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Does a continuous function in $W^1,p$ for $p<n$ lie in $W^1,n$?
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Let $Omega subseteq mathbbR^n$ be an open bounded domain. Suppose that $f in W^1,p(Omega)$ for some specific $1<p<n$, and that $f$ is continuous.
Is it true that $f in W^1,n_loc(Omega)$?
(In "almost" the other direction, we know that if $f in W^1,p$ for $p>n$ then it's continuous).
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
add a comment |Â
up vote
3
down vote
favorite
Let $Omega subseteq mathbbR^n$ be an open bounded domain. Suppose that $f in W^1,p(Omega)$ for some specific $1<p<n$, and that $f$ is continuous.
Is it true that $f in W^1,n_loc(Omega)$?
(In "almost" the other direction, we know that if $f in W^1,p$ for $p>n$ then it's continuous).
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $Omega subseteq mathbbR^n$ be an open bounded domain. Suppose that $f in W^1,p(Omega)$ for some specific $1<p<n$, and that $f$ is continuous.
Is it true that $f in W^1,n_loc(Omega)$?
(In "almost" the other direction, we know that if $f in W^1,p$ for $p>n$ then it's continuous).
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
Let $Omega subseteq mathbbR^n$ be an open bounded domain. Suppose that $f in W^1,p(Omega)$ for some specific $1<p<n$, and that $f$ is continuous.
Is it true that $f in W^1,n_loc(Omega)$?
(In "almost" the other direction, we know that if $f in W^1,p$ for $p>n$ then it's continuous).
real-analysis sobolev-spaces regularity-theory-of-pdes
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
asked Jul 31 at 7:56
Asaf Shachar
4,4573832
4,4573832
add a comment |Â
add a comment |Â
1 Answer
1
active
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up vote
4
down vote
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No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|sin (|x|^-a)$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $Bsetminus0$. The gradient of $u$ is
$$
nabla u(x) = fracxsin (|x|^-a) - fraca x^1+acos (|x|^-a) $$
which implies $|nabla u(x)| = O(|x|^-a)$, and $|nabla u(x)| sim |x|^-a$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $uin W^1,p(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.
answered Jul 31 at 15:08
user357151
13.7k31140
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|sin (|x|^-a)$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $Bsetminus0$. The gradient of $u$ is
$$
nabla u(x) = fracxsin (|x|^-a) - fraca x^1+acos (|x|^-a) $$
which implies $|nabla u(x)| = O(|x|^-a)$, and $|nabla u(x)| sim |x|^-a$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $uin W^1,p(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.
answered Jul 31 at 15:08
user357151
13.7k31140
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
add a comment |Â
up vote
4
down vote
accepted
No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|sin (|x|^-a)$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $Bsetminus0$. The gradient of $u$ is
$$
nabla u(x) = fracxsin (|x|^-a) - fraca x^1+acos (|x|^-a) $$
which implies $|nabla u(x)| = O(|x|^-a)$, and $|nabla u(x)| sim |x|^-a$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $uin W^1,p(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.
answered Jul 31 at 15:08
user357151
13.7k31140
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|sin (|x|^-a)$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $Bsetminus0$. The gradient of $u$ is
$$
nabla u(x) = fracxsin (|x|^-a) - fraca x^1+acos (|x|^-a) $$
which implies $|nabla u(x)| = O(|x|^-a)$, and $|nabla u(x)| sim |x|^-a$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $uin W^1,p(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.
answered Jul 31 at 15:08
user357151
13.7k31140
No, there is no such self-improvement. Continuity adds nothing for Sobolev regularity purposes.
Consider functions of the form $u(x)=|x|sin (|x|^-a)$, $u(0)=0$, where $a>0$. The domain can be the unit ball $B$.
The function $u$ is continuous on $B$ and smooth on $Bsetminus0$. The gradient of $u$ is
$$
nabla u(x) = fracxsin (|x|^-a) - fraca x^1+acos (|x|^-a) $$
which implies $|nabla u(x)| = O(|x|^-a)$, and $|nabla u(x)| sim |x|^-a$ most of the time, when the cosine term is bounded away from $0$.
Therefore, $uin W^1,p(B)$ if and only if $p<n/a$. Choose $a>0$ as you wish.
answered Jul 31 at 15:08
user357151
13.7k31140
answered Jul 31 at 15:08
user357151
13.7k31140
answered Jul 31 at 15:08
user357151
13.7k31140
answered Jul 31 at 15:08
user357151
13.7k31140
13.7k31140
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
add a comment |Â
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
Thanks, this is a great answer. Do you have an idea how to formally prove your map $u$ lies in a Sobolev space? I see naively that your calculation of the differential $nabla u$ is merely an application of the chain rule-$nabla(fg)=fnabla g+g nabla f$. However, is it clear this works in the weaker setting of weak derivatives? (in particular, both maps, $f=|x|,g=sin(|x|^-a)$ are not classically smooth at the origin, so this is not for instance, the case of a $C^1$ function multiplied by a Sobolev function. I wonder if one can phrase some more general claim here, or just "prove by hand" .
– Asaf Shachar
Jul 31 at 15:57
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
ACL characterization says it's okay to use pointwise derivatives as long as ACL condition holds. The condition holds here, as the function is Lipschitz on every horizontal segment except one.
– user357151
Jul 31 at 16:25
add a comment |Â
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