Riemann mapping from disk $mathbb{D}=B(0,1)$ to $mathbb{D} cup B(2-epsilon,1)$












0












$begingroup$


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







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  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 1:46







Dwagg

















asked Jan 7 at 1:39









DwaggDwagg

309111




309111








  • 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




    $begingroup$
    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










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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 $f(z)=z$.






share|cite|improve this answer









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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 $f(z)=z$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $f(z)=z$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $f(z)=z$.






        share|cite|improve this answer









        $endgroup$



        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 $f(z)=z$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 13:54









        David C. UllrichDavid C. Ullrich

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