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











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













$textbf{Problem}$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 16 at 11:25

























asked Nov 16 at 7:04









w.sdka

30919




30919












  • 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
1






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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000820%2fprove-that-w-01-p-is-a-banach-space%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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

536




536




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.





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


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000820%2fprove-that-w-01-p-is-a-banach-space%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Quarter-circle Tiles

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Mont Emei