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Thread: Analytic problem

  1. #1
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    Analytic problem

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

    I have no idea where to start. Thank you.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Analytic problem

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

    Thank you for you reply, I could do what you said but whats the next step? Thanks
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Analytic problem

    By hypothesis $\displaystyle |f(z)|\leq 1$ so, $\displaystyle |f(z)/z^2|\leq 1/r^2$ for $\displaystyle |z|=r$ . This equality is also valid for $\displaystyle |z|\leq r$ according to the Maximum Modulus Principle. If we fix $\displaystyle z$ in $\displaystyle |z|<1$ we have $\displaystyle |f(z)|\leq |z|^2/r^2$ for all $\displaystyle r\geq |z|$ and $\displaystyle <1$ . You can conclude.
    Last edited by FernandoRevilla; Dec 16th 2011 at 06:59 AM.
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    Re: Analytic problem

    Quote Originally Posted by FernandoRevilla View Post
    By hypothesis $\displaystyle |f(z)|\leq 1$ so, $\displaystyle |f(z)/z^2|\leq 1/r^2$ for $\displaystyle |z|=r$ . This equality is also valid for $\displaystyle |z|\leq r$ according to the Maximum Modulus Principle. If we fix $\displaystyle z$ in $\displaystyle |z|<1$ we have $\displaystyle |f(z)|\leq |z|^2/r^2$ for all $\displaystyle r\geq |z|$ and $\displaystyle <1$ . You can conclude.
    Sorry but I didn't get it. How can I deduce $\displaystyle |f(z)/z^2|\leq 1/r^2$for $\displaystyle |z|\leq r$ according to the Maximum Modulus Principle? Thanks!!!
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Analytic problem

    Quote Originally Posted by fareastmovement View Post
    Sorry but I didn't get it. How can I deduce $\displaystyle |f(z)/z^2|\leq 1/r^2$for $\displaystyle |z|\leq r$ according to the Maximum Modulus Principle? Thanks!!!
    What does the Maximum Modulus Principle say?
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    Re: Analytic problem

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

    but$\displaystyle |f(z)/z^2|\leq 1/r^2$. How can I make sure that $\displaystyle f(z_0)=1/r^2$? Or I am in the wrong direction? Thank you
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Analytic problem

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