6

4

Let $u$ be a nonconstant real-valued harmonic function defined in the open unit disk $D$. Suppose that $\Gamma\subset D$ is a smooth connected curve such that $u=0$ on $\Gamma$. Is there a universal upper bound for the length of $\Gamma$?

Remark: by the Hayman-Wu theorem, the answer is yes if $u$ is the real part of an injective holomorphic function; in fact, in this case there is a universal upper bound for the length of the entire level set in $D$. For general harmonic functions, level sets can have arbitrarily large length, e.g. $\Re z^n$.

flag
Leonid, so you should modify your question and say that your question is not about the whole curve $u=0$ but about connected componets of the curve $u=0$. Othervise the answer that I gave you is correct – Dmitri Jan 1 at 4:15
I this case this is a really cool question :)) – Dmitri Jan 1 at 4:29
1 
Can you give an example of smooth curve which can not be approximated by such level set? – Anton Petrunin Jan 1 at 4:51

1 Answer

8

It can get arbitrarily ugly. Indeed, approximate $1/z$ by a polynomial $p$ in the domain $K\subset\mathbb D$ whose complement is connected but goes from $0$ to the boundary along a long winding narrow path. Then each connected component of the set $\mbox{Re}p=A$ with large $A$ will have to escape the circle along essentially the same path and there are only finitely many $A$ for which we have branching points in these sets ($p'$ has finitely many roots).

link|flag
It won't help: just approximate the logarithm of the derivative of $1/z$ (which is still analytic in $K$), exponentiate, and integrate to get an approximation of $1/z$ without critical points. – fedja Jan 1 at 14:11
3 
This time I have a theorem up my sleeve (or, more precisely, up my friend's sleeve). It is true. The main idea is to relate the number of zeroes on a circle to the doubling exponent in the disk under the assumption that $u$ equals $0$ somewhere in the quarter-disk. If you allow to change the raius a bit, they essentially coincide. I don't know if it has ever been published anywhere but if you are really interested, let me know and I'll send you the notes. The details of the formal proof are a bit too long to post on MO (especially with the current limitations of TeX support). – fedja Jan 3 at 21:26

Your Answer

Get an OpenID
or

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