# Koenigs function

In mathematics, the Koenigs function is a function arising in complex analysis and dynamical systems. Introduced in 1884 by the French mathematician Gabriel Koenigs, it gives a canonical representation as dilations of a univalent holomorphic mapping, or a semigroup of mappings, of the unit disk in the complex numbers into itself.

## Existence and uniqueness of Koenigs function

Let D be the unit disk in the complex numbers. Let f be a holomorphic function mapping D into itself, fixing the point 0, with f not identically 0 and f not an automorphism of D, i.e. a Möbius transformation defined by a matrix in SU(1,1).
By the Denjoy-Wolff theorem, f leaves invariant each disk |z | < r and the iterates of f converge uniformly on compacta to 0: in fact for 0 < r < 1,
    $|f(z)|\leq M(r)|z|$
for |z | ≤ r with M(r ) < 1. Moreover f '(0) = λ with 0 < |λ| < 1.
Koenigs (1884) proved that there is a unique holomorphic function h defined on D, called the Koenigs function,
such that h(0) = 0, h '(0) = 1 and Schröder's equation is satisfied,
    $h(f(z))=f^{\prime }(0)h(z)~.$
The function h is the uniform limit on compacta of the normalized iterates, $g_{n}(z)=\lambda ^{-n}f^{n}(z)$.
Moreover, if f is univalent, so is h.
As a consequence, when f (and hence h) are univalent, D can be identified with the open domain U = h(D). Under this conformal identification, the mapping f becomes multiplication by λ, a dilation on U.

### Proof

* Uniqueness. If k is another solution then, by analyticity, it suffices to show that k = h near 0. Let
       $H=k\circ h^{-1}(z)$
    near 0. Thus H(0) =0, H'(0)=1 and, for |z | small,
       $\lambda H(z)=\lambda h(k^{-1}(z))=h(f(k^{-1}(z))=h(k^{-1}(\lambda z)=H(\lambda z)~.$
    Substituting into the power series for H, it follows that H(z) = z near 0. Hence h = k near 0.
* Existence. If $F(z)=f(z)/\lambda z,$ then by the Schwarz lemma
       $|F(z)-1|\leq (1+|\lambda |^{-1})|z|~.$
    On the other hand,
       $g_{n}(z)=z\prod _{j=0}^{n-1}F(f^{j}(z))~.$
    Hence g$_n$ converges uniformly for |z| ≤ r by the Weierstrass M-test since
       $\sum \sup _{|z|\leq r}|1-F\circ f^{j}(z)|\leq (1+|\lambda |^{-1})\sum M(r)^{j}<\infty .$
* Univalence. By Hurwitz's theorem, since each g$^n$ is univalent and normalized, i.e. fixes 0 and has derivative 1 there , their limit h is also univalent.

## Koenigs function of a semigroup

Let f$_t$ (z) be a semigroup of holomorphic univalent mappings of D into itself fixing 0 defined for t ∈ [0, ∞) such that
* $f_{s}$ is not an automorphism for s > 0
* $f_{s}(f_{t}(z))=f_{t+s}(z)$
* $f_{0}(z)=z$
* $f_{t}(z)$ is jointly continuous in t and z
Each f$_s$ with s > 0 has the same Koenigs function, cf. iterated function. In fact, if h is the Koenigs function of f = f$_1$, then h(f$_s$(z)) satisfies Schroeder's equation and hence is proportion to h.
Taking derivatives gives
    $h(f_{s}(z))=f_{s}^{\prime }(0)h(z).$
Hence h is the Koenigs function of f$_s$.

## Structure of univalent semigroups

On the domain U = h(D), the maps f$_s$ become multiplication by $\lambda (s)=f_{s}^{\prime }(0)$, a continuous semigroup. So $\lambda (s)=e^{\mu s}$ where μ is a uniquely determined solution of e $^μ$ = λ with Reμ < 0. It follows that the semigroup is differentiable at 0. Let
    $v(z)=\partial _{t}f_{t}(z)|_{t=0},$
a holomorphic function on D with v(0) = 0 and v'(0) = μ.
Then
    $\partial _{t}(f_{t}(z))h^{\prime }(f_{t}(z))=\mu e^{\mu t}h(z)=\mu h(f_{t}(z)),$
so that
    $v=v^{\prime }(0){h \over h^{\prime }}$
and
    $\partial _{t}f_{t}(z)=v(f_{t}(z)),\,\,\,f_{t}(z)=0~,$
the flow equation for a vector field.
Restricting to the case with 0 < λ < 1, the h(D) must be starlike so that
    $\Re {zh^{\prime }(z) \over h(z)}\geq 0~.$
Since the same result holds for the reciprocal,
    $\Re {v(z) \over z}\leq 0~,$
so that v(z) satisfies the conditions of Berkson & Porta (1978)
    $v(z)=zp(z),\,\,\,\Re p(z)\leq 0,\,\,\,p^{\prime }(0)<0.$
Conversely, reversing the above steps, any holomorphic vector field v(z) satisfying these conditions is associated to a semigroup f$_t$, with
    $h(z)=z\exp \int _{0}^{z}{v^{\prime }(0) \over v(w)}-{1 \over w}\,dw.$