Complex analysis/holomorphic/conformal map

[henry1964]

D is be a bounded domain in the complex plane. Suppose f : D -->D is a holomorphic automorphism (conformal bijection). Now define f_n(z) = f(f(f(f ..(z) (composed n times ).
Trying (and failing) to show:

(i) the sequence {f_n} has a subsequence that converges either to a constant
or to an automorphism of D

also

(ii) If the whole sequence {f_n} converges to g, then f(z)= z identically. $f(z)\equiv z

 

[Eynstone]

Welcome Henry.
I'll give a rough outline of a proof .
1. Let z0 be in D & f be non-constant. Since D is bounded, so is {f_n(z0)}. Choose a convergent subsequence of this , say {f_nk(z0)},k=1,2,...with f_nk(z0) -> w .
2. Choose zk ->z0 with f_nk(zk)=w(this is possible as the inverse of f is continuous & f_nk is obtained by finite iteration).
3. g_k(z) = f_nk(z)- w vanishes at zk & is bounded at z0. Choose a small neighbourhood of z0 in which a subsequence g_ki converges uniformly (to g(z),say) .
4. g is continuous. Since all f_nki are analytic, g can be continued analytically onto D.
5. Let g(z1) =g(z2) = s with distinct z1,z2. Let N be such that g_N(z1) is very close to s. Choose disjoint disks B1,B2 containing z1,z2 making g_N(B1) & g_N(B2) disjoing.Unless g = s identically, this is a contradiction.Thus, f_nk(z)-> w+s =constant.
6. If the range of g(z) +w didn't contain a disk in D, some f_nk would not be surjective.
Thus, g is surjective & by (5.) bijective.
For part (ii), note that f(g(z)) =g(z). If g is bijective, we are through.I
f not, f is constant, contrary to the very first supposition.
I hope this wasn't too hazy. If D is compact or a closed disk, the argument could be simplified ; f will have a fixed point by Brower's theorem.

 

[esisk]

Thank you Eynstone.
I am sure it all is there, but I am just trying to digest it in my own pace now.