at the usual proof that $|u| \in W^{1,p}$, it shows in particular that $\nabla u = 0$ a.e. in $\{ u=0 \}$. I added an argument that is a variation of this proof. Apr 27, 2017 at 23:18