Could anyone advise me on how to prove: If $g$ and $\text{log}|g|$ are harmonic in a simply connected domain $\Omega$, then $g \equiv$ constant on $\Omega.$

Hints will suffice, thank you very much.

  • $\begingroup$ Is $g$ assumed real-valued, or complex-valued? $\endgroup$ – Daniel Fischer Nov 2 '14 at 16:35
  • $\begingroup$ $g$ is assumed to be real-valued. $\endgroup$ – Alexy Vincenzo Nov 2 '14 at 16:37
  • $\begingroup$ $g$ can have no zeros (why?), so you can assume $g > 0$. Pedestrian: Compute $\Delta (\log g)$ using the Wirtinger derivatives. More elegant: mean value property and Jensen's inequality. $\endgroup$ – Daniel Fischer Nov 2 '14 at 16:45
  • $\begingroup$ @DanielFischer: I am actually trying to prove: If $f$ is holomorphic in $\mathbb{D}$ and $|f|$ is harmonic in $\mathbb{D},$ then $f \equiv$ constant. Hence, I thought that lemma in my main question might help. $\endgroup$ – Alexy Vincenzo Nov 2 '14 at 16:47
  • $\begingroup$ Setting $g = \lvert f\rvert$, your question gets you the desired result. (That $\lvert f\rvert$ cannot be harmonic if $f$ has zeros is hopefully clear.) $\endgroup$ – Daniel Fischer Nov 2 '14 at 16:52

If $\log |g| $ is harmonic then $$-\frac{1}{g^2} \left(\left(\frac{\partial g}{\partial x}\right)^2 +\left(\frac{\partial g}{\partial y}\right)^2\right)+\frac{1}{g} \left(\frac{\partial^2 g}{\partial x^2} +\frac{\partial^2 g}{\partial y^2}\right) =0,$$ if $g$ is harmonic then $$\frac{\partial^2 g}{\partial x^2} +\frac{\partial^2 g}{\partial y^2}=0$$ hence $$\left(\frac{\partial g}{\partial x}\right)^2 +\left(\frac{\partial g}{\partial y}\right)^2=0$$ but the last equality implies that $g$ is constant.

  • $\begingroup$ Thanks for your solution, may I know why the last equality implies $g$ is constant? $\endgroup$ – Alexy Vincenzo Nov 2 '14 at 16:58
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    $\begingroup$ If $\frac{\partial g}{\partial x} =0 $ then $g(x,y) = h(y) $ but also $\frac{\partial g}{\partial y} =0 $ hence $g(x,y) = v(x) $ therefore $v(x) =h(y) $ but this is possible only when $v(x) =h(y) =constant$ $\endgroup$ – user110661 Nov 2 '14 at 17:03
  • $\begingroup$ Sorry if I'm missing something simple, but how do we know $g_x^2+g_y^2=0$ implies $g_x=g_y=0$? Couldn't we just have $g_x=ig_y$ with neither identically zero? $\endgroup$ – Ruvi Lecamwasam Nov 2 '14 at 22:57

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