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Suppose $f(z)$ has Laurent series expansion in a neighborhood of $z=p$: $$f(z)=\sum^\infty_{k=-\infty}a_k(z-p)^k$$

A series I am interested in is $$S(z):=\sum^{-1}_{k=-\infty}a_{2k}\frac{(z-p)^{2k+1}}{2k+1}$$

Is it true that, for a fixed real $\theta$, $$\lim_{r\to 0^+}S(p+re^{i\theta})=\infty$$ ?

By series multisection, if $p=0$, $S(z)$ is equal to the principal part of the Laurent series of $$\int\frac{f(z)+f(-z)}2dz$$ However, in most cases $p\ne 0$, and I am not sure how does this help to prove the limit.

Any help will be appreciated.

Thanks in advance.

Szeto
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1 Answers1

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Counterexample: take $$f(z) = \sum_{k = -\infty}^{-1}\frac{z^{-2k+1}}{(-2k + 1)!},$$ convergent for every $z\ne 0$ and $S(z) = 0$ for all $z\in\Bbb C$.

In fact, when a finite number of $a_{2k}\ne 0$, $S$ has a pole and your condition is true. With a infinite number of $a_{2k}\ne 0$, $S$ has an essential singularity and the limit does not exists.

Martín-Blas Pérez Pinilla
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  • Can you find a counter example with not all coefficient of $z^{-2n}$ equal to zero? My intuition says no, but there may be exceptions. – Szeto Aug 23 '18 at 22:35
  • @Szeto, with a finite number of $a_{2k}\ne 0$, $S$ has a pole and your condition is true. With a infinite number of $a_{2k}\ne 0$, $S$ has a essential singularity and the limit $\not\exists$. – Martín-Blas Pérez Pinilla Aug 24 '18 at 13:39
  • Thank you. Please add that to your answer, and then I will accept it. – Szeto Aug 24 '18 at 13:57