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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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    $\begingroup$ 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 $n$. $\endgroup$ Mar 26, 2012 at 2:12
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    $\begingroup$ 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 $z_0 < 0$. $\endgroup$ Mar 26, 2012 at 2:20
  • $\begingroup$ @George: Thanks for this. I appreciate it. How did you come up with your example? $\endgroup$
    – Bey
    Mar 28, 2012 at 1:01

1 Answer 1

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The $\limsup $ formula for the radius of convergence implies that any power series with bounded coefficients has radius of convergence $R\ge 1$. So, a series with $R<1$ must have unbounded coefficients. The easiest way to make the radius of convergence small is to stick a singularity nearby.

Given $z_0\ne 0$, we pick a point $a$ just outside the closed unit disk, so that $|z_0-a|<1$. For example, $a=(1+|z_0|/2)z_0/|z_0|$ works. The function $f(z)=c/(z-a)$ is our example when $c$ is small enough, such as $c=|a|-1$. If you believe the arguments given above, you don't even have to look at the coefficients.

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