Let $D = \{ z \in \mathbb{C} ~|~ \text{Re}(z) > 1\}$ and $f : D \to \mathbb{C}$ be a holomorphic function such that $f(2z) = f(z) + 1$ for every $z \in D$. I now have to prove that $f$ has a analytic continuation on the right half-plane.
We know that $g: A \to \mathbb{C} : z \mapsto f(z) + 1$ is a holomorphic function where $A = \{ z \in \mathbb{C} ~|~ \text{Re}(z) > 2\}$.
So $g(z) - 1 = f(z)$ for all $z \in A$ and it then follows that $f(z)$ is an analytic continuation of $g(z)$ on $D$. But this would mean that $f(z)$ has an analytical continuation on $B =\{ z \in \mathbb{C} ~|~ \text{Re}(z) > 1/2 \}$.
But now I don't know how I can continue (or if this reasoning is correct). Can I just assume that $f(2z) = f(z) + 1$ also holds on $B$ and use the same argument over and over?
There is also an additional question where I have to show that $f$ has no analytic continuation on a set that contains zero but I have no idea how I can prove that either.