A simple bound for the Poisson kernel for the upper half plane.
$begingroup$
Let $P_z(t)=frac{1}{pi}Im(frac{1}{t-z})=frac{1}{pi} frac{y}{(x-t)^2+y^2}$
where $0<y=Im(z)$ and $x=Re(z)$
$P$ is the Poisson kernel for the upper half plane.
If $t in Bbb{R}$ how can we derive that $P_z(t) leq frac{c_z}{1+t^2}$ where $c_z$ is a constant depending on $z$?
Thank you in advance for your help.
real-analysis complex-analysis complex-numbers
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
$begingroup$
Let $P_z(t)=frac{1}{pi}Im(frac{1}{t-z})=frac{1}{pi} frac{y}{(x-t)^2+y^2}$
where $0<y=Im(z)$ and $x=Re(z)$
$P$ is the Poisson kernel for the upper half plane.
If $t in Bbb{R}$ how can we derive that $P_z(t) leq frac{c_z}{1+t^2}$ where $c_z$ is a constant depending on $z$?
Thank you in advance for your help.
real-analysis complex-analysis complex-numbers
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
$begingroup$
Let $P_z(t)=frac{1}{pi}Im(frac{1}{t-z})=frac{1}{pi} frac{y}{(x-t)^2+y^2}$
where $0<y=Im(z)$ and $x=Re(z)$
$P$ is the Poisson kernel for the upper half plane.
If $t in Bbb{R}$ how can we derive that $P_z(t) leq frac{c_z}{1+t^2}$ where $c_z$ is a constant depending on $z$?
Thank you in advance for your help.
real-analysis complex-analysis complex-numbers
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
Let $P_z(t)=frac{1}{pi}Im(frac{1}{t-z})=frac{1}{pi} frac{y}{(x-t)^2+y^2}$
where $0<y=Im(z)$ and $x=Re(z)$
$P$ is the Poisson kernel for the upper half plane.
If $t in Bbb{R}$ how can we derive that $P_z(t) leq frac{c_z}{1+t^2}$ where $c_z$ is a constant depending on $z$?
Thank you in advance for your help.
real-analysis complex-analysis complex-numbers
real-analysis complex-analysis complex-numbers
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
asked Jan 9 at 11:36
Marios GretsasMarios Gretsas
8,48011437
8,48011437
add a comment |
add a comment |
1 Answer
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$begingroup$
Certainly. Since your bound can depend on $z$ all that you are asking is if $frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is bounded in $t$. This is a continuous function which tends to $1$ as $|t| to infty$ so it is bounded.
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
Certainly. Since your bound can depend on $z$ all that you are asking is if $frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is bounded in $t$. This is a continuous function which tends to $1$ as $|t| to infty$ so it is bounded.
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
Certainly. Since your bound can depend on $z$ all that you are asking is if $frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is bounded in $t$. This is a continuous function which tends to $1$ as $|t| to infty$ so it is bounded.
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
Certainly. Since your bound can depend on $z$ all that you are asking is if $frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is bounded in $t$. This is a continuous function which tends to $1$ as $|t| to infty$ so it is bounded.
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
Certainly. Since your bound can depend on $z$ all that you are asking is if $frac {1+t^{2}} {(x-t)^{2}+y^{2}}$ is bounded in $t$. This is a continuous function which tends to $1$ as $|t| to infty$ so it is bounded.
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 9 at 11:54
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
add a comment |
add a comment |
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