1
$\begingroup$

Let $\Lambda\in \mathbf{C}$ be a discrete subset. We assume that $\mathrm{Re}(\lambda)<0$ for all the $\lambda\in \Lambda$. For $i\in \mathbf{N}$, $\lambda\in \Lambda$, let $m_{i,\lambda}\in \mathbf{Z}$.

We assume that the sum $$f_i(s)=\sum_{\lambda\in \Lambda} \frac{m_{i,\lambda}}{s-\lambda}$$ has only finite non zero terms.

If for $\mathrm{Re}(s)>0$, the sum $\sum_{i=0}^\infty f_i$ absolutely and uniformly converges to a holomorphic function $f$ defined on $\mathrm{Re}(s)>0$.

I would like to know

  1. Is $\sum_{i=1}^\infty m_{i,\lambda}$ finite? If so, we call it $m_\lambda$.
  2. Does $f$ have a meromorphic extension to $\mathbf{C}$ with pole at $\lambda$ of residue $m_\lambda$.
  3. Edit: If we suppose 1, could we get 2?

Thank you!

$\endgroup$
1
$\begingroup$

If all $m_{i,\lambda}\geq 0$, the answers to all these questions are "yes". The key observation is that $1/(s-\lambda)$ has positive real part when $s$ is in the right half-plane, while $\lambda$ is in the left half-plane. Therefore even if your series converges at ONE point $s$, then $\Re f_j(s)$ is a series with positive terms, from which all your statements immediately follow. Same if all $m_{i,\lambda}\leq 0$.

EDIT. In general, without positivity assumption, this is not true. The following example was given by user fedja.

For $n\geq 2$, let $p_n$ be rational functions with single (multiple) pole at $-n$, $p_n(z)\to \phi$ uniformly in the right half-plane. Here $\phi$ is an arbitrary bounded analytic function in the half-plane $\Re z>-1$.

Such functions $p_n$ are polynomials of $1/(z+n)$ and they exist by Runge's theorem (in fact one does not need Runge's theorem for approximating on a disk). For each $n$, let $Q_n$ be polynomials, partial sums of $e^z$, of sufficiently high degree $d_n$. The degree is such that the zeros of $Q_n\circ p_n$ are all in the disc $|z+n|<1/2$. This is possible because $Q_n(z)\to e^z$ as $d_n\to\infty$ uniformly on compact subsets of the plane, and $e^z$ has no zeros. (Zeros of $Q_n$ tend to $\infty$ but $p_n$ is close to $\infty$ only near the point $-n$.)

Consider now the rational functions $g_n=Q_n\circ p_n$. They converge to the function $e^\phi$ uniformly in the right half-plane.

Zeros and poles of $g_n$ are in the left half-plane, and they tend to $\infty$ as $n\to\infty$. We take the set of all these zeros and poles as our sequence $\Lambda$.

Finally, $$F_n(z):=g_n'(z)/g_n(z)= \sum_{\lambda\in\Lambda}\frac{m_{\lambda,n}}{z-\lambda},$$ where the sum is finite, and $m_{\lambda,n}$ are integers. This sequence tends to $\phi'$ uniformly in the right half-plane. But $\phi'$ was more or less arbitrary, so you can arrange it to have any singularities in the left half-plane. Now convert the sequence $F_n$ into a series, if you wish.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.