Given the set $D(\Omega)=\{ \phi \in C^\infty (\Omega):$ supp$(\phi):= \overline {\{x:\phi(x)\neq 0\} } \subset\Omega $ is compact$\},\, \, \Omega \subset R^n: $
In the case $\Omega $ is compact, can one say that $D(\Omega)= C^{\infty}(\Omega) $, since supp$(\phi)\subseteq \Omega$ ? In the case this is not true can you give an example illustrating this ?
In the case $\Omega $ is not compact but bounded can we still say $D(\Omega)= C^{\infty}(\Omega) $ ?
Let $\Omega \subset R$ be an open interval. For $f \in C^1 (\bar {\Omega})$ and $\phi \in D(\Omega)$ it follows after Integration by parts $$ \int _{\Omega} f'(x) \phi(x) \,\,dx=- \int_{\Omega} f(x) \phi'(x) \, \, dx.$$
One term of the integration by parts has disappeard because the support of $\phi$ is a compact subset of $\Omega$. Can somebody explain, why one term disappeared ? What I wrote is what is written in the book.
- Let $\Omega = (-1,1), f(x)=|x|, \,\, g(x)=\frac {x} {|x|} $ for $x\neq0,\,\,$ and $ g(0)=0.$ Then, $$<g,\phi>=-<f, \phi'> , \, \forall \phi \in D(\Omega), \,$$ where $<,>$ is the inner product in $L^2, $ which in our case is simply the integral of the product of the two functions.
Many thanks.
1) This is true in the sense that the restriction of the functions $\phi$ in $D(\Omega)$ are in $C^{\infty}(\Omega^{\circ})$ and $\text{supp}(\phi)$ are compact.
2) Consider $f(x)=1/x$ on $(0,1)$, then $f\in C^{\infty}(0,1)$ but $f\notin D(0,1)$.
3) Assume now in the one-dimensional, then $\Omega=(a,b)$ \begin{align*} &\int_{\Omega}f'(x)\phi(x)dx\\ &=\lim_{n\rightarrow\infty}\int_{a+1/n}^{b-1/n}f'(x)\phi(x)dx\\ &=\lim_{n\rightarrow\infty}f(b-1/n)\phi(b-1/n)-f(a+1/n)\phi(a+1/n)-\int_{a+1/n}^{b-1/n}f(x)\phi'(x)dx\\ &=-\int_{a}^{b}f(x)\phi'(x)dx\\ &=-\int_{\Omega}f(x)\phi'(x)dx. \end{align*}