Why is the laplacian a closed operator in $W^{2,p}(mathbb{R}^n)$?
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I have read that the laplacian is a closed operator in $W^{2,p}(Omega)$,(that is, $Delta : W^{2,p} to L^p$) where $Omega$ satisfies some conditions (I need the case $Omega = mathbb{R}^n$ so there shouldn't be any problems, since I think you don't need bounded domains). I need, in particular, to prove that the laplacian is closed in $W^{2,p}(mathbb{R}^n)$ with $p in (1,2)$. To do so, I would like t prove that the usual Sobolev norm and the norm defined by $|||u||| = ||u||_{L^p} +||Delta u||_{L^p}$ are equivalent, and so I would conclude by completeness of $W^{2,p}$. The fact is, when I have $p=2$ and $n=1$ I can see that, because I have:
$$int_{mathbb{R}} (u')^2 = int_{mathbb{R}} (u')(u') = - int_{mathbb{R}} u u'' leq ||u||_{L^2}^2 ||u''||_{L^2}^2 leq frac{1}{2} (||u||_{L^2}^2 + ||u''||_{L^2}^2)$$
and taking the square root I have:
$$||u'||_{L^2} leq frac{1}{sqrt{2}}(||u||_{L^2} + ||u''||_{L^2})$$
I think there shouldn't be any problem to generalise this to dimension $n$, if $p=2$. However, if $p neq 2$ I can't use the trick of writing $(u')^2$ as $(u')(u')$ and thus I don't know what to do.
real-analysis pde sobolev-spaces integral-inequality laplacian
asked Nov 17 at 11:47
tommy1996q
559413
|
show 3 more comments
up vote
1
down vote
favorite
I have read that the laplacian is a closed operator in $W^{2,p}(Omega)$,(that is, $Delta : W^{2,p} to L^p$) where $Omega$ satisfies some conditions (I need the case $Omega = mathbb{R}^n$ so there shouldn't be any problems, since I think you don't need bounded domains). I need, in particular, to prove that the laplacian is closed in $W^{2,p}(mathbb{R}^n)$ with $p in (1,2)$. To do so, I would like t prove that the usual Sobolev norm and the norm defined by $|||u||| = ||u||_{L^p} +||Delta u||_{L^p}$ are equivalent, and so I would conclude by completeness of $W^{2,p}$. The fact is, when I have $p=2$ and $n=1$ I can see that, because I have:
$$int_{mathbb{R}} (u')^2 = int_{mathbb{R}} (u')(u') = - int_{mathbb{R}} u u'' leq ||u||_{L^2}^2 ||u''||_{L^2}^2 leq frac{1}{2} (||u||_{L^2}^2 + ||u''||_{L^2}^2)$$
and taking the square root I have:
$$||u'||_{L^2} leq frac{1}{sqrt{2}}(||u||_{L^2} + ||u''||_{L^2})$$
I think there shouldn't be any problem to generalise this to dimension $n$, if $p=2$. However, if $p neq 2$ I can't use the trick of writing $(u')^2$ as $(u')(u')$ and thus I don't know what to do.
real-analysis pde sobolev-spaces integral-inequality laplacian
asked Nov 17 at 11:47
tommy1996q
559413
Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have read that the laplacian is a closed operator in $W^{2,p}(Omega)$,(that is, $Delta : W^{2,p} to L^p$) where $Omega$ satisfies some conditions (I need the case $Omega = mathbb{R}^n$ so there shouldn't be any problems, since I think you don't need bounded domains). I need, in particular, to prove that the laplacian is closed in $W^{2,p}(mathbb{R}^n)$ with $p in (1,2)$. To do so, I would like t prove that the usual Sobolev norm and the norm defined by $|||u||| = ||u||_{L^p} +||Delta u||_{L^p}$ are equivalent, and so I would conclude by completeness of $W^{2,p}$. The fact is, when I have $p=2$ and $n=1$ I can see that, because I have:
$$int_{mathbb{R}} (u')^2 = int_{mathbb{R}} (u')(u') = - int_{mathbb{R}} u u'' leq ||u||_{L^2}^2 ||u''||_{L^2}^2 leq frac{1}{2} (||u||_{L^2}^2 + ||u''||_{L^2}^2)$$
and taking the square root I have:
$$||u'||_{L^2} leq frac{1}{sqrt{2}}(||u||_{L^2} + ||u''||_{L^2})$$
I think there shouldn't be any problem to generalise this to dimension $n$, if $p=2$. However, if $p neq 2$ I can't use the trick of writing $(u')^2$ as $(u')(u')$ and thus I don't know what to do.
real-analysis pde sobolev-spaces integral-inequality laplacian
asked Nov 17 at 11:47
tommy1996q
559413
I have read that the laplacian is a closed operator in $W^{2,p}(Omega)$,(that is, $Delta : W^{2,p} to L^p$) where $Omega$ satisfies some conditions (I need the case $Omega = mathbb{R}^n$ so there shouldn't be any problems, since I think you don't need bounded domains). I need, in particular, to prove that the laplacian is closed in $W^{2,p}(mathbb{R}^n)$ with $p in (1,2)$. To do so, I would like t prove that the usual Sobolev norm and the norm defined by $|||u||| = ||u||_{L^p} +||Delta u||_{L^p}$ are equivalent, and so I would conclude by completeness of $W^{2,p}$. The fact is, when I have $p=2$ and $n=1$ I can see that, because I have:
$$int_{mathbb{R}} (u')^2 = int_{mathbb{R}} (u')(u') = - int_{mathbb{R}} u u'' leq ||u||_{L^2}^2 ||u''||_{L^2}^2 leq frac{1}{2} (||u||_{L^2}^2 + ||u''||_{L^2}^2)$$
and taking the square root I have:
$$||u'||_{L^2} leq frac{1}{sqrt{2}}(||u||_{L^2} + ||u''||_{L^2})$$
I think there shouldn't be any problem to generalise this to dimension $n$, if $p=2$. However, if $p neq 2$ I can't use the trick of writing $(u')^2$ as $(u')(u')$ and thus I don't know what to do.
real-analysis pde sobolev-spaces integral-inequality laplacian
real-analysis pde sobolev-spaces integral-inequality laplacian
asked Nov 17 at 11:47
tommy1996q
559413
asked Nov 17 at 11:47
tommy1996q
559413
asked Nov 17 at 11:47
tommy1996q
559413
asked Nov 17 at 11:47
tommy1996q
559413
asked Nov 17 at 11:47
tommy1996q
559413
559413
Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04
|
show 3 more comments
Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04
Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04
|
show 3 more comments
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Why do you want to show equivalence of these two norms? Isn't it enough to say that this is a bounded operator?
– Michał Miśkiewicz
Nov 18 at 23:41
Well, I know that if an operator is bounded then it is closed. I don’t know a proof of this fact, though. I would like the equivalence to conclude easily, but if the proof bounded $implies $ closed is easier I would use that, instead.
– tommy1996q
Nov 19 at 7:01
As far as I know, this is more of a tautology than a proof. What do you mean by closed here?
– Michał Miśkiewicz
Nov 19 at 9:28
Let $B_1$ and $B_2$ be Banach spaces. A linear operator $T: B_1 to B_2$ is closed if given a sequence ${x_n}$ converging in $B_1$ to $x$ such that ${Tx_n }$ converges in $B_2$ to $y$, then $x$ belongs to the domain of $T$ and $Tx=y$
– tommy1996q
Nov 19 at 13:04
Continuous implies closed should follow from the closed graph theorem, but I should prove boundedness then
– tommy1996q
Nov 19 at 13:04