# holomorphic function on unit disk with certain condition, Can we determine that function?

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.

• 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. – MPW Dec 30 '15 at 13:44
• If you want to use Schwarz lemma, then is it not that $f$ should be holomorphic from disk to disk? – p Groups Dec 30 '15 at 13:55
• @MPW why v is indentically 0? Can you give me some clue? – nicksohn Dec 30 '15 at 14:03
• @pGroups You are right. But Nothing comes to my mind except that. I'm quiet newb of Complex analysis. Thanks anyway. – nicksohn Dec 30 '15 at 14:03
• Anyway, Someone fix my text very beautifully. How can I write like such a elegant way? Can I learn somewhere? – nicksohn Dec 30 '15 at 14:12

$f(0)=r$ 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
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.
• Why $f$ is real means $f$ is real constant? – nicksohn Dec 30 '15 at 14:39