1
$\begingroup$

I'm learning complex analysis, specifically Laurent series, and need help with the following exercise:

If $f(z)$ is entire and for some $k \in \mathbb{N}$ the function $g(z) = z^kf(\frac{1}{z})$ is bounded in some annular region $\Delta(0, 0, \delta)$ with $\delta > 0$, show that $f(z)$ is a polynomial.

Note: the annulus $\Delta(0, 0, \delta)$ is defined as the region where $0 < |z| < \delta$.

Here's my work so far:

Since $f(z)$ is entire, by Taylor's theorem, we can expand $f(z)$ as a power series centered at $z_0 = 0$, and so

$$f(z) = \sum_{n = 0}^{\infty}a_n z^n \implies f\left(\frac{1}{z}\right) = \sum_{n = 0}^{\infty}a_n \frac{1}{z^n}.$$

Therefore

$$g(z) = z^kf(\frac{1}{z}) = a_0z^k + a_1z^{k-1} + a_2z^{k-2} + \ldots + a_{k-1}z + a_k + \frac{a_{k+1}}{z} + \frac{a_{k+2}}{z^2} + \ldots$$

This is as far as I got. I think the fact that $g(z)$ is bounded is key here but I don't know how to make use of this assumption for $g(z)$.

Edit: Considering the answer of Eric Wofsey, we have

$$g(z) = \sum_{n = 1}^{\infty}\frac{a_{n+k}}{z^n} + a_k + a_{k-1}z + \ldots + a_1z^{k-1} + a_0z^k.$$

However, since $z_0 = 0$ is a removable singularity of $g(z)$, by uniqueness of Laurent series, $g(z)$ is actually the Taylor series

$$g(z) = a_k + a_{k-1}z + \ldots + a_1z^{k-1} + a_0z^k.$$

It follows that

$$f(1/z) = \frac{g(z)}{z^k} = a_0 + \frac{a_1}{z} + \ldots + \frac{a_{k-1}}{z^{k-1}} + \frac{a_k}{z^k}$$

and so

$$f(z) = a_0 + a_1z + \ldots + a_{k-1}z^{k-1} + a_kz^k$$

which shows that $f(z)$ is a polynomial of degree $k$.

$\endgroup$
1
  • $\begingroup$ if $z^k f(1/z)$ is bounded on $0 < |z| < \delta$ then $f(z)$ is bounded by $C z^k$ on $|z| > 1/\delta$, and it is the Liouville theorem that if it is entire then it is a polynomial of degree $\le k$ $\endgroup$
    – reuns
    Commented Jun 29, 2016 at 3:14

1 Answer 1

1
$\begingroup$

If $g(z)$ is bounded on $\Delta(0,0,\delta)$, then its singularity at $0$ is removable, so it can be extended analytically to $0$. This means that $g(z)$ has a Taylor series around $0$, and so by uniqueness of Laurent series your Laurent series for $g(z)$ must actually be a Taylor series. You should be able to finish from here.

$\endgroup$
5
  • $\begingroup$ Thank you for your helpful explanation. I had forgotten that $z_0 = 0$ is a removable singularity of $g(z)$ if and only if $g(z)$ is bounded on some punctured disc centered at $0$. Although I understand your answer perfectly well, it is still unclear to me why $f(z)$ is a polynomial. Since we found that $g(z)$ is actually a Taylor series, then it must be of the form $g(z) = \sum_{n = 0}^{\infty}a_n z^n$. But I don't know how to get back to $f(z)$ and show that it is a polynomial. Can you help me with that? $\endgroup$
    – glpsx
    Commented Jun 29, 2016 at 1:28
  • 1
    $\begingroup$ @VonKar: Use the work you did before. You found a Laurent expansion for $g$ in terms of the Taylor expansion for $f$. Now equate that Laurent expansion to $g$'s Taylor expansion and compare terms. (The key fact here is that a function on an annulus can only have one Laurent expansion.) $\endgroup$ Commented Jun 29, 2016 at 1:33
  • $\begingroup$ I have edited my post with a complete answer. Could you tell me if my work is correct? $\endgroup$
    – glpsx
    Commented Jun 29, 2016 at 2:46
  • $\begingroup$ Looks good. Instead of editing the question, you should generally post the answer you have found as an answer. $\endgroup$ Commented Jun 29, 2016 at 3:01
  • $\begingroup$ Ok! I'll remember it next time. Thanks again for your helpful explanation, much appreciated. $\endgroup$
    – glpsx
    Commented Jun 29, 2016 at 3:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .