Koenigs function

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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,

${\displaystyle |f(z)|\leq M(r)|z|}$

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,

${\displaystyle h(f(z))=f^{\prime }(0)h(z)~.}$

The function h is the uniform limit on compacta of the normalized iterates, ${\displaystyle g_{n}(z)=\lambda ^{-n}f^{n}(z)}$.

Moreover, if f is univalent, so is h.^[1][2]

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.

• Uniqueness. If k is another solution then, by analyticity, it suffices to show that k = h near 0. Let

${\displaystyle H=k\circ h^{-1}(z)}$

near 0. Thus H(0) =0, H'(0)=1 and, for |z | small,

${\displaystyle \lambda H(z)=\lambda h(k^{-1}(z))=h(f(k^{-1}(z))=h(k^{-1}(\lambda z)=H(\lambda z)~.}$

Substituting into the power series for H, it follows that H(z) = z near 0. Hence h = k near 0.

• Existence. If ${\displaystyle F(z)=f(z)/\lambda z,}$ then by the Schwarz lemma

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

On the other hand,

${\displaystyle g_{n}(z)=z\prod _{j=0}^{n-1}F(f^{j}(z))~.}$

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

${\displaystyle \sum \sup _{|z|\leq r}|1-F\circ f^{j}(z)|\leq (1+|\lambda |^{-1})\sum M(r)^{j}<\infty .}$

• Univalence. By Hurwitz's theorem, since each gⁿ is univalent and normalized, i.e. fixes 0 and has derivative 1 there, their limit h is also univalent.

Let f_t (z) be a semigroup of holomorphic univalent mappings of D into itself fixing 0 defined for t ∈ [0, ∞) such that

• ${\displaystyle f_{s}}$ is not an automorphism for s > 0
• ${\displaystyle f_{s}(f_{t}(z))=f_{t+s}(z)}$
• ${\displaystyle f_{0}(z)=z}$
• ${\displaystyle f_{t}(z)}$ 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 = f₁, then h(f_s(z)) satisfies Schroeder's equation and hence is proportion to h.

Taking derivatives gives

${\displaystyle h(f_{s}(z))=f_{s}^{\prime }(0)h(z).}$

Hence h is the Koenigs function of f_s.