Mixing
A counter-example for integration by parts when there are “small” singularities
$begingroup$
I am looking for a "counter-example" to integration by parts of the following type:
$Omega subseteq mathbb R^n$ is an open, bounded, connected domain with smooth boundary. $u,v:bar Omega to mathbb R$ are real-valued functions, where $u$ is smooth and compactly supported in $Omega$, and $v$ is smooth on an open subset of $bar Omega$ whose complement is a closed subset of measure zero. I want $int_{Omega}(partial_iu)v neq -int_{Omega}u(partial_iv)$, i.e. to demonstrate failure of integration by parts.
Edit:
Does the answer change if we assume in addition that $v$ is continuous everywhere on $bar Omega$?. BigbearZzz gave here an example with a non-continuous $v$.
If $v$ was smooth on all $bar Omega$, then integration by parts would work. The point is that I am limiting the singular set to be closed and of measure zero. I guess integration by parts still cannot be saved, but I don't have a concrete example.
integration measure-theory definite-integrals geometric-measure-theory singularity
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
$endgroup$
add a comment |
$begingroup$
I am looking for a "counter-example" to integration by parts of the following type:
$Omega subseteq mathbb R^n$ is an open, bounded, connected domain with smooth boundary. $u,v:bar Omega to mathbb R$ are real-valued functions, where $u$ is smooth and compactly supported in $Omega$, and $v$ is smooth on an open subset of $bar Omega$ whose complement is a closed subset of measure zero. I want $int_{Omega}(partial_iu)v neq -int_{Omega}u(partial_iv)$, i.e. to demonstrate failure of integration by parts.
Edit:
Does the answer change if we assume in addition that $v$ is continuous everywhere on $bar Omega$?. BigbearZzz gave here an example with a non-continuous $v$.
If $v$ was smooth on all $bar Omega$, then integration by parts would work. The point is that I am limiting the singular set to be closed and of measure zero. I guess integration by parts still cannot be saved, but I don't have a concrete example.
integration measure-theory definite-integrals geometric-measure-theory singularity
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
$endgroup$
1
$begingroup$
I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05
add a comment |
$begingroup$
I am looking for a "counter-example" to integration by parts of the following type:
$Omega subseteq mathbb R^n$ is an open, bounded, connected domain with smooth boundary. $u,v:bar Omega to mathbb R$ are real-valued functions, where $u$ is smooth and compactly supported in $Omega$, and $v$ is smooth on an open subset of $bar Omega$ whose complement is a closed subset of measure zero. I want $int_{Omega}(partial_iu)v neq -int_{Omega}u(partial_iv)$, i.e. to demonstrate failure of integration by parts.
Edit:
Does the answer change if we assume in addition that $v$ is continuous everywhere on $bar Omega$?. BigbearZzz gave here an example with a non-continuous $v$.
If $v$ was smooth on all $bar Omega$, then integration by parts would work. The point is that I am limiting the singular set to be closed and of measure zero. I guess integration by parts still cannot be saved, but I don't have a concrete example.
integration measure-theory definite-integrals geometric-measure-theory singularity
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
$endgroup$
I am looking for a "counter-example" to integration by parts of the following type:
$Omega subseteq mathbb R^n$ is an open, bounded, connected domain with smooth boundary. $u,v:bar Omega to mathbb R$ are real-valued functions, where $u$ is smooth and compactly supported in $Omega$, and $v$ is smooth on an open subset of $bar Omega$ whose complement is a closed subset of measure zero. I want $int_{Omega}(partial_iu)v neq -int_{Omega}u(partial_iv)$, i.e. to demonstrate failure of integration by parts.
Edit:
Does the answer change if we assume in addition that $v$ is continuous everywhere on $bar Omega$?. BigbearZzz gave here an example with a non-continuous $v$.
If $v$ was smooth on all $bar Omega$, then integration by parts would work. The point is that I am limiting the singular set to be closed and of measure zero. I guess integration by parts still cannot be saved, but I don't have a concrete example.
integration measure-theory definite-integrals geometric-measure-theory singularity
integration measure-theory definite-integrals geometric-measure-theory singularity
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
edited Jan 9 at 12:14
Asaf Shachar
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
asked Jan 9 at 9:47
Asaf ShacharAsaf Shachar
5,47531141
5,47531141
1
$begingroup$
I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05
add a comment |
1
$begingroup$
I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05
1
1
$begingroup$
I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05
$begingroup$
I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I am a bit confused by the question, please tell me if I misread anything.
Consider the domain $Omega=B_1subsetBbb R^2$, i.e. the unit ball. Define $v$ to be
$$
v(x)=begin{cases}1 &; x<0 \
0 &; xge 0.
end{cases}
$$
We can see that $v$ is smooth on $overlineOmegabackslash l$, where $l={(x,y)inBbb R^2 : x=0}$ which is a closed set of measure zero.
Next, we take $varphiin C^infty_c(Omega)$. Since $nabla v(x)=0$ a.e., we have
$$
int_Omega varphi partial_i v = 0.
$$
On the other hand, we have
$$begin{align}
int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy
end{align}$$
since the distributional derivative of $v$ is the Hausdorff measure on $l$.
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
$endgroup$
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
|
show 1 more comment
$begingroup$
For a continuous counterexample, let's consider the case $Bbb R^n=Bbb R$ and $Omega = (0,1)$. Let $v$ be the Cantor function
It has the property that $v'(x)=0$ a.e., thus for any function $varphiin C^infty_c(0,1)$ we have
$$
int_0^1 varphi v' = 0.
$$
On the other hand, we have
$$begin{align}
int_0^1 varphi'v = -c_0int_{0}^1 varphi(t), dnu(t)
end{align}$$
where $nu=mathcal H^gammalvert_C$, the Hausdorff measure of dimension $gamma= ln 2 /ln 3$ restricted to the Cantor set $C$ (which is a compact set). Here $c_0$ is a normalizing constant that I don't remember.
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
I am a bit confused by the question, please tell me if I misread anything.
Consider the domain $Omega=B_1subsetBbb R^2$, i.e. the unit ball. Define $v$ to be
$$
v(x)=begin{cases}1 &; x<0 \
0 &; xge 0.
end{cases}
$$
We can see that $v$ is smooth on $overlineOmegabackslash l$, where $l={(x,y)inBbb R^2 : x=0}$ which is a closed set of measure zero.
Next, we take $varphiin C^infty_c(Omega)$. Since $nabla v(x)=0$ a.e., we have
$$
int_Omega varphi partial_i v = 0.
$$
On the other hand, we have
$$begin{align}
int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy
end{align}$$
since the distributional derivative of $v$ is the Hausdorff measure on $l$.
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
$endgroup$
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
|
show 1 more comment
$begingroup$
I am a bit confused by the question, please tell me if I misread anything.
Consider the domain $Omega=B_1subsetBbb R^2$, i.e. the unit ball. Define $v$ to be
$$
v(x)=begin{cases}1 &; x<0 \
0 &; xge 0.
end{cases}
$$
We can see that $v$ is smooth on $overlineOmegabackslash l$, where $l={(x,y)inBbb R^2 : x=0}$ which is a closed set of measure zero.
Next, we take $varphiin C^infty_c(Omega)$. Since $nabla v(x)=0$ a.e., we have
$$
int_Omega varphi partial_i v = 0.
$$
On the other hand, we have
$$begin{align}
int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy
end{align}$$
since the distributional derivative of $v$ is the Hausdorff measure on $l$.
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
$endgroup$
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
|
show 1 more comment
$begingroup$
I am a bit confused by the question, please tell me if I misread anything.
Consider the domain $Omega=B_1subsetBbb R^2$, i.e. the unit ball. Define $v$ to be
$$
v(x)=begin{cases}1 &; x<0 \
0 &; xge 0.
end{cases}
$$
We can see that $v$ is smooth on $overlineOmegabackslash l$, where $l={(x,y)inBbb R^2 : x=0}$ which is a closed set of measure zero.
Next, we take $varphiin C^infty_c(Omega)$. Since $nabla v(x)=0$ a.e., we have
$$
int_Omega varphi partial_i v = 0.
$$
On the other hand, we have
$$begin{align}
int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy
end{align}$$
since the distributional derivative of $v$ is the Hausdorff measure on $l$.
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
$endgroup$
I am a bit confused by the question, please tell me if I misread anything.
Consider the domain $Omega=B_1subsetBbb R^2$, i.e. the unit ball. Define $v$ to be
$$
v(x)=begin{cases}1 &; x<0 \
0 &; xge 0.
end{cases}
$$
We can see that $v$ is smooth on $overlineOmegabackslash l$, where $l={(x,y)inBbb R^2 : x=0}$ which is a closed set of measure zero.
Next, we take $varphiin C^infty_c(Omega)$. Since $nabla v(x)=0$ a.e., we have
$$
int_Omega varphi partial_i v = 0.
$$
On the other hand, we have
$$begin{align}
int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy
end{align}$$
since the distributional derivative of $v$ is the Hausdorff measure on $l$.
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
edited Jan 9 at 12:57
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
answered Jan 9 at 10:24
BigbearZzzBigbearZzz
8,58921652
8,58921652
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
|
show 1 more comment
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
Thanks. This is a nice example. I forgot to add the requirement that the singular function $v$ should be continuous on all $bar Omega$. Do you think that it is still possible to get a counter-example? (I will accept your answer, but I am waiting to see if someone will add a counter-example assuming $v$ is continuous everywhere, or prove that no such example exist).
$endgroup$
– Asaf Shachar
Jan 9 at 12:10
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
@AsafShachar The Cantor function is an example of a continuous function on $[0,1]$ that satisfies $f'(x)=0$ a.e. but has a (non-atomic) singular measure as its distributional derivative. This would be a counterexample you're looking for.
$endgroup$
– BigbearZzz
Jan 9 at 12:43
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
en.wikipedia.org/wiki/Cantor_function
$endgroup$
– BigbearZzz
Jan 9 at 12:44
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
Yes, this makes sense. Thank you. By the way, did you have a slick way to see that $int_Omega(partial_1varphi)v = int_{-1}^1 varphi(0,y), dy $ without computation? i.e. that "the distributional derivative of $v$ is the Hausdorff measure on $I$"? (By the way there is a small typo in your definition of $v$, it should be $x ge 0$). Thanks for your all help!
$endgroup$
– Asaf Shachar
Jan 9 at 12:57
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
$endgroup$
– BigbearZzz
Jan 9 at 13:02
$begingroup$
One way would be to observe that for a fixed $y$ we have $varphi(x,y) = int_{-infty}^x partial_x varphi(t,y) ,dt$. Since $v=1$ on the left half plane and $v=0$ otherwise, we have $varphi(0,y) = int_{-infty}^infty partial_x varphi(t,y) ,dt$. Then we use the fact that $varphi$ has compact support and Fubini's theorem.
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– BigbearZzz
Jan 9 at 13:02
|
show 1 more comment
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For a continuous counterexample, let's consider the case $Bbb R^n=Bbb R$ and $Omega = (0,1)$. Let $v$ be the Cantor function
It has the property that $v'(x)=0$ a.e., thus for any function $varphiin C^infty_c(0,1)$ we have
$$
int_0^1 varphi v' = 0.
$$
On the other hand, we have
$$begin{align}
int_0^1 varphi'v = -c_0int_{0}^1 varphi(t), dnu(t)
end{align}$$
where $nu=mathcal H^gammalvert_C$, the Hausdorff measure of dimension $gamma= ln 2 /ln 3$ restricted to the Cantor set $C$ (which is a compact set). Here $c_0$ is a normalizing constant that I don't remember.
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
$endgroup$
add a comment |
$begingroup$
For a continuous counterexample, let's consider the case $Bbb R^n=Bbb R$ and $Omega = (0,1)$. Let $v$ be the Cantor function
It has the property that $v'(x)=0$ a.e., thus for any function $varphiin C^infty_c(0,1)$ we have
$$
int_0^1 varphi v' = 0.
$$
On the other hand, we have
$$begin{align}
int_0^1 varphi'v = -c_0int_{0}^1 varphi(t), dnu(t)
end{align}$$
where $nu=mathcal H^gammalvert_C$, the Hausdorff measure of dimension $gamma= ln 2 /ln 3$ restricted to the Cantor set $C$ (which is a compact set). Here $c_0$ is a normalizing constant that I don't remember.
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
$endgroup$
add a comment |
$begingroup$
For a continuous counterexample, let's consider the case $Bbb R^n=Bbb R$ and $Omega = (0,1)$. Let $v$ be the Cantor function
It has the property that $v'(x)=0$ a.e., thus for any function $varphiin C^infty_c(0,1)$ we have
$$
int_0^1 varphi v' = 0.
$$
On the other hand, we have
$$begin{align}
int_0^1 varphi'v = -c_0int_{0}^1 varphi(t), dnu(t)
end{align}$$
where $nu=mathcal H^gammalvert_C$, the Hausdorff measure of dimension $gamma= ln 2 /ln 3$ restricted to the Cantor set $C$ (which is a compact set). Here $c_0$ is a normalizing constant that I don't remember.
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
$endgroup$
For a continuous counterexample, let's consider the case $Bbb R^n=Bbb R$ and $Omega = (0,1)$. Let $v$ be the Cantor function
It has the property that $v'(x)=0$ a.e., thus for any function $varphiin C^infty_c(0,1)$ we have
$$
int_0^1 varphi v' = 0.
$$
On the other hand, we have
$$begin{align}
int_0^1 varphi'v = -c_0int_{0}^1 varphi(t), dnu(t)
end{align}$$
where $nu=mathcal H^gammalvert_C$, the Hausdorff measure of dimension $gamma= ln 2 /ln 3$ restricted to the Cantor set $C$ (which is a compact set). Here $c_0$ is a normalizing constant that I don't remember.
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
answered Jan 9 at 13:22
BigbearZzzBigbearZzz
8,58921652
8,58921652
add a comment |
add a comment |
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I'm not sure if I understand the question correctly. For $n=2$, can't we just take the domain $Omega$ to be the unit disk and $v=1$ for $x>0$ and $v=0$ for $xle 0$ as a counterexample? Does this violate any of your assumptions?
$endgroup$
– BigbearZzz
Jan 9 at 10:05