Let $\Omega \subset \mathbb{R}^n$ be a open subset, and $p \in [1,\infty)$. Let $u \in W_0^{m,p} (\Omega)$ be a function and $\hat{u}: \mathbb{R}^n \rightarrow \mathbb{C}$ is a extention of $u$ with $\hat{u}(\mathbb{R} \backslash \Omega) = 0 $. To proof for all $|\alpha| \leq m$:
(1) $\hat{u} \in W^{m,p}(\mathbb{R}^n)$
(2) $\partial^{\alpha} \hat{u} = \hat{\partial^{\alpha} u}$
My answer is as follows:
For all $\varphi \in C_c^{\infty}(\Omega)$ satisfying the following condition:
$\int_{\Omega} \partial^{\alpha}u \varphi dx= (-1)^{|\alpha|} \int_{\Omega} u \partial^{\alpha} \varphi dx$ $ $ with $ $ $\partial^{\alpha} \in L^p$,$u \in L^p$.
So for the $\hat{u}$ satisfying this equlity:
$(-1)^{|\alpha|} \int_{\mathbb{R}^n} \hat{u} \partial^{\alpha} \varphi dx = (-1)^{|\alpha|} \int_{\Omega} u \partial^{\alpha} \varphi dx + (-1)^{|\alpha|} \int_{\mathbb{R}^n \backslash \Omega} \hat{u} \partial^{\alpha} \varphi dx = \int_{\Omega} \partial^{\alpha}u \varphi dx + \int_{\mathbb{R}^n \backslash \Omega} 0 \varphi dx.$
This equlity holds for all $\varphi \in C_{c}^{\infty}(\mathbb{R}^n)$. So we can define $\hat{\partial^{\alpha} u}$ as extension of $\partial^{\alpha} u$ and with the above equity follows $\partial^{\alpha} \hat{u} = \hat{\partial^{\alpha} u}$.
Q.R.D
Is this proof right? And how can I better formulate? Weil the testfunction changes it's domain, I don't know how to deal with this.
Some scripts define the support of a function with closure, some others not. But these make so many differences, so I feel very consumed, expecially when this definition be together used in the testfunction. Do you have any recommendations for books related to Testfunction?