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











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$textbf{Problem}$ 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_nVert_{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_kVert_{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, $uin W^{1,p}_0(Omega)$.



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



I want to know where my proof is wrong..



Any help is appreciated....



Thank you!










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  • 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.
    – Michał Miśkiewicz
    Nov 16 at 19:30

















up vote
1
down vote

favorite













$textbf{Problem}$ 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_nVert_{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_kVert_{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, $uin W^{1,p}_0(Omega)$.



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



I want to know where my proof is wrong..



Any help is appreciated....



Thank you!










share|cite|improve this question
























  • 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.
    – Michał Miśkiewicz
    Nov 16 at 19:30















up vote
1
down vote

favorite









up vote
1
down vote

favorite












$textbf{Problem}$ 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_nVert_{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_kVert_{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, $uin W^{1,p}_0(Omega)$.



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



I want to know where my proof is wrong..



Any help is appreciated....



Thank you!










share|cite|improve this question
















$textbf{Problem}$ 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_nVert_{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_kVert_{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, $uin W^{1,p}_0(Omega)$.



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



I want to know where my proof is wrong..



Any help is appreciated....



Thank you!







pde banach-spaces sobolev-spaces






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share|cite|improve this question













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edited Nov 16 at 11:25

























asked Nov 16 at 7:04









w.sdka

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  • 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.
    – Michał Miśkiewicz
    Nov 16 at 19:30




















  • 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.
    – Michał Miśkiewicz
    Nov 16 at 19:30


















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.
– Michał Miśkiewicz
Nov 16 at 19:30






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.
– Michał Miśkiewicz
Nov 16 at 19:30












1 Answer
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up vote
2
down vote



accepted










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.






share|cite|improve this answer








New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 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!!
    – w.sdka
    Nov 16 at 11:44












  • there is a same question: math.stackexchange.com/questions/2983140/…
    – chloe hj
    Nov 19 at 12:38











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










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.






share|cite|improve this answer








New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • 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!!
    – w.sdka
    Nov 16 at 11:44












  • there is a same question: math.stackexchange.com/questions/2983140/…
    – chloe hj
    Nov 19 at 12:38















up vote
2
down vote



accepted










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.






share|cite|improve this answer








New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • 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!!
    – w.sdka
    Nov 16 at 11:44












  • there is a same question: math.stackexchange.com/questions/2983140/…
    – chloe hj
    Nov 19 at 12:38













up vote
2
down vote



accepted







up vote
2
down vote



accepted






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.






share|cite|improve this answer








New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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.







share|cite|improve this answer








New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered Nov 16 at 11:39









chloe hj

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536




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chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • 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!!
    – w.sdka
    Nov 16 at 11:44












  • there is a same question: math.stackexchange.com/questions/2983140/…
    – chloe hj
    Nov 19 at 12:38


















  • 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!!
    – w.sdka
    Nov 16 at 11:44












  • there is a same question: math.stackexchange.com/questions/2983140/…
    – chloe hj
    Nov 19 at 12:38
















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!!
– w.sdka
Nov 16 at 11:44






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!!
– w.sdka
Nov 16 at 11:44














there is a same question: math.stackexchange.com/questions/2983140/…
– chloe hj
Nov 19 at 12:38




there is a same question: math.stackexchange.com/questions/2983140/…
– chloe hj
Nov 19 at 12:38


















 

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