A Short Note on Expanding Functions as Outer Infinite Compositions

A Short Note on Expanding Functions as Outer Infinite Compositions John Gill

April 2015

Suppose a complex function can be expanded as an infinite composition in the following way: (1) F ( z ) = lim ( f1
f2

fn ( z )) . n→∞

(See [1])

And suppose the inverses of the f k ( z ) exist: gk ( z ) = f k−1 ( z ) . Then, under suitable restrictions (2) G( z ) = F −1 ( z ) = lim ( gn
gn−1
 g1 ( z )) , outer or left compositions. n→∞

Conditions for convergence of (1) include the following variation on theorems in [2] Theorem 1 (Gill 2015) Suppose functions {f n (z)} are analytic in S0 = ( z < R 0 ) and ∞

f n (z) − z ≤ Cρn , 0 < σ = C∑ ρk < R 0 < ∞ in S0 . Then there exists R such that k =1

z < R < R + σ = R < R 0 and lim ( f1
f 2

f n (z) ) = F(z) for z < R . Convergence is uniform *

n →∞

on compact subsets. Sketch of Proof: Set Fn ,n+m ( z ) = fn
fn+1

fn+m ( z ) . Then it is easily determined that Fn ,n+m ( z ) < R + σ = R * . Now write fn ( z ) = z +ν n ( z ) , ν n ( z ) < C ρn . Set r = R0 − R * and use the C Cauchy Integral Formula to estimate the magnitude of ν n′( z ) : ν n′( z ) ≤ ρn . Therefore r C C   fn ( z ) − fn (ζ ) ≤ z − ζ + ν n ( z ) −ν n (ζ ) < z − ζ + ρn z − ζ = z − ζ  1 + ρn  by the r r   n C   Fundamental Theorem. Then F1,n ( z ) − F1,n (ζ ) < z − ζ ⋅ ∏  1 + ρ k  and r  k =1  n C   F1,n ( z ) − F1,n+m ( z ) = F1,n ( z ) − F1,n ( fn+1

fn+m ( z )) < z − fn+1

fn+m ( z ) ⋅ ∏  1 + ρ k  r  k =1  m −1

∞

k =1

k =1

Where fn+1

fn+m ( z ) − z ≤ ∑ fn+ k (Fn+ k +1,n+m ( z )) − Fn+ k +1,n+ m ( z ) + fn+ m ( z ) − z < ∑ C ρn+ k

Thus

n ∞ C   F1,n ( z ) − F1,n+m ( z ) < C ⋅ ∑ ρn+ k ⋅ ∏  1 + ρ k  → 0 as n → ∞ . r  k =1 k =1 

||

We have, in an inexact sense,

f n (z) − z ≤ Cρn ⇒ f n (z) = z + ν n (z) , ν n (z) ≤ Cρn , and z = g n (f n (z)) = g n (z + ν n (z)) = g n (ω) = ω − ν n (z) ⇒ g n (ω) − ω < Cρn And the following

Theorem 3 (Gill, 2011) Let {g n } be a sequence of complex functions defined on S=(|z| 0 . Then, for every z ∈ S0 = ( z < R 0 ) , 1

G n (z) = g n
g n −1

g1 (z) → G(z) , uniformly on compact subsets of S0

The following are simple topographical images. Example 1 :

F(z)

fn ( z ) =

z 1 + ρn z

, gn ( z ) = fn−1 ( z ) =

z 1 , ρn = 2 1 − ρn z n

G(z)

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