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This is a follow-up question to About the limit of the coefficient ratio for a power series over complex numbers.


Cauchy's estimation in complex analysis is a consequence of the Cauchy's integral formula, which can be stated as follow:

Theorem. If $f$ is holomorphic in an open set $\Omega$ that contains a closed disc $D$ centered at $w$ with radius $R$, then for all $n$

$\displaystyle{ \frac{|f^{(n)}(w)|}{n!} \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$, where $C$ is the boundary circle of $D$.

Note that if $f$ has power series expansion $\sum b_n (z-w)^n$ inside $D$, then $b_n = |f^{(n)}(w)|/n!$.

Consider the following question: If $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc except for a pole at $z_0$ on the unit circle, and $f$ has the power series expansion $\sum a_n z^n$ in the open unit disc. By the answer of Sivaram to the previous post, the coefficients $a_n$ of the expansion of $f$ will not converge to zero. I'm wondering if this can be proved by something that is a (partial) converse to the Cauchy's estimation, and apply it backwards we can ensure that $a_n$ will not go to zero.

I would like to know whether the following version exists:

Assume $f$ is holomorphic in an open set $\Omega$ that contains a closed unit disc except at $z_0$ on the unit circle, and $f$ has power series expansion $\sum a_n z^n$ inside the open unit disc. If for all $n$ large enough and all $R < 1$ we have

$\displaystyle{ |a_n| \leq \frac{\sup_{z \in C} |f(z)|}{R^n} }$,

where $D$ is a closed disc centered at $0$ with radius $R$ and $C$ is the boundary circle of $D$, then $f$ is also holomorphic at $z_0$.

This version may not work at all, but any reasonable modifications which can be used to prove that $a_n$ will be bounded away from zero are fine.

Question. Is there a reasonable version to the converse of the Cauchy's estimation?

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    Any function defined by power series is analytic in the interior of the radius of convergence of the series. On the boundary it depends and your $f$ might not even be well defined on the boundary, let alone permit talking about the supremum. Perhaps you would care to rephrase your question?2011-01-20
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    @Moron: Sure, I'll try to revise with a better form.2011-01-20

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