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Let $D$ denote the unit disk in $\mathbb{C}$. Suppose I have an analytic function $f:D\to D$. Then I can write down its series expansion at, say, $z_0\in D$: $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\,.$$

Now, we know that $|f(z)|\le 1$, since $f(z)\in D$ for all $z\in D$. Does this imply that the coefficients $a_n$ are bounded? My instinct is to say yes, but I can't find a particularly convincing reason.

Thanks!

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    No, they aren't bounded other than at $z_0=0$. Cauchy's integral formula gives a bound $a_n=\frac1{2\pi}\oint_\gamma\frac {f(z)}{(z-z_0)^{n+1}}dz$. So, $\vert a_n\vert\le(2\pi)^{-1}(1-\vert z_0\vert)^{-n-1}$. For $z_0\not=0$ this grows in $n$.2012-03-26
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    And taking $f(z)=c(1+z)\log(1+z)$ for example ($c$ is some constant to ensure that $f(z)\in D$), the coefficients will not be bounded if you expand about real $z_0 < 0$.2012-03-26
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    @George: Thanks for this. I appreciate it. How did you come up with your example?2012-03-28

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