Advanced Analysis

Dirichlet Characters and \(L\)-functions

Theorem
Suppose that \(n\) and \(d\) are relatively prime positive integers. Then there exist integers \(r_1\) and \(r_2\) such that \(r_1 n + r_2 d = 1\).
Proof
Let \(S\) be the set of integers expressible in the form \(r_1 n + r_2 d\) for integers \(r_1,r_2\). Let \(a\) be the smallest positive integer in \(S\). Clearly \(a\) is no greater than the minimum of \(n\) and \(d\). But also \(a\) must be a divisor of \(n\) and \(d\), since otherwise there would be some multiple \(k\) of \(a\) with \(ka < n < (k+1)a\), which would mean that \(0 < n - ka < a\) and \(n - ka = (1-r_1) n - r_2 d\), so \(n - ka \in S\) would be positive and smaller than \(a\). But \(1\) is the only common divisor, so \(a = 1\).
Corollary
Let \({\mathbb Z}^*_d\) be the set of equivalence classes modulo \(d\) of integers relatively prime to \(d\). Then \({\mathbb Z}^*_d\) is an abelian group.
Proof
The difficult part is showing the existence of inverses, but the Computation immediately above does exactly that: \([r_1][n] = [1]\).

Let \(\chi\) be a character on the multiplicative group \({\mathbb Z}^*_d\). We extend this to a function on the integers by letting \(\chi(n) := \chi([n])\) when \(n\) and \(d\) are relatively prime and \(\chi(n) = 0\) otherwise. All such functions \(\chi\) are known as Dirichlet characters and satisfy \(\chi(nm) = \chi(n)\chi(m)\) for all integers \(n\) and \(m\).
Definition
Given a Dirichlet character \(\chi\), the \(L\)-function associated to \(\chi\) is defined to equal
\[ L(s,\chi) := \sum_{n=1}^\infty \frac{\chi(n)}{n^s}. \]
If \(\chi\) is not trivial (i.e., takes values other than \(0\) and \(1\)), orthogonality of characters implies that
\[ \sum_{[n] \in {\mathbb Z}_d^*} \chi([n]) = 0. \]
Now every residue class modulo \(d\) occurs exactly once as \(n\) ranges from \((k-1)d+1\) to \(kd\). So in particular,
\[ \sum_{n=1}^{kd} \chi(n) = 0 \]
for all natural numbers \(k\). From this observation and the fact that \(|\chi| \leq 1\), it follows that
\[ \left| \sum_{n=1}^N \chi(n) \right| \leq d \]
for all natural numbers \(N\) when \(\chi\) is not trivial.
Theorem (Dirichlet-Type Convergence Test)
Suppose \(\{a_n\}_{n=1}^\infty\) is a sequence of complex numbers for which there is some positive \(C\) such that
\[ \left| \sum_{n=1}^N a_n \right| \leq C \]
for all natural numbers \(N\). Suppose also that \(\{f_n(x)\}_{n=1}^\infty\) is a sequence of continuous, complex-valued functions on some metric space \(X\) such that \(\sup_{x \in X} |f_n(x)| \rightarrow 0\) as \(n \rightarrow \infty\) and
\[ \sum_{n=1}^\infty \sup_{x \in X} |f_n(x) - f_{n+1}(x)| < \infty. \]
Then the series
\[ \sum_{n=1}^\infty a_n f_n(x) \]
converges uniformly to a continuous function on \(X\).
Proof
They key idea is summation by parts. Let \(\{b_n\}_{n=1}^\infty\) be any sequence. Let \(A_n := \sum_{k=1}^n a_k\) for each \(n \geq 1\) and let \(A_0 := 0\). Then
\[{}\sum_{n=1}^N a_n b_n{}\]
\[{}= \sum_{n=1}^N (A_n - A_{n-1}) b_n{}\]
\[{}= \sum_{n=1}^N A_n b_n - \sum_{n=1}^N A_{n-1} b_n{}\]
\[{}= \sum_{n=1}^N A_n b_n - \sum_{n=0}^{N-1} A_{n} b_{n+1}{}\]
\[{}= A_{N} b_{N+1}{}\]
\[{}+ \sum_{n=1}^N A_n (b_n - b_{n+1}).{}\]
Applying the identity pointwise gives that
\[{}S_N (x){}\]
\[{}:= \sum_{n=1}^N a_n f_n(x){}\]
\[{}= A_N f_{N+1}(x){}\]
\[{}+ \sum_{n=1}^N A_N (f_n(x) - f_{n+1}(x)).{}\]
Now for any \(N > M\), taking the difference of partial sums gives that
\[{}|S_N (x) - S_M(x)|{}\]
\[{}\leq |A_N f_{N+1}(x)| + |A_M f_{M+1} (x)|{}\]
\[{}+ \sum_{n=M+1} |A_n| |f_n(x) -f_{n+1}(x)|{}\]
so
\[{}\sup_{x \in X} |S_N (x) - S_M(x)|{}\]
\[{}\leq C \Bigg[ \sup_{x \in X} |f_{N+1}(x)| + \sup_{x \in X} |f_{M+1}(x)|{}\]
\[{}+ \sum_{n=M+1}^{N} \sup_{x \in X} |f_{n}(x) - f_{n+1}(x)| \Bigg].{}\]
The hypotheses guarantee that the right-hand side is smaller than \(\epsilon\) provided that \(M\) and \(N\) are sufficiently large.
Proposition
For any integer \(k\geq 0\) and \(s_0 > 0\),
\[ \frac{(\ln (n+1))^k}{(n+1)^s} - \frac{(\ln n)^k}{n^s} \leq 0 \]
for all \(n \geq e^{k/s_0}\) and all \(s \geq s_0\). Moreover, if \(n \geq e^{(k+1)/s_0}\) then
\[{}\sup_{s \geq s_0} \left|\frac{(\ln (n+1))^k}{(n+1)^s} - \frac{(\ln n)^k}{n^s}\right|{}\]
\[{}= \frac{(\ln n)^k}{n^{s_0}} - \frac{(\ln (n+1))^k}{(n+1)^{s_0}}.{}\]
Proof
Meta (Main Idea)
Use the Mean Value Theorem as a function of \(n\):
\[{}\frac{(\ln (n+1))^k}{(n+1)^s} - \frac{(\ln n)^k}{n^s}{}\]
\[{}= \frac{(\ln \xi)^{k-1} (k - s \ln \xi)}{\xi^{s+1}}{}\]
For some \(\xi\) between \(n\) and \(n+1\). When \(n \geq e^{k/s_0}\) and \(s \geq s_0\), \(\ln \xi \geq k/s_0\), so \(k - s \ln \xi \leq k (1 - ss_0^{-1}) \leq 0\).

Likewise, the function
\[{}\frac{(\ln (n+1))^k}{(n+1)^s} - \frac{(\ln n)^k}{n^s}{}\]
is increasing as a function of \(s\) on \([s_0,\infty)\) when \(n \geq e^{(k+1)/s_0}\) because its derivative in \(s\) is
\[{}-\frac{(\ln (n+1))^{k+1}}{(n+1)^s} + \frac{(\ln n)^{k+1}}{n^s}{}\]
which is nonnegative when \(s \geq s_0\) and \(n \geq e^{(k+1)/s_0}\).
Corollary
When \(\chi\) is not trivial, the \(L\)-function \(L(s,\chi)\) is defined on \(s > 0\) and is infinitely differentiable term-by-term.
Proof
Meta (Main Ideas)
Consider the functions \(f_n(s) := n^{-s}\). Differentiating (with respect to \(s\)) \(k\) times gives \(f_n^{(k)}(s) = (-\ln n)^k n^{-s}\). It suffices to show for each \(s_0 > 0\) that
\[ \sup_{s \geq s_0} | (\ln n)^k n^{-s}| \rightarrow 0 \]
as \(n \rightarrow \infty\) and that
\[{}\sum_{n=1}^\infty \sup_{s \geq s_0} \Big|(-\ln n)^k n^{-s}{}\]
\[{}- (-\ln (n+1))^k (n+1)^{-s}\Big|{}\]
\[{}<\infty.{}\]
For the latter fact, we use the proposition to bound the tail of the series by a telescoping sum.

1. To-Do List

To prove Dirichlet's Theorem, we need to:
  1. Establish that \(L(1,\chi) \neq 0\) when \(\chi\) is a nontrivial character.
  2. Establish for all nontrivial characters that \(L(1,\chi) \neq 0\) implies that
    \[ \sum_{p \text{ prime}} \frac{\chi(p)}{p^s}\]
    remains bounded as \(s \rightarrow 1^+\).
  3. Use the former fact to show that the sum of \(1/p\) over primes \(p\) congruent to \(r\) modulo \(d\) must be infinite by comparison to the argument for the Riemann Zeta Function (because we know that the sum of \(1/p\) over all primes does indeed diverge.) The computation above tells us that primes are in some sense equidistributed by remainder mod \(d\).