Uniform Convergence of General Dirichlet Series

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Theorem

Let $\map \arg z$ denote the argument of the complex number $z \in \C$.

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

Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.


Then $\map f s$ converges uniformly for all $s$ such that:

$\cmod {\map \arg {s - s_0} } \le a < \dfrac \pi 2$


Proof

Let $s = \sigma + i t$

Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.

Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$


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

\(\ds \map g s\) \(=\) \(\ds \map f {s + s_0}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \paren {s + s_0} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s_0} e^{-\lambda_n s}\)

Thus it suffices to show $\map g s$ converges uniformly for for $\cmod {\map \arg s} \le a < \frac \pi 2$

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 $m, n > N$:

$\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$

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

\(\ds \cmod {\sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0} e^{-\lambda_k s} }\) \(=\) \(\ds \cmod {\sum_{k \mathop = n}^m \paren {\map S {k, n} - \map S {k - 1, n} } e^{-\lambda_k s} }\)
\(\ds \) \(=\) \(\ds \cmod {\map S {m, n} e^{-\lambda_m s} + \sum_{k \mathop = n}^{m - 1} \map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }\)
\(\ds \) \(\le\) \(\ds \cmod {\map S {m, n} e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }\) Triangle Inequality

Because $\map S {k, j}$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j > N$:

$\ds \cmod {\map S {k, j} } < \frac {\epsilon \cos a} 3$

which gives us:

\(\ds \cmod {\map S {m, n} e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }\) \(\le\) \(\ds \frac {\epsilon \cos a} 3 \paren {\cmod {e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } } }\)

We see that:

\(\ds \cmod {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} }\) \(=\) \(\ds \cmod {\int_{\lambda_k}^{\lambda_{k + 1} } -s e^{-x s} \rd x}\)
\(\ds \) \(\le\) \(\ds \int_{\lambda_k}^{\lambda_{k + 1} } \cmod {-s e^{-x s} } \rd x\) Modulus of Complex Integral
\(\ds \) \(=\) \(\ds \int_{\lambda_k}^{\lambda_{k + 1} } \cmod s e^{-x \sigma} \rd x\)
\(\ds \) \(=\) \(\ds \cmod s \int_{\lambda_k}^{\lambda_{k + 1} } e^{-x \sigma} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\cmod s} \sigma \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }\)
\(\ds \) \(=\) \(\ds \map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }\)

From Shape of Secant Function, we have that on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:

$\cmod {\map \arg s} \le a \implies \map \sec {\map \arg s} \le \sec a$

which gives us:

$\map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} } \le \sec a \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$

Hence:

\(\ds \frac {\epsilon \cos a} 3 \paren {\cmod {e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } } }\) \(\le\) \(\ds \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m \sigma} + \sec a \sum_{k \mathop = n}^{m - 1} e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }\)
\(\ds \) \(=\) \(\ds \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m \sigma} + \sec a \paren {e^{-\lambda_n \sigma} - e^{-\lambda_m \sigma} } }\) Telescoping Sum

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

$e^{-\lambda_k \sigma} < 1 \le \sec a$

which gives us:

\(\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} }\) \(\le\) \(\ds \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m\ sigma} + \sec a \paren {e^{-\lambda_n \sigma} - e^{-\lambda_{m}\sigma} } }\)
\(\ds \) \(\le\) \(\ds \frac {\epsilon \cos a} 3 \paren {\sec a + \sec a \paren {1 + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\epsilon \cos a} 3 \paren {3 \sec a}\)
\(\ds \) \(=\) \(\ds \epsilon\)

$\blacksquare$


Sources

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