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.

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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: if 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.

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 Schroeder'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. [1][2]

As a consequence, when f (and hence h) are univalent, D ca 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

\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.
|F(z) - 1|\le (1%2B|\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%2B|\lambda|^{-1}) \sum M(r)^j <\infty.

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\in [0,\infty) such that

Each f_s with s > 0 has the same Koenigs 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

h(f_s(z)) =f_s^\prime(0) h(z).

Hence h is the Koenigs function of fs.

Structure of univalent semigroups

On the domain U = h(D), the maps fs 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^\mu=\lambda 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 Berkson & Porta (1978)

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 ft, with

h(z)= z \exp \int_0^z {v^\prime(0) \over v(w)} -{1\over w} \, dw.

Notes

  1. ^ Carleson & Gamelin 1993, p. 28-32
  2. ^ Shapiro 1993, p. 90-93

References