MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question regarding harmonic maps from all of ${\Bbb R}^2$ into a domain in ${\Bbb R}^2$. Before stating my question in full generality, let me ask a special case of the question first. Is it possible to find two non-constant harmonic functions $u$ and $v$ on ${\Bbb R}^2$ such that $u>v^3$ at every point? My guess is that the answer is negative.

share|cite|improve this question
Some motivation to this question would be welcome. – Marcin Kotowski Oct 1 '11 at 17:53
Actually, I am interested in showing that the tube domain in ${\Bbb C}^2$ with base given by $y>x^3$ is Kobayashi-hyperbolic. If the question that I have asked had a positive answer, it would mean that the domain is Brody-hyperbolic, which is a somewhat weaker statement. – Alexander Isaev Oct 2 '11 at 6:17
up vote 7 down vote accepted

The answer is, indeed, negative. WLOG $v(0)=0$. Take the intersection of the region $-A<v<2A$ with a huge disk (more precisely, take the connected component $\Omega$ of this intersection containing the origin). It is simply connected by the maximum principle. The nice thing about the plane is that once we have a curve on the boundary that passes not too far from the origin, we can make the harmonic measure of the circle piece of the boundary arbitrarily small by choosing big enough radius. Now, the harmonic measures of the pieces $v=-A$ and $v=2A$ are balanced essentially as $2:1$, so the piece $v=2A$ has harmonic measure about $1/3$. Also $u$ is at least $-A^3$ everywhere in $\Omega$ and at least $8A^3$ on the piece $v=2A$. So, $u(0)\ge -\frac 23 A^3+\frac 13 8A^3=2A^3$. Since it is true for all $A>0$, $u(0)=+\infty$.

share|cite|improve this answer
Thank you very much for your solution. Your argument proves that there exists no non-constant harmonic map from the plane into the domain in the plane lying above the cubic parabola $y>x^3$. Consider now the region defined as follows: $y>tx^3$ for $x\ge 0$ and $y>sx^3$ for $x<0$, where $s,t>0$. If one applies your argument to this more general domain, it seems that the answer depends on $s$ and $t$. However, I believe that there is no non-constant map into this domain for any $s$ and $t$. Can your proof be adopted to the case of general $s$ and $t$? Alex Isaev. – Alexander Isaev Oct 2 '11 at 6:13
In fact, I think I know how to adapt your proof to this more general situation. One has to consider the region where $-kA<v<nA$ for suitable $k$, $n$. – Alexander Isaev Oct 2 '11 at 8:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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