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Suppose $f$ is a harmonic function on a connected open set $\Omega$ in the complex plane, and suppose also that $f$ is holomorphic on some open subset $U$ of $\Omega$. Prove that $f$ is holomorphic on all of $\Omega$.

Unfortunately, I do not know how to start this problem. There are only a few ways I know how to show a function is holomorphic - using the "limit definition", Morera's Theorem, showing there is some sequence $f_n\subset H(\Omega)$ converging normally to $f$, etc. However, I can't see how to solve this problem with these methods. Could I receive some suggestions on what to be thinking about? Thanks.

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    $\begingroup$ Are you aware of the fact that if two homomorphic agree on a subset of $\Omega$ which contains a limit point in $\Omega$ then the functions are identical on $\Omega$. $\endgroup$ Dec 13 '14 at 0:05
  • $\begingroup$ Yes, I know this fact for holomorphic functions. $\endgroup$
    – user122916
    Dec 13 '14 at 0:11
  • $\begingroup$ Is $f:\Omega\rightarrow\Bbb R$ or $f:\Omega\rightarrow\Bbb C$? Harmonic functions are usually real-valued while homomorphic functions are complex-valued. $\endgroup$ Dec 13 '14 at 0:31
  • $\begingroup$ Complex-valued, and harmonic meaning it satisfies Laplace's equation. $\endgroup$
    – user122916
    Dec 13 '14 at 1:33
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Let $$\partial_z := \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right) \qquad \text{and} \qquad \partial_{\bar{z}} := \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$$

Since $f$ is harmonic on $\Omega$, $\partial_z \partial_{\bar z}f = 0$ on $\Omega$. Hence $\partial_{\bar{z}} f$ is constant on $\Omega$. Since $f$ is holomorphic on $U$, $\partial_{\bar{z}} f = 0$ on $U$. Since $U$ has a point of accumulation, $\partial_{\bar{z}} f = 0$ on $\Omega$ by the identity theorem for holomorphic functions. Hence, $f$ is holomorphic on $\Omega$.

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  • $\begingroup$ How can I use the identity theorem for holomorphic functions (the uniqueness principle) on all of $\Omega$ when I don't know $f$ is holomorphic there? $\endgroup$
    – user122916
    Dec 13 '14 at 1:35
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    $\begingroup$ You're using it on $\partial_{\bar{z}} f$, not $f$. Since $f = u + iv$ is harmonic, $u$ and $v$ have continuous second partials, so $\partial_{\bar{z}} f$ is $C^1$ on $\Omega$. Further, $\partial_{\bar{z}} f$ is complex-differentiable with derivative $0$. So $\partial_{\bar{z}} f$ is holomorphic on $\Omega$. $\endgroup$
    – kobe
    Dec 13 '14 at 1:41
  • $\begingroup$ Of course! Nice and concise. Thank you. $\endgroup$
    – user122916
    Dec 13 '14 at 1:49

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