Limit conditions of a subharmonic function imply that it is constant Let $u$ be a subharmonic function on $\mathbb{C}$. Suppose that $$\limsup_{z\to \infty} \frac{u(z)}{\log|z|}=0$$
I'm trying to prove that this implies $u(z)$ is constant. I have a feeling that it may have to do with Hadamard's Three Circles Theorem and/or the maximum principle for sub/superharmonic functions, but I'm not getting anywhere.
 A: Answer using Three Circles and Maximum principle:
For any $r>0$, let $m(r) = \sup_{|z| = r} u(z)$. Fix $0<a<b<r$. The three circles theorem says that  $$m(b)\leq \frac{\log r - \log b}{\log r - \log a}m(a) + \frac{\log b - \log a}{\log r - \log a}m(r).$$ Taking the $\limsup$ as $r\to \infty$ of both sides of this inequality, I get that $m(b)\leq m(a)$, since by your assumption $\limsup_{r\to\infty} m(r)/\log(r) = 0$. By the maximum principle, we conclude that $m(a) = m(b)$, and that $u$ is constant on $\{|z|\leq b\}$. But $a$ and $b$ were arbitrary, so $u$ is constant on $\mathbb{C}$.
Answer using Jensen's formula:
Jensen's formula for subharmonic functions says the following.

Let $u\colon \mathbb{C}\to \mathbb{R}\cup\{-\infty\}$ be a subharmonic function such that $u(0) \neq -\infty$. Let $\mu = \Delta u$ be the Laplacian of $u$ (so it is a positive Radon measure on $\mathbb{C}$). Then $$\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta = u(0) + \frac{1}{2\pi}\int_0^r\frac{\mu(\mathbb{D}_t)}{t}\,dt,$$ where $\mathbb{D}_t$ is the open disk of radius $t$ around the origin.

Let $u$ be a subharmonic function satisfying your $\limsup$ condition. Up to translating it, you may assume $u(0) \neq -\infty$. Unless $u$ is harmonic, the measure $\mu = \Delta u$ is nonzero, and hence there is some radius $R>0$ for which $\mu(\mathbb{D}_R)>0$. It follows from Jensen's formula that for $r>R$ $$\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta =u(0) + \frac{1}{2\pi}\int_0^r\frac{\mu(\mathbb{D}_t)}{t}\,dt\geq u(0) + \frac{\mu(\mathbb{D}_R)}{2\pi}\int_R^r\frac{dt}{t}$$$$ = u(0) + \frac{\mu(\mathbb{D}_R)}{2\pi}\log(r/R)\geq \mathrm{const}\cdot\log r.$$ On the other hand, because of your assumption on $u$, for any $M>0$ one has $$\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta = \frac{1}{2\pi}\int_0^{2\pi}\log(r)\frac{u(re^{i\theta})}{\log r}\,d\theta\leq M\log r$$ when $r\gg 1$. These two inequalities give a contradiction. It follows that $u$ must be a harmonic function. Dealing with this case should be easier.
