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Let $f = u + iv$ be a holomorphic function on unit (open) disk $\mathbb D^2$.

and Suppose $v$ is nonnegative and $v(0)=0$.

Can we determine all such $f$?

I wanted to use schwarz lemma, but I couldn't catch anything about bound condition.

I think it is about liouville's theorem.

Is it related to harmonic function?

Thanks.

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    $\begingroup$ I would guess that if $v(0)=0$ then $v$ can't be nonnegative unless it is constant, so $v\equiv 0$ which would mean $f$ is a real constant. But I'm just thinking nonrigorously. $\endgroup$ – MPW Dec 30 '15 at 13:44
  • $\begingroup$ If you want to use Schwarz lemma, then is it not that $f$ should be holomorphic from disk to disk? $\endgroup$ – p Groups Dec 30 '15 at 13:55
  • $\begingroup$ @MPW why v is indentically 0? Can you give me some clue? $\endgroup$ – nicksohn Dec 30 '15 at 14:03
  • $\begingroup$ @pGroups You are right. But Nothing comes to my mind except that. I'm quiet newb of Complex analysis. Thanks anyway. $\endgroup$ – nicksohn Dec 30 '15 at 14:03
  • $\begingroup$ Anyway, Someone fix my text very beautifully. How can I write like such a elegant way? Can I learn somewhere? $\endgroup$ – nicksohn Dec 30 '15 at 14:12
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$ f(0)=r $ some real number. Then open mapping theorem says that if $f$ is non-constant , then there is a open nbd of $0$ which maps onto an open nbd of $r$ ... which implies there exists a point lets say $ r_1 -ir_2$ where $r_1, r_2$ +ve real number, is in the image. But $v$ is non-negative. contadiction. So $f$ is constant

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If $0<r<1$ then $$0=v(0)=\frac1{2\pi}\int_0^{2\pi}v(re^{it})dt.$$Since $v\ge0$ (and $v$ is continuous) this shows that $v=0$. So $f$ is real, hence constant.

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  • $\begingroup$ Why $f$ is real means $f$ is real constant? $\endgroup$ – nicksohn Dec 30 '15 at 14:39
  • $\begingroup$ Open Mapping Theorem. Or the Cauchy-Riemann equations. $\endgroup$ – David C. Ullrich Dec 30 '15 at 14:56

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