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\begin{fclemmastar}[]
  \glsref[mult]{Multiplicative} \glspl{af} form an abelian group
  with $\conv$. Moreover, for \gls{cmult} $f$, the
  Dirichlet inverse is $\mu f$.
\end{fclemmastar}

\begin{proof}
  For the first part, suffices to show closedness.
  Let $f, g : \Nbb \to \Cbb$ be \gls{mult}, and
  $m, n$ coprime.

  If $mn = ab$, we can write $a = a_1 a_2$,
  $b = b_1 b_2$ where $a_1 = (a, m)$, $a_2 = (a,
  n)$, $b_1 = (b, m)$ and $b_2 = (b, n)$.
  Therefore we have:
  \begin{align*}
    (f \conv g)(mn)
    &= \sum_{mn = ab} f(a) g(b) \\
    &= \sum_{\substack{m = a_1 b_1 \\ n = a_2
    b_2}} f(a_1 a_2) g(b_1 b_2)
    &&\text{(above observation)} \\
    &= \sum_{\substack{m = a_1 b_1 \\ n = a_2
    b_2}} f(a_1) f(a_2) g(b_1) g(b_2)
    &&\text{(by multiplicativity)} \\
    &= (f \conv g)(m) (f \conv g)(n)
  \end{align*}
  (also need to check that inverses are \gls{mult}).

  Now, remains to show that for \gls{cmult} $f$,
  $f \conv \mu f = \dId$.

  Note that $f \conv \mu f$ is \gls{mult} by the first
  part. So enough to show $(f \conv \mu f)(p^k) =
  \dId(p^k)$ for prime powers $p^k$. Calculate:
  \begin{align*}
    f \conv \mu f(p)
    &= f(p) + \mu f(p) \\
    &= f(p) - f(p) \\
    &= 0 \\
    &= \dId(p)
  \end{align*}
  and for $k \ge 2$:
  \begin{align*}
    f \conv \mu f(p^k)
    &= f(p^k) + \mu f(p) f(p^{k - 1}) \\
    &= f(p^k) - f(p) f(p^{k - 1}) \\
    &= f(p)^k - f(p)f(p)^{k - 1} \\
    &= 0 \\
    &= \dId(p^k)
    \qedhere
  \end{align*}
\end{proof}

\begin{example*}
  $\tauf_k(n) = \sum_{n = n_1 \cdots n_k} 1$. Then
  \[
    \tauf_k = \ub{1 \conv 1 \conv \cdots \conv 1}_{\text{$k$
    times}}
  .\]
  So $\tauf_k$ is \gls{mult} by the previous
  result.
\end{example*}

\begin{fcdefnstar}[von Mangoldt function]
  \glssymboldefn{vman}%
  \glsnoundefn{vman}{von Mangoldt function}{NA}%
  The \emph{von Mangoldt function} $\Lambda : \Nbb
  \to \Rbb$ is
  \[
    \Lambda(n) = \begin{cases}
      \log p & \text{if $n = p^k$ for some prime
      $p$} \\
      0 & \text{otherwise}
    \end{cases}
  \]
  Then $\log = 1 \conv \Lambda$ since
  \[
    \log n
    = \log \prod_{p \mid n} p^{\alpha_p(n)}
    = \sum_{p \mid n} \alpha_p(n) \log p
    = 1 \conv \Lambda(n)
  .\]
  Since $\mob$ is the inverse of $1$, we have
  \[
    \log \conv \mob
    = 1 \conv \Lambda \conv \mob
    = \Lambda \conv 1 \conv \mob
    = \Lambda \conv \dId
    = \Lambda
  .\]
\end{fcdefnstar}

\subsection{Dirichlet Series}

For a sequence $(a_n)_{n \in \Nbb}$, we want to
associate a generating function that gives
information of $(a_n)_{n \in \Nbb}$. Might
consider
\[
  (a_n)_{n \in \Nbb}
  \leftrightarrow \sum_{n = 1}^{\infty} a_n x^n
.\]
If we do this, then $\sum_p x^p$ is hard to
control. So this is not very useful.

The following series has nicer number-theoretic
properties:
\begin{fcdefnstar}[Formal series]
  \glssymboldefn{Df}%
  For $f : \Nbb \to \Cbb$, define a (formal)
  series
  \[
    D_f(s) = \sum_{n = 1}^{\infty} f(n) n^{-s}
  \]
  for $s \in \Cbb$.
\end{fcdefnstar}

\begin{fclemmastar}[]
  Assuming:
  - $f : \Nbb \to \Cbb$ satisfying $|f(n)| \le
  n^{o(1)}$
  Then:
  $\Df_f(s)$ converges absolutely for $\Re(s) > 1$
  and defines an analytic function for $\Re(s) >
  1$.
\end{fclemmastar}

\begin{proof}
  Let $\eps > 0$ (fixed), $n \in \Nbb$ and $\Re(s) > 1 +
  2\eps$.

  Then
  \begin{align*}
    \sum_{n = N}^{\infty} |f(n) n^{-s}|
    &= \sum_{n = N}^\infty |f(n)| n^{-\Re(s)} \\
    &\vll \sum_{n = N}^\infty n^{\eps - \Re(s)} \\
    &\le \sum_{n = N}^\infty n^{-1 - \eps} \\
    &\le N^{-1 - \eps} + \sum_{n = N + 1}^\infty
    \int_{n - 1}^n t^{-1 - \eps} \dd t \\
    &\le N^{-1 - \eps} + \int_N^\infty t^{-1 -
    \eps} \dd t \\
    &\vll N^{-\eps}
  \end{align*}
  Hence we have absolute convergence for $\Re(s) >
  1 + 2\eps$. Also, $\Df_f(s)$ is a uniform limit
  of the functions $\sum_{n = 1}^N f(n) n^{-s}$.
  From complex analysis, a uniform limit of
  analytic functions is analytic. Hence $\Df_f(s)$
  is analytic for $\Re(s) > 1 + 2\eps$.

  Now let $\eps \to 0$.
\end{proof}