2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Oh~~~really? Could you please show me a counterexample? $\endgroup$ – No One Apr 15 '16 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 '16 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 '16 at 15:03
3
$\begingroup$

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 $(1)$.

$\endgroup$
  • $\begingroup$ +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 '16 at 7:00
  • 1
    $\begingroup$ @frank000: You are right, and even $\liminf$ should be sufficient. Thanks for the feedback. $\endgroup$ – Martin R Jun 1 '16 at 7:43
  • $\begingroup$ A related thread I just posted. $\endgroup$ – user175968 Jun 1 '16 at 7:48
1
$\begingroup$

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 $r\to0$.)

$\endgroup$
  • $\begingroup$ 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 '16 at 22:05
  • $\begingroup$ @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? $\endgroup$ – David C. Ullrich Apr 19 '16 at 23:01
  • $\begingroup$ Why is $v$ necessarily subharmonic? $\endgroup$ – helloworld112358 May 11 '17 at 5:00
  • $\begingroup$ @helloworld112358 Fubini... $\endgroup$ – David C. Ullrich May 14 '17 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.