Bound on gradient positive harmonic function
My question is essentially the same as An inequality concerning an harmonic function , however I did not find the answer given satisfactory. To restate it, I would like to solve the following:
Let $h$ be a positive harmonic function on $K(0, rho) = { z in mathbb{C} vert vert z vert < rho }$. Show that
$$
vert nabla h(0) vert leq frac{2}{rho} h(0)
$$
and deduce that
$$
vert nabla h(z) vert leq frac{2rho}{rho^2- vert z vert^2} h(z) qquad quad , vert z vert < rho
$$
Showing that $
vert nabla h(0) vert leq frac{2}{rho} h(0)
$ was rather easy using Harnacks inequality, my issue comes when trying to prove the last bit. Supposedly we should be able to define an automorphism $varphi: K(0,rho) to K(0,rho)$ such that $varphi(0)= z$ and hence use the first result on $h circ varphi$, since if $varphi$ is holomorphic $hcirc varphi$ is again positive harmonic. Now my problem is that I fail to see why doing so wouldn't just imply that
$$
vert nabla h(z) vert leq frac{2}{rho} h(z)
$$
which is not quite what we want.
complex-analysis inequality harmonic-functions
asked yesterday
MikkelD
1
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My question is essentially the same as An inequality concerning an harmonic function , however I did not find the answer given satisfactory. To restate it, I would like to solve the following:
Let $h$ be a positive harmonic function on $K(0, rho) = { z in mathbb{C} vert vert z vert < rho }$. Show that
$$
vert nabla h(0) vert leq frac{2}{rho} h(0)
$$
and deduce that
$$
vert nabla h(z) vert leq frac{2rho}{rho^2- vert z vert^2} h(z) qquad quad , vert z vert < rho
$$
Showing that $
vert nabla h(0) vert leq frac{2}{rho} h(0)
$ was rather easy using Harnacks inequality, my issue comes when trying to prove the last bit. Supposedly we should be able to define an automorphism $varphi: K(0,rho) to K(0,rho)$ such that $varphi(0)= z$ and hence use the first result on $h circ varphi$, since if $varphi$ is holomorphic $hcirc varphi$ is again positive harmonic. Now my problem is that I fail to see why doing so wouldn't just imply that
$$
vert nabla h(z) vert leq frac{2}{rho} h(z)
$$
which is not quite what we want.
complex-analysis inequality harmonic-functions
asked yesterday
MikkelD
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday
add a comment |
My question is essentially the same as An inequality concerning an harmonic function , however I did not find the answer given satisfactory. To restate it, I would like to solve the following:
Let $h$ be a positive harmonic function on $K(0, rho) = { z in mathbb{C} vert vert z vert < rho }$. Show that
$$
vert nabla h(0) vert leq frac{2}{rho} h(0)
$$
and deduce that
$$
vert nabla h(z) vert leq frac{2rho}{rho^2- vert z vert^2} h(z) qquad quad , vert z vert < rho
$$
Showing that $
vert nabla h(0) vert leq frac{2}{rho} h(0)
$ was rather easy using Harnacks inequality, my issue comes when trying to prove the last bit. Supposedly we should be able to define an automorphism $varphi: K(0,rho) to K(0,rho)$ such that $varphi(0)= z$ and hence use the first result on $h circ varphi$, since if $varphi$ is holomorphic $hcirc varphi$ is again positive harmonic. Now my problem is that I fail to see why doing so wouldn't just imply that
$$
vert nabla h(z) vert leq frac{2}{rho} h(z)
$$
which is not quite what we want.
complex-analysis inequality harmonic-functions
asked yesterday
MikkelD
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My question is essentially the same as An inequality concerning an harmonic function , however I did not find the answer given satisfactory. To restate it, I would like to solve the following:
Let $h$ be a positive harmonic function on $K(0, rho) = { z in mathbb{C} vert vert z vert < rho }$. Show that
$$
vert nabla h(0) vert leq frac{2}{rho} h(0)
$$
and deduce that
$$
vert nabla h(z) vert leq frac{2rho}{rho^2- vert z vert^2} h(z) qquad quad , vert z vert < rho
$$
Showing that $
vert nabla h(0) vert leq frac{2}{rho} h(0)
$ was rather easy using Harnacks inequality, my issue comes when trying to prove the last bit. Supposedly we should be able to define an automorphism $varphi: K(0,rho) to K(0,rho)$ such that $varphi(0)= z$ and hence use the first result on $h circ varphi$, since if $varphi$ is holomorphic $hcirc varphi$ is again positive harmonic. Now my problem is that I fail to see why doing so wouldn't just imply that
$$
vert nabla h(z) vert leq frac{2}{rho} h(z)
$$
which is not quite what we want.
complex-analysis inequality harmonic-functions
complex-analysis inequality harmonic-functions
asked yesterday
MikkelD
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
MikkelD
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
MikkelD
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
MikkelD
1
asked yesterday
MikkelD
1
1
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MikkelD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday
add a comment |
You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday
You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday
add a comment |
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MikkelD is a new contributor. Be nice, and check out our Code of Conduct.
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You forgot to apply the chainrule when computing the gradient
– TheOscillator
yesterday
Of course! Doing so gave me the result that I wanted. I don't know how I missed that, but thanks a lot :-)
– MikkelD
yesterday