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I am facing the following question:

Let $f(z)$ be a meromorphic function on $\mathbb{C}$ with simple pole at $a_1,a_2,\dots$ such that $$0<|a_1|\leq |a_2|\leq |a_3|\leq\dots$$ and let $ A_n \equiv \underset{\,\,z=a_n}{\operatorname{Res}}f(z)$, and $C_m$ be a sequence of simple closed contours.

Prove that if the following are true:

  1. $C_m$ does not pass through any $a_n$;
  2. $\lim_{m\to\infty}r_m=\infty$, where $r_m=d(C_m,0)$;
  3. there is a constant $\beta$ such that $L_m=$Length of $C_m\leq \beta r_m$;
  4. $max_{z\in C_m}|f(z)|=\circ(r_m)$.

then $$f(z)=f(0)+\sum_{n=1}^{\infty} A_n(\frac{1}{z-a_n}+\frac{1}{a_n}).$$

I tried to show the convergence of the above summation first and then by Mittag-Leffler Theorem in $\mathbb{C}$ we can get $f(z)=\sum_{n=1}^{\infty} A_n(\frac{1}{z-a_n}+\frac{1}{a_n})+g(z)$ where $g$ is entire.

But I can't show the convergence since in these assumptions, $C_m$ may encircle no poles, so they don't give any information for convergence. I think we may assume more. For example, we assume $C_m$ contains the origin. Then how to show the convergence and if we can show the convergence, how to estimate $g$ ?

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Koma
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1 Answers1

1

We want to show that $$ \tag 1 f(z) = f(0) + \sum_{n=1}^\infty \alpha_n \left( \frac{1}{z-z_n} + \frac{1}{z_n} \right), $$ where

  1. $f$ is a meromorphic function with simple poles at $z_n, \, n \in \mathbb{N}$, holomorphic at $z=0$,
  2. $\alpha_n$ is the residue of $f$ at the pole $z_n$,
  3. $f$ is uniformly bounded on every circle $\mathcal C_n$ with radius $R_n$ containing all poles $z_k$ with $k \le n$, i.e. $$ \tag 2\exists A > 0 | \,\, \forall (n \in \mathbb{N}, z \in \mathcal C_n), \,\, |f(z)| \le A $$

To do that, suppose without loss of generality that the poles $z_n$ are ordered monotonically as in $$ 0 < |z_1| \le |z_2| \le \dots, $$ and fix some $z_0 \in \mathbb{C}$ such that $f$ is holomorphic at $z_0$.

Consider the integral $$ \tag 3 I_n(z_0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z-z_0}, $$ which applying Residue's theorem becomes $$ \tag 4 I_n(z_0) = f(z_0) + \sum_{k=1}^n \frac{\alpha_k}{z_k-z_0}. $$ In particular for $z_0=0$ we have $$ \tag 5 I_n(0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z} = f(0) + \sum_{k=1}^n \frac{\alpha_k}{z_k}.$$ Taking the difference between (3) and (5) (in the integral form) we see that: $$ \tag 6 |I_n(z_0) - I_n(0) | \le \oint_{\mathcal C_n} \frac{|dz|}{2\pi} \left| \frac{z_0 f(z)}{(z-z_0) z} \right| \le \frac{Az_0}{R_n - |z_0|} \to 0, \text{when } n \to \infty.$$ However, we also know that $$ \tag 7 I_n(z_0)-I_n(0) = f(z_0) - f(0) - \sum_{k=1}^n \left( \frac{1}{z_0-z_k} + \frac{1}{z_k} \right), $$ which taking the limit $n\to \infty$ gives the wanted result (1).

For an example of application of (1) see e.g. here.

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glS
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