# Uniform Convergence of General Dirichlet Series

## Theorem

Let $\arg \left({z}\right)$ denote the argument of the complex number $z \in \C$.

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

Let $f \left({s}\right)$ converge at $s_0 = \sigma_0 + i t_0$.

Then $f \left({s}\right)$ converges uniformly for all $s$ such that:

$\left\vert{\arg \left({s - s_0}\right)}\right\vert \le a < \dfrac \pi 2$

## Proof

Let $s = \sigma+it$

Let $s_0$ be such that $f \left({s_0}\right)$ converges.

Let $S\left({m, n}\right) = \displaystyle \sum_{ k\mathop = n}^m a_k e^{-\lambda_k s_0}$

We may create a new Dirichlet series that converges at 0 by writing:

 $\displaystyle g \left({s}\right)$ $=$ $\displaystyle f \left({s+s_0}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \left({s + s_0}\right)}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_ns_0} e^{-\lambda_ns}$

Thus it suffices to show $g \left({s}\right)$ converges uniformly for for $\left\vert \arg\left(s\right)\right\vert \le a < \frac \pi 2$

By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$ there exists an $N$ independent of $s$ such that for all $m,n>N$:

$\left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert < \epsilon$

By Abel's Lemma: Formulation 2 we may write:

 $\displaystyle \left\vert \sum_{k \mathop = n}^m a_k e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert$ $=$ $\displaystyle \left\vert \sum_{k \mathop = n}^m \left(S\left(k,n\right) - S\left(k-1,n\right)\right) e^{-\lambda_ks} \right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert S\left(m,n\right)e^{-\lambda_ms} + \sum_{k \mathop = n}^{m-1} S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert$ $\displaystyle$ $\le$ $\displaystyle \left\vert S\left(m,n\right) e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert$ Triangle Inequality

Because $S\left(k,j\right)$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j>N$

$\displaystyle \left\vert S\left(k,j\right) \right\vert < \frac {\epsilon \cos\left(a\right)}{3}$

which gives us:

 $\displaystyle \left\vert S\left(m,n\right) e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert S\left(k,n\right) \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert$ $\le$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left(\left\vert e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert \right)$

We see that:

 $\displaystyle \left\vert e^{-\lambda_ks} - e^{-\lambda_{k+1}s} \right\vert$ $=$ $\displaystyle \left\vert \int_{\lambda_k}^{\lambda_{k+1} } -se^{-xs} \rd x \right\vert$ $\displaystyle$ $\leq$ $\displaystyle \int_{\lambda_k}^{\lambda_{k+1} } \left\vert-se^{-xs}\right\vert \rd x$ Modulus of Complex Integral $\displaystyle$ $=$ $\displaystyle \int_{\lambda_k}^{\lambda_{k+1} } \left\vert s \right\vert e^{-x\sigma} \rd x$ $\displaystyle$ $=$ $\displaystyle \left\vert s \right\vert \int_{\lambda_k}^{\lambda_{k+1} } e^{-x\sigma} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {\left\vert s \right\vert} {\sigma} \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$ $\displaystyle$ $=$ $\displaystyle \sec\left(\arg\left(s\right)\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$

From Shape of Secant Function, we have that on the interval $\left( -\frac \pi 2 , \frac \pi 2\right)$:

$\left\vert \arg\left(s\right) \right\vert \le a \implies \sec\left(\arg\left(s\right)\right) \le \sec\left(a\right)$

which gives us:

$\sec\left(\arg\left(s\right)\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right) \le \sec\left(a\right) \left( e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$

Hence:

 $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left(\left\vert e^{-\lambda_ms} \right\vert + \sum_{k \mathop = n}^{m-1} \left\vert \left(e^{-\lambda_ks} - e^{-\lambda_{k+1}s}\right) \right\vert \right)$ $\le$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \sum_{k \mathop = n}^{m-1} e^{-\lambda_k\sigma} - e^{-\lambda_{k+1}\sigma} \right)$ $\displaystyle$ $=$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \left( e^{-\lambda_n\sigma} - e^{-\lambda_{m}\sigma} \right) \right)$ Telescoping Sum

Because $\sigma>0$, we have that $- \lambda_k \sigma <0$ and hence:

$e^{-\lambda_k \sigma} < 1 \le \sec\left(a\right)$

which gives us:

 $\displaystyle \left\vert \displaystyle \sum_{k \mathop = n}^m a_n e^{-\lambda_ks_0} e^{-\lambda_ks} \right\vert$ $\le$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left( e^{-\lambda_m\sigma} + \sec\left(a\right) \left( e^{-\lambda_n\sigma} - e^{-\lambda_{m}\sigma} \right) \right)$ $\displaystyle$ $\le$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left( \sec\left(a\right) + \sec\left(a\right) \left( 1 + 1 \right) \right)$ $\displaystyle$ $=$ $\displaystyle \frac {\epsilon \cos\left(a\right)}{3} \left( 3 \sec\left(a\right) \right)$ $\displaystyle$ $=$ $\displaystyle \epsilon$

$\blacksquare$