6

Let $u$ be a non-constant harmonic function of two variables defined, say, in the unit disk (or on the half plane for example). It is known that $u$ can vanish on some lines, as it discussed in here. But can $u$ vanish on a set of positive measure ?

Community
  • 1
Mher
  • 5,101
  • 1
  • 21
  • 35

1 Answers1

4

Here's a way to fill in the gap in @Yiorgos' solution. The problem is local, so we may as well work on a simply connected domain.

Let $v$ be a harmonic conjugate of $u$, so that $f = u+iv$ is holomorphic. It follows from Cauchy-Riemann's equations that $f' = u'_x -iu'_y$. In particular, the gradient of $u$ vanishes exactly at the points where $f' = 0$. But $f'$ is also holomorphic, so the zero set of $f'$ is a discrete set (unless $f' \equiv 0$, but this only happens when $u$ is constant).

Hence, removing the discrete (thus measure zero) set of points where $f'=0$, the rest of the zero set is a countable union of graphs of $C^1$-functions.

(To see this last claim, take a countable exhaustion $K_1 \subset K_2 \subset \cdots$ of the unit disc minus $f^{-1}(0)$ by compact sets. For each point in $K_j$, use the implicit function theorem to see that the zero set of $u$ locally is the graph of a $C^1$ function. Use compactness to select a finite number of such graphs to cover the entire zero set in $K_j$, and take the union over all $j$.)

mrf
  • 44,697
  • We are in dimension $n \geq 2$. Hence, the zero set of a holomorphic function (or of the derivative of a holomorphic function) is not necessarily discrete, take e.g. $f(z_1, z_2) = z_1 \cdot z_2$. Or am I missing something? – PhoemueX Oct 21 '14 at 16:45
  • @PhoemueX I interpreted the question as two real variables. (Seems supported by "unit disk" and "upper halfplane" as well.) – mrf Oct 21 '14 at 17:14
  • But then just take $f(z_1, z_2) = z_1^2 \cdot z_2$. Then $f'(z_1, z_2) = (2 z_1 z_2, z_1^2)$ which vanishes on the set where $z_1 = 0$. But this set is not discrete. – PhoemueX Oct 21 '14 at 19:52
  • 1
    @PhoemueX Two real variables $(x,y)$, one complex $z=x+iy$. – mrf Oct 21 '14 at 20:42
  • Oh, of course. I somehow thought about the way we can (locally) extend an analytic function $f : \Bbb{R}^n \to \Bbb{C}^n$ to a holomorphic function $U \to \Bbb{C}^n$ with $U \subset \Bbb{C}^n$ open. Because this "doubles" the number of real variables. But in this case, we are doing something very different. Thanks for the clarification. – PhoemueX Oct 21 '14 at 20:48