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.



  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 $3/4$.

  • $\begingroup$ Brilliant hint. An ideal stackexchange answer. $\endgroup$ – 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. $z_0\in A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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