The integral of the Poisson kernel is a constant function in $zeta$?

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For the ball of radius 1, $B$, in $mathbb R$, the Poisson kernel takes the form

$$P(x,zeta) = fracxomega _n-1$$

where $xin B$, $zetain mathbb S$ (the surface of $B$), and $omega _n-1$ is the surface area of the unit n−1-sphere.

I would like to know why, for $x=|x|omegain [0,1[times ,mathbb S$, $,$ the function
$$zeta mapsto int_mathbb S P(|x|w,zeta) , dsigma(omega)$$
is constant ?

complex-analysis fourier-analysis harmonic-analysis harmonic-functions

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edited Aug 6 at 18:14

user357151

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asked Aug 6 at 11:16

Z. Alfata

878413

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up vote
0
down vote

favorite

For the ball of radius 1, $B$, in $mathbb R$, the Poisson kernel takes the form

$$P(x,zeta) = fracxomega _n-1$$

where $xin B$, $zetain mathbb S$ (the surface of $B$), and $omega _n-1$ is the surface area of the unit n−1-sphere.

I would like to know why, for $x=|x|omegain [0,1[times ,mathbb S$, $,$ the function
$$zeta mapsto int_mathbb S P(|x|w,zeta) , dsigma(omega)$$
is constant ?

complex-analysis fourier-analysis harmonic-analysis harmonic-functions

share|cite|improve this question

edited Aug 6 at 18:14

user357151

13.9k31140

asked Aug 6 at 11:16

Z. Alfata

878413

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

For the ball of radius 1, $B$, in $mathbb R$, the Poisson kernel takes the form

$$P(x,zeta) = fracxomega _n-1$$

where $xin B$, $zetain mathbb S$ (the surface of $B$), and $omega _n-1$ is the surface area of the unit n−1-sphere.

I would like to know why, for $x=|x|omegain [0,1[times ,mathbb S$, $,$ the function
$$zeta mapsto int_mathbb S P(|x|w,zeta) , dsigma(omega)$$
is constant ?

complex-analysis fourier-analysis harmonic-analysis harmonic-functions

share|cite|improve this question

edited Aug 6 at 18:14

user357151

13.9k31140

asked Aug 6 at 11:16

Z. Alfata

878413

For the ball of radius 1, $B$, in $mathbb R$, the Poisson kernel takes the form

$$P(x,zeta) = fracxomega _n-1$$

where $xin B$, $zetain mathbb S$ (the surface of $B$), and $omega _n-1$ is the surface area of the unit n−1-sphere.

I would like to know why, for $x=|x|omegain [0,1[times ,mathbb S$, $,$ the function
$$zeta mapsto int_mathbb S P(|x|w,zeta) , dsigma(omega)$$
is constant ?

complex-analysis fourier-analysis harmonic-analysis harmonic-functions

share|cite|improve this question

edited Aug 6 at 18:14

user357151

13.9k31140

asked Aug 6 at 11:16

Z. Alfata

878413

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 18:14

user357151

13.9k31140

edited Aug 6 at 18:14

user357151

13.9k31140

edited Aug 6 at 18:14

user357151

13.9k31140

13.9k31140

asked Aug 6 at 11:16

Z. Alfata

878413

asked Aug 6 at 11:16

Z. Alfata

878413

asked Aug 6 at 11:16

Z. Alfata

878413

878413

add a comment | 

add a comment | 

1 Answer
1

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1
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accepted

If $xi$ is another point of $mathbbS$, there exists a matrix $Min SO(n)$ mapping $zeta$ to $xi$. Note that
$$
P(x, zeta) = P(Mx, Mzeta)
$$
because $|Mx|=|x|$ and $|Mx-Mzeta| = |x-zeta|$. So,

$$
int_mathbb S P(|x|omega,zeta) , dsigma(omega)
= int_mathbb S P(|x|M omega , xi) , dsigma(omega)
= int_mathbb S P(|x|omega', xi) , dsigma(omega')
$$
by the orthogonal change of variable $omega'=Momega$ (the Jacobian determinant of which is $1$.)

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

If $xi$ is another point of $mathbbS$, there exists a matrix $Min SO(n)$ mapping $zeta$ to $xi$. Note that
$$
P(x, zeta) = P(Mx, Mzeta)
$$
because $|Mx|=|x|$ and $|Mx-Mzeta| = |x-zeta|$. So,

$$
int_mathbb S P(|x|omega,zeta) , dsigma(omega)
= int_mathbb S P(|x|M omega , xi) , dsigma(omega)
= int_mathbb S P(|x|omega', xi) , dsigma(omega')
$$
by the orthogonal change of variable $omega'=Momega$ (the Jacobian determinant of which is $1$.)

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

add a comment | 

up vote
1
down vote



accepted

If $xi$ is another point of $mathbbS$, there exists a matrix $Min SO(n)$ mapping $zeta$ to $xi$. Note that
$$
P(x, zeta) = P(Mx, Mzeta)
$$
because $|Mx|=|x|$ and $|Mx-Mzeta| = |x-zeta|$. So,

$$
int_mathbb S P(|x|omega,zeta) , dsigma(omega)
= int_mathbb S P(|x|M omega , xi) , dsigma(omega)
= int_mathbb S P(|x|omega', xi) , dsigma(omega')
$$
by the orthogonal change of variable $omega'=Momega$ (the Jacobian determinant of which is $1$.)

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

If $xi$ is another point of $mathbbS$, there exists a matrix $Min SO(n)$ mapping $zeta$ to $xi$. Note that
$$
P(x, zeta) = P(Mx, Mzeta)
$$
because $|Mx|=|x|$ and $|Mx-Mzeta| = |x-zeta|$. So,

$$
int_mathbb S P(|x|omega,zeta) , dsigma(omega)
= int_mathbb S P(|x|M omega , xi) , dsigma(omega)
= int_mathbb S P(|x|omega', xi) , dsigma(omega')
$$
by the orthogonal change of variable $omega'=Momega$ (the Jacobian determinant of which is $1$.)

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

If $xi$ is another point of $mathbbS$, there exists a matrix $Min SO(n)$ mapping $zeta$ to $xi$. Note that
$$
P(x, zeta) = P(Mx, Mzeta)
$$
because $|Mx|=|x|$ and $|Mx-Mzeta| = |x-zeta|$. So,

$$
int_mathbb S P(|x|omega,zeta) , dsigma(omega)
= int_mathbb S P(|x|M omega , xi) , dsigma(omega)
= int_mathbb S P(|x|omega', xi) , dsigma(omega')
$$
by the orthogonal change of variable $omega'=Momega$ (the Jacobian determinant of which is $1$.)

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 18:14

user357151

13.9k31140

answered Aug 6 at 18:14

user357151

13.9k31140

answered Aug 6 at 18:14

user357151

13.9k31140

13.9k31140

add a comment | 

add a comment | 

 

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