# Prove that $W_0^{1,p}$ is a Banach space

$$\textbf{Problem}$$ Prove that $$W_0^{1,p}(\Omega)$$ is a Banach space where $$\Omega$$ be an open and bounded set in $$\mathbb{R}^n$$

$$\textbf{Proof}$$ $$\quad$$Let $$\{u_n\}$$ be the Cauchy Sequence in $$W_0^{1,p}(\Omega)$$. Then, $$\{u_n\}$$ be also the Cacuhy Sequence in $$W^{1,p}(\Omega)$$. Since $$W^{1,p}(\Omega)$$ is a Banach space, there exists $$u \in W^{1,p}(\Omega)$$ such that $$\Vert u-u_n \Vert _{W^{1,p}(\Omega)} \rightarrow 0$$ as $$n \rightarrow \infty$$. We suffices to show that $$u \in W_0^{1,p}(\Omega)$$.

Since $$u_n \in W_0^{1,p}(\Omega)$$, there exists $$\phi_{n_j} \in C^{\infty}_{0}(\Omega)$$ such that $$\Vert u_n - \phi_{n_j}\Vert _{W^{1,p}(\Omega)} \rightarrow 0$$ as $$n_j \rightarrow 0$$.

Thus, \begin{align*} \Vert u - \phi_{n_j} \Vert_{W^{1,p}(\Omega)}\leq \Vert u-u_k \Vert_{W^{1,p}(\Omega)}+\Vert u_k-u_n\Vert_{W^{1,p}(\Omega)}+\Vert u_n-\phi_{n_j}\Vert_{W^{1,p}(\Omega)} \end{align*} (i) There exists $$N_1>0$$ such that \begin{align*} \Vert u-u_k\Vert_{W^{1,p}(\Omega)}<\epsilon/3 \end{align*} for $$k>N_1$$. ($$u_n$$ converge to $$u$$ in $$W^{1,p}(\Omega)$$)

(ii) There exists $$N_2>0$$ such that \begin{align*} \Vert u_k-u_n \Vert_{W^{1,p}(\Omega)}<\epsilon/3 \end{align*} for $$n,k>N_2$$. ($$u_n$$ Cauchy sequence in $$W^{1,p}(\Omega)$$)

(iii) There exists $$N_3>0$$ such that \begin{align*} \Vert u_n-\phi_{n_j}\Vert _{W^{1,p}(\Omega)}<\epsilon/3 \end{align*} for $$n_j>N_3$$. ($$u_n \in W^{1,p}_0(\Omega)$$)

Consequently, $$\phi_{n_j} \in C^{\infty}_0(\Omega)$$ converge to $$u$$ in $$W^{1,p}(\Omega)$$. i.e, $$u\in W^{1,p}_0(\Omega)$$.

I'm not sure my proof is right....

I want to know where my proof is wrong..

Any help is appreciated....

Thank you!

• It seems that what you're proving here is the following general fact: if $B$ is a metric space and $A \subseteq B$, then the closure $\bar{A}$ (defined as the set of all possible limits of sequences in $A$) is a closed set. Nov 16, 2018 at 19:30

I think your proof is right. But according to the definition of $$W^{1,p}_0{\Omega}$$, which is $$W^{1,p}_0{\Omega}$$ is the completion of $$C_c^{\infty}(\Omega)$$ in $$W^{1,p}(\Omega)$$. So I think it's no need to proof $$W^{1,p}_0(\Omega)$$ is a Banach space, because it's natural.
• I also agree with your answer... But, in our homework, we can't use that $W^{1,p}_0(\Omega)$ is the completion of $C^{\infty}_c(\Omega)$ and the closed subspace of Banach space is a Banach space... Anyway, thank you!!