# Are the singular points of harmonic function on the disk always of measure zero?

Let $$f : \mathbb D^2 \to \mathbb R$$ be a smooth function with no singular points, i.e. $$df \neq 0$$ on $$\mathbb D^2$$. (Here $$\mathbb D^2$$ is the closed unit disk in $$\mathbb R^2$$). Let $$\omega:\mathbb D^2 \to \mathbb R$$ be the harmonic function corresponding to the Dirichlet problem imposed by $$f$$, i.e $$\omega|_{\partial \mathbb D^2}=f|_{\partial \mathbb D^2}$$.

Is it true that $$d\omega \neq 0$$ on a set of full measure in $$M$$?

*By harmonic I mean $$\Delta \omega=\omega_{xx}+\omega_{yy}=0$$.

Sine $$\omega$$ is harmonic, it is real-analytic. In particular its partial derivatives $$\omega_x,\omega_y:\mathbb D^2 \to \mathbb R$$ are real-analytic. Define $$A=\{ p\in \mathbb D^2 \, | \, \omega_x(p)=\omega_y(p)=0\}$$. We shall prove $$A$$ has measure zero. Indeed, $$A$$ is closed. So, if it had positive measure, then it would contain an accumulation point in $$\operatorname{Int} \mathbb D^2$$ (see the lemma below). By the identity theorem, this would imply $$\omega_x=\omega_y=0$$ on $$\operatorname{Int} \mathbb D^2$$, which would imply $$\omega$$ is constant on $$\mathbb D^2$$. However, $$f$$ cannot be constant* on $$\partial \mathbb D^2$$ which is a contradiction to $$f|_{\partial \mathbb D^2}=\omega|_{\partial \mathbb D^2}$$.
• $$f$$ is not constant on $$\partial \mathbb D^2$$: Indeed, we assumed that $$df \neq 0$$ everywhere, so $$\min f,\max f$$ are both obtained on $$\partial \mathbb D^2$$, and are distinct, since $$f$$ is not constant on the whole disk.
Lemma: A closed subset $$A \subseteq \mathbb D^2$$ of positive measure contains an accumulation point in $$\operatorname{Int} \mathbb D^2$$.
Assume by contradiction that $$A$$ does not contain an accumulation point in $$\operatorname{Int} \mathbb D^2$$. This implies (by compactness), that for any natural $$n$$, $$A \cap B_{1-1/n}$$ is finite, where $$B_r$$ is the closed disk of radius $$r$$. Thus, $$A \cap \operatorname{Int} \mathbb D^2=\cup_n A \cap B_{1-1/n}$$ is countable, hence of measure zero. Thus $$A$$ has measure zero, contradiction.