Let $g: \mathbb{C}\to\mathbb{C}$ be a complex analytic function so that $g(z^2) = g(z)+g(z-1)$ for all $z\in \mathbb{C}$. Prove that g is identically zero.

Plugging in $z=0$ shows that $g(-1) = 0$. Plugging in $z=1$ shows that $g(0) = 0$. And plugging in $z=-1$ shows that $g(1) = g(-2)$. Suppose one can show that there exists some $M > 0$ and $d > 0$ so that $|g(z)| \leq M | z|^d$ for all z. I think this implies that g must be a polynomial, but I'm not sure why. I know that if f is a polynomial, we can write $f(x)$ as $f(x) = \sum_{i=0}^n f^{(i)}(a)/i! (x-a)^i$ for any real number $a$. If g is a polynomial, then we see that in order for the leading coefficients of both sides to be equal, g must be a constant polynomial. Then it immediately follows that g is identically zero.


1 Answer 1


Denote by $D(R)$ the disk centered in zero with radius $R$. Let $M(R)$ be the maximum modulus of $g$ on the closed disk $D(R)$.

Let $K$ be $M(10)$.

Then for all $w\in D(81)$ we find a $z$ in $D(9)$ with $w=z^2$. Because of $z,z-1\in D(10)$ we obtain $|g(w)|\le |g(z)|+|g(z-1)|\le 2M(10)=2K$, so $M(81)\le 2M(10)$. With the same argument, $M(81^2-1)\le2M(81)\le 4M(10)=4K$. And we repeat the argument. Inductively, one can see that the sequence defined recursively by $a_0=10$, $a_{n+1}=a_n^2-1$ satisfies $a_n\ge 10\cdot 3^n$.

We obtain $M(10\cdot 3^n)\le M(a_n)\le K\cdot 2^n$. This shows that the function $h(z)=g(z)/z$, which is also entire, no pole in zero, is bounded. For instance, for a $z$ in the region $3^n\le |z|\le 3^{n+1}$, $n\ge 1$, we have: $$ g(z)\le M(3^{n+1})\le K\cdot 2^{n+1}\le 2K\cdot 3^n\le 2K\;|z|\ . $$ So $h(z)=g(z)/z$ is bounded for $z\ge 3$ by $2K$, and we adjust this upper bound to $\max(2K, M(3))$ to also cover $D(3)$.

It is thus a constant. But only the constant zero for $h$ matches the given functional equation of $g$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .