Hadamard three-circle theorem

Remarks

Proof idea: choose a real $\alpha$ such that $r_1^{\alpha} M(r_1) = r_2^{\alpha} M(r_2)$. For rational $\alpha = p/q$ with $q > 0$, the maximum modulus principle for $z^p f(z)^q$ on the annulus gives $r^{\alpha} M(r) \le r_1^{\alpha} M(r_1)$. A limit over rational $\alpha$, or the subharmonicity of $\log \abs{f(z)} + \alpha \log \abs{z}$, handles irrational $\alpha$, and the bound rearranges to the inequality. Example: for $f(z) = z^n$, $\log M(r) = n \log r$ is linear in $\log r$, so equality holds. If $M(r) = 0$ for some $r$, then $f$ vanishes on a circle, so $f \equiv 0$ on the annulus and the inequality is trivial.

Going deeper

For a student

The three-circle theorem is a quantitative form of the maximum modulus principle. It is the circle version of the Hadamard three-lines theorem; the map $z \mapsto e^{z}$ carries a vertical strip onto an annulus. It is used in the theory of entire functions to study the growth of $M(r)$, and it extends to subharmonic functions.