$\textbf{Problem}$ 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!

  • $\begingroup$ 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. $\endgroup$ Nov 16, 2018 at 19:30

1 Answer 1


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.

  • $\begingroup$ 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!! $\endgroup$
    – user453447
    Nov 16, 2018 at 11:44
  • $\begingroup$ there is a same question: math.stackexchange.com/questions/2983140/… $\endgroup$
    – chloe
    Nov 19, 2018 at 12:38

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