Existence of Abscissa of Absolute Convergence

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Theorem

Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a Dirichlet series.

Let the series $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert { a_n n^{-s} } \right\vert$ not converge for all $s \in \C$, or diverge for all $s \in \C$.


Then there exists a real number $\sigma_a$ such that $f \left({s}\right)$ converges absolutely for all $s = \sigma + it$ with $\sigma > \sigma_a$, and does not converge absolutely for all $s$ with $\sigma < \sigma_a$.

We call $\sigma_a$ the abscissa of absolute convergence of the Dirichlet series.


Proof

Let $S$ be the set of all complex numbers $s$ such that $f \left({s}\right)$ converges absolutely.

By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $f \left({s_0}\right)$ converges absolutely, so $S$ is not empty.

Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lemma that $f \left({s}\right)$ converges absolutely for all $s \in \C$, a contradiction of our assumptions.

Therefore the infimum:

$\sigma_a = \inf \left\{{\sigma: s = \sigma + i t \in S}\right\} \in \R$

is well defined.


Now if $s = \sigma + it$ with $\sigma > \sigma_a$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $f \left({s'}\right)$ is absolutely convergent.

Then it follows from Dirichlet Series Absolute Convergence Lemma that $f \left({s}\right)$ is absolutely convergent.


If $s = \sigma + it$ with $\sigma < \sigma_a$, and $f \left({s}\right)$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$.


Therefore, $\sigma_a$ has the claimed properties.

$\blacksquare$


Note

It is conventional to set $\sigma_a = -\infty$ if the series $f \left({s}\right)$ is absolutely convergent for all $s \in \C$, and $\sigma_a = \infty$ if the series converges absolutely for no $s \in \C$.

Therefore, allowing $\sigma_a$ to be an extended real number, $\sigma_a$ is defined for all Dirichlet series.


Also see


Sources