# 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.

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## 1 Answer

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.

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Thank you! Both solutions were very helpful. – Connor May 14 '12 at 18:47
@froggie Hadamard's three circles theorem should be $\log m(a)$,$\log m(b)$ in your first solution,then when $r\to\infty$,the second term would not cancel. – Daniel S. May 4 '14 at 7:22
Moreover,three circles theorem requires analytic function,here $u$ is just subharmonic,not analytic. – Daniel S. May 4 '14 at 7:31
@DanielS.: There is a form of the three circles theorem for subharmonic functions, which is that the function $m(r)$ is a convex function of $\log r$. For a reference, see for instance Hormander's Notions of Convexity Corollary 3.2.22, or just google "three circles theorem subharmonic". To derive the statement you are used to seeing about holomorphic functions, all you have to do is use that for a holomorphic function $f$, we have $\log|f|$ is subharmonic. – froggie May 4 '14 at 8:00