3
$\begingroup$

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!

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ 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 $n$. $\endgroup$ – George Lowther Mar 26 '12 at 2:12
  • 2
    $\begingroup$ 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 $z_0 < 0$. $\endgroup$ – George Lowther Mar 26 '12 at 2:20
  • $\begingroup$ @George: Thanks for this. I appreciate it. How did you come up with your example? $\endgroup$ – Bey Mar 28 '12 at 1:01
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.