Smoothness of level curves Lets $f:\mathbb{R}^2 \to \mathbb{R} $, where $f$ is harmonic, continuous and non-constant. How do I go about showing that the level curves of $f$ are smooth?
Thanks!
 A: This isn't actually true... let $f(x,y) = x^2 - y^2$ for example. Then the level set $f(x,y) = 0$ consists of two lines which isn't a smooth curve at the origin. The issue is that $\nabla f$ is zero at the origin.
In general a harmonic function $f(x,y)$ can be written locally as $Re(F(z))$ where $F(z) = f(x,y) + ig(x,y)$ is analytic. Suppose $\nabla f(x_0,y_0)$ is nonzero. Then in a neighborhood of $(x_0,y_0)$, you can look at $F(x,y) = (f(x,y), g(x,y))$ as a function from a region in ${\bf R}^2$ to ${\bf R}^2$. The Jacobian determinant of $F$ is given by $$ {\partial f \over \partial x}{\partial g \over \partial y} - {\partial f \over \partial y}{\partial g \over \partial x}$$ By the Cauchy-Riemann equations this is ${\displaystyle \left(\partial f \over \partial x\right)^2 + \left(\partial f \over \partial y\right)^2} = ||\nabla f||^2$ which is assumed to be nonzero. So by the inverse function theorem, the level sets of $f(x,y)$, which are just the inverse images of sets $x = c$ under the map $F(x,y)$, are smooth near $(x_0,y_0)$.
It is also true that if $\nabla f(x_0,y_0) = 0$ (and $f$ is not a constant function), then locally the level set of $f(x,y)$ containing $(x_0,y_0)$ is the union of finitely many smooth curves; to show this you use the fact that locally $F(z) = c + g(z)^n$ for some constant $c$, some positive integer $n$, and some analytic $g(z)$ with $g'(x_0 + iy_0) \neq 0$. 
A: Hint: The level curves are perpendicular to the gradient of a function. What condition on $f$ or $\nabla f$ might insure that the direction of $\nabla f$ is smooth?
