# Are entire functions “essentially” determined by their maximum modulus function?

(Note: This has been asked on Math SE, but without an answer after almost two years and one offered bounty.)

For an entire function $$f$$ let $$M(r,f)=\max_{|z|=r}|f(z)|$$ be its maximum modulus function. $$M(r, f)$$ does not change

• if $$f$$ is replaced by $$e^{i\varphi}f(e^{i\theta}z)$$,
• if $$f$$ is replaced by by $$\overline{f(\overline z)}$$,
• or combinations thereof.

My question is:

What can be said about entire functions $$f$$ and $$g$$ which satisfy $$M(r, f) = M(r, g)$$ for all $$r > 0$$?

Are $$f$$ and $$g$$ necessarily related by the above listed transformations?

Some thoughts:

• If $$f(0) = 0$$ with multiplicity $$k$$ at the origin then $$M(r, f) \sim r^k$$ for $$r \to 0$$, so that $$g(0) = 0$$ with the same multiplicity. We can divide both functions by $$z^k$$ and therefore assume that $$|f(0)| = |g(0)| \ne 0$$. After multiplying both functions with suitable factors $$e^{i\varphi}$$ we can assume that $$f(0) = g(0) = 1$$.
• If $$f(z) = 1 + a_m z^m + \cdots$$ at the origin with $$a_m \ne 0$$ then $$M(r, f) \sim 1 + |a_m| r^m$$ for $$r \to 0$$, so $$g(z) = 1 + b_m z^m + \cdots$$ with $$|b_m| = |a_m|$$. After suitable rotations $$z \to e^{i\theta}z$$ in the argument we can assume that $$a_m = b_m > 0$$.

After this normalization the question is whether necessarily $$g = f$$ or $$g(z) = \overline{f(\overline z)}$$, or if no such conclusion can be drawn.