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Let $f: \Omega \to \mathbb{R}$ be a harmonic function, where $\Omega \subset \mathbb{R}^2$ is an open subset.

What can be said about the points where $$\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=0? $$(Is it discrete, empty, or on the boundary of $\Omega$?).


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This is taken from Rudin's "Real and complex analysis". – user48900 Jan 9 '13 at 6:02
$f(x,y) =x^2-y^2$ is harmonic, $(0,0)$ is the only point where the derivative vanishes, so depending on how you choose $\Omega$ it may be empty or on the boundary or inside. If $f$ is harmonic, then $f = \text{Re } \phi$, where $\phi$ is analytic. Then $\phi' = f_x - i f_y$. If the set of points for which $f_x = f_y = 0$ has an accumulation point, then $\phi'$ is zero on that connected component of $\Omega$, and hence $f$ is constant on that component. – copper.hat Jan 9 '13 at 6:26
@user48900 Adding relevant tags might attract subscribers. Added (harmonic-functions). – AD. Jan 9 '13 at 6:33
@copper A more streamlined argument is to observe that $f_x-if_y$ is holomorphic by directly checking that the Cauchy-Riemann equations hold. // Also this is an answer... – user53153 Jan 9 '13 at 6:51
@PavelM: Nice observation. My complex/harmonic analysis was never very smooth... – copper.hat Jan 9 '13 at 7:02
up vote 1 down vote accepted

The question was settled in the comments: since the $\partial/\partial z$ derivative of a harmonic function is holomorphic, it follows that the zero set is discrete. It may or may not be empty. It does not make much sense to talk about it being on the boundary since $f$ is not assumed to be defined there, let alone differentiable.

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