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 $n$...


I think that the affirmation does not hold. Let's have some $\varphi \in H^{1}_{0} (\Omega)$ and sequence of functions $\varphi_i$ converging to $\varphi$ in $H^1(\Omega)$. Assume that $\varphi_i(x),\varphi(x)\geq 0$ for all $x \in \Omega$

Than functions $$\psi_i(x) = \varphi_i(x)+\frac{e^{-||x||^2}}{i}$$

converges to $\varphi$ in $H^1$ and support of $\psi_i$ is $\Omega$

  • $\begingroup$ Are you sure that the support of $\psi_i$ is $\Omega$? $\endgroup$ – AndreasT Mar 11 '13 at 22:19
  • $\begingroup$ if you assume $\varphi_i(x) \geq 0$ $\endgroup$ – tom Mar 11 '13 at 22:25
  • $\begingroup$ Why should $\psi_i(x)=0$ for $x\notin\Omega$ then? $\endgroup$ – AndreasT Mar 11 '13 at 22:27
  • $\begingroup$ Ohh I forgot that $\psi$ has to be zero on boundary. $\endgroup$ – tom Mar 11 '13 at 22:30
  • $\begingroup$ Yep, more than that: its support must be compactly contained in $\Omega$... $\endgroup$ – AndreasT Mar 11 '13 at 22:32

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