Existence of Abscissa of Convergence
Extract the definition of the Abscissa of Convergence into a separate page, and rename this proof of existence to something like Existence of Abscissa of Convergence.
Until this has been finished, please leave {{refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can remove this instance of {{MissingLinks}} from the code.
Contents
Theorem
Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.
Let the series $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
Then there exists a real number $\sigma_c$ such that $\map f s$ converges for all $s = \sigma + it$ with $\sigma > \sigma_c$, and does not converge for all $s$ with $\sigma < \sigma_c$.
We call $\sigma_c$ the abscissa of convergence of the Dirichlet series.
General
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
Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges.
By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges, so $S$ is not empty.
Which hypothesis?
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.
To discuss this proof in more detail, feel free to use the talk page.
If you are able to fix this page, then when you have done so you can remove this instance of {{Questionable}} from the code.
If you would welcome a second opinion as to whether your correction is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Convergence Lemma that $\map f s$ converges for all $s \in \C$, a contradiction of our assumptions.
Therefore the infimum:
- $\sigma_c = \inf \set {\sigma: s = \sigma + i t \in S} \in \R$
is well defined.
Now if $s = \sigma + it$ with $\sigma > \sigma_c$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $\map f {s'}$ is convergent.
Then it follows from Dirichlet Series Convergence Lemma that $\map f s$ is convergent.
If $s = \sigma + it$ with $\sigma < \sigma_c$, and $\map f s$ is convergent then $s$ contradicts the definition of $\sigma_c$.
Therefore, $\sigma_c$ has the claimed properties.
$\blacksquare$
Note
It is conventional to set $\sigma_c = -\infty$ if the series $\map f s$ is convergent for all $s \in \C$, and $\sigma_c = \infty$ if the series converges for no $s \in \C$.
Therefore, allowing $\sigma_c$ to be an extended real number, $\sigma_c$ is defined for all Dirichlet series.
Also see
Sources
- 1976: Tom M. Apostol: Introduction to Analytic Number Theory: $\S 11.6$: Theorem $11.9$
