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Must an entire function be bounded in any open disk around the origin? I think the answer is yes and this is my attempt at a proof:

Let $f$ be an entire function and $D$ be an open disk of radius $r$ centered at the origin. Since $f$ is entire, it can be represented by a power series $f(z) = \sum^{\infty}_{n=0}c_nz^n$ with radius of convergence $r$ for every $z \in D$. Since the power series converges in $D$, $f$ must be bounded in $D$.

Is this correct?

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  • $\begingroup$ Try a convergence radius strictly greater than $r$. Or the maximum modulus principle. $\endgroup$ – Hagen von Eitzen Nov 16 '13 at 23:20
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    $\begingroup$ Shorter: $\overline{D}$ is compact. $\endgroup$ – Daniel Fischer Nov 16 '13 at 23:21
  • $\begingroup$ An entire function can be represented by a power series with infinite radius of convergence at every point, in fact. $\endgroup$ – zibadawa timmy Nov 16 '13 at 23:21
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Quite a bit more is true: Every analytic function is continuous, and continuous functions are necessarily bounded on all compact sets; this is true because the continuous image of a compact set is compact. Noting that the closure of the disk

$$\overline{D(0, R)}$$

is compact for all finite $R$, the result follows.

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  • $\begingroup$ I see. Thanks! +1 and accetped. $\endgroup$ – Jean Valjean Nov 17 '13 at 16:21

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