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Fall 2010 - Problem 11

harmonic functions, Wirtinger derivatives

Let Ω⊆C\Omega \subseteq \C be a non-empty open connected set. If f ⁣:Ω→C\func{f}{\Omega}{\C} is harmonic and f2f^2 is also harmonic, show that either ff or f‾\conj{f} is holomorphic on Ω\Omega.
Solution.

Recall the Wirtinger derivatives:

∂f∂z=12(∂f∂x−i∂f∂y)∂f∂z‾=12(∂f∂x+i∂f∂y).\begin{aligned} \pder{f}{z} &= \frac{1}{2}\p{\pder{f}{x} - i\pder{f}{y}} \\ \pder{f}{\conj{z}} &= \frac{1}{2}\p{\pder{f}{x} + i\pder{f}{y}}. \end{aligned}

From a quick calculation,

∂2f∂z∂z‾=∂2f∂z‾∂z=Δf4.\frac{\partial^2 f}{\partial z \partial \conj{z}} = \frac{\partial^2 f}{\partial \conj{z} \partial z} = \frac{\Delta{f}}{4}.

where Δ\Delta is the Laplacian. By assumption on ff, we have

∂2f∂z‾∂z=0and∂2f2∂z‾∂z=0\frac{\partial^2 f}{\partial \conj{z} \partial z} = 0 \quad\text{and}\quad \frac{\partial^2 f^2}{\partial \conj{z} \partial z} = 0

Since the Wirtinger derivatives are derivations, we have

∂f2∂z=2f∂f∂z  ⟹  ∂2f2∂z‾∂z=2∂f∂z‾∂f∂z+2f∂2f∂z‾∂z=2∂f∂z‾∂f∂z.\begin{aligned} \pder{f^2}{z} = 2f \pder{f}{z} \implies \frac{\partial^2 f^2}{\partial \conj{z} \partial z} &= 2\pder{f}{\conj{z}} \pder{f}{z} + 2f \frac{\partial^2 f}{\partial \conj{z} \partial z} \\ &= 2\pder{f}{\conj{z}} \pder{f}{z}. \end{aligned}

Since the left-hand side vanishes, we are left with

∂f∂z‾∂f∂z=0.\pder{f}{\conj{z}} \pder{f}{z} = 0.

Now suppose ff is not holomorphic. In particular, ff is harmonic hence C1C^1, which implies that ff does not satisfy the Cauchy-Riemann equations everywhere on Ω\Omega. Thus, there exists z0∈Ωz_0 \in \Omega such that ∂f∂z‾(z0)≠0\pder{f}{\conj{z}}\p{z_0} \neq 0. By continuity, this means that ∂f∂z‾≠0\pder{f}{\conj{z}} \neq 0 on a neighborhood of z0z_0. By the equation above, this implies that ∂f∂z\pder{f}{z} vanishes on a non-empty open set. Since Ω\Omega is connected and ∂f∂z\pder{f}{z} is harmonic, ∂f∂z\pder{f}{z} is identically 00. Hence,

0=∂f∂z‾=12(∂f∂x‾+i∂f∂y‾)=12(∂f‾∂x+i∂f‾∂y)=∂f‾∂z‾,0 = \conj{\pder{f}{z}} = \frac{1}{2} \p{\conj{\pder{f}{x}} + i\conj{\pder{f}{y}}} = \frac{1}{2} \p{\pder{\conj{f}}{x} + i\pder{\conj{f}}{y}} = \pder{\conj{f}}{\conj{z}},

so f‾\conj{f} satisfies the Cauchy-Riemann equations, hence holomorphic.