# Harmonic function vanishing on a set of positive measure.

I'm preparing for a qualifying exam, and came across a question I couldn't figure out:

If $\Omega$ is a region and $h:\Omega\to \mathbb{R}$ is a harmonic function vanishing on a set of positive measure, then $h\equiv 0$.

It gives a hint to consider $\nabla h$, but I haven't managed to figure out a way to use it. Of course, since $\Omega$ is connected it suffices to show that $A=\{z\in \Omega: h(z)=0\}$ is open. It seems right to consider a Lebesgue point of density for $A$; but all that's managed to do for me is to confirm that $h$ vanishes on $A$, but that's about it.

Strategy:

1. The gradient of $h$ vanishes at every Lebesgue point of density of $A$.
2. The function $f(z)=h_x-ih_y$ is holomorphic (check the Cauchy-Riemann equations directly).
3. A holomorphic function vanishing on a non-discrete set (in particular on a set of positive measure) is identically zero.
4. Hence, $h$ is constant, and the constant must be zero.

Suggestion for step 1: Suppose $h(z_0)=0$ and $\nabla h(z_0)\ne 0$. The definition of differentiability implies that $h(z)\ne 0$ if $|z-z_0|$ is small enough and the angle between $\nabla h(z_0)$ and $z-z_0$ is at most $45$ degrees. But then the density of $\{h=0\}$ at $A$ is at most $3/4$.

• Brilliant hint. An ideal stackexchange answer. – user135671 Mar 23 '14 at 0:54

An alternative way to prove step 1 for the above solution:

Let $z_0\in A$ such that $z_0$ is in the Lebesgue set of $h^2$, i.e. $$\frac{1}{\pi r^2} \int_{D(z_0,r)}|h(w)|^2dA(w)\to 0$$ as $r\to 0$. The partial derivative $\frac{\partial h}{\partial x}(z)$ is also harmonic, therefore by the mean value propery and Green's theorem we have $$\frac{\partial h}{\partial x}(z_0)=\frac{1}{\pi r^2}\int_{D(z_0,r)} \frac{\partial h}{\partial x}(x,y)dxdy= \frac{1}{\pi r^2} \int_{\partial D(z_0,r)}hdy$$ $$= \frac{1}{\pi r^2} \int_0^{2\pi} h(z_0+re^{it})r \cos tdt$$ This implies $$r^2|\frac{\partial h}{\partial x}(z_0)|\leq \frac{1}{\pi}\int_0^{2\pi} |h(z_0+re^{it})|rdt$$ and integrating from $0$ to $R$ yields $$\frac{R^3}{3}|\frac{\partial h}{\partial x}(z_0)|\leq \frac{1}{\pi}\int_{D(z_0,R)}h(w)dA(w)$$ By Cauchy Schwarz we have $$|\frac{\partial h}{\partial x}(z_0)|\leq\frac{3}{\pi R^3} (\pi R^2)^{1/2} \int_{D(z_0,R)} |h(w)|^2dA(w)=\frac{3}{\sqrt{\pi}R^2}\int_{D(z_0,R)}|h(w)|^2dA(w)$$ where the latter converges to $0$ as $R\to 0$. So you have $h_x(z_0)=0$ for a.e. $z_0\in A$. Similarly the same holds for $h_y(z_0)$, a.e. $z_0\in A$.