Given a non-zero holomorphic function $f$ fixing $0$ which isn't a Mobius transform, the Koenigs function of $f$, which we'll call $h$, is the function which linearizes $f$ in the sense that $$ h(f(z)) = f'(0)h(z). $$ I am interested in finding an expression or estimates for $h$ in the case where $f$ is a polynomial of low degree with real coefficients. (I'd be pretty happy with degree 3)

I have heard that there is a way of guessing the power series of $h$. Does anyone know of any references which explain how to do this?

  • Can't you just write down a power series for $h$ and equate coefficients, starting $h(z)=z+h''(0)z^2/2+\ldots$? – user25199 Feb 11 '14 at 18:19
  • @Carl I'm not sure what you mean, what am I equating the coefficients of the power series of $h$ with? – D. Kelleher Feb 11 '14 at 18:46
  • @Carl, I think I see what you mean now, I have been able to use it to get a recurrence relation for the coefficients. Thank you. – D. Kelleher Feb 12 '14 at 5:12
up vote 5 down vote accepted

You do not tell the crucial thing: how large is $|f'(0)|$. There is no simple expression for coefficients or any other simple expression for $h$, even when $f$ is quadratic polynomial $\lambda z+z^2$.

However the global behaviour of $h$ has been be studied a lot, with remarkable results.

In the following description, I assume for simlicity that $f$ is a polynomial.

If $|f'(0)|<1$, $h$ is analytic in a neighborhood of $0$. It has an analytic continuation to the immediate domain of attraction of $0$, the boundary of this domain is the natural boundary for $h$, and the boundary behaviour of $h$ is relatively well understood.

If $|f'(0)|$ is a root of unity, the equation has no solution, even as a formal series.

If $|f'(0)|=1$ but not a root of unity, a formal power series $h$ exists. Let $f'(0)=e^{2\pi i\alpha}$, then everything depends on the fine Diophantine properties of $\alpha$. For quadratic polynomial $f$, there is a necessary and sufficient condition in terms of $\alpha$ for $h$ to be analytic in a neighborhood of $0$ (Yoccoz got a Fileds medal for this). When analytic, its maximal domain of analyticity is the so-called Siegel disc, and $h$ maps this Siegel disc onto a round disc bijectively.

When $|f'(0)|>1$, $h$ exists but its analytic continuation is not single valued. However the inverse of $h$ has a nice property: it is an entire function of finite order. This result is due to Poincare and Valiron.

For the introduction to the properties of this function see P. Fatou, Sur les equations fonctionnelles, in 3 parts, available on and G. Valiron, Fonctions analytiques, Paris, Press Universitaires, 1954. For the case $|f'(0)|=1$, see the work of Yoccoz.

It follows from this brief description that there can be no simple expression for $h$ (whatever polynomial $f$ of degree at least $2$ you use). It is easy to obtain a recurrence for coefficients, but it is very complicated, and can hardly be used to study $h$. One exception is the work of Siegel on the case $|f'(0)|=1$. he was able to show that the radius of convergence of the power series is positive for certain $\alpha$ by a kind of direct analysis of this recurrence.

  • Thanks for the answer, this is very interesting and very useful. For the family of polynomials I'm working with, $0<f'(0)<1$, so they are relatively well behaved. – D. Kelleher Feb 12 '14 at 5:32

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.