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Is there any simple (general) way to find the complex solution of a functional-equation?

For example, it is given $ f(z) = f\left(\frac{z^2+4}{3}\right) $, and the question is to find all the functions $f(z)$ witch satisfy the given condition.

In an another topic I have seen a interesting solution, where another equation ($f(2z) = f^2 (z) $) was solved by differentiating, defining a new function, and using the identity theorem. This solution is elegant, but I think that it doesn't help in the case above, and in many other cases. So, what is "the usual way" (if it exists) of solving this?

$$ f(z) = f\left(\frac{z^2+4}{3}\right) $$

I am happy to assume $f$ is an entire function, so that you can differentiate freely.

Mohsen Shahriari
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Icarus 369
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    If you differentiate $f$ you are assuming that $f$ is analytic. This is a very strong condition that you have not included in your problem. – ajotatxe Jul 12 '16 at 17:30
  • Of course, this condition isn't given in my problem (but I think that it was in the similar topic here), so I probably would lose some of the solutions by taking that $ f $ is a holomorphic function. (I tried to derive looking for some solutions, I understand that it is not necessary the whole solutions set) – Icarus 369 Jul 12 '16 at 17:44
  • A typical functional equation of this sort will have no nonconstant analytic solutions, so it makes a huge difference whether you require that. – Eric Wofsey Jul 12 '16 at 18:13
  • Assume that $f$ is an entire function (holomorphic over the whole complex plane), so the differentiation is possible and allowed. – Icarus 369 Jul 12 '16 at 18:24
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    https://math.stackexchange.com/questions/4904571/entire-function-f-such-that-fz-fpz-where-p-is-a-polynomial-of-degre – Riemann Apr 24 '24 at 11:43

2 Answers2

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Let me assume you are interested in a functional equation of the form $f(z)=f(g(z))$ where $g(z)$ is some entire function and you want solutions $f(z)$ which are entire. In this case, for most choices of $g(z)$, there will be no nonconstant solutions. Indeed, suppose that there is some $a\in\mathbb{C}$ such that $g(a)=a$. If $|g'(a)|<1$, then if you start with a point $z_0$ near $a$ and iterate $g$, you will get closer and closer to $a$. Since $f$ must take the same value at all these points, we find that $f$ takes the same value at a set of points that accumulates at $a$, so by the identity theorem $f$ must be constant. On the other hand, if $|g'(a)|>1$, then $g$ has a holomorphic inverse $g^{-1}$ near $a$ such that $f(z)=f(g^{-1}(z))$ for all $z$ near $a$ and $|(g^{-1})'(a)|<1$, and we can apply the argument above with $g^{-1}$ to again find that $f$ must be constant.

So if $g$ has a fixed point $a$ at which $|g'(a)|\neq 1$, then any entire solution to $f(z)=f(g(z))$ must be constant. In particular, for instance, it is easy to verify that this condition holds for $g(z)=\frac{z^2+4}{3}$. If $g$ has a fixed point $a$ with $|g'(a)|=1$, I'm not sure whether you can always find a grand orbit of $g$ that accumulates at $a$; probably this is known and someone who knows complex dynamics better than me could answer that question.

Eric Wofsey
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  • Thank you, your answer was very helpful - most of the equations that I had to solve are of the type $ f(z) = f(g(z)) $. I still have only one question, what to do if there is a situation $ f(h(z)) \pm f(z)= g(z) $? For example, if we have $f(z) - f(z^2) = z$, what is the usual aproach? I tried something (again with derivinig), but my logic looks circular, so I would be grateful for an advice. – Icarus 369 Jul 13 '16 at 07:28
  • If $h(0)=0$, one technique will probably often be useful is looking at the Taylor series of $f$. You can write each side of the equation as a power series, and then equate them term by term, which will give you a bunch of equations relating the Taylor coefficients of $f$. If it is particularly simple to write down the Taylor series of $g(z)$ and $h(z)$ (as in your example), these equations will be easy to understand. – Eric Wofsey Jul 13 '16 at 07:35
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    In an answer to the related https://math.stackexchange.com/q/4904571/96384, one (simple) example as asked for in your last paragraph came up: For $g(z) = \zeta_n \cdot z$ with $\zeta_n$ an $n$-th root of unity, e.g. any $f(z) = z^m$ with $k|m$ is a non-constant entire solution to $f(z) = f(g(z))$. These $g$ have fixed points but only finite orbits; if we replace $\zeta_n$ with an element of the unit circle of infinite order, again $f$ is necessarily constant. – Torsten Schoeneberg Apr 24 '24 at 19:54
  • See added answer for a generalization of my previous comment. – Torsten Schoeneberg May 13 '24 at 04:35
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Inspired by Eric Wofsey's answer, I'd like to add that one can completely answer the question which entire functions satisfy $f(z) = f(P(z))$ for any complex polynomial $P$. In most cases, only constant functions satisfy this, but there are two interesting exceptions.

In fact, unless $P$ is linear $P(z) = az +b$ with $|a|=1$, $f$ must be constant. That's trivial for constant $P$, and both for $\deg(P) \ge 2$ and for $P(z) =az+b$ with $|a| \neq 1$ the methods in the answers to Entire function $f$ such that $f(z)=f(P(z))$, where $P$ is a polynomial of degree at least $2$. show what we want. In fact, the argument with the maximum modulus principle can be applied far more generally to any function $h(z)$ instead of $P(z)$ for which there exists some domain $D$ with boundary $\partial D$ such that either $h$ maps some point of $\partial D$ inside $D$, or every point of $\partial D$ has a preimage under $h$ inside $D$. For the mentioned polynomials, it is easy to find disks which satisfy one or the other of these.

So now assume $P(z) =az+b$ with $|a|=1$.

In case $a=1$, we are asking about periodic entire functions $f(z) = f(z+b)$ with period $b\neq 0$ (otherwise the condition is empty). These are well-known to be in one-to-one correspondence with analytic functions $g$ on the punctured plane $\mathbb C \setminus \{0\}$, via defining $g(w) = f\left(\dfrac{b}{2\pi i}\log(w)\right)$; that is, with $g(w) = \sum_{n=-\infty}^\infty a_nw^n$, $f$ can be written as a Laurent series in $e^{\frac{2\pi i}{b}}$, $$f(z)=\sum_{n=-\infty}^\infty a_n e^{\frac{2\pi i}{b} n}$$ Compare e.g. Show this formula holds for an analytic, periodic function in each half plane or Periodic holomorphic function on the right half-plane or Entire "periodic" function and links from there about more statements to be made once one imposes growth conditions.

In case $a \neq 1$, but there is a (say minimal) $k$ with $a^k=1$, i.e. $a$ is a primitive $k$-th root of unity, the following interesting thing happens: $P(z)=az+b$ has the fixed point $p:= \frac{b}{1-a}$, and a quick calculation shows $$(P(z)-p) = a(z-p)$$ So if we develop $f$ into its Taylor Series around $p$, $f(z)=\sum_{n\ge 0} a_n(z-p)^n$, the condition $f(P(z))=f(z)$ becomes, via the identity principle for power series, $$a_n=0 \text{ for all } n \not \equiv 0 \pmod k $$ Or in words, the power series around $p$ for $f(z)$ contains only terms where the exponent is a multiple of $k$. E.g. for $p=0$, $k=2$ these are the "even" functions $f(z)= f(-z)$.

Finally, in case $|a|=1\neq a$ and $a$ is not a unit root (i.e. of infinite multiplicative order), again $f$ must be constant, as can be seen by the exact argument as before (fixed point $p$, identity of power series) but now noting that $a^n \neq 1$ for all $n$ so all $a_n$ except $a_0$ have to be $0$.

Torsten Schoeneberg
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