Hadamard three-circle theorem for subharmonic functions

Remarks

The maximum $M(r)$ exists because an upper semicontinuous function attains its maximum on a compact circle. Proof idea: choose real $a, b$ such that $h(z) = a + b \log\abs{z}$ equals $M(r_1)$ on $\abs{z} = r_1$ and $M(r_2)$ on $\abs{z} = r_2$. The function $h$ is harmonic on $\C \setminus \set{0}$, so $u - h$ is subharmonic on the open annulus, and its upper limits at the two boundary circles are at most $0$. The maximum principle for subharmonic functions gives $u \le h$ on $A$, so $M(r) \le a + b \log r$, which is the stated inequality. Applying the same bound to every sub-annulus gives convexity in $\log r$.

Example: for $u(z) = \log\abs{z}$, $M(r) = \log r$ is linear in $\log r$ and equality holds. For a holomorphic $f$, the function $\log\abs{f}$ is subharmonic, and the theorem becomes the classical Hadamard three-circle theorem for $\log \max_{\abs{z}=r} \abs{f(z)}$.

Going deeper

For a student

The proof is a direct comparison with the harmonic functions $a + b\log\abs{z}$, which are the radial harmonic functions in the plane. The same argument in $\R^n$ with $n \ge 3$ shows that the maximum over spheres is convex as a function of $r^{2-n}$. The result is a standard tool for growth estimates of holomorphic and subharmonic functions, together with the Phragmén–Lindelöf principle.