Is this a "folk theorem" about analytic functions of a complex variable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:53:43Z http://mathoverflow.net/feeds/question/110552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110552/is-this-a-folk-theorem-about-analytic-functions-of-a-complex-variable Is this a "folk theorem" about analytic functions of a complex variable? Garabed Gulbenkian 2012-10-24T15:56:37Z 2012-10-24T19:07:29Z <p>In a comment on question 110345 I made a claim that might be incorrect. I claimed that if f(z) is a non-constant analytic function defined by a power series whose circle of convergence C has a positive radius, then f(z) cannot be bounded at all points in the interior of C. But is this really true? Or am I just imagining that I learned it somewhere. I could not come up with any simple counter-examples. It sounds like some weird generalization of Liouville's theorem.</p> http://mathoverflow.net/questions/110552/is-this-a-folk-theorem-about-analytic-functions-of-a-complex-variable/110555#110555 Answer by quid for Is this a "folk theorem" about analytic functions of a complex variable? quid 2012-10-24T16:09:48Z 2012-10-24T16:09:48Z <p>This is not true. For a counterexample take </p> <p><code>$$\sum_{n=1}^{\infty} \frac{x^n}{n^2}$$</code></p> <p>The radius of convergence is one and this is bounded by </p> <p><code>$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$$</code></p> <p>You are likely confusing this with the <a href="http://en.wikipedia.org/wiki/Maximum_modulus_principle" rel="nofollow">maximum modulus principle</a> saying that there must be no local maximu (in absolute value) in the interior if the function is nonconstant.</p> http://mathoverflow.net/questions/110552/is-this-a-folk-theorem-about-analytic-functions-of-a-complex-variable/110575#110575 Answer by Johannes Ebert for Is this a "folk theorem" about analytic functions of a complex variable? Johannes Ebert 2012-10-24T19:07:29Z 2012-10-24T19:07:29Z <p>You are probably misunderstanding the following folk theorem: If $D$ is the convergence disc of a power series converging to $f$, then there must be some singularity of $f$ on $\partial D$. In other words, you cannot continue $f$ analytically onto a larger disc. A counterexample that is more explicit than quids example is the power series expansion of $\sqrt{1+z}$ around $z=0$, which has convergence radius $1$. The singularity at $z=-1$ is not a pole.</p> <p>A hint to the proof: if $D \subset U$ is a disc in the domain of definition of a holomorphic function $f$, then the Taylor expansion around the midpoint of $D$ converges in $D$.</p>