\section{Introduction}  \label{sec:intro}

Let $\mu(n)$ denote the Moebius arithmetic function and $M(x)$ its sum
function. Von Mangoldt first established the estimate $M(x) = o(x)$,
essentially going through the steps used in proving the Prime Number
Theorem (PNT).  A few years later, Landau showed by relatively simple
real variable arguments that this and several other estimates followed
from the PNT \cite[\S150]{landau}.  Similarly, these relations imply each
other and the PNT; thus they are said to be ``equivalent'' to the PNT.

In this article we consider an analog of the $M$ bound for Beurling
generalized (g-)numbers.  A Beurling g-prime system is an unbounded
sequence of real numbers ${\mathcal{P}} =\{p_i\}$ satisfying $1 < p_1 \le p_2
\le \ldots\,$, and the multiplicative semigroup generated by $\mathcal{P}$ and
1 is called the associated collection $\mathcal{N}$ of g-integers
(cf.~\cite{BD69}, \cite{diamond-zhangbook}, \cite{MV07}).  The
counting function $N(x)$ of $\mathcal{N}$ is the number of elements of $\mathcal{N}$
not exceeding $x$.  The g-Chebyshev function of $\mathcal{N}$ is
\[
\psi(x) = \sum_{p_i^{\alpha_i} \le x} \log p_i.
\]
(Here and below we give our Beurling functions the same names used in
classical number theory.)

The PNT-related assertions are somewhat different  
for g-numbers: not all implications between the several corresponding
assertions hold unconditionally, see e.g.~\cite{d-vPNTequiv2016},
\cite[Chap.~14]{diamond-zhangbook}, \cite{zhang1987}.  Here we study
the Beurling version of the assertion $M(x) = o(x)$ and show this can
be deduced (a) from the PNT under an O-boundedness condition
$N(x)=O(x)$ or (b) from a Chebyshev-type upper bound assuming
$N(x)\sim ax$ (for $a>0$) and the integral condition \eqref{L1cond}
(below).  At the end, we give examples in which $M$ estimates fail.


A word about the Beurling version of $M(x)$.  The characteristic
property of the Moebius function is its being the (multiplicative)
convolution inverse of the $1$ function.  For g-numbers, we define the
measure $\mathrm{d}M$ as the convolution inverse of $\mathrm{d}N$; by familiar Mellin
transform properties (cf. \cite{BD04}, \cite{diamond-zhangbook},
\cite{MV07}), 
\[
 \int_{1-}^\infty u^{-s} \mathrm{d}M(u) = 1/\zeta(s) \ \, {\rm with} \ \,
\zeta(s) = \zeta_{\mathcal{N}}(s) = \int_{1-}^\infty u^{-s} \mathrm{d}N(u),
\]
the Beurling number version of the Riemann zeta function. Note that
these definitions make sense even when $\mathrm{d}N$ is not discrete or when
factorization into primes is not unique.
\smallskip

We thank Wen-Bin Zhang for his useful comments.




\section{PNT hypothesis}  \label{sec:PNT}

In this section we show that the PNT together with the ``O-density''
condition $N(x) \ll x$ implies $M(x) = o(x)$.
\begin{theorem}  \label{th:M}
Let $\mathcal{N}$ be a g-number system for which the PNT holds and $N(x) \ll
x$.   Then $M(x) = o(x)$. 
\end{theorem}

Other sufficient conditions for an $M$ estimate are known, 
e.g.~\cite[ Prop.~14.10]{diamond-zhangbook}, if the PNT holds for a
g-number system and the integer counting function satisfies the
logarithmic density condition
\begin{equation}
\label{log-density}
\int_{1^{-}}^{x}\frac{\mathrm{d}N(t)}{t}\sim a\log x,
\end{equation}
then $M(x) = o(x)$.  The present result differs from the other in that
log-density and O-density are conditions that do not imply one another;
also, the proofs are very different.


The key to our argument is the following relation.
\begin{lemma} \label{lem:key}
Under the hypotheses of the theorem,
\begin{equation}  \label{eq:main-ident}
\frac{M(x)}{x} = \frac {-1}{\log x}\int_1^x \!\frac{M(t)}{t^2}\,
\mathrm{d}t + o(1). 
\end{equation}
\end{lemma}


\begin{proof}[Proof of the lemma]
A variant of Chebyshev's identity for primes reads
\[
L\mathrm{d}M =  -\mathrm{d}M * \mathrm{d}\psi,
\]
where $L$ is the operator of multiplication by $\log t$ and $*$ is
multiplicative convolution.  This can be verified (in the classical or
in the Beurling case) via a Mellin transform.  Note that this
transform carries convolutions into pointwise products and the $L$
operator into differentiation.  The equivalent Mellin formula is the
identity
\[
\Big\{\frac{1}{\zeta(s)} \Big\}' = -\frac{1}{\zeta(s)} \cdot
\frac {\zeta'(s)}{\zeta(s)}.
\]


Now add and subtract the term $(\delta_1+\mathrm{d}t)*\mathrm{d}M$ in
the variant of the
Chebyshev relation, with $\delta_1$ the point mass at 1 and $\mathrm{d}t$
the Lebesgue measure on $(1,\, \infty)$.  Integrating, we find
\begin{equation}  \label{eq:Cheb'}
\int_{1-}^x L\,\mathrm{d}M =  \int_{1-}^x\mathrm{d}M *
(\delta_1+\mathrm{d}t - \mathrm{d}\psi) - \int_{1-}^x\mathrm{d}M 
* (\delta_1+\mathrm{d}t).
\end{equation}

Integrating by parts the left side of the last formula, we get
\[
M(x) \log x - \int_1^x \frac{M(t)}{t}\, \mathrm{d}t = M(x) \log x + O(x),
\]
since $|\mathrm{d}M| \le \mathrm{d}N$, and hence $|M(t)| \le N(t) \ll
t$.  If we evaluate the convolution integrals by the iterated integral
formula
\[
 \int_{1-}^x\mathrm{d}A * \mathrm{d}B = \int\!\!\!\int_{st \le x}\!
 \mathrm{d}A(s)\, \mathrm{d}B(t) = \int_{1-}^xA(x/t)\, \mathrm{d}B(t)
\]
and use the PNT, we find the first term on the right side of
\eqref{eq:Cheb'} to be 
\[
\int_{1-}^x \{x/t-\psi(x/t)\}\,\mathrm{d}M(t) = 
\int_{1-}^x o(x/t)\,\mathrm{d}N(t) = o(x \log x).
\]
Also, the last term of \eqref{eq:Cheb'} is, upon integrating by parts,
\[
\int_{1-}^x \frac xt \, \mathrm{d}M(t) = M(x) + x\!\int_1^x
\!\frac{M(t)}{t^2}\,\mathrm{d}t. 
\]

Finally, we combine the bounds for the terms of \eqref{eq:Cheb'},
divide through by $x \log x$, and note that $M(x) \ll x$ to obtain
\eqref{eq:main-ident}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th:M}]
First, we can assume that $M(x)$ has an infinite number of sign
changes.  Otherwise, there is some number $z$ such that $M(x)$ is of
one sign for all $x \ge z$.  By \eqref{eq:main-ident}, as $x \to \infty$,
\[
\frac{M(x)}{x}
= \frac {-1}{\log x}\, \int_1^z
\!\frac{O(t)}{t^2}\,\mathrm{d}t -\frac{1}{\log x}\int_z^x
\!\frac{M(t)}{t^2}\,\mathrm{d}t  + o(1).
\]
Thus we have
\[
\frac{M(x)}{x} 
+ \frac{1}{\log x}\int_z^x \!\frac{M(t)}{t^2}\,\mathrm{d}t
=  \frac {O(\log z)}{\log x} +o(1) = o(1).
\]
Since $M(x)$ and the integral are of the same sign, $M(x)/x \to 0$ 
as $x \to \infty$, and this case is done.

Now suppose that $M(x)$, which we regard as a right-continuous
function, changes sign at $x$.  We show that $M(x)/x = o(1)$.  If
$M(x)/x = 0$, there is nothing more to say here.  If, on the other
hand, $M(x) > 0$, then there is a number $y \in (x-1,\ x)$ with $M(y)
\le 0$.  If we apply \eqref{eq:main-ident} again, we find
\begin{align*}
\frac{M(x)}{x}
&= \frac {-\log y}{\log x}\, \frac{1}{\log y}\int_1^y
\!\frac{M(t)}{t^2}\,\mathrm{d}t -\frac{1}{\log x}\int_y^x
\!\frac{M(t)}{t^2}\,\mathrm{d}t  + o(1) \\
&= \frac {\log y}{\log x}\,\frac{M(y)}{y} + 
\frac{1}{\log x} \int_y^x \frac{O(t)\,\mathrm{d}t}{t^2} +o(1). 
\end{align*}
Thus $M(x)/x - M(y)/y  = o(1)$, and since $M(x)/x> 0 \ge M(y)/y$,
each is $o(1)$.  A similar story holds if $M(x) < 0$.

Finally, suppose that $M(t)$ changes sign at $t=y$ (so that
$M(y)/y = o(1)$) and  $M(t)$ is of one sign 
for $y < t \le z$.  By yet another application of
\eqref{eq:main-ident},  we find for any $x \in (y,\, z]$
\[
\frac{M(x)}{x}
= \frac {-\log y}{\log x}\, \frac{1}{\log y}\int_1^y
\!\frac{M(t)}{t^2}\,\mathrm{d}t -\frac{1}{\log x}\int_y^x
\!\frac{M(t)}{t^2}\,\mathrm{d}t  + o(1)
\]
or
\[
\frac{M(x)}{x} + \frac{1}{\log x}\int_y^x \!\frac{M(t)}{t^2}\,\mathrm{d}t 
= \frac {\log y}{\log x}\,\frac{M(y)}{y} +o(1) = o(1)
\]
as $y \to \infty$.  Since $M(x)/x$ and the integral are of the same
sign, it follows that $M(x)/x = o(1)$.
\end{proof}





\section{Chebyshev hypothesis}  \label{sec:cheb}

What happens if the PNT hypothesis of the last theorem is weakened to
just a Chebyshev upper bound? In the examples of Section \ref{Sect
  examples} we show that even two-sided Chebyshev estimates by
themselves are not strong enough to ensure that $M(x) = o(x)$ holds.
Furthermore, this bound could fail even if, in addition to Chebyshev
estimates, one also assumes that $N$ satisfies both
(\ref{log-density}) and $N(x)\ll x$, see Example \ref{ex1C-M}. We
shall show, however, that if the regularity hypothesis on $N$ is
slightly augmented, then one can indeed deduce the desired $M$ bound.


\begin{theorem} \label{thCheby-moebius} Suppose that a g-number
 system satisfies the following conditions: \begin{enumerate}
 \item [(a)] a Chebyshev upper bound, that is,  
\begin{equation}
\label{chebyeq1} \limsup_{x\to\infty} \frac{\psi(x)}{x} < \infty,
\end{equation}
\item [(b)]  for some positive constant $a, \ N(x) \sim ax$,
\item [(c)] for some $\b \in (0,1/2)$ and all $\sigma \in (1,2)$
\begin{equation}
\label{L1cond} \int_{1}^{\infty} \frac{|N(x)-ax|}{x^{\sigma+1}}\:
\mathrm{d}x \ll (\sigma-1)^{-\b}\,.
\end{equation} 
 \end{enumerate}
Then, $M(x)=o(x)$ holds.
\end{theorem}

As a simple consequence we have an improvement of a result of
W.-B.~Zhang \cite[Cor.~2.5]{zhang1987}:
\begin{corollary}  If a g-number system satisfies \eqref{chebyeq1} and
\[
N(x) - ax \ll x \log ^{-\gamma} x
\]
for some $\gamma > 1/2$, then $M(x)=o(x)$ holds.
\end{corollary}

 W.-B. Zhang had conjectured\footnote{Oral communication to the authors.} that a Chebyshev
 bound along with the $L^{1}$ bound 
\begin{equation}
\label{strongL1cond} \int_{1}^{\infty} \frac{|N(x)-ax|}{x^{2}}
\mathrm{d}x <\infty
\end{equation} 
imply $M(x) = o(x)$.  Naturally, (\ref{strongL1cond}) is included in
(\ref{L1cond}) and $N(x)\sim ax$. 

It is easy to show that $M(x) = o(x)$ is always implied by
\begin{equation} \label{eqlandausharprelation}
 m(x) =\int_{1^{-}}^{x}\frac{\mathrm{d}M(u)}{u}=

 o(1),
\end{equation}
but the converse implication is not true in general
\cite{diamond-zhangbook}.  On the other hand, it has recently been
shown \cite[Thm. 2]{d-vPNTequiv2016} that the assertions become
equivalent under the additional hypothesis (\ref{strongL1cond}).  So, Theorem
\ref{thCheby-moebius} and the quoted result yield at once:

\begin{corollary} \label{corCheby-moebius} Under (\ref{chebyeq1}) and
  (\ref{strongL1cond}), relation (\ref{eqlandausharprelation})
  holds as well. 
\end{corollary}

We can also strengthen another result of W.-B. Zhang
\cite[Thm. 2.3]{zhang1987}.  

\begin{corollary} \label{cor2Cheby-moebius} The condition
(\ref{strongL1cond})
and 
\begin{equation}\label{avcondpointwiseeq}
\int_{1}^{x}\frac{(N(t)-at)\log t}{t}\:\mathrm{d}t\ll x
\end{equation}
imply (\ref{eqlandausharprelation}).
\end{corollary}
\begin{proof} Chebyshev bounds are known to hold under the hypotheses
  (\ref{strongL1cond}) and (\ref{avcondpointwiseeq})
  \cite[Thm. 11.1]{diamond-zhangbook}. The rest follows from Corollary
  \ref{corCheby-moebius}. 
\end{proof}

We shall prove Theorem \ref{thCheby-moebius} using several lemmas.
Our method is inspired by W.-B. Zhang's proof of a H{a}l\'{a}sz-type
theorem for Beurling primes \cite{zhang1987}.  Our first step is to
replace $M(x)=o(x)$ by an equivalent asymptotic relation.

\begin{lemma}\label{lem1C-M} Suppose that the integer counting
  function of a g-number system has a positive density, i.e.,
  $N(x)\sim a x$ for some $a>0$. Then, $M(x)=o(x)$ if and only if
\begin{equation}
\label{MoebiusAv} f(x)=\int_{1}^{x}\left(\int_{1}^{u}\log t\:
  \mathrm{d}M(t) \right)\frac{\mathrm{du}}{u}=o(x\log x). 
\end{equation}
\end{lemma}
\begin{proof}The direct implication is trivial. For the converse, we
  set 
\[
g(x)=\int_{1^{-}}^{x} \log t\, (\mathrm{d}N(t)+\mathrm{d}M(t)). 
\]
Notice that 
\[
\mathrm{d}N+\mathrm{d}M=2\sum_{n=0}^{\infty}\,
\mathrm{d}\Pi^{\ast 2n}/(2n)!
\]
is a non-negative measure, so that $g$ is non-decreasing. The
hypotheses (\ref{MoebiusAv}) and $N(x)\sim ax$ give
$$
\int_{1}^{x}\frac{g(u)}{u}\,\mathrm{d}u\sim  a x\log x.
$$
Hence,
$$
\int_{1}^{x}g(u)\,\mathrm{d}u= x \int_{1}^{x}\frac{g(u)}{u} \,\mathrm{d}u
- \int_{1}^{x}\!\left(\int_{1}^{t}\frac{g(u)}{u} \,\mathrm{d}u\right)
\mathrm{d}t \sim \frac{a}{2} x^{2}\log x.  
$$
Since $g$ is a non-decreasing function, we see by a simple
differencing argument (see e.g. \cite[p. 34]{korevaarbook}) that
$g(x)\sim a x\log x$. But also we have $\int_{1^{-}}^{x}\log u\,
\mathrm{d}N(u)\sim a x\log x $; consequently, $\int_{1^{-}}^{x}\log
u\, \mathrm{d}M(u) = o( x\log x)$.  Integration by parts now yields
$$
\log x\, M(x)= \int_{1}^{x}\log u\, \mathrm{d}M(u) 
+\int_{1}^{x} \frac{M(u)}{u}\, \mathrm{d}u = o(x\log x)+O(x),
$$
and the result then follows by dividing by $\log x$.
\end{proof}

The next lemma provides a crucial analytic estimate.
\begin{lemma}\label{lem3CC-M} Suppose that $N(x)\sim a x$, with $a>0$. Then, 
\begin{equation}
\label{invzeta}
\frac{1}{\zeta(\sigma+it)}=o\left(\frac{1}{\sigma-1}\right), \quad
\sigma \to 1^{+},
\end{equation}
uniformly for $t$ on compact intervals.
\end{lemma}
\begin{proof} We first show that (\ref{invzeta}) holds pointwise,
i.e., for each fixed $t\in \mathbb{R}$, without the uniformity
requirement. If $t=0$ this is clear because $\zeta(\sigma)\sim
a/(\sigma-1)$ and thus $1/\zeta(\sigma)=o(1)$. Note that $N(x)\sim a x$ 
implies  
$$
\zeta(s)=\frac{a}{s-1}+o\left(\frac{|s|}{\sigma-1}\right)
$$
uniformly. If $t\neq 0$, we obtain, $\zeta(\sigma+2it)=o_{t}(1/(\sigma-1))$. 
Applying the 3-4-1 inequality, we conclude that 
$$
1\leq |\zeta(\sigma)|^{3}|\zeta(\sigma+it)|^{4} |\zeta(\sigma+2it)|
=|\zeta(\sigma+it)|^{4} \,o_{t}\!\left(\frac{1}{(\sigma-1)^{4}}\right), 
$$
which shows our claim. Equivalently, we have 
$$
\exp\left(-\int_{1}^{\infty}x^{-\sigma}(1+\cos(t\log x))
\,\mathrm{d}\Pi(x)\right)
=\frac{1}{|\zeta(\sigma)\zeta(\sigma+it)|}=o_{t}(1).
$$
The left-hand side of the last formula is a net of continuous
monotone functions in the variable $\sigma$ that tend pointwise to 0 as $\sigma\to 1^{+}$;
Dini's theorem then asserts that it must also converge to 0 uniformly
for $t$ on compact sets, as required. 
\end{proof}

The next lemma is very simple but useful.


\begin{lemma}\label{lem3C-M}Let $F$ be a right continuous function of
  local bounded variation with support in $[1,\infty)$ satisfying the
  bound $F(x)=O(x)$. Set $\widehat{F}(s)=\int_{1^{-}}^{\infty}x^{-s}
  \mathrm{d}F(x)$, $\Re s>1$. Then,  
$$
\int_{\Re s=\sigma} \left| \frac{\widehat{F}(s)}{s}\right|^{2}
|\mathrm{d}s|\ll \frac{1}{\sigma-1} \,.
$$
\end{lemma}
\begin{proof}
Indeed, by the Plancherel theorem,
$$
\int_{\Re s=\sigma}  \left| \frac{\widehat{F}(s)}{s}\right|^{2}
|\mathrm{d}s|=2\pi\! \int_{0}^{\infty}\!\! e^{-2\sigma x}
|F(e^{x})|^{2}\,\mathrm{d}x\ll \int_{0}^{\infty}\!\! e^{-2(\sigma-1) x}
\,\mathrm{d}x= \frac{1}{2(\sigma-1)}\,. \qedhere
$$
\end{proof}


If a g-number system satisfies a Chebyshev upper bound, then Lemma
\ref{lem3C-M} implies  
\begin{equation}
\label{L2Chevy}
\int_{\Re s=\sigma} \left|\frac{\zeta'(s)}{s\zeta(s)}\right|^{2} 
|\mathrm{d}s|\ll \frac{1}{\sigma-1}.
\end{equation}
Similarly, $N(x)=O(x)$ yields
\begin{equation}
\label{L2zeta}
\int_{\Re s=\sigma} \left|\frac{\zeta(s)}{s}\right|^{2} |\mathrm{d}s|
\ll \frac{1}{\sigma-1}.
\end{equation}

Also, we shall need the following version of the Wiener-Wintner theorem
\cite{bateman-diamond2000,montgomeryBook1970}.  
\begin{lemma}\label{lem4C-M} Let $F_{1}$ and $F_{2}$ be right
continuous functions of local bounded variation with support in
$[1,\infty)$. Suppose that their Mellin-Stieltjes transforms
$\widehat{F}_{j}(s)=\int_{1^{-}}^{\infty}x^{-s} \mathrm{d}F_{j}(x)$
are convergent on $\Re s>\alpha$, that $F_{1}$ is non-decreasing, and
$|\mathrm{d}F_{2}|\leq \mathrm{d}F_{1}$. Then, for all
$b\in\mathbb{R}$, $c>0$, and $\sigma>\alpha$,
$$
\int_{b}^{b+c}|\widehat{F}_{2}(\sigma+it)|^{2}\,\mathrm{d}t
\leq 2 \int_{-c}^{c}|\widehat{F}_{1}(\sigma+it)|^{2}\,\mathrm{d}t.
$$
\end{lemma}


We are ready to present the proof of Theorem \ref{thCheby-moebius}.
\begin{proof}[Proof of Theorem \ref{thCheby-moebius}] 
In view of Lemma
  \ref{lem1C-M}, it suffices to show (\ref{MoebiusAv}).  
The Mellin-Stieltjes transform of the function $f$ is 
$$
-\frac{1}{s}\left(\frac{1}{\zeta(s)}\right)'.
$$
Given $x > 1$, it is convenient to set $\ss=1+1/\log x$.  By the
Perron inversion formula, the Cauchy-Schwarz inequality, and
(\ref{L2Chevy}), we have
\begin{align*}
\nonumber
\frac{|f(x)|}{x}&=\frac{1}{2\pi}\left|\int_{\Re s=\ss}  
\frac{x^{s-1}\zeta'(s)}{s^{2}\zeta^{2}(s)}\, \mathrm{d}s
\right|\leq \frac{e}{2\pi} \int_{\Re s=\ss}
\left|\frac{\zeta'(s)}{s^{2}\zeta^{2}(s)}\right| |\mathrm{d}s|  \\
&
\ll \log^{1/2}x \left( \int_{\Re s=\ss} \left|
\frac{1}{s\zeta(s)}\right|^{2} |\mathrm{d}s| \right)^{1/2}.
\end{align*}

Next, we take a large number $\lambda$, arbitrary but fixed. We
split the integration line $\{\Re s=\ss\}$ of the last integral into two
parts, $\{\ss+it:\: |t|\geq \lambda\} $ and $\{\ss+it:\: |t|\leq
\lambda\} $, and we denote the corresponding integrals over these sets
as $I_{1}(x)$ and $I_{2}(x)$ respectively, so that
\begin{equation}
\label{eq1proofM}
\frac{|f(x)|}{x}\ll \big((I_{1}(x))^{1/2}+(I_{2}(x))^{1/2}\big) \log^{1/2}x.
\end{equation}

To estimate $I_{1}(x)$, we apply Lemma \ref{lem4C-M} to
$|\mathrm{d}M|\leq \mathrm{d}N$ and employ  (\ref{L2zeta}), 
$$
\left(\int_{-\lambda -m-1}^{-\lambda-m} +\int_{\lambda+m}^{\lambda+m+1}
\right)\left|\frac{1}{\zeta(\ss+it)}\right|^{2} \mathrm{d}t\leq 4
\int_{-1}^{1} \left|\zeta(\ss+it)\right|^{2} \mathrm{d}t \ll \log
x. 
$$
Therefore, we have the bound
\begin{equation}
\label{eq2proofM}
I_{1}(x)\ll \log x \sum_{m=0}^{\infty} \frac{1}{1+m^{2}
+\lambda^{2}}\ll \frac{\log x}{\lambda}.
\end{equation}

To deal with $I_{2}(x)$, we need to derive further properties of the
zeta function. Using the hypothesis \eqref{L1cond}, we find that
\begin{align*}
\zeta(s)-\frac{a}{s-1}&= s \!\int_{1}^{\infty}\!\!
x^{-s}\frac{N(x)-ax}{x}\:\mathrm{d}x +a \\
&\ll |s|\!\int_{1}^{\infty}\!\! x^{-\sigma}\frac{|N(x)-ax|}{x}\,
\mathrm{d}x =O_t\!\left(\frac{1}{(\sigma-1)^{\beta}}\right). 
\end{align*}
Hence, we obtain
\begin{equation}
\label{eq3proofM}
\zeta(\ss+it)= \frac{a}{\ss-1+it}
+ O(\log^{\beta} x)
\end{equation}
for some number $\beta \in (0,\,1/2)$, uniformly for $t$ on
compact sets. We are ready to estimate $I_{2}(x)$. Set
$\eta=(1-2\beta)/(1-\beta)$ and note that $\eta \in (0,\, 1)$.  Then,
using Lemma~\ref{lem3CC-M},
$$
I_{2}(x)\leq \int_{-\lambda}^{\lambda} \left|
\frac{1}{\zeta(\ss+it)}\right|^{2}\mathrm{d}t\leq 
\left(\int_{-\lambda}^{\lambda} \left|
\frac{1}{\zeta(\ss+it)}\right|^{2-\eta}\mathrm{d}t\right) 
 o_{\lambda}(\log^{\eta}x).
$$   
On the other hand, applying  Lemma \ref{lem4C-M} to 
$\mathrm{d}F_{1}=\exp^{\ast}(-(1-\eta/2)\,\mathrm{d}\Pi)$ and
$\mathrm{d}F_{2}=\exp^{\ast}((1-\eta/2)\,\mathrm{d}\Pi)$
and using  (\ref{eq3proofM}), we find 
\begin{align*}
\int_{-\lambda}^{\lambda}\left|\frac{1}{\zeta(\ss+it)}\right|^{2-\eta}\mathrm{d}t
&
\leq 4\int_{-\lambda}^{\lambda} \left|\zeta(\ss+it)\right|^{2-\eta}\mathrm{d}t
\\
&
 \ll  \int_{-\lambda}^{\lambda} \frac{\mathrm{d}t}{ ((\ss-1)^{2}+t^{2})^{1-\eta/2}}
+\lambda\log^{(2-\eta)\beta}x\ll \lambda \log^{1-\eta}x,
\end{align*}
which implies $I_{2}(x)= o_{\lambda}(\log
x)$. Inserting this and the bound (\ref{eq2proofM}) into
(\ref{eq1proofM}), we arrive at 
$
|f(x)|/(x\log x)\ll \lambda^{-1/2} +o_{\lambda}(1).$ Taking first the limit
superior as $x\to\infty$ and then $\lambda\to\infty$, we have shown
that 
$$
\lim _{x\to\infty} \frac{f(x)}{x\log x}=0.
$$
By Lemma~\ref{lem1C-M}, $M(x) = o(x)$.
This completes the proof of Theorem \ref{thCheby-moebius}.
\end{proof}



\section{Three examples}\label{Sect examples}
The examples of this section center on the importance of
$N(x)$ being close to $ax$ in Theorem \ref{thCheby-moebius}.  In
the first example, $N(x)/x$ has excessive wobble and in the second
one, excessive growth; in both cases $M(x) = o(x)$ fails. The third
example shows that condition~\eqref{L1cond} is not sufficient to
insure the convergence of $N(x)/x$, whence the introduction of this
hypothesis. 

In preparation for treating the first two examples, we give a
necessary condition for $M(x) = o(x)$. An analytic function $G(s)$ on
the half-plane $\{s:\:\Re s>\alpha\}$ is said to have a right-hand zero of
order $\beta>0$ at $s=it_{0}+\alpha$ if $\lim_{\sigma\to\alpha^{+}}
(\sigma -\alpha)^{-\beta}G(it_{0}+\sigma)$ exists and is non-zero.
Our examples violate the following necessary condition:
\begin{lemma}
\label{lexC-M}
If $M(x)=o(x)$, then $\zeta(s)$ does not have any right-hand zero of
order $\geq 1$ on $\{\Re s=1\}$. 
\end{lemma}
\begin{proof} We must have
$$
\frac{1}{\zeta(\sigma+it)}=s\!\int_{1}^{\infty} x^{-it-\sigma}
\frac{M(x)}{x}\:\mathrm{d}x= o\!\left(\frac{1}{\sigma-1}\right), \quad
\sigma\to 1^{+},  
$$
uniformly for $t$ on compact intervals.
\end{proof}

\begin{example}
\label{ex1C-M}
We consider
$$
\Pi (x)=\sum_{2^{k+1/2}\leq x} \frac{2^{k+1/2}}{k}\,.
$$
This satisfies the Chebyshev bounds: we have
\[
\psi(x)=  2^{\lfloor \log x/\log 2 + 1/2\rfloor +1/2}\log 2 +O(x/\log x).
\]
Thus
$$
\liminf_{x\to\infty} \frac{\psi(x)}{x}=\log 2 \quad \mbox{and} \quad
\limsup_{x\to\infty} \frac{\psi(x)}{x}=2 \log 2. 
$$
Further, the zeta function of this g-number system can be
explicitly computed:  
$$
\log \zeta(s)= 2^{-(s-1)/2}\sum_{k=1}^{\infty} \frac{2^{-k(s-1)}}{k} = - 2^{-(s-1)/2}\log(1-2^{-(s-1)}).
$$
We conclude that 
$$\zeta(\sigma)\sim \frac{1}{(\sigma-1)\log 2}, \quad \sigma\to1^{+};$$
therefore, by the Hardy-Littlewood-Karamata Tauberian Theorem
\cite{diamond-zhangbook}, \cite{korevaarbook}, $N$ has logarithmic density 
\begin{equation}
\label{log-densityex1}
\int_{1^{-}}^{x} \frac{\mathrm{d}N(u)}{u}\sim \frac{\log x}{\log 2}.
\end{equation}
Furthermore, $\zeta(s)$  has infinitely many right-hand zeros of order
1 at the points $s=1\pm i 2\pi (2n+1) /\log 2$, $n\in\mathbb{N}$, because
\[
\zeta\left(\sigma \pm i\frac{2\pi(2n+1)}{\log 2}\right)= \frac{1}{
  \zeta (\sigma)}\sim (\sigma-1)\log 2. 
\]
It follows, by Lemma \ref{lexC-M}, that
$$M(x)=\Omega(x).$$ 


To show the wobble of $F(x) = N(x)/x$, we apply an idea of Ingham
\cite{ingham1942} that is based on a finite form of the Wiener-Ikehara
method.  We use (essentially) the result given in \cite[Thm.~11.12]{BD04}.
For $\Re s > 0$,
\[
\int_1^\infty x^{-s-1} F(x)\,\mathrm{d}x = \frac{\zeta(s+1)}{s+1} = G(s).
\]
The discontinuities of $G(s)$ on the line segment $(-8\pi i / \log 2,
8\pi i / \log 2)$, which provide a measure of the wobble, occur at
$s=0$ and $s = \pm 4\pi i / \log 2$.

We analyze the behavior of $G$ near these points.  Let $s$ be a
complex number with $\Re s \ge 0$ (to avoid any logarithmic fuss)
and $0 <|s| \le 1/2$ (so $\log s \ll \log |1/s|$ is valid).
 For $n = 0, \, \pm 1$, a small calculation shows that
\[
\zeta(1+ s + 4 \pi n i/\log 2) = \frac{1}{s \log 2}
+ O(\log |1/s|).
\]
For $n = -1,\,0, \, 1$ set
\[
\gamma_n =  4 \pi n/\log 2, \quad \  \a_n = 1/(\log 2+ 4 \pi in).
\]
Take $T$ a number between $4\pi/\log 2$ and $8\pi/\log 2$, e.g.~$T=36$,
and set
\[
G^*(s) =  \sum_{-1\le n \le 1}\frac{\a_{n}}{s-i \gamma_n}, \quad 
F_T^*(u) = \sum_{-1\le n \le 1}\!\!\a_{n}\Big(1 - \frac{|\gamma_{n}|}{T}
\,\Big)\, e^{i \gamma_nu}.
\]

Now $G - G^*$ has a continuation to the closed strip $\{s\!: \sigma
\ge 0, \, |t| \le T\}$ as a function that is continuous save
logarithmic singularities at ${\gamma}_{-1},\,{\gamma}_0,\,{\gamma}_1$.  In particular,
$G - G^*$ is integrable on the imaginary segment $(-iT, \, iT)$.  The
result in \cite{BD04} is stated for an extension that is continuous at
all points of such an interval, but integrability is a sufficient 
condition for the result to hold.


We find that 
\[
\liminf_{u \to \infty} F(u) \le \inf_u F_T^*(u) < \sup_uF_T^*(u) 
\le \limsup_{u \to \infty} F(u).
\]
By a little algebra,
\begin{align*}
\sup_uF_T^*(u) &= \frac 1{\log 2} + \frac 2{|\log 2 - 4 \pi i|}\Big(1 -
\frac{4 \pi}{36 \log 2}\Big) > 1.52, \\
\inf_uF_T^*(u) &= \frac 1{\log 2} - \frac 2{|\log 2 - 4 \pi i|}\Big(1 -
\frac{4 \pi}{36 \log 2}\Big) < 1.37.
\end{align*}
Thus $N(x)/x$ has no asymptote as $x \to \infty$.


It is interesting to note that $\zeta$ is $4\pi i/\log 2$ periodic.
So $\zeta(s)$ has an analytic continuation to $\{s\!: \Re s=1,\,
s\neq1+4n\pi i/\log 2,\: n\in\mathbb{N}\}$. One could show a larger
oscillation by a more elaborate analysis exploiting additional
singularities.
\medskip

Finally, we discuss $m(x)=\int_{1^{-}}^{x}u^{-1} \mathrm{d}M(u)$ for
this example.  Since $M(x)=\Omega(x)$, we necessarily have
$m(x)=\Omega(1)$. We will prove that
\begin{equation}
\label{m O-bound}
m(x)=O(1).
\end{equation}
This shows that, in general, having Chebyshev bounds, log-density
(\ref{log-density}), $N(x)\ll x $, and (\ref{m O-bound}) together do
not suffice to deduce the estimate $M(x)=o(x)$.

To prove (\ref{m O-bound}), we first need to improve (\ref{log-densityex1}) to  
\begin{equation}
\label{2log-densityex1}
\int_{1^{-}}^{x} \frac{\mathrm{d}N(u)}{u}= \frac{\log x}{\log 2}
-\frac{\log \log x}{2}+O(1). 
\end{equation}
 This estimate can be shown by applying a Tauberian theorem of
 Ingham-Fatou-Riesz type \cite{debruyne-vindasCT}
 (cf.~\cite{ingham1935,korevaarbook}). In fact, the Laplace transform
 of the non-decreasing function $\tau_{1}(x)=\int_{1^{-}}^{e^{x}}
 u^{-1} \mathrm{d}N(u)$ is analytic on $\Re s>0$ and 
\begin{align*}
G(s)&=\mathcal{L}\{\tau_{1}; s\}- \frac{1}{s^{2}\log 2} 
+\frac{\log (1/s)}{2s}=\frac{\zeta(s+1)}{s}-\frac{1}{s^{2}\log 2} 
+\frac{\log (1/s)}{2s} \\ 
&
= \frac{1+\log\log2}{2s}+O(\log^{2} |1/s|), \quad |s|<1/2,
\end{align*}
as a small computation shows. In the terminology of
\cite{debruyne-vindasCT}, $G(s)$ has local pseudo\-measure boundary
behavior on the imaginary segment $(-i/2,i/2)$; in fact, its boundary
value on that segment is the sum of a pseudomeasure and a locally
integrable function.  One then deduces (\ref{2log-densityex1})
directly from \cite[Thm. 3.7]{debruyne-vindasCT}.  
Similarly, we use the fact that $(s\,\zeta(s+1))^{-1}$ has a continuous
extension to the same imaginary segment and we apply the same
Tauberian result to the non-decreasing function
$\tau_{2}(x)=\int_{1^{-}}^{e^{x}} u^{-1}
(\mathrm{d}M(u)+\mathrm{d}N(u))$, whose Laplace transform is   
$\mathcal{L}\{\tau_{2}; s\}= \mathcal{L}\{\tau_{1}; s\}+(s\zeta(s+1))^{-1}.$
The conclusion is again the asymptotic formula $\tau_{2}(\log x)= \log
x/\log 2-(\log \log x)/2+O(1)$.  One then obtains (\ref{m O-bound}) upon
subtracting  (\ref{2log-densityex1}) from this formula. 
\end{example}


\begin{example}
\label{ex2C-M} As a second example, we consider a modification of the
Beurling-Diamond examples from \cite{beurling1937,diamond1970} (see
also \cite{d-s-v}), namely, the continuous prime measure
$\mathrm{d}\Pi_{B}$ given by 
$$
\Pi_{B}(x)= \int_{1}^{x}\frac{1-\cos (\log u)}{\log u}\: \mathrm{d}u  
$$
and the discrete g-prime system 
$$q_{k}=\Pi^{-1}_{B}(k), \quad k=1,2,\dots,$$
with g-prime and g-integer counting functions $\pi_{D}(x)$ and $N_{D}(x)$.

We study here the continuous prime measure
$\mathrm{d}\Pi_{C}=2\mathrm{d}\Pi_{B}$ and the discrete $g$-primes
formed by taking each $q_{k}$ twice, that is, the g-prime system  
$$\mathcal{P}= \{q_{1},\, q_{1},\, q_{2},\, q_{2},\, q_{3},\,
q_{3},\,\dots\}.$$ 
The associated number-theoretic functions will be denoted as
$\pi_{\mathcal{P}}(x)$, $\Pi_{\mathcal{P}}(x)$, $N_{\mathcal{P}}(x)$,
$M_{\mathcal{P}}(x)$, and $\zeta_{\mathcal{P}}(s)$, and those
corresponding to $\mathrm{d}\Pi_{C}$ we denote by $N_{C}(x)$, $M_{C}(x)$, and
$\zeta_{C}(s)$.

It is easy to verify that 
\begin{equation*}
\Pi_{C}(x)=  \frac{x}{\log x}\left(2-\sqrt{2}\cos \left(
\log x-\frac{\pi}{4}\right)\right) +O\left(\frac{x}{\log^{2} x}\right)
\end{equation*}
and, since $\pi_{\mathcal{P}}(x)=2\pi_{D}(x)= 2\lfloor\Pi_{B}(x)\rfloor= \Pi_{C}(x)+ O(1)$,
\begin{equation}
\label{asympdiamondexeq1}
\pi_{P}(x)=\frac{x}{\log x}\left(2-\sqrt{2}\cos \left(
\log x-\frac{\pi}{4}\right)\right)  +O\left(\frac{x}{\log^{2}
x}\right) ,
\end{equation}
whence both $\Pi_{C}$ and $\pi_{P}$ satisfy lower and upper Chebyshev
bounds. The zeta function $\zeta_{C}(s)$ can be explicitly computed:
$$
\log\zeta_{C}(s)= -2\log(s-1)+\log (s-1-i)+\log(s-1+i),
$$
and so
$$
\zeta_{C}(s)= \frac{(s-1)^{2}+1}{(s-1)^{2}}= 1+\frac{1}{(s-1)^{2}}\,.
$$

Now $\zeta_{C}(s)$ has right-hand zeros of order 1 located at $1\pm i$,
and so Lemma \ref{lexC-M} implies that $M_{C}(x)=\Omega(x)$. For the
discrete g-number system, we have
$$
\zeta_{\mathcal{P}}(s)=\zeta_{C}(s)\exp(G(s)),
$$
where $G(s)=\int_{1}^{\infty}x^{-s}\mathrm{d}(\Pi_{\mathcal{P}}-\Pi_{C})(x)$ is
analytic on $\Re s>1/2$ because 
$$\Pi_{\mathcal{P}}(x)=\pi_{\mathcal{P}}(x)+O(\sqrt x)=\Pi_{C}(x)+O(\sqrt x).$$
Thus, the same argument yields $M_{\mathcal{P}}(x)=\Omega(x)$.
 
Note that $\zeta_{C}$ is the Mellin transform of the measure
$\mathrm{d}N_{C}= \delta_{1}+\log u\, \mathrm{d}u$, and therefore we
have the exact formula
$$
N_{C}(x)= x\log x- x+2, \quad x\geq 1.
$$
We now show that $N_{\mathcal{P}}$ satisfies a lower bound of a
similar type.   It is well known \cite{d-s-v,diamond1970} that  
$N_{D}(x)= cx +O(x\log^{-3/2}x)$
with $c>0$. Thus $N_D(x) \ge c'x$ for some $c'>0$ and all 
$x\ge 1$ and so
\begin{align*}
N_{\mathcal{P}}(x) &= \int_{1}^x \mathrm{d}N_D*\mathrm{d}N_D
= \int_{1}^x N_D(x/t)\,\mathrm{d}N_D(t) 
\ge \int_{1}^x c'x/t \,\mathrm{d}N_D(t) \\
&\ge c'\!\!\int_{1}^x N_D(x/t)\,\mathrm{d}t
\ge c'^{\, 2}\!\!\int_{1}^x x/t \,\mathrm{d}t = c'^{\, 2}x \log x \ne O(x).
\end{align*}

Applying the Dirichlet hyperbola method, one can actually obtain the
sharper asymptotic estimate
\begin{equation}
\label{eqasympex2}
N_{\mathcal{P}}(x)=c^{2} x\log x+ bx+ O\left(\frac{x}{\log^{1/2}x}\right)
\end{equation}
with certain constants $b, \, c \in\mathbb{R}$. We leave the
verification of (\ref{eqasympex2})  to the reader. 
\end{example} 
 
\begin{example}  \label{ex:3C-M}
There exists a non-decreasing function $N$ on $[1, \, \infty)$ for which
\[
\int_{1}^{\infty} \frac{|N(x)-x|}{x^{\sigma + 1}}\,
\mathrm{d}x \ll (\sigma -1)^{-1/3}\, \tag{\ref{L1cond} bis}.
\]
holds for $1 < \sigma < 2$, but  $\limsup_{x \to \infty} N(x)/x = \infty$.
\end{example}

To see this, set $f(n) = e^{e^n}$ and take
\[
N(x) = \begin{cases}
x, \ x \ge 1, & x \not\in \displaystyle\bigcup_{m=1}^\infty \big[f(m),
\,e^{m/3}f(m)\big] \\
e^{n/3}f(n), & f(n) \le x \le e^{n/3}f(n).
\end{cases}
\]
Clearly, $N\big(e^{e^n}\big)/ e^{e^n} = e^{n/3} \to \infty.$

On the other hand, for each $n$ we have
\begin{align*}
\int_{f(n)}^{\,e^{n/3}f(n)} \frac{|N(x)-x|}{x^{\sigma + 1}}\,\mathrm{d}x 
&<e^{n/3}f(n)\int_{f(n)}^{\,e^{n/3}f(n)} \frac{\mathrm{d}x}{x^{\sigma + 1}} \\
&< e^{n/3}f(n)\int_{f(n)}^\infty \frac{\mathrm{d}x}{x^{\sigma + 1}}
< e^{n/3} f(n)^{-(\sigma -1)}.
\end{align*}
Instead of summing $ e^{n/3} f(n)^{-(\sigma -1)}$, we calculate the
corresponding integral:  
\[
\int_0^\infty e^{u/3} e^{-(\sigma -1) e^u} \mathrm{d}u 
= \int_0^\infty v^{1/3} e^{-(\sigma -1) v} \frac{\mathrm{d}v}{v} =\Gamma(1/3)  
(\sigma -1)^{-1/3}.
\]
Also, $\{e^{n/3} f(n)^{-(\sigma -1)}\}$ is a unimodal sequence whose maximal
term is of size at most $(3e(\sigma -1))^{-1/3} \ll (\sigma-1)^{-1/3}$. 
Thus (\ref{L1cond} bis) holds. 





