Riemann mapping from disk $mathbb{D}=B(0,1)$ to $mathbb{D} cup B(2-epsilon,1)$
$begingroup$
If $epsilon>0$ and $f_{epsilon}$ is the Riemann mapping from $mathbb{D}$ (unit disk) to the union of $mathbb{D}$ and the unit disk centered at $2-epsilon$, made unique by specifying $f_{epsilon}(0)=0$ and $f_{epsilon}'(0) >0$, then is it true that $f_{epsilon} to $ identity as $epsilon to 0$ in some sense of "convergence"?
Intuitively, I want to show that $f_{epsilon}$ is the identity on most of $mathbb{D}$ and "blows up" at $1$. In particular this would show that the image of $lim f_{epsilon}$ is not the limit if the image of $f_{epsilon}$.
I am not sure how to proceed. Any help or pointers to relevant theorems would be appreciated.
complex-analysis conformal-geometry
asked Jan 7 at 1:39
DwaggDwagg
312111
$endgroup$
add a comment |
$begingroup$
If $epsilon>0$ and $f_{epsilon}$ is the Riemann mapping from $mathbb{D}$ (unit disk) to the union of $mathbb{D}$ and the unit disk centered at $2-epsilon$, made unique by specifying $f_{epsilon}(0)=0$ and $f_{epsilon}'(0) >0$, then is it true that $f_{epsilon} to $ identity as $epsilon to 0$ in some sense of "convergence"?
Intuitively, I want to show that $f_{epsilon}$ is the identity on most of $mathbb{D}$ and "blows up" at $1$. In particular this would show that the image of $lim f_{epsilon}$ is not the limit if the image of $f_{epsilon}$.
I am not sure how to proceed. Any help or pointers to relevant theorems would be appreciated.
complex-analysis conformal-geometry
asked Jan 7 at 1:39
DwaggDwagg
312111
$endgroup$
1
$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32
add a comment |
$begingroup$
If $epsilon>0$ and $f_{epsilon}$ is the Riemann mapping from $mathbb{D}$ (unit disk) to the union of $mathbb{D}$ and the unit disk centered at $2-epsilon$, made unique by specifying $f_{epsilon}(0)=0$ and $f_{epsilon}'(0) >0$, then is it true that $f_{epsilon} to $ identity as $epsilon to 0$ in some sense of "convergence"?
Intuitively, I want to show that $f_{epsilon}$ is the identity on most of $mathbb{D}$ and "blows up" at $1$. In particular this would show that the image of $lim f_{epsilon}$ is not the limit if the image of $f_{epsilon}$.
I am not sure how to proceed. Any help or pointers to relevant theorems would be appreciated.
complex-analysis conformal-geometry
asked Jan 7 at 1:39
DwaggDwagg
312111
$endgroup$
If $epsilon>0$ and $f_{epsilon}$ is the Riemann mapping from $mathbb{D}$ (unit disk) to the union of $mathbb{D}$ and the unit disk centered at $2-epsilon$, made unique by specifying $f_{epsilon}(0)=0$ and $f_{epsilon}'(0) >0$, then is it true that $f_{epsilon} to $ identity as $epsilon to 0$ in some sense of "convergence"?
Intuitively, I want to show that $f_{epsilon}$ is the identity on most of $mathbb{D}$ and "blows up" at $1$. In particular this would show that the image of $lim f_{epsilon}$ is not the limit if the image of $f_{epsilon}$.
I am not sure how to proceed. Any help or pointers to relevant theorems would be appreciated.
complex-analysis conformal-geometry
complex-analysis conformal-geometry
asked Jan 7 at 1:39
DwaggDwagg
312111
asked Jan 7 at 1:39
DwaggDwagg
312111
edited Jan 7 at 1:46
Dwagg
asked Jan 7 at 1:39
DwaggDwagg
312111
asked Jan 7 at 1:39
DwaggDwagg
312111
asked Jan 7 at 1:39
DwaggDwagg
312111
312111
1
$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32
add a comment |
1
$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32
1
1
$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32
add a comment |
1 Answer
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I was about to say "Montel" when I saw the comment above...
Normalize $f_epsilon$ so $f_epsilon(0)=0$, $f_epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_epsilon'(0)>1$.
Now say $f$ is the limit of $f_epsilon$ (or some subsequence). Since $f(Bbb D)$ is open and connected it's clear that $f(Bbb D)subset Bbb D$. And $f(0)=0$, $f'(0)ge1$ now shows that $f(z)=z$.
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
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$begingroup$
I was about to say "Montel" when I saw the comment above...
Normalize $f_epsilon$ so $f_epsilon(0)=0$, $f_epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_epsilon'(0)>1$.
Now say $f$ is the limit of $f_epsilon$ (or some subsequence). Since $f(Bbb D)$ is open and connected it's clear that $f(Bbb D)subset Bbb D$. And $f(0)=0$, $f'(0)ge1$ now shows that $f(z)=z$.
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
$begingroup$
I was about to say "Montel" when I saw the comment above...
Normalize $f_epsilon$ so $f_epsilon(0)=0$, $f_epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_epsilon'(0)>1$.
Now say $f$ is the limit of $f_epsilon$ (or some subsequence). Since $f(Bbb D)$ is open and connected it's clear that $f(Bbb D)subset Bbb D$. And $f(0)=0$, $f'(0)ge1$ now shows that $f(z)=z$.
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
add a comment |
$begingroup$
I was about to say "Montel" when I saw the comment above...
Normalize $f_epsilon$ so $f_epsilon(0)=0$, $f_epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_epsilon'(0)>1$.
Now say $f$ is the limit of $f_epsilon$ (or some subsequence). Since $f(Bbb D)$ is open and connected it's clear that $f(Bbb D)subset Bbb D$. And $f(0)=0$, $f'(0)ge1$ now shows that $f(z)=z$.
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
$endgroup$
I was about to say "Montel" when I saw the comment above...
Normalize $f_epsilon$ so $f_epsilon(0)=0$, $f_epsilon'(0)>0$. "Subordination" (ie the Schwarz lemma applied to a certain composition) shows that in fact $f_epsilon'(0)>1$.
Now say $f$ is the limit of $f_epsilon$ (or some subsequence). Since $f(Bbb D)$ is open and connected it's clear that $f(Bbb D)subset Bbb D$. And $f(0)=0$, $f'(0)ge1$ now shows that $f(z)=z$.
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
answered Jan 7 at 13:54
David C. UllrichDavid C. Ullrich
59.3k43893
59.3k43893
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$begingroup$
Use the fact that ${f_{epsilon}}$ is a normal family.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 5:43
$begingroup$
@KaviRamaMurthy Ok, so since ${f_{epsilon}}$ is a normal family, it has a convergent subsequence $f_nto f$. And we are to conclude that this subsequence converges to the identity?
$endgroup$
– Dwagg
Jan 7 at 13:32