Let $W_0^{k,p}(\Omega)$ denote the closure of all testfunctions with compact support in $\Omega\subseteq \mathbb{R}^d$.
Would it be right to identify any function in $u\in W_0^{k,p}(\Omega)$ as a function in $u\in W_0^{k,p}(\mathbb{R}^n)$ when identifying $u$ with its zero extention (only in one direction) such that $W_0^{k,p}(\Omega)\subseteq W_0^{k,p}(\mathbb{R}^d)$?
I think yes, and can provide a proof for, but I'm interested in your independent opinion to exclude wether I made a mistake or not.
Yes, since $u\in W^{k,p}(\Omega)$ has zero trace, you can extend it by zero in $\mathbb{R}^d\setminus\Omega$ to obtain a function $\tilde{u}\in W^{k,p}(\mathbb{R}^d)$ such that $\tilde{u} = u$ in $\Omega$ and $\tilde{u} = 0$ in $\mathbb{R}^d\setminus\Omega$. The proof of the integrability is straightforward, while for the weak differentiability take any $\phi\in C_c^\infty(\mathbb{R}^d)$. Then $$ \int_{\mathbb{R}^d}\tilde{u}\partial^\alpha\phi\,dx = \int_\Omega u\partial^\alpha\phi\, dx = (-1)^{\vert\alpha\vert} \int_\Omega \partial^\alpha u \phi\,dx = (-1)^{\vert\alpha\vert}\int_{\mathbb{R}^d}v\phi\, dx$$ for all multiindices $\alpha\in\mathbb{N}^d_0$ such that $\vert\alpha\vert\leq k$, where $$ v = \begin{cases} \partial^\alpha u, &&\text{in } \Omega, \\ 0, &&\text{in } \mathbb{R}^d\setminus\Omega. \end{cases}$$ Note that the space is $W^{k,p}(\mathbb{R}^d)$ and $\textbf{not}$ $W^{k,p}_0(\mathbb{R}^d)$, as the latter isn't defined.