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,
 * $$ |f(z)|\le M(r) |z|$$

for |z | ≤ r with M(r ) < 1. Moreover $f$ '(0) = $λ$ with 0 < |$λ$| < 1.

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


 * $$|F(z) - 1|\le (1+|\lambda|^{-1})|z|~.$$


 * On the other hand,
 * $$ g_n(z) = z\prod_{j=0}^{n-1} F(f^j(z))~.$$


 * Hence gn converges uniformly for |z| ≤ r by the Weierstrass M-test since


 * $$ \sum \sup_{|z|\le r} |1 -F\circ f^j(z)| \le (1+|\lambda|^{-1}) \sum M(r)^j <\infty.$$


 * Univalence. By Hurwitz's theorem, since each gn 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 Re$μ$ < 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
 * $$\Re {zh^\prime(z)\over h(z)} \ge 0 ~.$$

Since the same result holds for the reciprocal,
 * $$ \Re {v(z)\over z}\le 0 ~,$$

so that $v(z)$ satisfies the conditions of
 * $$ v(z)= z p(z),\,\,\, \Re p(z) \le 0, \,\,\, p^\prime(0) < 0.$$

Conversely, reversing the above steps, any holomorphic vector field $v(z)$ satisfying these conditions is associated to a semigroup  $f_{t}$, with
 * $$ h(z)= z \exp \int_0^z {v^\prime(0) \over v(w)} -{1\over w} \, dw.$$