# Prove that $g(z)$ is identically zero

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.

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 $$g$$.