Suppose $\Omega\subset\mathbb{R}^n$ is a bounded domain. Let $u\in H_0^1(\Omega)$ and $\tilde{u}\in L^2(\Omega)$ satisfies $$\int_\Omega\nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

Suppose that there exist a set $\Gamma$ of positive measure such that $\nabla u=0, a.e.\ x\in\Gamma$. How can one show that $\tilde{u}=0,\ a.e.\ x\in\Gamma$?

Any help is appreciate. I have to solve this problem without considering the fact that $\tilde{u}=-\Delta u$. I know that i have to take some "nice" $v\in C_0^{\infty}(\Omega)$, but what $v$?

Edit 1: Why am i trying to solve this problem? Let $p\geq 1$. Suppose $u\in W_0^{1,p}(\Omega)$ and $\tilde{u}\in L^q(\Omega)$ $(\frac{1}{p}+\frac{1}{q}=1)$ satisfies $$\int_\Omega|\nabla u|^{p-2} \nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

In this case we dont have enough regularity to show that $\tilde{u}=-\Delta_p u$, so i need a more direct aproach, and consequently i think that this aproach is the same for both cases.

Lastly we can have the $\Phi$-laplacian too and considering the spaces where it makes sense, we have $$\int_\Omega\Phi(|\nabla u|)\nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

Edit 2: In general, consider $\sigma$ a vector valued function. Suppose $\sigma$ is in a convenient space and $$\int_\Omega \sigma\cdot\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

where, $\cdot$ stands for inner product. Now i want to conclude the same thing for all the cases above.

Note: $\tilde{u}$ is called weak divergence of $\sigma$.

Edit 3: In any open set $U$ contained in $\Gamma$, we can take functions with compact supoort in $U$ and conclude that $\tilde{u}=0,\ a.e.\ x\in U$, but in the general case, how to proceed? Any opinion?

The question is: If closure of $\Gamma$ have interior empty, what we have to do?


  • $\begingroup$ Take a small ball around an arbitrary point in $\Gamma$. Then the integral must be small by Lebesgue differentiation theorem? $\endgroup$
    – timur
    Oct 30, 2012 at 17:31

1 Answer 1


If $\Gamma$ is a bounded domain then considering $v \in C_0^\infty(\Gamma) \subset C_0^\infty(\Omega)$ we have $$ 0= \int_\Gamma \nabla u \cdot \nabla v =\int_\Gamma \tilde u v$$ Therefore $\tilde u=0$ in $L^2(\Gamma)$. I don't know what to do if $\Gamma$ is less regular.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.