I'm trying to prove that $W_0^{k,p}(\Omega)$ is Banach space by showing it is complete space.
So, here is what i got so far. Let $\{u_m\}$ be a Cauchy sequence in $W_0^{k,p}(\Omega)$, then, since $W_0^{k,p}(\Omega)\subset W^{k,p}(\Omega)$ and $W^{k,p}(\Omega)$ is complete, we have limit $u$ of $u_m$ in $W^{k,p}(\Omega)$.
I want to show that this $u$ is indeed in $W_0^{k,p}(\Omega)$. So, i need to show that there exists $\{\xi_j\}_{j=1}^{\infty}\subset C_0^\infty(\Omega)$ which converges to $u$ in $W^{k,p}(\Omega)$.
How can i construct such sequence of smooth functions? Maybe construct from each smooth sequences in $u_m$ approximation?
Thanks in advance.
For each $m\in \mathbb{N}$, let $(u_{m, k})_{k}\subseteq C_0^\infty$ be a sequence convergin to $u_m$ in $W^{1,p}$.
Now, for each $n\in \mathbb{N}$, we may find $m\in\mathbb{N}$ such that $$ \lVert u - u _m \rVert < \frac{1}{n} $$ We may then find $k\in\mathbb{N}$ such that $$ \lVert u_m - u_{m, k}\rVert < \frac{1}{n} $$ So, define $v_n = u_{m,k}$. Then notice that $(v_n)$ is a sequence in $C_0^\infty$ converging to $u$ in $W^{1,p}$.
EDIT: As the comment below mentions, suppose you have a complete metric space $X$ and a subset $A\subseteq X$. Then we define the closure $\overline{A}$ of $A$ to be the collection of all $x\in X$ such that $x = \lim x_n$ for some sequence $(x_n)_n\subseteq A$.The above method allows one to show that $\overline{A}$ is indeed closed (in the topological sense).