Mixing
Liouville's theorem for harmonic functions
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I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.
The proof -
Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function
calculus integration multivariable-calculus harmonic-functions
asked Jan 5 at 15:45
ChikChakChikChak
757418
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add a comment |
$begingroup$
I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.
The proof -
Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function
calculus integration multivariable-calculus harmonic-functions
asked Jan 5 at 15:45
ChikChakChikChak
757418
$endgroup$
add a comment |
$begingroup$
I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.
The proof -
Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function
calculus integration multivariable-calculus harmonic-functions
asked Jan 5 at 15:45
ChikChakChikChak
757418
$endgroup$
I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.
The proof -
Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function
calculus integration multivariable-calculus harmonic-functions
calculus integration multivariable-calculus harmonic-functions
asked Jan 5 at 15:45
ChikChakChikChak
757418
asked Jan 5 at 15:45
ChikChakChikChak
757418
asked Jan 5 at 15:45
ChikChakChikChak
757418
asked Jan 5 at 15:45
ChikChakChikChak
757418
asked Jan 5 at 15:45
ChikChakChikChak
757418
757418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
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add a comment |
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The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
$endgroup$
add a comment |
$begingroup$
Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
$endgroup$
add a comment |
$begingroup$
Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
$endgroup$
Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.
The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
answered Jan 5 at 16:13
WojowuWojowu
17.4k22666
17.4k22666
add a comment |
add a comment |
$begingroup$
The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
add a comment |
$begingroup$
The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
add a comment |
$begingroup$
The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
answered Jan 5 at 16:03
José Carlos SantosJosé Carlos Santos
156k22125227
156k22125227
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
add a comment |
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
$begingroup$
So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
$endgroup$
– mouthetics
Jan 5 at 16:54
1
1
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
$begingroup$
@mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
$endgroup$
– Wojowu
Jan 5 at 18:12
add a comment |
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