Harmonic functions on the plane - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:15:37Z http://mathoverflow.net/feeds/question/76470 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76470/harmonic-functions-on-the-plane Harmonic functions on the plane Alexander Isaev 2011-09-27T04:52:45Z 2011-10-01T16:37:35Z <p>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. </p> http://mathoverflow.net/questions/76470/harmonic-functions-on-the-plane/76933#76933 Answer by fedja for Harmonic functions on the plane fedja 2011-10-01T16:37:35Z 2011-10-01T16:37:35Z <p>The answer is, indeed, negative. WLOG $v(0)=0$. Take the intersection of the region <code>$-A&lt;v&lt;2A$</code> 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 $A>0$, $u(0)=+\infty$.</p>