Existence of Abscissa of Convergence/General

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $s = \sigma + i t$

Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s}$ be a general Dirichlet series.

Then there exists a extended real number, $\sigma_0$, such that

$(1): \quad$ For $\sigma < \sigma_0$, $\map f s$ diverges
$(2): \quad$ For $\sigma > \sigma_0$, $\map f s$ converges.


Proof

If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is vacuously true

If there does exist such an $s_0$, let $\sigma_0$ be the infimum of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below.

  • It is clear that $\map f s$ diverges if $\sigma < \sigma_0$ by construction of $\sigma_0$ as the infimum of all convergent $s$
  • As $\sigma_0$ is the infimum of the the real part of the set all s such that $\map f s$ converges, for any $\epsilon > 0$, there exists an $s$ such that $\map f s$ converges and $\sigma_0 \le \sigma < \sigma_0+ \epsilon$ as otherwise $\sigma_0 + \epsilon$ would be the infimum of the set.
Therefore, for any $s$ such that $\sigma > \sigma_0$, we can pick $\epsilon = \sigma - \sigma_0$ and find a $q$ such that $\map f q$ converges and $\map \Re q < \sigma$.
By Dirichlet Series Convergence Lemma, We therefore have that $\map f s$ converges.

$\blacksquare$