I am trying to prove if $u\in W^{1,p}(\Omega)$ has support compactly inside $\Omega$, then $u\in W^{1,p}_0(\Omega)$, where $\Omega\subset \mathbb R^N$ is open.
Intuitively this is true. Assume $V:=\text{spt}(u)$, since $u\equiv 0$ outside $V$ and there always has some room between $\partial \Omega$ and $\partial V$, hence $u$ should have $0$ trace. But I met some difficulties to understand the proof given by H.Brezis book.
Here is the argument. we choose another set $W$ such that $V\subset\subset W\subset\subset \Omega$ and choose $\alpha\in C_c^\infty(W)$ such that $\alpha\equiv 1$ over $V$. Hence $\alpha u = u$.
Next, we could have a sequence $(u_n)\subset C_c^\infty(\mathbb R^N)$ such that $u_n\to u$ in $L^p(\Omega)$ and $\nabla u_n\to \nabla u$ over $(L^p(W))^N$. So far so good.
Then the book states that $\alpha u_n\to \alpha u$ in $W^{1,p}(\Omega)$. I feel very uncomfortable with this sentence. Of course we have $\alpha u_n\to\alpha u$ in $W^{1,p}(W)$ but why in $\Omega$. Yes, I know it should be just $0$ outside $W$ but I feel uncomfortable if I use this idea. Because if I use this idea, then I can just say outside $V$ $u$ is just $0$ and hence $0$ trace and I done. No need for this argument at all.
So, please help me to understand how sentence marked with bold is true. Thank you!
Notice that $\text{supp}(\alpha) \subset W$, so that both $\alpha u_n$ and $\alpha u$ vanish outside $W$, and then so do their gradients. Therefore $$\| \alpha u_n -\alpha u\|_{W^{1,p}(W)}=\| \alpha u_n -\alpha u\|_{W^{1,p}(\Omega)}.$$
Be careful, trace zero only works when the domain is nice enough. In a general domain you have to show that $u$ is approximable by test functions in $\Omega$. But yes, it's a pretty 'obvious' result.