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!
pde banach-spaces sobolev-spaces
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up vote
1
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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!
pde banach-spaces sobolev-spaces
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
add a comment |
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!
pde banach-spaces sobolev-spaces
$textbf{Problem}$ 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
pde banach-spaces sobolev-spaces
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
add a comment |
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
add a comment |
1 Answer
1
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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.
New contributor
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
add a comment |
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.
New contributor
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
add a comment |
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.
New contributor
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
add a comment |
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.
New contributor
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.
New contributor
New contributor
answered Nov 16 at 11:39
chloe hj
536
536
New contributor
New contributor
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
add a comment |
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
add a comment |
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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