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Suppose we have a harmonic function $f$ such that $e^f$ is also harmonic. I want to prove $f$ is either holomorphic or antiholomorphic.

I tried using the fact that $\Delta u= 4\partial_z(\partial_{\overline{z}} u)=0$ but I didn't get anywhere with that and I assumed that does not imply $\partial_{\overline{z}}=0$ or $\partial_z =0$.

I actually have the same problem but with $f^2$ instead of $e^f$ and I suppose it is similar, so I hope I can do one knowing how to do the other one.

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1 Answer 1

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Let $D \subset \Bbb C$ be a domain and $f: D \to \Bbb C$ a function with continuous derivatives up to order $2$. $\exp$ is holomorphic, i.e. $\partial_{\overline{z}} e^z = 0$, therefore the chain rule for Wirtinger derivatives gives $$ \partial_z (e^f) = e^f \partial_z f \, ,\\ \partial_{\overline{z}} (e^f) = e^f \partial_{\overline{z}}f \, . $$

It follows that $$ \Delta e^f= 4\partial_z(\partial_{\overline{z}} e^f) = 4\partial_z( e^f \partial_{\overline{z}} f) = 4 \left(e^f (\partial_z f)(\partial_{\overline{z}} f) + e^f \partial_z(\partial_{\overline{z}} f \right) = 4 e^f (\partial_z f)(\partial_{\overline{z}} f) + e^f \Delta f \, . $$ In particular, if $f$ and $e^f$ are harmonic then $$ (\partial_z f(z))(\partial_{\overline{z}} f(z)) = 0 $$ for all $z \in D$. Both $\partial_z f(z)$ and $\partial_{\overline{z}} f$ are harmonic in $D$, therefore one of them must be identically zero in $D$ (see for example product of harmonic functions), which means that $f$ is anti-holomorphic or holomorphic.

The same approach works if $f$ and $f^2$ are harmonic, or even for $f$ and $\phi \circ f$, if $\phi$ is holomorphic in $\Bbb C$ and not linear.

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