I have the following question: Let $f$ be an analytic function satisfying the functional equation: $f(z)=u(z)f(az)$ where $a$ is a real constant. Let $g$ be another function satisfying the same functional equation. In this case I asking if $f=g$. Thank you in advance.
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In the solution by Robert Israel, the uniqueness question (which was asked) is not addressed. Of course one cannot conclude that $f=g$, because if $f$ satisfies the equation, and $p$ is any function such that $p(z)=p(az)$ then $fp$ satisfies the equation. And this is the general description of all solutions because if $f$ and $g$ satisfy the equation, then their ratio has the property that $p(z)=p(az)$. 


A necessary condition for a nonzero solution is $u(z) u(az) = 1$. If that is true and $u$ is entire, it has an entire square root. Now $\sqrt{u(z)} \sqrt{u(az)} = u(a/2) = \pm 1$: if it is $1$, then $f(z) = \sqrt{u(z)}$ is a solution. If it is $1$, then $f(z) = (z  a/2) \sqrt{u(z)}$ is a solution. 

