Koenigs function

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 | : |f(z)|\le M(r) |z| for |z | ≤ r with M(r )

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 . 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 near 0. Hence 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 g**n 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

• 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 , 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 , 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 with Reμ : 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 :\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)

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.

Notes

References

• ASIN: B0006BTAC2

References

1. {{harvnb. Carleson. Gamelin. 1993
2. {{harvnb. Shapiro. 1993