Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a holomorphic function from $H=\{z \in \mathbb C : \operatorname{Im}(z)>0\}$ to $D=\{z \in \mathbb C : |z|<1\}$. Suppose that $f(z_0)=f'(z_0)=0$.

Show that: \[ |f(z)| \leq \frac{|z-z_0|^2}{|z-\overline{z}_0|^2} \quad\text{and}\quad |f''(z_0)| \leq \frac{1}{2|\operatorname{Im}(z_0)|^2} \]

Thanks everyone for your help!!

share|improve this question
2  
What have you tried? In my experience, questions that show some effort beyond a reproduction of an exercise get more attention. If you've tried something and it didn't work, it's okay and even desirable to mention that. My gut reaction is to use some conformal mapping to get a map $D \to D$ and then apply Schwarz or Shwarz–Pick, but I don't have time to pursue this at the moment. Hopefully someone else will. Welcome, by the way! –  Dylan Moreland Jun 27 '12 at 18:35
    
Hey Dylan! First of all thanks for the greeting! I just tried to call $g(z)=f(z)/|z-z_0|^2$ and using the maximum modulus principle as when you demonstrate that $|f(z)|<|z|$ when f goes from D to D...it worked for demonstrating that $|f(z)|<|z|^2$, but I can't solve this one... –  Nico Jun 27 '12 at 18:41
1  
Should the $z$ in your second inequality be $z_0$? –  Robert Israel Jun 27 '12 at 19:02
    
yep, I mistyped –  Nico Jun 27 '12 at 19:35
    
And by the way solving the first inequality the second becomes pretty obvious to me...the problem is the first one! –  Nico Jun 27 '12 at 19:37

1 Answer 1

Hint: let $g = f \circ \varphi^{-1}$ where $\varphi(z) = \dfrac{z - z_0}{z - \overline{z_0}}$, and note that $g(w)/w^2$ has a removable singularity at $0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.