Analytic problem

**fareastmovement** · December 16th 2011, 03:13 AM

Let $f(z)$ be analytic in the disk $|z| <1$ . If $f(z)$ has a zero of order 2 at the origin and $|f(z)| \le 1$ in that disk. Prove that $|f(z)|\le|z|^2$ in $|z|<1$

I have no idea where to start. Thank you.

**FernandoRevilla** · December 16th 2011, 03:35 AM

Hint: In the series expansion $f(z)=\sum_{n=0}^{+\infty}a_nz^n$ we have $a_0=a_1=0$ , so $f(z)/z^2$ is analytic in $|z|<1$ .

**fareastmovement** · December 16th 2011, 05:07 AM

Thank you for you reply, I could do what you said but whats the next step? Thanks

**FernandoRevilla** · December 16th 2011, 06:31 AM

By hypothesis $|f(z)|\leq 1$ so, $|f(z)/z^2|\leq 1/r^2$ for $|z|=r$ . This equality is also valid for $|z|\leq r$ according to the Maximum Modulus Principle. If we fix $z$ in $|z|<1$ we have $|f(z)|\leq |z|^2/r^2$ for all $r\geq |z|$ and $<1$ . You can conclude.

**fareastmovement** · December 16th 2011, 07:17 AM

Originally Posted by FernandoRevilla

By hypothesis $|f(z)|\leq 1$ so, $|f(z)/z^2|\leq 1/r^2$ for $|z|=r$ . This equality is also valid for $|z|\leq r$ according to the Maximum Modulus Principle. If we fix $z$ in $|z|<1$ we have $|f(z)|\leq |z|^2/r^2$ for all $r\geq |z|$ and $<1$ . You can conclude.

Sorry but I didn't get it. How can I deduce $|f(z)/z^2|\leq 1/r^2$ for $|z|\leq r$ according to the Maximum Modulus Principle? Thanks!!!

**FernandoRevilla** · December 16th 2011, 07:25 AM

Originally Posted by fareastmovement

Sorry but I didn't get it. How can I deduce $|f(z)/z^2|\leq 1/r^2$ for $|z|\leq r$ according to the Maximum Modulus Principle? Thanks!!!

What does the Maximum Modulus Principle say?

**fareastmovement** · December 16th 2011, 07:48 AM

Suppose $f$ is analytic on $|z-z_0|<\epsilon$ . If $|f(z)| \le |f(z_0)|$ for $z$ on this region then $f(z) \equiv f(z_0)$ on the region.

but $|f(z)/z^2|\leq 1/r^2$ . How can I make sure that $f(z_0)=1/r^2$ ? Or I am in the wrong direction? Thank you

**FernandoRevilla** · December 16th 2011, 11:34 AM

Better use this version: Let $D \subset \mathbb{C}$ be a bounded domain, and let $f$ be a continuous function on the closed set $\overline{D}$ that is analytic on $D$ . Then the maximum value of $|f|$ on $\overline{D}$ (which always exists) occurs on the boundary $\partial D$ . So, in our case, it is not possible $|f(z)/z^2|>1/r^2$ if $|z|<r$ .

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