I noticed this post and this paper, which gives a version of Liouville's theorem for subharmonic functions and the reference of its proof, but I think there must be an easier proof for the following version of Liouville's theorem with a stronger condition.

A subharmonic function that is bounded above on the complex plane $\mathbb C$ must be constant

I think we may need to use the fact that the maximum of a subharmonic function cannot be achieved in the interior of its domain unless the function is constant(MVP). But how do we prove that a bounded-above subharmonic function on the complex plane $\mathbb C$ can achieve its maximum at a certain point of $\mathbb C$?(Maybe we don't need to use MVP for proof)

Thanks in advance!

  • $\begingroup$ Oh~~~really? Could you please show me a counterexample? $\endgroup$
    – No One
    Apr 15, 2016 at 14:46
  • $\begingroup$ Oh~~~I am confused, isn't my conjecture a special case of the theorem 8 of the following paper? dm.unibo.it/~arcozzi/subharmonic.pdf $\endgroup$
    – No One
    Apr 15, 2016 at 14:51
  • $\begingroup$ See my comment to your answer~~~So you believe that the theorem in paper I linked is not correct? $\endgroup$
    – No One
    Apr 15, 2016 at 15:03

2 Answers 2


If $v$ is subharmonic in the complex plane $\Bbb C$ then $$ \tag 1 v(z) \le \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} M(r_1, v) + \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, v) $$ for $0 < r_1 < |z| < r_2$, where $$ M(r, v) := \max \{ v(z) : |z| = r \} \quad . $$ That is the "Hadamard three-circle theorem" for subharmonic functions, and follows from the fact that the right-hand side of $(1)$ is a harmonic function which dominates $v$ on the boundary of the annulus $\{ z : r_1 < |z| < r_2 \}$ .

(Remark: It follows from $(1)$ that $M(r, v)$ is a convex function of $\log r$.)

Now assume that $v(z) \le K$ for all $z \in \Bbb C$. Then $M(r_2, v) \le K$, and $r_2 \to \infty$ in the inequality $(1)$ gives $$ \tag 2 v(z) \le M(r_1, v) $$ for $0 < r_1 < |z|$. It follows that $$ v(z) \le \limsup_{r_1 \to 0} M(r_1, v) = v(0) $$ because $v$ is upper semi-continuous. Thus $v$ has a maximum at $z=0$ and therefore is constant.

Remark: As noted in the comments, the condition “$v$ is bounded above” can be relaxed to $$\liminf_{r \to \infty} \frac{M(r, v)}{\log r} = 0 $$ which is still sufficient to conclude $(2)$ from $(1)$.

  • $\begingroup$ +1. The inequality given basically says the real value function $f(s)=M(e^s, v)$ concave up, and clearly $f(s)$ is increasing too. I think with these conditions and additional assumption $\displaystyle lim_{s\to \infty}f(s)/s=0$ we can prove f(s) is constant. That means in the original question the assumption can be relaxed to $\limsup_{z\to \infty} \frac{v(z)}{\log|z|}=0$. $\endgroup$
    – user175968
    Jun 1, 2016 at 7:00
  • 2
    $\begingroup$ @frank000: You are right, and even $\liminf$ should be sufficient. Thanks for the feedback. $\endgroup$
    – Martin R
    Jun 1, 2016 at 7:43
  • $\begingroup$ A related thread I just posted. $\endgroup$
    – user175968
    Jun 1, 2016 at 7:48

Yes it's true - sorry about my stupidity earlier.

I haven't looked at that paper - you can decide whether what's below is simpler.

Say $u$ is sh in the plane and bounded above. Define $$v(z)=\frac1{2\pi}\int_0^{2\pi}u(e^{it}z)\,dt.$$Then $v$ is a radial sh function, which is to say there exists $\phi$ with $v(z)=\phi(|z|)$. Since $u$ is sh, $\phi$ is non-decreasing, so if $\phi$ is non-constant there exist $r_1<r_2$ with $\phi(r_1)<\phi(r_2)$. Choose $a$ and $b$ with $$\phi(r_j)=a+b\log(r_j)\quad(j=1,2).$$Note that $b>0$.

Now $v$ sh shows that $$\phi(r)\ge a+b\log(r)\quad(r>r_2).$$ Because if $r>r_2$ but $\phi(r)<a+b\log(r)$ then the harmonic function that equals $\phi(|z|)$ on the boundary of $\{r_1<|z|<r\}$ would be smaller than $a+b\log(r_2)$ for $|z|=r_2$, contradicting the subhamonicity of $v$.

So $b>0$ shows that $u$ is not bounded above, contradiction. So $\phi$ is constant.

So the averages of $u$ on circles centered at the origin are constant. Hence the average over disks centered at the origin are constant: $$\frac{1}{\pi r^2}\int_{|z|<r}u(z)\,dxdy=c.$$

The same applies to the bounded-above subharmonic function $u(z-p)$ for every $p$: $$\frac{1}{\pi r^2}\int_{|z-p|<r}u(z)\,dxdy=c_p.$$

But if $p$ is fixed and $r\to\infty$ we have $$\frac{m(D(p,r)\cap D(0,r))}{m(D(0,r))}\to1$$(where $D(p,r)$ is a disk and $m$ is Lebesgue measure). So letting $r\to\infty$ shows that $c_p=c_0$ for every $p$, and hence $u$ is constant (since $u$ sh shows that the average of $u$ over $D(p,r)$ tends to $u(p)$ as $r\to0$.)

  • $\begingroup$ Thank you for the answer. But I can't see why $\frac{m(D(p,r)\cap D(0,r))}{m(D(0,r))}\to1$ implies the convergence of the integral. $u$ is just bounded above, not necessarily bounded in the absolute value. The difference of $c_p$ and $c$ is hard to estimate. $\endgroup$
    – No One
    Apr 19, 2016 at 22:05
  • 1
    $\begingroup$ @TiWen Hmm, I see your point. Ok, try this: Assume for typing convenience that $u\le0$ and $|p|=1$. Now $D(p,r)\subset D(0,r+1)$, so $$c_p\ge\frac1{\pi r^2}\int_{D(0,r+1)}u=\left(\frac{r+1}{r}\right)^2c.$$Letting $r\to\infty$ shows $c_p\ge c$. Similarly $c\ge c_p$. Believe that one? $\endgroup$ Apr 19, 2016 at 23:01
  • $\begingroup$ Why is $v$ necessarily subharmonic? $\endgroup$ May 11, 2017 at 5:00
  • $\begingroup$ @helloworld112358 Fubini... $\endgroup$ May 14, 2017 at 14:26

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