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Does there exist a holomorphic function $f\in H^\infty$ (i.e. holomorphic and bounded on the open unit disc) such that its power series expansion at $0$ doesn't converge anywhere in $\{|z|=1\}$?

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Well, it is pretty clear you cannot have $f^{(n)}(0)$ monotonic decreasing, since the alternating series test would have convergence for $z=1$. Using summation by parts, you might be able to show convergence for $|z|=1$, $z\neq 1$ as well if it is monotonic decreasing. – nayrb Aug 10 '12 at 18:28
up vote 2 down vote accepted

To every $f \in H^\infty$ there corresponds $f^\star \in L^\infty({\mathbb T})$ such that $f^\star(\omega) = \lim_{r \to 1-} f(r \omega)$ almost everywhere. The functions $f^\star$ thus obtained are the members of $L^\infty$ whose Fourier coefficents $c_n = (2 \pi)^{-1} \int_{0}^{2\pi} f^\star(e^{it}) e^{-int}\ dt$ are $0$ for integers $n < 0$, and the Maclaurin series of $f$ on $\mathbb T$ is the Fourier series of $f^\star$. But by Carleson's theorem the Fourier series of any function in $L^2$ converges almost everywhere.

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+1 But I'd like to see a proof without Carleson ... there ought to be a tauberian theorem for that. – user31373 Aug 11 '12 at 5:12
Me too.. By the way, thank you for your clear answer! The fact that the Fourier coefficients of $f^*$ are those of the power series follows from an application of the Dominated convergence theorem, right? – Mizar Aug 11 '12 at 13:30

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