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Assume that $f$ is an analytic function on the unit disc $\mathbb{D}$ and continuous up to the closure. Therefore $f(z)=\sum\limits_{n=0}^\infty c_nz^n$ for all $z \in \mathbb{D}$. If $f$ have $m$ zeros in $\mathbb{D}$ how can you prove that $$ \min\{|f(z)| : |z|=1\}\leq |c_0|+\ldots+|c_m| $$

To begin with, the minimum of the funcion in $|z|=1$ shold be equal to the minimun on the whole $\overline{\mathbb{D}}$. Then I tried using the mean value principle around the zeros, or the expansion near a zero, but I couldn't do it.

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The minimum principle holds only if $f\ne 0$ in the domain. So, if $m=0$, you can use $\min_{\partial D} |f| = \min_{D} |f|\le |f(0)|=|c_0|$. When $m\ge 1$, the minimum on the boundary is not in general the same as the minimum in $D$, the latter being 0. – user31373 May 15 '12 at 22:52
@balestrav: I'm deleting my answer since there was a mistake in the application of Rouche (the hypothesis were not quite satisfied), and I don't see a clear way to patch it up without appealing to the same argument given by Leonid, so you should probably accept his answer. – Jose27 May 17 '12 at 3:03
@balestrav: Hm, apparently I can't delete an accepted answer. Please un-accept it. – Jose27 May 17 '12 at 3:04
up vote 2 down vote accepted

Let $P(z)=c_0+\dots +c_m z^m$ and $g(z)=f(z)-P(z)$. Suppose that $|P(z)|<|f(z)|$ for all $z\in\partial D$. Then the Rouche theorem says that $g$ has $m$ zeros in $D$. This is absurd; $g$ has a zero of order $m+1$ at the origin. Therefore, $|P(z)|\ge |f(z)|$ for some $z\in\partial D$.

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