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Let f(x) ve given by a power series everywhere where ir converges, thus the domain of f(x) consists of the open circle of convergence plus some boundary points of it (posibbly none). Then is f continuous? I know that if f is real valued then yes by abel's limit theorem, but what about complex valued f? I've looked online and found something about a thing called 'Stoltz sector', I didn't really understand it, plus it posibbly doesn't answer my question (therefore I won't quote it).

Manuel Ocaña
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  • Related https://math.stackexchange.com/questions/119926/how-does-a-complex-power-series-behave-on-the-boundary-of-the-disc-of-convergenc – D.R. Mar 06 '22 at 04:46

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Ah, I have found the exact answer to your question here: https://mathoverflow.net/questions/110345/does-a-power-series-converging-everywhere-on-its-circle-of-convergence-define-a. The answer provides a 1916 paper from Sierpinski, in which "[he] produces an example where the function converges everywhere on the unit circle but is discontinuous (in fact unbounded) on the circle."

D.R.
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  • I see you linked a site similar to MSE, what makes that site different to MSE? – Manuel Ocaña Mar 06 '22 at 05:16
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    @ManuelOcaña https://math.meta.stackexchange.com/questions/41/differences-between-mathoverflow-and-math-stackexchange (it's essentially the "older brother" to this site) – D.R. Mar 06 '22 at 05:18