Mixing
composition of functions in sobolev space
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If we have $uin H^s(Omega)$ where $Omega$ is a domain of $mathbb R^n,s>0$ and $ g : mathbb R^n to mathbb R^n$, what is the least condition on $g$ to ensure $g(u) in H^s.$ I think $gin C^infty$ is enough, but it seems not optimal. Need some references. Thanks!
real-analysis sobolev-spaces
asked Dec 29 '18 at 7:11
user13676user13676
13
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add a comment |
$begingroup$
If we have $uin H^s(Omega)$ where $Omega$ is a domain of $mathbb R^n,s>0$ and $ g : mathbb R^n to mathbb R^n$, what is the least condition on $g$ to ensure $g(u) in H^s.$ I think $gin C^infty$ is enough, but it seems not optimal. Need some references. Thanks!
real-analysis sobolev-spaces
asked Dec 29 '18 at 7:11
user13676user13676
13
$endgroup$
add a comment |
$begingroup$
If we have $uin H^s(Omega)$ where $Omega$ is a domain of $mathbb R^n,s>0$ and $ g : mathbb R^n to mathbb R^n$, what is the least condition on $g$ to ensure $g(u) in H^s.$ I think $gin C^infty$ is enough, but it seems not optimal. Need some references. Thanks!
real-analysis sobolev-spaces
asked Dec 29 '18 at 7:11
user13676user13676
13
$endgroup$
If we have $uin H^s(Omega)$ where $Omega$ is a domain of $mathbb R^n,s>0$ and $ g : mathbb R^n to mathbb R^n$, what is the least condition on $g$ to ensure $g(u) in H^s.$ I think $gin C^infty$ is enough, but it seems not optimal. Need some references. Thanks!
real-analysis sobolev-spaces
real-analysis sobolev-spaces
asked Dec 29 '18 at 7:11
user13676user13676
13
asked Dec 29 '18 at 7:11
user13676user13676
13
edited Dec 29 '18 at 7:49
user13676
asked Dec 29 '18 at 7:11
user13676user13676
13
asked Dec 29 '18 at 7:11
user13676user13676
13
asked Dec 29 '18 at 7:11
user13676user13676
13
13
add a comment |
add a comment |
1 Answer
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If $Omega$ is unbounded, then $C^{infty}$ is not sufficient without growth conditions on $g.$ For example taking $Omega = mathbb R$ and $g(x) = 1$ (which is smooth) we get $g circ u equiv 1$ which is not integrable regardless of what $u$ is.
For the case $s = 1,$ we have the following result.
Theorem: If $Omega subset mathbb R^n$ is bounded and $g : mathbb R rightarrow mathbb R$ is globally Lipschitz continuous, then $g circ u in H^1(Omega)$ whenever $u in H^1(Omega).$ For general $Omega,$ the same holds if $g(0) = 0.$
The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n in C^{infty}(Omega) cap H^1(Omega)$ such that $u_n rightarrow u$ in $H^1(Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.
The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).
By iterative differentiation we get for $u in H^k$ with $k$ integer, we need $g in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.
For general $s>0$ I do not know what happens.
answered Jan 2 at 18:09
ktoiktoi
2,3861616
$endgroup$
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $Omega$ is unbounded, then $C^{infty}$ is not sufficient without growth conditions on $g.$ For example taking $Omega = mathbb R$ and $g(x) = 1$ (which is smooth) we get $g circ u equiv 1$ which is not integrable regardless of what $u$ is.
For the case $s = 1,$ we have the following result.
Theorem: If $Omega subset mathbb R^n$ is bounded and $g : mathbb R rightarrow mathbb R$ is globally Lipschitz continuous, then $g circ u in H^1(Omega)$ whenever $u in H^1(Omega).$ For general $Omega,$ the same holds if $g(0) = 0.$
The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n in C^{infty}(Omega) cap H^1(Omega)$ such that $u_n rightarrow u$ in $H^1(Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.
The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).
By iterative differentiation we get for $u in H^k$ with $k$ integer, we need $g in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.
For general $s>0$ I do not know what happens.
answered Jan 2 at 18:09
ktoiktoi
2,3861616
$endgroup$
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
add a comment |
$begingroup$
If $Omega$ is unbounded, then $C^{infty}$ is not sufficient without growth conditions on $g.$ For example taking $Omega = mathbb R$ and $g(x) = 1$ (which is smooth) we get $g circ u equiv 1$ which is not integrable regardless of what $u$ is.
For the case $s = 1,$ we have the following result.
Theorem: If $Omega subset mathbb R^n$ is bounded and $g : mathbb R rightarrow mathbb R$ is globally Lipschitz continuous, then $g circ u in H^1(Omega)$ whenever $u in H^1(Omega).$ For general $Omega,$ the same holds if $g(0) = 0.$
The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n in C^{infty}(Omega) cap H^1(Omega)$ such that $u_n rightarrow u$ in $H^1(Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.
The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).
By iterative differentiation we get for $u in H^k$ with $k$ integer, we need $g in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.
For general $s>0$ I do not know what happens.
answered Jan 2 at 18:09
ktoiktoi
2,3861616
$endgroup$
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
add a comment |
$begingroup$
If $Omega$ is unbounded, then $C^{infty}$ is not sufficient without growth conditions on $g.$ For example taking $Omega = mathbb R$ and $g(x) = 1$ (which is smooth) we get $g circ u equiv 1$ which is not integrable regardless of what $u$ is.
For the case $s = 1,$ we have the following result.
Theorem: If $Omega subset mathbb R^n$ is bounded and $g : mathbb R rightarrow mathbb R$ is globally Lipschitz continuous, then $g circ u in H^1(Omega)$ whenever $u in H^1(Omega).$ For general $Omega,$ the same holds if $g(0) = 0.$
The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n in C^{infty}(Omega) cap H^1(Omega)$ such that $u_n rightarrow u$ in $H^1(Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.
The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).
By iterative differentiation we get for $u in H^k$ with $k$ integer, we need $g in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.
For general $s>0$ I do not know what happens.
answered Jan 2 at 18:09
ktoiktoi
2,3861616
$endgroup$
If $Omega$ is unbounded, then $C^{infty}$ is not sufficient without growth conditions on $g.$ For example taking $Omega = mathbb R$ and $g(x) = 1$ (which is smooth) we get $g circ u equiv 1$ which is not integrable regardless of what $u$ is.
For the case $s = 1,$ we have the following result.
Theorem: If $Omega subset mathbb R^n$ is bounded and $g : mathbb R rightarrow mathbb R$ is globally Lipschitz continuous, then $g circ u in H^1(Omega)$ whenever $u in H^1(Omega).$ For general $Omega,$ the same holds if $g(0) = 0.$
The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n in C^{infty}(Omega) cap H^1(Omega)$ such that $u_n rightarrow u$ in $H^1(Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.
The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).
By iterative differentiation we get for $u in H^k$ with $k$ integer, we need $g in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.
For general $s>0$ I do not know what happens.
answered Jan 2 at 18:09
ktoiktoi
2,3861616
answered Jan 2 at 18:09
ktoiktoi
2,3861616
answered Jan 2 at 18:09
ktoiktoi
2,3861616
answered Jan 2 at 18:09
ktoiktoi
2,3861616
2,3861616
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
add a comment |
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
$begingroup$
Thanks，I have find a result for $mathbb R^n$ in R. Danchin's notes "Fourier Analysis Methods for PDE" (prop.1.48) where ,for $H^s$, $g$ should in the space $W^{s+1,infty}.$
$endgroup$
– user13676
Jan 3 at 10:49
add a comment |
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