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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?


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How are the level curves of a function $f:\mathbb{R}^2 \to \mathbb{R}^2$ defined? If you mean the set of arguments where the function takes on a specific value, this will typically be a point, not a curve. Did you mean $f:\mathbb{R}^2 \to \mathbb{R}$? – joriki Sep 30 '11 at 10:51
Even for $f:\mathbb{R}^2 \to \mathbb{R}$ the level curves may not be smooth. For instance, $f(x,y)=y-|x|$. – lhf Sep 30 '11 at 10:54
Yes sorry I meant, $f:\mathbb{R}^2 \to \mathbb{R}$. – Alexander Sep 30 '11 at 10:57
Note: lhf's comment was made when the requirement that $f$ is harmonic was an option in parentheses. If $f$ is harmonic, it is smooth. In particular, that means that the requirement of continuity is now redundant, since harmonic functions are necessarily continuous. – joriki Sep 30 '11 at 11:00

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?

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All partial derivatives of a harmonic function is also a harmonic function? And harmonic functions are twice differentiable, therefore $\nabla f$ has smooth elements? Sorry; taking a stab in the dark. – Alexander Sep 30 '11 at 11:09
The statement was changed to require harmonicity after I started writing my answer. However, you still need to show that $\frac{\nabla f}{|\nabla f|}$ (the direction of $\nabla f$) is smooth. – robjohn Sep 30 '11 at 11:17
Is it sufficient to show that the elements of $\nabla f$ are smooth? – Alexander Sep 30 '11 at 11:28
@Alexander: One has to be careful when $\nabla f=0$. – robjohn Sep 30 '11 at 11:39
When $\nabla f = 0$, $|\nabla f|=0$ so the direction is still defined in the limit? Sorry I'm unsure and thanks for the help so far, much appreciate it! – Alexander Sep 30 '11 at 11:55

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 $g'(x_0 + iy_0) \neq 0$.

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