Let $f$ be holomorphic function defined on a domain which contains the closed unit disk $\overline {D(0,1)}.$ Suppose $f$ maps $\overline {D(0,1)}$ into open unit disk $D(0,1).$ Could anyone advise me on how to prove there exists exactly one $w \in D(0,1)$ such that $f(w)=w \ ?$

Hints will suffice, thank you.


Hint: Prove that $f|_{\overline{D(0,1)}}$ is a contraction, use Banach's fixed point theorem.

  • $\begingroup$ Thanks for the suggestion. Is it possible to address this problem with only ideas from Complex analysis? $\endgroup$ – Alexy Vincenzo Sep 7 '14 at 18:54
  • $\begingroup$ Yes. Hint: What can you say about $|f(w)/w|$ in the punctured unit disk? $\endgroup$ – Yiannis Galidakis Sep 7 '14 at 19:33
  • $\begingroup$ I am sorry, I do not get you. Do I use Rouché theorem to show $f(z)-z$ has exactly one zero ? $\endgroup$ – Alexy Vincenzo Sep 7 '14 at 19:55
  • 1
    $\begingroup$ $|f(z) -z -(-z)| =|f(z)| < 1, \forall z$ such that $|z|=1.$ So $f(z) -z$ has same number of zeros(counting multiplicity) as $-z$ on $D(0,1).$ $\endgroup$ – Alexy Vincenzo Sep 7 '14 at 20:09

Here I have a some what similar result on finite Blaschke products. Some times it may helps you.

$B(z)≠z$ Blaschke product can have at most one fixed point in $D$

Let $z_0$ be a fixed point of Blaschke product $B(z)$ in $D.$ Define $f:C \to C$ by $$f(z)=φ_{z_0 } (B(φ_{-z_0 } (z) ) ) ,∀z∈C$$ Where $φ_{z_0 } (z)=\dfrac{z-z_0}{1-¯z_0 z} ,∀z∈C$
Now $f(0)=0$
Note that $f:D \to D$ and analytic in $D.$
Suppose $B$ has another fixed point $z_1$ in $D$ and $φ_{-z_0 } (z_2 )=z_1.$
Then $φ_{z_0 } (z_1 )=z_2.$ Here $z_1,z_2∈D.$
$$f(z_2 )=φ_{z_0 } (B(φ_{-z_0 } (z_2 ) ) )$$ $$f(z_2 )=φ_{z_0 } (B(z_1 ) )$$ $$f(z_2 )=φ_{z_0 } (z_1 )$$ $$f(z_2 )=z_2$$ By the Schwarz lemma, $$f(z)=z ,∀z∈D$$ $$φ_{z_0 } (B(φ_{-z_0 } (z) ) )=z$$
$$B(φ_{-z_0 } (z) )=φ_{-z_0 } (z)$$
$$B(z)=z ,∀z∈D$$
Hence non trivial Blaschke product can have at most one fixed point in the open unit disk $D.$
Since $|B(z)|=1$ for $|z|=1,$ we know that $B(z) ¯B (1/¯z)=1.$
Thus if $z_0≠0$ is a fixed point of $B$ then $\dfrac{1}{ ¯z_o}$ is also a fixed point of $B.$
Therefore for each non zero fixed point of $B$ in $D,$ we have corresponding fixed point in $C-¯D={z∈C│|z|>1}.$
Since $B(z)≠z$ Blaschke product can have at most one fixed point in $D,$ we have $B(z)≠z$ Blaschke product can have at most two fixed points in $C-∂D.$

$B(0)=0$ Blaschke products have no other fixed points in $C-∂D.$

If this Blaschke product have other fixed point in $D,$ it has at least three fixed points in $C-∂D.$ This is a contradiction.


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