let $\Omega \subset R^n$ a open set .let $\varphi \in H^{1}_{0} (\Omega)$ . Suppose that the suport of $\varphi$ is compact. By definition , there exists a sequence of functions $ \varphi_i , i \in N$ in $C_{0}^{\infty}(\Omega)$ converging to $\varphi$ in $H^{1}(\Omega)$ .
Consider an open set ${\Omega}^{'}$ satisfying (this set exists because the suport of $\varphi$ is compact) : $ \operatorname{supp}\varphi \subset {\Omega}^{'} $
and $\overline{{\Omega}^{'}} \subset \Omega$
The affirmation " there exists $i_0 \in N$ such that $i \geq i_0 $ implies $ \operatorname{supp}\varphi_i \subset {\Omega}^{'}$ " is true ?
My professor said that the affirmation is true . the hint of my professor is: Suppose the affirmation is not true. since $\varphi_i , i \in N$ converges to $\varphi$ in $H^{1}(\Omega)$, then $\varphi_i \rightarrow \varphi$ a.e. the negation of your thesis contradicts this $"\varphi_i \rightarrow \varphi$ a.e." I dont know how to prove the affirmation..... someone can give me a hint ?
Thank you ( my english is terrible , sorry )
I am thinking that the claim is false...
Let $\Omega=(-2,2)\subset\mathbb R$ and $\Omega'=(-\frac12,\frac12)$. Choose $\varphi=0\in H_0^1(\Omega)$ and $$ \varphi_n(x) ~=~ \frac1n e^{-\frac{1}{x^2-1}} \chi_{(-1,1)}(x) $$ where $\chi_A$ is the characteristic function of the set $A$, i.e. $$ \varphi_n(x) ~=~ \begin{cases} \frac1ne^{-\frac{1}{x^2-1}} & \text{if }|x|< 1 \\[4pt] 0 & \text{if }|x|\geq 1 \end{cases} $$ Now, clearly $\varphi_n\in C^\infty_c(\Omega)$ converges to $0$ in $L^2(\Omega)$. Besides, $$ \varphi_n'(x) ~=~ -\frac1n \frac{2x}{(x^2-1)^2} e^{-\frac{1}{x^2-1}} \chi_{(-1,1)}(x) $$ so $\varphi_n'$ too converges to $0=\varphi'$ in $L^2(\Omega)$. (To see this, just notice that you have convergence in $L^\infty$ too in both cases).
So, $\varphi_n\to0$ in $H^1(\Omega)$ but ${\sf supp}\varphi_n = [-1,1]$ for every $n$...