I am pretty stuck on a homework problem on harmonic functions, or rather subharmonic functions (which for us are allowed to take the value $-\infty$). The statement is as follows:
Supper $u$ is subharmonic on $\mathbb{C}$ and twice-differentiable. Let $M(r)$ be the mean value of $u$ on the circle of radius $r$. Then
$$\lim_{r \to \infty} \frac{M(r)}{\log(r)}$$ exists, although is possibly infinite. We know the basic theorems like max mod, Poisson Kernel, Harnack, Lindelof etc. but this theorem seems to defy me haha
Can somebody give me an idea of maybe just the first step? Thanks a lot!
The main idea is that the circular mean of a subharmonic function is a convex function of the radius: that is, $M(r) = \phi(\log r)$ where $\phi$ is convex. Indeed, the inequality $\Delta u\ge 0$, written in polar coordinates, implies $$ M''(r)+\frac{1}{r}M'(r)\ge 0 $$ Writing $t=\log r$, we have $$ \phi''(t) = M(e^t)'' =(e^t M'(e^t))' = e^{2t}M''(e^t) + e^t M'(e^t) = r^2 M''(r)+rM'(r) \ge 0 $$ proving the convexity of $\phi$.
The existence of the limit $\lim_{t\to\infty}\phi(t)/t$ for a convex function $\phi$ can be shown as follows: $(\phi(t)-\phi(0))/t$ is increasing with $t$.