# If $u=v$ on $A \subset \Omega$, then $\nabla u = \nabla v$ on $A$ too

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $A \subset \Omega$ be measure nonzero. For $u, v \in H^1(\Omega)$, if $u=v$ (a.e) on $A$, how to prove that $\nabla u = \nabla v$ on $A$?

This is problematic since this is the weak gradient.

Reference also appreciated.

• What is $A$? Does it have reasonable properties? Because you are asking for the implication "$u=0$ a.e. $\Rightarrow$ $\nabla u=0$". – Siminore Jun 9 '14 at 14:09
• What does that mean? That $A \neq \emptyset$, or that the measure of $A$ is positive? – Siminore Jun 9 '14 at 14:15
• That means the latter – assa888 Jun 9 '14 at 14:15
• You can use Stampacchia's theorem. Take a look, for example, in Heinonen's book, pages: 18,19 and 20. amazon.com/Nonlinear-Potential-Degenerate-Equations-Mathematics/… – Tomás Jun 9 '14 at 14:17
• @Siminore, there is no restriction. If $A$ has positive measure and $u=0$ a.e. in $A$ then, $\nabla u=0$ a.e. in $A$. So, in general, this result can be stated as follows: Assume that $A\subset \Omega$ and $u=0$ a.e. in $A$ then, $\nabla u=0$ a.e. in $A$. – Tomás Jun 9 '14 at 14:20

The following is more or less the same as Evans, Partial Differential Equations, Chapter 5, Exercise 18.

Let $u \in H^1(\Omega)$. For $\varepsilon > 0$ let

$$F_\varepsilon (z) := \begin{cases} \left(z^{2}+\varepsilon^{2}\right)^{1/2}-\varepsilon, & z\geq0,\\ 0, & z<0. \end{cases}$$

Show $F_\varepsilon\circ u \in H^1(\Omega)$, $(F_{1/n} \circ u)_n$ is Cauchy in $H^1(\Omega)$ and $F_1/n \circ u \rightarrow u^+$ a.e., where $u^+$ is the positive part of $u$.

Furthermore, show that

$$\partial_j (F_\varepsilon \circ u)(x) \xrightarrow[\varepsilon \downarrow 0]{} \begin{cases} \left(\partial_{j}u\right)\left(x\right), & u\left(x\right)>0,\\ 0, & u\left(x\right)\leq0 \end{cases}$$

almost everywhere.

This will yield $u^+ \in H^1(\Omega)$ with weak derivative

$$\partial_j u^+(x) = \begin{cases} \left(\partial_{j}u\right)\left(x\right), & u\left(x\right)>0,\\ 0, & u\left(x\right)\leq0 \end{cases}$$

Do the same for $u^-$ instead of $u^+$.

This will finally yield

$$\partial_j u = \partial_j u^+ - \partial_j u^-,$$ a.e., where we used that $u = u^+ - u^-$.

But the above identity implies $\partial_j u = 0$ a.e. on $\{x \mid u(x) = 0\}$.

Finally, apply the above to $u-v$.

• Doing the same for $\,u^{-}\,$ instead of $\,u^{+}\,$ yields just $\,\partial_j u = \partial_j u^{+} + \partial_j u^{-}\,$ a.e. on $\Omega\backslash \{x \mid u(x) = 0\}$, and nothing else. – mkl314 Jun 10 '14 at 10:18
• Why do you exclude the set where $u$ vanishes? If you show $\partial u^- (x) = 0$ on the set where $u(x) \geq 0$, then (depending on the sign that you choose for $u^-$), you get $\partial_j u = \partial_j u^+ \pm \partial_j u^-$ a.e. on $\Omega$. Now note that on the set where $u$ vanishes, both $\partial_j u^+$ and $\partial_j u^-$ vanish. – PhoemueX Jun 10 '14 at 11:26
• I just mean to say that the proof of $\partial_j u =\partial_j u^+ \pm \partial_j u^-$ a.e. on Ω is missing. – mkl314 Jun 10 '14 at 12:06
• That directly follows by linearity of $\partial_j$ and because $u = u^+ \pm u^-$. I edited my post to include that point. – PhoemueX Jun 10 '14 at 12:38
• OK, that'll do. But this question concerns the basics of Real Analysis which is why it is not a good idea to indicate unimportant details while leaving out most important. – mkl314 Jun 10 '14 at 13:54

Obviously, it suffices to show that $\,u=0\,$ a.e. on $A$ implies $\,\nabla u=0\,$ a.e. on $A$. It is clear that the approximate derivative $ap\,\nabla u=0\,$ a.e. on $\{x\in\Omega\colon\, u(x)=0\}$ — for details see theorem 3 in section 6.1.3 of "Measure theory and fine properties of functions" by L.C. Evans and R.F. Gariepy. Hence by remark $(ii)$ in theorem 4 (Ibid.), the weak derivative $\nabla u=ap\,\nabla u=0\,$ a.e. on a subset $A\subset \{x\in\Omega\colon\,u(x)=0\}$.   Q.E.D.