Existence of a $C_c^{\infty}$ function $\Phi$ s.t. $\int \nabla \Phi \neq 0.$

Question

Let $\Omega\subseteq \mathbb R^n$ be open, bounded and connected set with $\mathcal{L^n}(\Omega)>0$. Let $\Omega=\Omega_1\cup\Omega_2$ where $\Omega_1\cap\Omega_2=\emptyset$ with $\mathcal{L^n}(\Omega_1)>0$ and $\mathcal{L^n}(\Omega_2)>0.$ Also $\Omega_1,$ $\Omega_2$ are $\mathcal L^n$-measurable sets. Prove that $\exists$ $\Phi\in C_c^{\infty}(\Omega)$ such that $\displaystyle \int_{\Omega_1}\nabla\Phi(x)dx\neq 0.$

Here $\Phi$ is real valued function and $C_c^{\infty}(\Omega)$ consists of all $C^{\infty}(\Omega)$ functions which have compact support in $\Omega.$

I tried to show by contradiction and wished to have $\mathcal{L^n}(\Omega_1)=0 $ but couldn't be able to do that.

Any help is appreciated. Thank you.

Answer 1

If possible, let for all $\Phi\in C_c^{\infty}(\Omega),$ $\int\limits_{\Omega_1}\nabla\Phi=0.$

Then \begin{align} \int\limits_{\Omega_1}\nabla\Phi=0&\implies \int\limits_{\Omega_1}\frac{\partial \Phi}{\partial x_i}=0\ \ \ \ \ \ \forall\ \ 1\leq i\leq n\ \ \textrm{ and }\ \forall\ \Phi \in C_c^{\infty}(\Omega)\\ &\implies\int\limits_{\Omega}\chi_{\Omega_1} \frac{\partial \Phi}{\partial x_i}=0 \ \ \ \ \ \ \forall\ \ 1\leq i\leq n\ \ \textrm{ and }\ \forall\ \Phi \in C_c^{\infty}(\Omega)\\ &\implies T_{\chi_{\Omega_1}}\left(\frac{\partial \Phi}{\partial x_i}\right)=0\ \ \ \ \ \ \forall\ \ 1\leq i\leq n\ \ \textrm{ and }\ \forall\ \Phi \in C_c^{\infty}(\Omega), \end{align} where $T_{\chi_{\Omega_1}}$ is a distribution on $\Omega$ defined by $T_{\chi_{\Omega_1}}(\Phi)=\int\limits_{\Omega}\chi_{\Omega_1} \Phi \ \ \ \ \forall \ \ \Phi \in C_c^{\infty}(\Omega),\ $ i.e., $T_{\chi_{\Omega_1}}: C_c^{\infty}(\Omega)\to \mathbb R.$

From the property of distribution^*, $$T_{\chi_{\Omega_1}}\left(\frac{\partial \Phi}{\partial x_i}\right)=-\frac{\partial T_{\chi_{\Omega_1}}}{\partial x_i}(\Phi)$$

So $\frac{\partial T_{\chi_{\Omega_1}}}{\partial x_i}=0 \ \ \ \ \forall 1\leq i\leq n.$

$\implies T_{\chi_{\Omega_1}}=T_k=$ constantly distributed operator, where $k\in \mathbb R$.

$\implies \int\limits_{\Omega}\chi_{\Omega_1}\Phi=\int\limits_{\Omega} k\Phi \ \ \ \ \forall \ \Phi \in C_c^{\infty}(\Omega)$

$\implies \int\limits_{\Omega}\left(\chi_{\Omega_1}-k\right)\Phi=0 \ \ \ \ \forall \ \Phi \in C_c^{\infty}(\Omega)$

$\implies \chi_{\Omega_1}=k$ a.e. $\Omega$ $\ \ \ $[By Fundamental Theorem of Calculus of Variations]

$\implies \mathcal L^n(\Omega_2)=0,$ which is a contradiction.

Therefore, we can conclude that there exists a $C_c^{\infty}$ function $\Phi$ on $\Omega$ such that $\int\limits_{\Omega_1}\nabla \Phi\neq 0.$

Note: Definition of distribution can be seen at the definition $1.2.2$ in the book 'Topics in Functional Analysis and Applications' by S. Kesavan.

Definition(distribution): A linear functional $T$ on $C_c^{\infty}(\Omega)$ is said to be distribution on $\Omega$ if whenever $\Phi_m\to 0$ in $C_c^{\infty}(\Omega)$ we have $T(\Phi_m)\to 0.$

^{*_{It's proved in the last part of the comment $3.$ Fundamental Solutions (Before the exercise section) in the first chapter 'Distribution' of the book 'Topics in Functional Analysis and Applications' by S. Kesavan. Also it's on the exercise $1.2$}}

Answer 2

Hint: Assuming $\Omega_1$ and $\Omega_2$ are measurable, then there exists compact set $F\subset \Omega_1$ such that $\mathcal{L}^n(F)+\varepsilon=\mathcal{L}^n(\Omega_1)$. Likewise, there exists open set $G\supset \Omega_1$ such that $\mathcal{L}^n(G)-\varepsilon=\mathcal{L}^n(\Omega_1)$. Hence $\mathcal{L}^n(G\cap\Omega_2)$ is small.

Then by smooth Urysohn's lemma there exists $\Phi$ smooth such that $\Phi \equiv 1$ on $F$ and $\Phi\equiv 0$ on $G^c$.