I was trying to prove that if a holomorphic function $f$ (non constant) satisfies $$ f(x+y)=f(x)f(y)$$
for all $x, y, x+y$ are in a neighborhood of $0$, then $f(z) = \exp(bz)$ for some $b$.
My idea was to transform the product into Cauchy functional equation by using $\log$. This works if $f(0)\not=0$ : we could choose a smaller neighborhood of $0$ with isolated singularity theorem and prove that $\log(f)$ is linear. However, this doesn't work for the case $f(z)=0$.
Instead, I try to directly use the original equation by choosing a point $z_0$ s.t. $f(z_0) = \exp(bz_0)\not= 0$ for some $b$, and consider the sequence $z_n = 2^{-n}z_0$. I want to show that $\forall n,\ f(z_n) = \exp(bz_n)$, but I can't get rid of the root $f(z_n) = -\exp(bz_n)$ from $f(z^{n+1})^2 = f(z^n)$. Could someone give me a hint?
Best Answer
Suppose that the property $f(z+w)=f(z)f(w)$ holds for all $z,w\in D(0,r)$ for some $r>0$. Let's suppose that $f$ has a root $z_0$ in the complex plane; We will show that $f\equiv 0$ in the disk $D(0,r/2)$ and by the Identity principle we will have that $f=0$, a contradiction since $f$ is not constant.
First we will need the following: let $z\in D(0,r)$; then $f(z)=f(z/n)^n$ for all $n$. Indeed, since $kz/n\in D(0,r)$ for all $n\in\mathbb{N}$ and $k<n$ we can write $f(z)=f((n-1)z/n+z/n)=f((n-1)z/n)f(z/n)$ and use induction, since for $n=2$ we have $f(z)=f(z/2+z/2)=f(z/2)f(z/2)=f(z/2)^2$.
Now back to our task: let $z\in D(0,r/2)$; we can pick an integer $n$ large enough such that $z_0/n\in D(0,r/2)$. Then we can write $z=z_0/n +(z-z_0/n)$ and now $z_0/n\in D(0,r/2)\subset D(0,r)$ and $|z-z_0/n|\leq|z|+|z_0/n|<r/2+r/2=r$. Therefore, $f(z)=f(z_0/n+(z-z_0/n))=f(z_0/n)f(z-z_0/n)$. But it is $f(z_0)=f(z_0/n)^n$, therefore $f(z_0/n)^n=0$ which is of course true only if $f(z_0/n)=0$; hence $f(z)=0$.
This proves that $f$ has no zeros. Now you can apply your method with $\log$.