Do Sobolev spaces contain nowhere differentiable functions?Are functions of bounded variation a.e. differentiable?Derivable functions & Sobolev spacesA (non trivial) continuous map on a Banach space which is nowhere Frechet differentiableCompactly supported functions and Sobolev spaces on manifoldsCompact embedding and fractional Sobolev spaces in unbounded domainSobolev space compact embeddingsIs the composition of two nowhere differentiable functions still nowhere differentiable?
Do Sobolev spaces contain nowhere differentiable functions?
Are functions of bounded variation a.e. differentiable?Derivable functions & Sobolev spacesA (non trivial) continuous map on a Banach space which is nowhere Frechet differentiableCompactly supported functions and Sobolev spaces on manifoldsCompact embedding and fractional Sobolev spaces in unbounded domainSobolev space compact embeddingsIs the composition of two nowhere differentiable functions still nowhere differentiable?
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Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?
fa.functional-analysis sobolev-spaces
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
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add a comment |
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Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?
fa.functional-analysis sobolev-spaces
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
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Not for $n=1$, of course...
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– Nate Eldredge
6 hours ago
4
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"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
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– Dirk
5 hours ago
add a comment |
$begingroup$
Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?
fa.functional-analysis sobolev-spaces
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
$endgroup$
Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?
fa.functional-analysis sobolev-spaces
fa.functional-analysis sobolev-spaces
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
edited 5 hours ago
Arnold Neumaier
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
asked 8 hours ago
Arnold NeumaierArnold Neumaier
1,5887 silver badges27 bronze badges
1,5887 silver badges27 bronze badges
$begingroup$
Not for $n=1$, of course...
$endgroup$
– Nate Eldredge
6 hours ago
4
$begingroup$
"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
$endgroup$
– Dirk
5 hours ago
add a comment |
$begingroup$
Not for $n=1$, of course...
$endgroup$
– Nate Eldredge
6 hours ago
4
$begingroup$
"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
$endgroup$
– Dirk
5 hours ago
$begingroup$
Not for $n=1$, of course...
$endgroup$
– Nate Eldredge
6 hours ago
$begingroup$
Not for $n=1$, of course...
$endgroup$
– Nate Eldredge
6 hours ago
4
4
$begingroup$
"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
$endgroup$
– Dirk
5 hours ago
$begingroup$
"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
$endgroup$
– Dirk
5 hours ago
add a comment |
1 Answer
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As Nate Eldredge pointed out, $W^1,2(mathbbR)$ functions are absolutely continuous on $mathbbR$, and therefore differentiable a.e., and so the answer is no.
For $ngeq 2$, the answer is yes.
When n=2, this is a classical result of L. Cesari (Ann. Sc. Norm. Super. Pisa Cl. Sci., 1941); see also an explicit construction applicable to $W^1,n(mathbbR^n)$ in J. Serrin, Arch. Ration. Mech. Anal., 1961. The paper of Cesari also contains a positive result for $W^1,p$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder'on in Riv. Mat. Univ. Parma, 1951.
For $n>2$, it looks like a suitable construction is given at Are functions of bounded variation a.e. differentiable?.
answered 3 hours ago
AnonymousAnonymous
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New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
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I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
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– Wojowu
3 hours ago
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
As Nate Eldredge pointed out, $W^1,2(mathbbR)$ functions are absolutely continuous on $mathbbR$, and therefore differentiable a.e., and so the answer is no.
For $ngeq 2$, the answer is yes.
When n=2, this is a classical result of L. Cesari (Ann. Sc. Norm. Super. Pisa Cl. Sci., 1941); see also an explicit construction applicable to $W^1,n(mathbbR^n)$ in J. Serrin, Arch. Ration. Mech. Anal., 1961. The paper of Cesari also contains a positive result for $W^1,p$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder'on in Riv. Mat. Univ. Parma, 1951.
For $n>2$, it looks like a suitable construction is given at Are functions of bounded variation a.e. differentiable?.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
add a comment |
$begingroup$
As Nate Eldredge pointed out, $W^1,2(mathbbR)$ functions are absolutely continuous on $mathbbR$, and therefore differentiable a.e., and so the answer is no.
For $ngeq 2$, the answer is yes.
When n=2, this is a classical result of L. Cesari (Ann. Sc. Norm. Super. Pisa Cl. Sci., 1941); see also an explicit construction applicable to $W^1,n(mathbbR^n)$ in J. Serrin, Arch. Ration. Mech. Anal., 1961. The paper of Cesari also contains a positive result for $W^1,p$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder'on in Riv. Mat. Univ. Parma, 1951.
For $n>2$, it looks like a suitable construction is given at Are functions of bounded variation a.e. differentiable?.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
add a comment |
$begingroup$
As Nate Eldredge pointed out, $W^1,2(mathbbR)$ functions are absolutely continuous on $mathbbR$, and therefore differentiable a.e., and so the answer is no.
For $ngeq 2$, the answer is yes.
When n=2, this is a classical result of L. Cesari (Ann. Sc. Norm. Super. Pisa Cl. Sci., 1941); see also an explicit construction applicable to $W^1,n(mathbbR^n)$ in J. Serrin, Arch. Ration. Mech. Anal., 1961. The paper of Cesari also contains a positive result for $W^1,p$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder'on in Riv. Mat. Univ. Parma, 1951.
For $n>2$, it looks like a suitable construction is given at Are functions of bounded variation a.e. differentiable?.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As Nate Eldredge pointed out, $W^1,2(mathbbR)$ functions are absolutely continuous on $mathbbR$, and therefore differentiable a.e., and so the answer is no.
For $ngeq 2$, the answer is yes.
When n=2, this is a classical result of L. Cesari (Ann. Sc. Norm. Super. Pisa Cl. Sci., 1941); see also an explicit construction applicable to $W^1,n(mathbbR^n)$ in J. Serrin, Arch. Ration. Mech. Anal., 1961. The paper of Cesari also contains a positive result for $W^1,p$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder'on in Riv. Mat. Univ. Parma, 1951.
For $n>2$, it looks like a suitable construction is given at Are functions of bounded variation a.e. differentiable?.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
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Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
answered 3 hours ago
AnonymousAnonymous
411 bronze badge
411 bronze badge
New contributor
Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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Check out our Code of Conduct.
1
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
add a comment |
1
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
1
1
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
$begingroup$
I'll be honest, my first thought after reading this answer was "what about $1<n<2$?"
$endgroup$
– Wojowu
3 hours ago
add a comment |
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$begingroup$
Not for $n=1$, of course...
$endgroup$
– Nate Eldredge
6 hours ago
4
$begingroup$
"The Sobolev space"? Maybe you need to be a little more precise about the exponents. Maybe you want just some Sobolev space $W^{m,p)$ with $m>0$? Or $m=1$? Oh, and: Good question!
$endgroup$
– Dirk
5 hours ago