A problem on fixed point 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. 
 A: Hint: Prove that $f|_{\overline{D(0,1)}}$ is a contraction, use Banach's fixed point theorem.
A: Here I have a some what similar result on finite Blaschke products. Some times it may helps you.
Theorem
$B(z)≠z$ Blaschke product can have at most one fixed point in $D$  
Proof
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 $C-∂D.$  
Corollary
$B(0)=0$ Blaschke products have no other fixed points in $C-∂D.$  
Proof
If this Blaschke product have other fixed point in $D,$ it has at least three fixed points in $C-∂D.$ This is a contradiction.
