Extremal distance
$begingroup$
I'm reading Conformally Invariant Processes in the Plane by Lawner and I have a doubt, he takes $R_L=(0,L)times(0,ipi) subset mathbb C$ and define $partial_1=[0,ipi]$, $partial_2 = [L,L+ipi]$, then define $f(z)=2 min {mathbb P_z(B_tau in partial_1),P_z(B_tau in partial_2)}$, where $B$ is a Brownian motion starting at $z$, $tau=inf{tge 0colon B(t)in partial (R_L)}$, and
$$
Omega(R_L,partial_1,partial_2)= sup{f(z) colon z in R_L }.
$$
Then he takes a Jordan domain D (a.k.a. a domain bounded with boundary a jordan curve) and 4 points in the boundary ordered clockwise ($z_1,z_2,z_3,z_4$).
If $A_1,A_2$ are the arcs between $z_1,z_2$ and $z_3,z_4$ then he define $Omega(D,A_1,A_2)$ in the same way as above.
At least he define the $pi$-extremal distance $L(D; A_1,A_2)$ as the only $L$ such that $Omega(R_L,partial_1,partial_2)=Omega(D,A_1,A_2)$.
I think this $L$ is the only number in $mathbb R$ such that exists a conformal map $fcolon D to R_L$ such that $f(z_1)=ipi$, $f(z_2)=0$, $f(z_3)=L$, $f(z_4)=L+ipi$.
My question is: How can i prove that this $L$ exists and how can i prove that such $f$ exist too?
complex-analysis probability-theory conformal-geometry
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
$endgroup$
add a comment |
$begingroup$
I'm reading Conformally Invariant Processes in the Plane by Lawner and I have a doubt, he takes $R_L=(0,L)times(0,ipi) subset mathbb C$ and define $partial_1=[0,ipi]$, $partial_2 = [L,L+ipi]$, then define $f(z)=2 min {mathbb P_z(B_tau in partial_1),P_z(B_tau in partial_2)}$, where $B$ is a Brownian motion starting at $z$, $tau=inf{tge 0colon B(t)in partial (R_L)}$, and
$$
Omega(R_L,partial_1,partial_2)= sup{f(z) colon z in R_L }.
$$
Then he takes a Jordan domain D (a.k.a. a domain bounded with boundary a jordan curve) and 4 points in the boundary ordered clockwise ($z_1,z_2,z_3,z_4$).
If $A_1,A_2$ are the arcs between $z_1,z_2$ and $z_3,z_4$ then he define $Omega(D,A_1,A_2)$ in the same way as above.
At least he define the $pi$-extremal distance $L(D; A_1,A_2)$ as the only $L$ such that $Omega(R_L,partial_1,partial_2)=Omega(D,A_1,A_2)$.
I think this $L$ is the only number in $mathbb R$ such that exists a conformal map $fcolon D to R_L$ such that $f(z_1)=ipi$, $f(z_2)=0$, $f(z_3)=L$, $f(z_4)=L+ipi$.
My question is: How can i prove that this $L$ exists and how can i prove that such $f$ exist too?
complex-analysis probability-theory conformal-geometry
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
$endgroup$
add a comment |
$begingroup$
I'm reading Conformally Invariant Processes in the Plane by Lawner and I have a doubt, he takes $R_L=(0,L)times(0,ipi) subset mathbb C$ and define $partial_1=[0,ipi]$, $partial_2 = [L,L+ipi]$, then define $f(z)=2 min {mathbb P_z(B_tau in partial_1),P_z(B_tau in partial_2)}$, where $B$ is a Brownian motion starting at $z$, $tau=inf{tge 0colon B(t)in partial (R_L)}$, and
$$
Omega(R_L,partial_1,partial_2)= sup{f(z) colon z in R_L }.
$$
Then he takes a Jordan domain D (a.k.a. a domain bounded with boundary a jordan curve) and 4 points in the boundary ordered clockwise ($z_1,z_2,z_3,z_4$).
If $A_1,A_2$ are the arcs between $z_1,z_2$ and $z_3,z_4$ then he define $Omega(D,A_1,A_2)$ in the same way as above.
At least he define the $pi$-extremal distance $L(D; A_1,A_2)$ as the only $L$ such that $Omega(R_L,partial_1,partial_2)=Omega(D,A_1,A_2)$.
I think this $L$ is the only number in $mathbb R$ such that exists a conformal map $fcolon D to R_L$ such that $f(z_1)=ipi$, $f(z_2)=0$, $f(z_3)=L$, $f(z_4)=L+ipi$.
My question is: How can i prove that this $L$ exists and how can i prove that such $f$ exist too?
complex-analysis probability-theory conformal-geometry
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
$endgroup$
I'm reading Conformally Invariant Processes in the Plane by Lawner and I have a doubt, he takes $R_L=(0,L)times(0,ipi) subset mathbb C$ and define $partial_1=[0,ipi]$, $partial_2 = [L,L+ipi]$, then define $f(z)=2 min {mathbb P_z(B_tau in partial_1),P_z(B_tau in partial_2)}$, where $B$ is a Brownian motion starting at $z$, $tau=inf{tge 0colon B(t)in partial (R_L)}$, and
$$
Omega(R_L,partial_1,partial_2)= sup{f(z) colon z in R_L }.
$$
Then he takes a Jordan domain D (a.k.a. a domain bounded with boundary a jordan curve) and 4 points in the boundary ordered clockwise ($z_1,z_2,z_3,z_4$).
If $A_1,A_2$ are the arcs between $z_1,z_2$ and $z_3,z_4$ then he define $Omega(D,A_1,A_2)$ in the same way as above.
At least he define the $pi$-extremal distance $L(D; A_1,A_2)$ as the only $L$ such that $Omega(R_L,partial_1,partial_2)=Omega(D,A_1,A_2)$.
I think this $L$ is the only number in $mathbb R$ such that exists a conformal map $fcolon D to R_L$ such that $f(z_1)=ipi$, $f(z_2)=0$, $f(z_3)=L$, $f(z_4)=L+ipi$.
My question is: How can i prove that this $L$ exists and how can i prove that such $f$ exist too?
complex-analysis probability-theory conformal-geometry
complex-analysis probability-theory conformal-geometry
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
asked Dec 23 '18 at 14:29
Claudio DelfinoClaudio Delfino
43
43
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