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\hypersetup{pdftitle={A quantitative refinement of Lu's elementary prime number theorem},pdfauthor={Rajat Bhattacharjee}}
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\title{A quantitative refinement of Lu's elementary prime number theorem\\[5pt]
\large Balog's identity and a polylogarithmic loss in the square-root exponent}
\author{Rajat Bhattacharjee}
\date{19 October 2026}
\begin{document}
\maketitle
\begin{abstract}
We refine Lu's elementary method for the prime number theorem to
obtain, for every fixed $K>400$,
\[
 \psi(X)-X=O_K\!\left(
 X\exp\!\left\{-\frac{\sqrt{\log X}}{(\log\log X)^K}\right\}\right)
 \qquad(X\to\infty),
\]
where $\psi$ is Chebyshev's function. The main ingredient is a
comparison estimate that remains uniform as the degree grows and
the short-interval parameter shrinks. Combining this estimate with
exponential weights in Balog's identity, a single finite energy step
and positivity-based interpolation yields the stated bound, retaining
the full degree tail and all arithmetic correction terms.
The resulting remainder strengthens Lu's fixed-parameter estimates
and Goz\'e's quantitative refinement within the real-variable framework;
stronger remainder estimates are known by other methods.
\end{abstract}
\tableofcontents

\section{Introduction}
All logarithms are natural. Sums with real cutoffs include their
integer upper endpoints, and $0^0=1$ in logarithmic polynomials.
Let $\mu$ be the M\"obius function, $\Lambda$ the von Mangoldt function,
$\psi(x)=\sum_{n\le x}\Lambda(n)$, and $\gamma$ Euler's constant.
We also write $\pi(x)=\#\{p\le x:p\text{ prime}\}$,
$\vartheta(x)=\sum_{p\le x}\log p$, $\tau(n)=\sum_{d\mid n}1$,
and $\varphi(n)=\#\{1\le a\le n:(a,n)=1\}$.

\begin{theorem}\label{main:theorem}
For every fixed real $K>400$ there are constants $C_K,X_K>0$ such that
\[
 |\psi(X)-X|\le C_KX\exp\{-\Phi_K(\log X)\}\quad(X\ge X_K),
 \qquad \Phi_K(L)=\frac{\sqrt L}{(\log L)^K}\quad(L>1).
\]
In particular, $K=512$ is admissible. Neither this value nor the
threshold is asserted to be optimal.
\end{theorem}

\subsection{Contribution and relation to earlier work}
\label{intro:literature}
The elementary remainder method was developed by
Bombieri~\cite{Bombieri} and Wirsing~\cite{Wirsing}, who obtained
arbitrary fixed logarithmic savings, and by
Diamond and Steinig~\cite{DiamondSteinig} and
Lavrik and Sobirov~\cite{LavrikSobirov}, who obtained stronger
exponential remainders. These results provide the historical setting
for Lu's work. The fixed-power preliminary estimate used below belongs
to this earlier theory; the growing-degree estimates are treated
separately in the present proof.

Write $L=\log X$. Lu~\cite{Lu} proves, by elementary means, a
remainder with decay profile $L^{1/2-\varepsilon}$ for every fixed
$\varepsilon>0$. We use the same arithmetic objects: the centered
remainder $R$, the generalized von Mangoldt functions $\Lambda_i$,
their smooth approximations $m_i$, and Balog's identity. The point of
the present argument is quantitative uniformity as both the degree
and the short-interval parameter vary. Its final parameter is
\[
 \lambda=K\frac{\log\log\log X}{\log\log X}.
\]
This choice is justified by the estimates below; it cannot be inferred
by substituting a varying parameter into Lu's fixed-parameter theorem.
For fixed $K$ and $\varepsilon>0$,
\[
 \frac{\sqrt L/(\log L)^K}{L^{1/2-\varepsilon}}
 =\frac{L^\varepsilon}{(\log L)^K}\longrightarrow\infty.
\]
Thus the stated profile is asymptotically stronger than that family
of fixed-parameter estimates.

There is recent work on precisely this quantitative question.
Goz\'e's 2024 thesis~\cite[Introduction, p.~4]{Goze} proves that there
are absolute constants $c,c'>0$ such that
\[
 |R_1(X)|\le X\exp\{-(\log X)^{1/2-\varepsilon(X)}\},\qquad
 \varepsilon(X)=\frac{c'}{\log\log\log X},\qquad
 \log\log\log X\ge c,
\]
where $R_1(X)=\psi(X)-\lfloor X\rfloor+2\gamma$. Its difference from
$\psi(X)-X$ is bounded, so this gives the same asymptotic error profile
for $\psi$. The result was also presented by Balazard and
Martin~\cite[p.~3]{BM2025}. In terms of $L$, the profile is
\[
 \Phi_G(L)=L^{1/2-c'/\log\log L}
          =\sqrt L\exp\!\left(-\frac{c'\log L}{\log\log L}\right).
\]
For every fixed $K$ and $c'>0$,
\[
 \log\frac{\Phi_K(L)}{\Phi_G(L)}
 =\frac{c'\log L}{\log\log L}-K\log\log L
 \longrightarrow\infty.
\]
This compares asymptotic profiles; it does not compare explicit
thresholds. Goz\'e~\cite[\S20.3, p.~132]{Goze} also identifies the
bound for the coefficients $B_i$ in his Proposition~17.6 as a source
of the restriction on the shrinking parameter. A detailed comparison
with that estimate is needed to assess the originality of the
uniform comparison used here.
A March 2026 seminar announcement by Balazard~\cite{Balazard2026}
also describes ongoing joint work with Goz\'e and Martin on simplifying
the proofs of Lu and Goz\'e. It does not specify a further remainder.
Consequently, neither the first use of a shrinking parameter nor
priority for simplifying this method is claimed here.

A different comparison is essential. Koukoulopoulos~\cite[\S1,
equation~(1.1) and Theorem~1.1]{Koukoulopoulos} proves the stronger
Korobov--Vinogradov profile
\[
 \psi(X)-X=O\!\left(X\exp\!\left\{-c_0
       \frac{L^{3/5}}{(\log L)^{1/5}}\right\}\right),\qquad c_0>0,
\]
using elementary estimates and Fourier inversion. That paper calls
the method largely elementary and distinguishes it from proofs using
no complex variables. Its profile dominates the present one since
\[
 \frac{c_0L^{3/5}/(\log L)^{1/5}}{\Phi_K(L)}
 =c_0L^{1/10}(\log L)^{K-1/5}\longrightarrow\infty.
\]
Accordingly, this paper makes no claim to a new best numerical PNT
remainder, or to a new best remainder among all methods described as
elementary. Its scope is the real-variable Balog--Lu argument and the
uniform estimates needed for the displayed profile. This distinction
concerns the proof method, not an improvement over Koukoulopoulos's
theorem. A focused literature review completed on 19 October 2026
did not establish historical priority for this restricted contribution.

\subsection{Organization and proof dependencies}
The proof follows Lu's progression from short-interval estimates to
smoothing and a contraction for the prime remainder. Balog's identity
relates that remainder to the generalized von Mangoldt coefficients
and the centered divisor average. Its complete exponential generating
form supplies the final forcing identity, with the factorial source
summed explicitly.

We cite Lu's precise forms of the classical fixed-power preliminary
estimates, the elementary interval sieve and the finite Wirsing
inequality. Their hypotheses and normalizations are recorded where
used. The finite summation formulas in Section~\ref{arith:inputs}
prove the required degree-uniform M\"obius bounds, and the prime-power,
floor and unit corrections remain explicit. These inputs come from
earlier elementary work and do not depend on Lu's final remainder
theorem or on the present theorem. The energy parameter $N_0$ below
is one sufficiently large absolute integer, fixed before the degree
or interval margin varies.

Positivity transfers the terminal seed to a common domain, and the
order-sensitive finite comparison in Lemma~\ref{grow:comparison-lemma}
controls all degrees needed by the exponential series. A single fixed
smoothing cutoff and one finite energy step give the coefficient
$3/4$ at degree one. The higher cutoff contribution costs at most
$1/64$; the remaining normalized degree errors vanish, leaving room
for the sufficient $7/8$ slope. The common shrinking-parameter
schedule~\eqref{local:outer-schedule} and the full exponential tail
are essential: a fixed-parameter estimate is not substituted at a
varying parameter.

The last step centers at $1/\zeta(1+\sigma)$ and retains the complete
exponential-average correction. Positive block mass controls the
chord error with both integer endpoints. The mean retains its self
term, and the finite envelope closes with coefficient $28/29$.
No further energy iteration is used. All constants described as
absolute are independent of the growing degree and the shrinking
interval margin.

The choices made here do not establish the endpoint
$\Phi(L)=\sqrt L$; $K=512$ is a convenient, unoptimized example.
The term elementary refers to finite divisor identities, an
elementary sieve and real-variable arguments, including the cited
classical preliminary estimate. No complex zero-free region or
Fourier inversion is used. The comparisons in
Section~\ref{intro:literature} delimit the contribution, whose
historical priority remains unresolved.

\section{The path of the proof and Balog's identity}
\label{path:section}
Dirichlet convolution is $(f*g)(n)=\sum_{d\mid n}f(d)g(n/d)$.
Write $\mathbf1(n)=1$ and $e_0(n)=\mathbf1_{n=1}$; hence
$\mu*\mathbf1=e_0$. Prime factorization gives, independently,
\begin{equation}\label{main:divisor-algebra}
 \mathbf1*\Lambda=\log,
 \qquad \sum_{d\mid n}\Lambda(d)=\log n.
\end{equation}
For $j\ge1$ define
\[
 \Lambda_j=\mu*(\log^j),\quad
 m_j(t)=\sum_{d\le t}\frac{\mu(d)}d\log^j(t/d),\quad
 m(t)=m_0(t)=\sum_{d\le t}\frac{\mu(d)}d.
\]
Here $\log^j$ is an ordinary arithmetic function, not a convolution
power. Put, for all $t>0$,
\[
 \omega(n)=\Lambda(n)-1,\qquad
 R(t)=\sum_{n\le t}\omega(n)+2\gamma.
\]
Thus $R(t)=2\gamma$ below $1$, $\omega(1)=-1$, and
$R(t)=\psi(t)-\lfloor t\rfloor+2\gamma$ above $1$.
For $j\ge2$ set
\[
 R_j(t)=\sum_{n\le t}\bigl(\Lambda_j(n)-j\log^{j-1}n\bigr),
 \quad R_1=R,\quad Z_1=R-2\gamma,\quad Z_j=R_j\ (j\ge2).
\]
All $Z_j$ vanish below $1$. Finally set
\[
 u_j(n)=\Lambda_j(n)-m_j(n),\qquad
 U_j(t)=\sum_{n\le t}u_j(n).
\]

\begin{lemma}[Positivity and the centered divisor average]
\label{main:divisor-average}
For $j\ge1$,
\[
 \Lambda_{j+1}(n)=(\log n)\Lambda_j(n)
                  +(\Lambda*\Lambda_j)(n)\ge0,
 \qquad \Lambda_j(1)=0.
\]
Moreover, for $t\ge1$,
\begin{equation}
 \Rtilde(t):=\sum_{n\le t}R(t/n)
 =\sum_{n\le t}\log n-\sum_{n\le t}\tau(n)
       +2\gamma\lfloor t\rfloor=O(\sqrt t).
 \label{main:divisoridentity}
\end{equation}
Consequently, for $1\le y<x$,
\begin{equation}
 \sum_{n\le x}[R(x/n)-R(y/n)]
 =\Rtilde(x)-\Rtilde(y)
     -2\gamma(\lfloor x\rfloor-\lfloor y\rfloor).
 \label{main:below-one}
\end{equation}
\end{lemma}
\begin{proof}
The operator $\mathcal D f(n)=(\log n)f(n)$ is a derivation for
convolution. Differentiating $\mu*\mathbf1=e_0$ gives
$\mathcal D\mu=-\mu*\Lambda$. Applying this to
$\mu*\log^j$ proves the recurrence; induction from $\Lambda_1=\Lambda$
proves positivity. Expanding the finite divisor average gives
\eqref{main:divisoridentity}. To bound it, let $N=\lfloor\sqrt t\rfloor$.
The elementary hyperbola formula is
\[
 \sum_{n\le t}\tau(n)=2\sum_{n\le N}\lfloor t/n\rfloor-N^2
 =t\log t+(2\gamma-1)t+O(\sqrt t),
\]
using $\sum_{n\le N}1/n=\log N+\gamma+O(1/N)$.
Integral comparison gives $\sum_{n\le t}\log n=t\log t-t+O(\log(2t))$.
All linear terms cancel, including $2\gamma t$. The terms with
$y<n\le x$ in the second sum of \eqref{main:below-one} have
$R(y/n)=2\gamma$, proving its floor correction.
\end{proof}

\begin{lemma}[Balog's identity and its complete source]\label{main:balog-lemma}
For every integer $k\ge0$ and every real $X\ge1$,
\begin{equation}\label{main:balog}
 \boxed{R(X)(\log X)^k+
 \sum_{i=1}^k\binom ki\sum_{n\le X}
 \Lambda_i(n)\log^{k-i}(X/n)R(X/n)=\mathcal B_k(X),}
\end{equation}
where the complete right side is
\begin{equation}\label{main:balog-source}
 \mathcal B_k(X)=\sum_{d\le X}\mu(d)\log^k(X/d)\Rtilde(X/d).
\end{equation}
If $|\Rtilde(t)|\le C_0\sqrt t$ for $t\ge1$, then, uniformly in $k$,
\begin{equation}\label{main:balog-source-bound}
 |\mathcal B_k(X)|\le C_0 X\,2^{k+1}k!.
\end{equation}
\end{lemma}
\begin{proof}
For $k\ge1$, \eqref{main:balog} is Balog's identity as stated in
Lu~\cite[\S6, Lemma~20, p.~1024]{Lu}. For $k=0$, finite M\"obius
inversion gives $R(X)=\sum_{d\le X}\mu(d)\Rtilde(X/d)$, since
$\sum_{d\mid n}\mu(d)=e_0(n)$. Thus $\Lambda_0=e_0$ accounts for
the degree-zero term. All upper cutoffs are inclusive, and the
surviving unit $n=d=1$ contributes $R(X)$.

The bound on $\Rtilde$ was proved in
Lemma~\ref{main:divisor-average}. Since
$t^{-1/2}\log^k(X/t)$ is nonnegative and decreasing on $(0,X]$,
\[
 |\mathcal B_k(X)|
 \le C_0\sqrt X\sum_{d\le X}d^{-1/2}\log^k(X/d)
 \le C_0\sqrt X\int_0^X t^{-1/2}\log^k(X/t)\dd t
 =C_0 X\,2^{k+1}k!.
\]
The integral over $(0,1]$ includes the first lattice term. No
degree-dependent constant is hidden in this estimate.
\end{proof}

The task in~\eqref{main:balog} is to control the coefficients
$\Lambda_i$ on short intervals. Following Lu, we separate them as
$\Lambda_i=m_i+u_i$. The smooth moments $m_i$ provide the mean;
the local sums of $u_i$ control the interpolation error. A direct
two-lattice estimate gives the high-degree input. The positive recurrence
for $\Lambda_i$ then transfers that input through the lower degrees,
and a finite real-convolution energy step gives a strict degree-one
bound. Finally, we sum~\eqref{main:balog} with weights $\sigma^k/k!$,
keep its full source and infinite tail, and close a finite-envelope
contraction. The following lemma supplies the first local estimate.

\begin{lemma}[One local estimate, uniform in the degree]
\label{main:two-lattice}
Let $1\le y<x$, $\Delta=x-y\ge\max(1,\sqrt x)$, and
$h=\log(x/\Delta)$. For every integer $i\ge1$,
\begin{align}
 |U_i(x)-U_i(y)|&\le\Delta E_i(h),\label{main:latticebound}\\
 E_i(h)&=4Q_i(h)+(3h+4)h^i,\qquad
 Q_i(h)=\int_0^\infty e^{-z}(h+z)^i\dd z
       =i!\sum_{r=0}^i\frac{h^r}{r!}.\notag
\end{align}
If $h\ge\max(1,2i)$, then $Q_i(h)\le2h^i$ and
$E_i(h)\le15h^{i+1}$.
\end{lemma}
\begin{proof}
Extend $f_i(t)=\log^i t$ by zero for $t<1$. Finite rearrangement gives
the exact identity
\begin{equation}
 U_i(x)-U_i(y)=\sum_{d\le x}\mu(d)
 \left\{\sum_{y/d<a\le x/d}f_i(a)
       -\frac1d\sum_{y<n\le x}f_i(n/d)\right\}.
 \label{main:twolatticeexact}
\end{equation}
For $d\le\Delta$, the two lattices have the same continuous integral.
Monotone rectangle comparison bounds the sum of their two endpoint
errors by $4\log^i(x/d)$. Its sum is at most
\[
 4\int_0^\Delta\log^i(x/t)\dd t=4\Delta Q_i(h).
\]
For $d>\Delta$, reverse the first lattice and use $a\le x/\Delta$.
The number of available $d$ for each $a$ is at most $\Delta/a+1$,
and $f_i(a)\le h^i$. Since $x/\Delta\le\Delta$, this costs at most
$\Delta(h+2)h^i$. In the second lattice there are at most
$\Delta+1\le2\Delta$ integers $n$, and
$\sum_{\Delta<d\le n}1/d\le h+1$. Its cost is at most
$2\Delta(h+1)h^i$. This proves \eqref{main:latticebound}, with both
integer-count corrections retained. Finally
\[
 Q_i(h)=h^i\sum_{r=0}^i\frac{i(i-1)\cdots(i-r+1)}{h^r}
       \le h^i\sum_{r\ge0}(i/h)^r\le2h^i.
\]
No constant here depends on $i$.
\end{proof}



\section{The preliminary estimates and their exact normalizations}
\label{arith:inputs}

The preliminary estimate needed in the subsequent proof is
\begin{equation}\label{arith:prior}
 r(x):=\sum_{n\le x}\frac{1-\Lambda(n)}n-2\gamma
       =O((\log x)^{-3})\qquad(x\ge2).
\end{equation}
We take this classical fixed-power estimate from
Lu~\cite[\S6, p.~1026, equation~(114) and the proof of Lemma~21]{Lu},
with the preliminary exponent fixed at $3$ once and for all. Lu attributes
that estimate to the earlier elementary results of
Bombieri~\cite{Bombieri}, Wirsing~\cite{Wirsing}, and
Diamond--Steinig~\cite{DiamondSteinig}; its use does not depend on
Lu's final theorem or on the present remainder. Equation~(114) has
exactly the centering $-2\gamma$ and inclusive cutoff displayed above.

For the fixed M\"obius estimates we also cite Lu's Lemma~21,
retaining the exact normalization identities in
Lemma~\ref{arith:lem-mobius}. The interval sieve and finite Wirsing
inequality are cited separately below. All imported exponents and
constants are fixed before the degree or interval parameter varies;
the degree-uniform consequences are proved here.

We shall freely use finite summation by parts in the form
\[
 \sum_{n\le x}a_nf(n)=A(x)f(x)-\int_1^x A(t)f'(t)\,dt,
 \qquad A(t)=\sum_{n\le t}a_n.
\]
The atom at $1$ is included because $A(1-)=0$.
For the elementary preliminary estimates only, write
\[
 H_j(x)=\sum_{n\le x}\frac{(\log n)^j}{n}\qquad(j=0,1).
\]
The harmonic and factorial estimates are
\[
 H_0(x)=\log x+\gamma+O(x^{-1}),\qquad
 \sum_{n\le x}\log n=x\log x-x+O(\log(2x)).
\]
The first follows by summing
$0\le1/n-\log(1+1/n)\le1/(2n^2)$ and bounding its tail;
the second follows from the upper and lower integral sums for $\log t$.

The binomial coefficient $\binom{2N}{N}\le4^N$ gives
$\vartheta(2N)-\vartheta(N)\le2N\log2$. Dyadic summation, followed by
$\psi(x)=\sum_{j\ge1}\vartheta(x^{1/j})$, proves
$\psi(x)\ll x$. Summing $\mathbf1*\Lambda=\log$ and retaining the floors
then gives
\begin{equation}\label{arith:mertens}
 P(x):=\sum_{n\le x}\frac{\Lambda(n)}n=\log x+O(1),
 \qquad \sum_{n\le x}\frac{\Lambda(n)-1}{n}=O(1),
\end{equation}
because the error in replacing $\lfloor x/n\rfloor$ by $x/n$ is
at most $\psi(x)\ll x$.

The exact centered divisor average and its below-one floor correction
were proved in Lemma~\ref{main:divisor-average}; in particular
$\widetilde R(x)=0$ for $x<1$, whereas $R(x)=2\gamma$ there.

\subsection{A complete interval sieve}

\begin{lemma}\label{arith:lem-sieve}
Uniformly for real $A\ge0$ and $H\ge3$,
\[
 \pi(A+H)-\pi(A)\le\frac{2H}{\log H}
 \left(1+O\left(\frac{\log\log(3H)}{\log H}\right)\right).
\]
In particular, for each fixed $0<\eta<1$,
\begin{equation}\label{arith:initial-local}
 |R(x)-R(y)|\le (x-y)\{1+O_\eta((\log x)^{-\eta})\}
\end{equation}
uniformly when $x/2\le y<x$ and
$x-y\ge x\exp[-(\log x)^{1-\eta}]$.
\end{lemma}
\begin{proof}
The first assertion is Selberg's elementary interval-sieve estimate
in Lu~\cite[\S3, p.~985, equation~(17)]{Lu}; the error there is
explicitly independent of both the starting point and the interval
length. Replacing $\log\log H$ by $\log\log(3H)$ and increasing the
absolute constant covers bounded $H\ge3$. The case $A=0$ follows by
a right limit. We record the conversion to $R$ because its prime-power
and integer corrections are used later.

Proper prime powers contribute at most $O(\sqrt x\log^2x)$, so for
$H=x-y\ge3$,
\[
 \psi(x)-\psi(y)\le
 \frac{2H\log x}{\log H}
 \left(1+O\left(\frac{\log\log(3H)}{\log H}\right)\right)
 +O(\sqrt x\log^2x).
\]
This also proves $\psi(x)-\psi(y)\ll H$ whenever $H\ge x^{2/3}$.
On the range in \eqref{arith:initial-local},
$\log H=\log x+O((\log x)^{1-\eta})$ and the prime-power and floor
errors are $O_\eta(H(\log x)^{-\eta})$; the same is true of the
error caused by the sieve correction because $\log\log x/\log x=O_\eta((\log x)^{-\eta})$. Subtracting
$\lfloor x\rfloor-\lfloor y\rfloor=H+O(1)$ from the nonnegative
prime sum proves both sides of \eqref{arith:initial-local}.
\end{proof}

\subsection{The two M\"obius estimates, with the unit atoms retained}

\begin{lemma}\label{arith:lem-mobius}
The classical fixed-exponent M\"obius estimates give
\[
 m(x)=O((\log x)^{-3}),\qquad
 m_1(x)=1+O((\log x)^{-2}).
\]
Consequently
\begin{equation}\label{arith:K}
 K_m:=\int_1^\infty|m(t)|\frac{dt}{t}<\infty,
 \qquad |m_j(x)|\le K_m j(\log x)^{j-1}\quad(j\ge1,x>1),
\end{equation}
where the last constant is independent of $j$. Put $J=\int_1^\infty|m_1(t)-1|\,dt/t<\infty$. For every $j\ge2$,
\begin{equation}\label{arith:fixed-mj}
 |m_j(x)-j(\log x)^{j-1}|\le Jj(j-1)(\log x)^{j-2}\qquad(x>1).
\end{equation}
\end{lemma}
\begin{proof}
The first two estimates follow from the transfer recorded in
Lu~\cite[\S6, p.~1026, proof of Lemma~21]{Lu}, taking exponent $3$
in his equation~(114): it gives $m=O((\log x)^{-3})$ and
$m_1-1=O((\log x)^{-3}\log\log x)$, hence the weaker displayed
bound. Thus $K_m$ and $J$ are finite, including the compact initial
interval. We record the exact normalization identities of that
transfer before deriving the degree-uniform bounds.

If $F,G$ are right-continuous functions of locally bounded variation,
zero below $1$, and $XY=x$, their multiplicative hyperbola formula is
\begin{equation}\label{arith:hyperbola}
 \int_{[1,x]}F(x/t)\,dG(t)
 =\int_{[1,X]}F(x/t)\,dG(t)
 +\int_{[1,Y]}\{G(x/t)-G(X)\}\,dF(t).
\end{equation}
It follows by splitting the finite-variation product measure on $uv\le x$
into $u\le X$ and $u>X$. At $v=Y$ the subtracted difference is zero;
all atoms, including the atoms at $1$, are counted once.

Let $F(t)=\log t-P(t)-\gamma$ for $t\ge1$ and zero below $1$.
Then $F=r+O(t^{-1})=O((\log t)^{-3})$ and direct expansion, using
$\mu*\Lambda=-\mu\log$, gives
\begin{equation}\label{arith:mobius-exact}
 \int_{[1,x]}F(x/t)\,dm(t)=(\log x-\gamma)m(x).
\end{equation}
In particular $dF=dt/t-\sum\Lambda(n)\delta_n/n-\gamma\delta_1$;
the atom $-\gamma\delta_1$ is not dropped.

Now let $q(t)=\log t+\gamma-H_0(t)$ for $t\ge1$, zero below $1$.
We have $q(t)=O(t^{-1})$, $|dq|([1,Y])\ll\log(2Y)$, and exactly
\begin{equation}\label{arith:mone-exact}
 \int_{[1,x]}q(x/t)\,dm(t)=m_1(x)+\gamma m(x)-1.
\end{equation}
The constant $-1$ is the unit convolution $\mu*\mathbf1$.
These identities retain the centering terms in the cited estimates;
no atom is absorbed by a change of normalization.

Finite-sum integration gives, including the atom $m(t)=1$ on $[1,2)$,
\[
 m_j(x)=j\int_1^x m(t)\log^{j-1}(x/t)\frac{dt}{t}\quad(j\ge1),
\]
which proves \eqref{arith:K}. For $j\ge2$ it also gives
\[
 m_j(x)=j(j-1)\int_1^x m_1(t)\log^{j-2}(x/t)\frac{dt}{t}.
\]
The constant part $m_1=1$ contributes $j\log^{j-1}x$, and
$\int_1^\infty|m_1(t)-1|dt/t<\infty$ bounds the remainder by
$Jj(j-1)\log^{j-2}x$. The same $J$ works for every degree.
\end{proof}



\section{A finite descent uniform in the degree}
\label{grow:section}

This section keeps the degree variable.  Its constants come only from
the elementary estimates established in Section~\ref{arith:inputs},
with its fixed cited preliminary input; in particular they do not
depend on a degree or on a short-interval parameter.  Put
\[
 a_i(n)=i(\log n)^{i-1},\qquad r_i(n)=\Lambda_i(n)-a_i(n),
 \qquad A_i(s,t)=\sum_{t<n\le s}a_i(n),
\]
with $a_1(1)=1$.  Differences of $R_i$ are sums of $r_i$ also when
$i=1$, since the constant $2\gamma$ in $R_1$ cancels in a difference.
The finite convolution identities give
\begin{align}
 \Lambda_{i+1}&=(\log)\Lambda_i+\Lambda*\Lambda_i,
       &\Lambda_i&\ge0,\quad \Lambda_i(1)=0,
       \label{grow:positive-recurrence}\\
 r_{i+1}&=(\log)r_i+\Lambda*r_i+\mathfrak c_i,
       &\mathfrak c_i&=i\Lambda*\log^{i-1}-\log^i.
       \label{grow:centered-recurrence}
\end{align}
The first identity was proved by logarithmically differentiating
$\mu*\mathbf1=e_0$; subtracting the integer main rows proves the
second.  Only $\mathfrak c_1$ vanishes.  For example,
$\mathfrak c_i(p)=-(\log p)^i$ for a prime $p$ and $i\ge2$.

Fix constants $C_M,C_\psi\ge1$ and a fixed lower threshold so that
\eqref{arith:mertens} and Lemma~\ref{arith:lem-sieve} give
\begin{equation}
 |P(t)-\log t|\le C_M\quad(t\ge1),\qquad
 \psi(s)-\psi(t)\le C_\psi(s-t)
       \quad(s-t\ge s^{2/3}).
 \label{grow:source-constants}
\end{equation}
The second assertion is required only above that fixed threshold.
Both assertions include prime powers and real endpoints.  Also put
\[
 J=\int_1^\infty |m_1(t)-1|\frac{dt}{t}<\infty.
\]
The integral formula proved with \eqref{arith:fixed-mj} gives the
explicit all-order estimate
\begin{equation}
 |m_i(t)-i(\log t)^{i-1}|
       \le J i(i-1)(\log t)^{i-2}\qquad(i\ge2, t\ge1).
 \label{grow:uniform-mj}
\end{equation}
Indeed the constant part of
$i(i-1)\int_1^t m_1(v)\log^{i-2}(t/v)\,dv/v$ is
$i(\log t)^{i-1}$; the absolute integral of the remaining factor is
at most $J$.  At the unit endpoint the assertion follows by continuity
when necessary.  Thus no fixed-degree $O_i$ is being used for a
growing degree.

\begin{lemma}[A positive harmonic bound with its integer endpoints]
\label{grow:harmonic}
For every real $y\ge1$ and integer $i\ge1$,
\begin{equation}
 \sum_{d\le y}\Lambda_i(d)\lfloor y/d\rfloor
       =\sum_{n\le y}(\log n)^i,
 \qquad
 \sum_{d\le y}\frac{\Lambda_i(d)}d\le2(\log y)^i.
 \label{grow:harmonic-bound}
\end{equation}
\end{lemma}
\begin{proof}
The exact identity is the finite sum of
$\Lambda_i*\mathbf1=\log^i$.  For $d\le y$ we have
$\lfloor y/d\rfloor\ge y/(2d)$.  Positivity and
$\sum_{n\le y}(\log n)^i\le y(\log y)^i$ prove the inequality.
The term at $d=1$ is zero, and the floors were retained before taking
the inequality.
\end{proof}

Here is a finite form of the comparison that will be used below. Its
normalization $\mathcal D_i$ measures the local remainder on a common
interval class.

\begin{lemma}[Uniform finite comparison]
\label{grow:certificate}
\label{grow:comparison-lemma}
There are fixed constants $V_0,C_a,C_H>0$ with the following property.
Let $q\in[1/2,1)$, $\kappa=\log(1/(1-q))$, $T\ge3$ be an integer,
$0<\eta<1/4$, $0<\theta\le1/6$, and $u\ge V$.  Suppose
\begin{align}
 &\kappa\le\min(V^{1-\eta},V/6),\qquad
 V^{1-\eta}\ge2T,\qquad \eta(T+1)>3,
       \label{grow:hypotheses-a}\\
 &V\ge\max\{V_0,100(1+J)T^2,12C_MT\},\qquad
 \theta V\ge\kappa+\log u,\qquad W:=V/\theta\le u.
       \label{grow:hypotheses-b}
\end{align}
For $V\le S\le u$ put $z=e^S$, $H=(1-q)z$, and
\[
 \mathcal D_i(S)=
 \frac{|R_i(z)-R_i(qz)|}{HS^{i-1}}.
\]
Set $G_T=15+2JT(T-1)$ and $H_T=C_H(1+J)T^4$.  On the entire common
domain $V\le S\le u$,
\begin{equation}
 \mathcal D_T(S)\le\frac{G_T}{S},\qquad
 \mathcal D_i(S)\le8T\quad(1\le i\le T).
 \label{grow:seed-coarse}
\end{equation}
On $W\le S\le u$, for every $2\le i\le T$,
\begin{equation}
 \boxed{\displaystyle
 \mathcal D_i(S)\le C_a\theta^i+\frac{H_T}{V}
   +\frac{16T^3}{i(i-1)}
                 \left(\frac WS\right)^{(i-1)/2}.}
 \label{grow:comparison}
\end{equation}
All constants are independent of $q,T,\eta,\theta,V,u$.
\end{lemma}

\begin{proof}
We give the arithmetic recurrence and its corrections before making
the comparison.

\smallskip\noindent
\emph{The terminal seed and a common coarse bound.}
The hypotheses imply $H=e^{S-\kappa}\ge e^{5S/6}$ and $qz\ge1$.
In Lemma~\ref{main:two-lattice} the actual logarithmic interval
parameter is $\kappa\le S^{1-\eta}$.  Since $E_T(h)$ there is an
increasing polynomial for $h\ge0$ and $S^{1-\eta}\ge2T$,
\[
 |U_T(z)-U_T(qz)|\le15H S^{(1-\eta)(T+1)}.
\]
After division by $HS^{T-1}$ this is at most $15/S$, because
$2-\eta(T+1)<-1$.  There are at most $H+1\le2H$ integers in the
interval.  Equation~\eqref{grow:uniform-mj} therefore bounds the
remaining smooth-row correction by
$2JT(T-1)/S$ in this normalization.  This proves the terminal estimate
in \eqref{grow:seed-coarse}, including both counting errors in the
two-lattice estimate.

Positivity in \eqref{grow:positive-recurrence} gives, for $n>1$,
\[
 \Lambda_i(n)\le
       \frac{\Lambda_T(n)}{(\log n)^{T-i}}\quad(1\le i\le T).
\]
As $\log n\ge S-\log2$ on the present interval, its positive mass
satisfies
\begin{equation}
 \frac{\sum_{qz<n\le z}\Lambda_i(n)}{HS^{i-1}}
 \le\left(\frac S{S-\log2}\right)^{T-i}
                 \left(2T+\frac{G_T}{S}\right)\le5T.
 \label{grow:positive-coarse}
\end{equation}
Here the $2T$ retains the possible extra integer in the main row;
$V\ge100(1+J)T^2$ makes $G_T/S\le1$ and the displayed power at most
$2$.  The $i$th integer main row costs at most $2i\le2T$.
Consequently $\mathcal D_i\le8T$ on the same domain, without any
successive extension of that domain.  No interval here contains the
unit, but the original definitions at the unit have not been changed.

\smallskip\noindent
\emph{The complete arbitrary-cut recurrence.}
Write $b=e^{(1-\theta)S}$ and
$\Delta R_i(s,t)=R_i(s)-R_i(t)$.  For every $i\ge1$, finite summation
and a split of the convolution in \eqref{grow:positive-recurrence}
give the exact identity
\begin{align}
 \Delta R_{i+1}(z,qz)={}&S\Delta R_i(z,qz)
  +\sum_{d\le b}\Lambda(d)\Delta R_i(z/d,qz/d)
  +F_i+\mathsf T_i+M_i,                 \label{grow:exact-recurrence}\\
 F_i={}&\sum_{qz<n\le z}(\log n-S)\Lambda_i(n),\notag\\
 \mathsf T_i={}&\sum_{m\le e^{\theta S}}\Lambda_i(m)
             \sum_{\max(b,qz/m)<d\le z/m}\Lambda(d),\notag\\
 M_i={}&S A_i(z,qz)
       +\sum_{d\le b}\Lambda(d)A_i(z/d,qz/d)-A_{i+1}(z,qz).
       \notag
\end{align}
Every lower endpoint in these integer sums is strict and every upper
endpoint is inclusive.  These three rows reorganize the correction
$\mathfrak c_i$ in \eqref{grow:centered-recurrence}; they do not drop
that correction.

The hypotheses imply
\[
 \frac Hb=e^{\theta S-\kappa}\ge u\ge S.
\]
For $m\le e^{\theta S}$, put $s=z/m$.  Since
$\log s\ge(1-\theta)S\ge5S/6$ and $\kappa\le S/6$,
$H/m=s e^{-\kappa}\ge s^{2/3}$.  All these upper endpoints exceed
the fixed sieve threshold once $V_0$ is fixed.  Positivity permits the
inner interval in $\mathsf T_i$ to be enlarged to $(qz/m,z/m]$.
Lemma~\ref{grow:harmonic} and \eqref{grow:source-constants} thus give
\begin{equation}
 0\le\mathsf T_i\le
 C_\psi H\sum_{m\le e^{\theta S}}\frac{\Lambda_i(m)}m
       \le2C_\psi H(\theta S)^i.
 \label{grow:tail-row}
\end{equation}
Also, with the exact integer main row retained,
\begin{equation}
 |F_i|\le(\log2)
       \{ |\Delta R_i(z,qz)|+i(H+1)S^{i-1}\}.
 \label{grow:frozen-row}
\end{equation}

\smallskip\noindent
\emph{The missing mean and all endpoint terms.}
Let $E_P(t)=P(t)-\log t$, so $E_P(1)=0$ and $|E_P(t)|\le C_M$.
For $i\ge2$ define the positive operator
\[
 (\mathcal P_i f)(S)=\frac1{S^i}
  \sum_{d\le b}\frac{\Lambda(d)}d
       (S-\log d)^{i-1}f(S-\log d).
\]
For $0\le\xi<i-1$, direct partial summation gives
\begin{align}
 \frac{\mathcal P_i[(W/S)^\xi]}{(W/S)^\xi}
   &=\frac{1-\theta^{i-\xi}}{i-\xi}
                         +\Xi_{i,\xi}(S),
                            \label{grow:mass-exact}\\
 \Xi_{i,\xi}(S)
   &=S^{\xi-i}\left\{(\theta S)^{i-1-\xi}E_P(b)
       +(i-1-\xi)\int_0^{(1-\theta)S}
          E_P(e^v)(S-v)^{i-2-\xi}\,dv\right\},\notag\\
 |\Xi_{i,\xi}(S)|&\le\frac{C_M}{S}.
                            \label{grow:mass-error}
\end{align}
For the last bound, the endpoint plus the integral of the absolute
derivative of $(S-v)^{i-1-\xi}$ is exactly $S^{i-1-\xi}$.
The lower endpoint term vanishes because $E_P(1)=0$; the upper cutoff
term $E_P(b)$ is displayed explicitly.  Thus this bound is uniform
also when $i$ and $\xi$ grow.
Put $\Xi_i=\Xi_{i,0}$ for $i\ge2$, and put
$\Xi_1=E_P(b)/S$; then the corresponding mass identity at $i=1$ is
$P(b)/S=1-\theta+\Xi_1$.

Define the exact integer errors
\[
 \epsilon_i(s,t)=A_i(s,t)-i(s-t)(\log s)^{i-1}.
\]
Substituting these definitions and the mass identity in $M_i$ gives,
for every $i\ge1$,
\begin{align}
 \frac{M_i}{HS^i}={}&-\theta^i+i\Xi_i(S)\notag\\
 &+\frac{S\epsilon_i(z,qz)
   +\sum_{d\le b}\Lambda(d)\epsilon_i(z/d,qz/d)
   -\epsilon_{i+1}(z,qz)}{HS^i}.
 \label{grow:main-exact}
\end{align}
In particular the omitted mean is exactly $-\theta^i$, not zero.

Here are bounds for the displayed endpoint terms.  If $s/2\le t<s$,
$L_s=\log s\ge1$, and $h=s-t\ge L_s$, then, for $i\ge2$,
\begin{equation}
 |\epsilon_i(s,t)|
 \le iL_s^{i-1}
       +i(i-1)(\log2)(h+1)L_s^{i-2}
 \le2i^2hL_s^{i-2}.
 \label{grow:integer-errors}
\end{equation}
Indeed $|\lfloor s\rfloor-\lfloor t\rfloor-h|\le1$ and
the mean value theorem bounds the change in $(\log n)^{i-1}$.
For $i=1$ retain instead the exact floor error
$\epsilon_1(s,t)=\lfloor s\rfloor-\lfloor t\rfloor-(s-t)$,
of absolute value at most $1$.
The hypotheses $H/d\ge H/b\ge S\ge\log(z/d)$ verify every condition
of \eqref{grow:integer-errors} at every daughter.  For $i\ge2$,
\[
 \sum_{d\le b}\Lambda(d)|\epsilon_i(z/d,qz/d)|
 \le2i^2H S^{i-2}P(b)\le C i^2HS^{i-1}.
\]
The two other integer errors in \eqref{grow:main-exact} have the
same normalized bound.  At $i=1$ the convolution of floor errors is
at most $\psi(b)\ll b\le H/S$, and the other two errors are covered
by the same conclusion.  Thus for all $i\ge1$ the absolute value of
the second line of \eqref{grow:main-exact} is at most
$C(i+1)^2/S$, with a fixed $C$.

Transpose \eqref{grow:exact-recurrence}, divide by $HS^i$, and use
\eqref{grow:seed-coarse} for the frozen row
$\mathcal D_i(S)/S$ in \eqref{grow:frozen-row}.  Equations
\eqref{grow:tail-row}--\eqref{grow:main-exact} prove, simultaneously
for $2\le i<T$,
\begin{equation}
 \mathcal D_i(S)\le\mathcal D_{i+1}(S)
                 +(\mathcal P_i\mathcal D_i)(S)+E_i,
 \qquad E_i=C\theta^i+\frac{CT^2}{V}.
 \label{grow:comparison-row}
\end{equation}
Here and below $C$ is one fixed sufficiently large constant.
The inequality is used only for $S\ge W$, so all daughters lie in
the established common domain: a nonzero term has $d\ge2$ and
\begin{equation}
 V\le\theta S\le S-\log d\le S-\log2.
 \label{grow:daughter-domain}
\end{equation}
In particular the frozen row has not been silently absorbed into a
degree-dependent constant or assumed small before the coarse bound.

\smallskip\noindent
\emph{A finite simultaneous comparison.}
Set $\varepsilon=C_M/V$, $\xi_i=(i-1)/2$, and
$J_i(S)=(W/S)^{\xi_i}$.  The exact masses imply the upper bounds
\[
 \mathcal P_i1\le q_i:=\frac1i+\varepsilon,
 \qquad
 \mathcal P_iJ_i\le q_i^*J_i,
 \qquad q_i^*:=\frac2{i+1}+\varepsilon.
\]
They are strictly below $1$.  For $2\le i\le k\le T-1$, the ideal
resolvent products telescope:
\[
 \prod_{j=i}^k(1-1/j)^{-1}=\frac{k}{i-1},\qquad
 \prod_{j=i}^k(1-2/(j+1))^{-1}
             =\frac{k(k+1)}{i(i-1)}.
\]
Each logarithmic perturbation caused by $\varepsilon$ is at most
$6\varepsilon$.  This follows from $1-2/(j+1)\ge1/3$ and
$\varepsilon\le1/(12T)$.  Since $6T\varepsilon\le1/2<\log2$,
each perturbed product is at most twice its displayed ideal value.

Define nonnegative numbers backwards by
\[
 a_T=G_T/V,\quad b_T=8T,\qquad
 a_i=\frac{a_{i+1}+E_i}{1-q_i},\quad
 b_i=\frac{b_{i+1}}{1-q_i^*}\quad(2\le i<T).
\]
We claim
\begin{equation}
 \mathcal D_i(S)\le a_i+b_iJ_i(S)\qquad(W\le S\le u).
 \label{grow:barrier}
\end{equation}
The terminal degree follows from the seed.  Descend one degree at a
time, and at each degree induct over the consecutive half-open strips
$[W+n\log2,W+(n+1)\log2)$.  Every same-degree daughter lies in an
earlier strip by \eqref{grow:daughter-domain}.  Daughters in $[V,W)$
are covered by $\mathcal D_i\le8T\le b_iJ_i$.
For the already treated upper degree,
$J_{i+1}(S)\le J_i(S)$ when $S\ge W$.  Substitution of these bounds
in \eqref{grow:comparison-row} gives precisely
$a_i+b_iJ_i(S)$ by the definitions of $a_i,b_i$.
These finitely many strips cover every finite upper endpoint $u$;
at a strip boundary its daughters are still in earlier strips.
This proves the comparison without an infinite recursion or a
regularity assumption on the step functions $R_i$.

Finally, unfolding the finite recurrences and using the telescoping
products gives
\[
 a_i\le\frac2{i-1}
       \left\{\frac{(T-1)G_T}{V}
                  +\sum_{j=i}^{T-1}jE_j\right\},\qquad
 b_i\le\frac{16T^3}{i(i-1)}.
\]
For $0<\theta\le1/2$ and $i\ge2$,
\[
 \frac1{i-1}\sum_{j=i}^\infty j\theta^j
  =\frac{\theta^i}{i-1}
       \left\{\frac{i}{1-\theta}
                       +\frac{\theta}{(1-\theta)^2}\right\}
  \le6\theta^i.
\]
It follows that
$a_i\le12C\theta^i+(2TG_T+CT^4)/V$.
Choose $C_a=12C$ and a fixed $C_H$ with
$2TG_T+CT^4\le C_H(1+J)T^4$.  Together with
\eqref{grow:barrier} this proves \eqref{grow:comparison}; the
terminal degree obeys the same enlarged bound.  All sums and
comparisons in this proof are finite, and each endpoint correction
was retained before it was bounded.
\end{proof}



\section{Fixed-cutoff smoothing and one signed energy step}
\label{local:section}

We prove only the strict degree-one coefficient needed in the final
exponential sum.  The cutoff fraction in this section is one fixed
number; neither an iteration of strict prime estimates nor a local
decay theorem is used.  All constants denoted by $C$ below are absolute,
unless their dependence on this fixed cutoff is displayed.

\subsection{The raw interval bounds and the integer interpolant}

The proof of Lemma~\ref{arith:lem-sieve}, before any fixed-margin
specialization, gives the following useful uniform form.  If
$z=e^s$, $q\in[1/2,1)$, $\kappa=\log(1/(1-q))\le s/3$ and
$H=(1-q)z\ge3$, then
\begin{equation}\label{local:raw-sieve}
 \frac{|R(z)-R(qz)|}{H}
 \le1+C\left\{\frac\kappa s+\frac{\log s}s
       +s^2e^{\kappa-s/2}+e^{\kappa-s}\right\}.
\end{equation}
Indeed, the displayed prime-counting sieve there bounds the prime
sum divided by $H$ by
$2+C\{\kappa/s+\log s/s+s^2e^{\kappa-s/2}\}$.
The proper-prime-power term is the third term here.  Subtracting
$\lfloor z\rfloor-\lfloor qz\rfloor=H+O(1)$, and using the
nonnegativity of the prime sum for the opposite inequality, proves
\eqref{local:raw-sieve}.  Thus its constant is independent of a
shrinking margin.  We also use the absolute bound
$|R(z)-R(qz)|\le CH$ when $H\ge z^{2/3}$, and the elementary
bounds proved above:
\begin{equation}\label{local:elementary}
\begin{gathered}
 R(z)=O(z),\quad
 \sum_{n\le z}\frac{|\omega(n)|}{n}\le C(1+\log z),\quad
 \sum_{n\le z}|\omega(n)|\le Cz,\\
 \left|\sum_{n\le z}\frac{\omega(n)}n\right|\le C,
 \qquad \omega(n)=\Lambda(n)-1.
\end{gathered}
\end{equation}
Here $\omega(1)=-1$.

For $11/12\le p<1$, take integer knots
\[
 b_0=3,\qquad b_{j+1}=b_j+\lfloor b_je^{-(\log b_j)^p}\rfloor,
\]
and let $S_0$ interpolate $(b_j,R(b_j))$ linearly.  Its derivative
at a knot is the left derivative.  All increments are positive, and
$b_{j+1}\le2b_j$.  In particular, for every integer $n\ge4$,
\begin{equation}\label{local:cells}
 S_0'(n)=S_0(n)-S_0(n-1),\qquad
 S_0'(t)=S_0'(n)\quad(n-1<t\le n).
\end{equation}
If $Z=\log b_j$ and $b_je^{-Z^p}\ge2$, the interval length
$\Delta=b_{j+1}-b_j$ obeys
$\Delta\ge b_je^{-Z^p}/2$.  Consequently its relative logarithmic
width is at most $Z^p+\log4$.  Applying
\eqref{local:raw-sieve} to the whole interval, provided $Z^{p-1}\le1/8$ and $Z$ exceeds a fixed absolute lower
bound, gives
\begin{equation}\label{local:raw-mesh}
 |S_0'(t)|\le1+C\left\{Z^{p-1}+\frac{\log Z}{Z}
                                      +Z^2e^{-Z/3}\right\}
 \quad(b_j<t\le b_{j+1}).
\end{equation}
This statement has no fixed-$p$ implied constant; its stated conditions
will be checked on the common schedule below.

There is also a pointwise chord bound requiring no prime asymptotic:
\begin{equation}\label{local:chord}
 E(z):=R(z)-S_0(z)
   =O\left(ze^{-(\log(z/2))^p}\log z\right)\quad(z>3).
\end{equation}
On the block containing $z$, both deviations from $R(b_j)$ are
bounded by $\sum_{b_j<n\le b_{j+1}}(\Lambda(n)+1)$, which is
at most $\Delta(\log b_{j+1}+1)$.  Since $b_j\ge z/2$, this
proves \eqref{local:chord} with a constant independent of $p$.

\subsection{The complete fixed-cutoff smoothing identity}

Fix $u>v>1$, $q\in[1/2,1)$ and $0<\theta\le1/6$.  Set
\[
 f(t)=\mathbf1_{(v,u]}(t)
       \frac{R(e^t)-R(qe^t)}{(1-q)e^t},\qquad
 g(t)=\mathbf1_{(v,u]}(t)S_0'(e^t),\qquad
 h(t)=\frac1t\int_0^t f(t-s)g(s)\,\dd s,
\]
with $h(0)=0$.  Suppose $|f|\le A_f$, $|g|\le A_g$, where
$A_f,A_g\le2$.  At an equation endpoint $t\in[w,u]$, use
\[
 x=e^t,\quad H=(1-q)x,\quad a=e^{\theta t},\quad B=x/a,
 \qquad \Delta_nR=R(x/n)-R(qx/n).
\]
Assume $\theta t\ge2v$, $a\ge4$,
$\kappa\le(1-\theta)t/3$, and the fixed elementary lower bounds.
Then, with an absolute constant independent of $p,q,\theta$, the
EXACT equation has the form
\begin{equation}\label{local:smoothed}
 f(t)+h(t)=d_2(t,q)+r_\theta(t,q),\qquad
 d_2(t,q)=\frac{R_2(e^t)-R_2(qe^t)}{(1-q)e^t t},
\end{equation}
where the complete remainder satisfies
\begin{align}
 |r_\theta(t,q)|\le C\bigg\{&(1+A_fA_g)\theta+\frac1t
 +\frac{(1+A_g)e^{\kappa-\theta t}}t
 +\frac{e^{\kappa-t/2}}t
 +\frac{e^{\kappa+\theta t-t}}t+e^{\kappa-t}\notag\\
 &\hspace{25mm}+t\exp\bigl[\kappa-(\theta t-\log4)^p\bigr]
                         \bigg\}.\label{local:complete-error}
\end{align}
We prove this, recording the exact differences before bounding them.

The $i=1$ instance of \eqref{grow:exact-recurrence} is
\begin{equation}\label{local:arithmetic}
 t\Delta_1R+\sum_{n\le B}\Lambda(n)\Delta_nR
   =\Delta R_2-F_1-T_1-M_1.
\end{equation}
Its three correction rows retain the original strict lower and
inclusive upper endpoints.  Their bounds can be seen directly here.
The frozen-log row is at most $(\log2)\sum_{qx<n\le x}\Lambda(n)
=O(H)$, by the absolute interval sieve.  In the tail, $m\le a$ and
every full inner interval has length $H/m\ge(x/m)^{2/3}$.
Its prime sum is at most $CH/m$, even when the actual interval is
clipped.  The exact harmonic bound
$\sum_{m\le a}\Lambda(m)/m\le2\log a$ therefore gives
$T_1\le CH\theta t$.
For the main row, \eqref{grow:main-exact} retains the missing mean
$-\theta Ht$ and the harmonic endpoint $H D_P(B)$, where
$|D_P|\le C_M$.  To check its remaining integer terms directly, write
$\epsilon_i(s,z)=A_i(s,z)-i(s-z)(\log s)^{i-1}$.
Then $|\epsilon_1|\le1$, and for $s/2\le z<s$,
$|\epsilon_2(s,z)|\le C\{s-z+\log s\}$.
Consequently their unnormalized contribution is at most
$C(H+t+B)$.  These facts give
\begin{equation}\label{local:arithmetic-error}
 \frac{|F_1+T_1+M_1|}{Ht}
 \le C\left\{\theta+\frac1t+e^{\kappa-t}
                           +\frac{e^{\kappa-\theta t}}t\right\}.
\end{equation}
Thus neither the missing mean nor the weighted-mass endpoint has
been removed by an unproved signed cancellation.

For constant centering the exact identity is
\begin{equation}\label{local:below-one}
 \sum_{n\le x}\Delta_nR
 =\Rtilde(x)-\Rtilde(qx)
                   -2\gamma(\lfloor x\rfloor-\lfloor qx\rfloor).
\end{equation}
The last term uses $R(z)=2\gamma$ below one.  In particular,
\begin{align}
 C_{\rm cen}:={}&\sum_{n\le B}\Lambda(n)\Delta_nR
                   -\sum_{a<n\le B}\omega(n)\Delta_nR\notag\\
 ={}&\sum_{n\le a}\omega(n)\Delta_nR
 +\Rtilde(x)-\Rtilde(qx)
 -2\gamma(\lfloor x\rfloor-\lfloor qx\rfloor)
 -\sum_{B<n\le x}\Delta_nR.\label{local:centering-exact}
\end{align}
The first sum is at most $CH(1+\theta t)$: every daughter satisfies
the absolute interval sieve, and \eqref{local:elementary} applies.
For the last sum, expand the jumps first; then
\[
 \sum_{B<n\le x}|\Delta_nR|
 \le\sum_{m\le a}|\omega(m)|(H/m+1)
 \le C\{H(1+\theta t)+a\}.
\]
The displayed $+1$ is the retained integer-count error.  Using
$\Rtilde(x)=O(\sqrt x)$ and the floor term in
\eqref{local:centering-exact} proves
\begin{equation}\label{local:centering-error}
 \frac{|C_{\rm cen}|}{Ht}
 \le C\left\{\theta+\frac1t+
 \frac{e^{\kappa+\theta t-t}}t+\frac{e^{\kappa-t/2}}t
                                        +e^{\kappa-t}\right\}.
\end{equation}
In particular the unit in the first sum is present.

Put $c(n)=\omega(n)-S_0'(n)$ and
$Q_a=\exp[-(\log(a/4))^p]$.  The exact cumulative coefficient error is
\begin{equation}\label{local:floor-chords}
 K_a(d):=\sum_{a<n\le d}c(n)
       =E(\lfloor d\rfloor)-E(\lfloor a\rfloor)
       =O(dQ_at)\quad(a<d\le x).
\end{equation}
Set $K_a(d)=0$ when $d\le a$.  Both floor arguments are retained;
if they coincide the difference is zero.  Otherwise they are at least
$a/2$, and \eqref{local:chord} proves the bound.  For $z=x,qx$ the
finite jump expansion is exactly
\[
 \sum_{a<n\le B}c(n)R(z/n)
 =\sum_{m\le z/a}\omega(m)K_a(\min(B,z/m))+2\gamma K_a(B).
\]
The constant is displayed before subtracting the two values of $z$.
It follows from \eqref{local:elementary} that
\begin{equation}\label{local:coefficient-error}
 C_{\rm coef}:=\sum_{a<n\le B}
       [\omega(n)-S_0'(n)]\Delta_nR=O(xQ_at^2).
\end{equation}

There is a separate, smaller discrete-to-integral difference
\begin{equation}\label{local:cell-error}
 C_{\rm cell}:=\sum_{a<n\le B}S_0'(n)\Delta_nR
       -\int_a^B S_0'(z)\{R(x/z)-R(qx/z)\}\,\dd z
       =O(A_gx/a).
\end{equation}
Indeed \eqref{local:cells} is exact on every complete unit cell.
The change of $R(z/n)$ on that cell is bounded by
$\sum_{z/n<m\le z/(n-1)}|\omega(m)|$.
These half-open reciprocal intervals are disjoint as $n$ varies,
so Chebyshev bounds their total by $Cx/a$ for $z=x,qx$.
The two incomplete endpoint cells have the same bound, by
$R(z)=O(z)$.  Their logarithms lie in the support where $A_g$ applies,
since $\theta t\ge2v$.  No step function is differentiated.

Finally substitution $z=e^s$ in the integral gives its normalized
value
\[
 \frac1t\int_{\theta t}^{(1-\theta)t}g(s)f(t-s)\,\dd s.
\]
Both factors there lie above their lower support.  Completing this
to $h(t)$ adds the EXACT term
\[
 C_{\rm end}=\frac1t
 \left\{\int_v^{\theta t}+\int_{(1-\theta)t}^{t-v}\right\}
                    g(s)f(t-s)\,\dd s,
 \qquad |C_{\rm end}|\le2A_fA_g\theta.
\]
Thus the precise remainder in \eqref{local:smoothed} is
\[
 r_\theta=-\frac{F_1+T_1+M_1+C_{\rm cen}+C_{\rm coef}
                              +C_{\rm cell}}{Ht}+C_{\rm end}.
\]
Equations \eqref{local:arithmetic-error},
\eqref{local:centering-error}--\eqref{local:cell-error} prove
\eqref{local:complete-error}.  It includes both completed end ranges,
both floor chords, the constant $2\gamma$, all reciprocal jumps,
all prime powers and every count correction.

\subsection{A finite real-convolution inequality}

\begin{lemma}[Finite Wirsing inequality; Lu's Lemma~3]
\label{lem:s049-wirsing}
There are absolute constants $C_W>0$ and an integer $N_W\ge2$
with the following property. Let $X>2Y>0$ and let real measurable
$f,g$ on $[0,X]$ satisfy
\[
 \int_0^t f^2\le Ft,\qquad \int_0^t g^2\le Gt
       \quad(0<t\le X),\qquad F,G>0.
\]
Put $h(t)=t^{-1}\int_0^t f(t-s)g(s)\,ds$. For an integer
$N\ge N_W$ write $\epsilon=1/N$. If
\[
 \left|\int_0^t h(s)\,ds\right|
 \le\frac{\epsilon^3}{4}\sqrt{FG}\,t
       \quad(Y\le t\le X),
\]
then
\begin{equation}\label{local:wirsing-result}
 \int_Y^X h^2\le(\tfrac12+C_W/N)FG(X-Y).
\end{equation}
The constants are independent of $N,X,Y,F,G,f,g$.
\end{lemma}
\begin{proof}[Source and normalization]
This is Lu~\cite[\S4, pp.~991--992, Lemma~3]{Lu}, based on
Wirsing~\cite[proof of Lemma~1, pp.~2--6]{Wirsing}. Lu states the
result for reciprocal sufficiently large integers and explicitly
makes the error constant absolute. The source's large-$X,Y$
convention causes no restriction here: replacing $f(t),g(t)$ by
$f(t/a),g(t/a)$ on $[0,aX]$ gives $h(t/a)$, and multiplies every
integral and its corresponding interval length by $a$. Apply the
source at large $aY$ and divide by $a$. The initial-interval energy
hypotheses and the factor $X-Y$ are unchanged.

The initial strip in the source's geometric-block argument can be
kept explicit. Write its absolute block constant as $C_b$ (Lu's
equation~(36)). Complete blocks of ratio $1+1/N$ ending at $X$ leave
a first point $a$ with $Y\le a<(1+1/N)Y$. Cauchy--Schwarz gives
$|h|\le\sqrt{FG}$ from the initial-interval hypotheses, so
\[
 \int_Y^Xh^2\le FG\{(a-Y)+(\tfrac12+C_b/N)(X-a)\}
 \le(\tfrac12+(C_b+1)/N)FG(X-Y),
\]
using $a-Y\le Y/N$ and $Y<X-Y$. Thus one may take $C_W=C_b+1$;
the unpaired strip is included, and no short-interval energy bound
is assumed.
\end{proof}

Fix once and for all an integer
\begin{equation}\label{local:fixed-energy-parameter}
 N_0\ge\max\{N_W,2,\lceil400C_W/9\rceil\}.
\end{equation}
Thus $1/2+C_W/N_0\le209/400$. This integer is absolute and does not
vary with the growing degree, the interval margin or the outer $X$.

\subsection{The signed primitive and a quantitative single step}

The primitive uses the arithmetic signs, not an absolute variation
bound.  Before truncation put
$F_q(t)=[R(e^t)-R(qe^t)]/[(1-q)e^t]$ for $t\ge0$.
Finite jump integration gives, for every $T\ge0$,
\begin{equation}\label{eq:s049-raw-primitive}
 \int_0^T F_q(t)\,\dd t
 =\sum_{n\le qe^T}\frac{\omega(n)}n
 +\frac1{1-q}\sum_{qe^T<n\le e^T}
                       \omega(n)(1/n-e^{-T}).
\end{equation}
To verify it, integrate the contribution of each jump over
$\log n\le t<\log(n/q)$, clipped to $[0,T]$; the integral of
$e^{-t}/(1-q)$ is $1/n$ for a complete jump and is the displayed
coefficient for a partial one.  This includes $n=1$ at the lower
endpoint.  The partial coefficient is between zero and $1/n$.
The first sum is bounded by \eqref{local:elementary}; the absolute
partial sum is bounded on $(qe^T,e^T]$ by Chebyshev, since
$q\ge1/2$.  Therefore the whole primitive is at most an absolute
$C_*$ in magnitude.  For the truncated $f$,
\begin{equation}\label{local:truncated-primitive}
 F(T):=\int_0^T f(t)\,\dd t
 =\mathbf1_{T>v}\left\{\int_0^{\min(T,u)}F_q(t)\,\dd t
                              -\int_0^v F_q(t)\,\dd t\right\},
 \qquad |F(T)|\le K_f:=2C_*.
\end{equation}
Thus the lower-support subtraction has not been dropped.

Let $C(T)=\int_0^T(g*f)(t)\,\dd t$.  Finite Fubini yields
$C(T)=\int_0^Tg(s)F(T-s)\,\dd s$.  Both factors vanish below $v$,
so $C(T)=0$ for $T\le2v$, and
$|C(T)|\le A_gK_f(T-2v)$ otherwise.  Integration by parts retains
the upper endpoint:
\begin{align}
 \int_0^T h(t)\,\dd t
 &=\frac{C(T)}T+\int_{2v}^T\frac{C(t)}{t^2}\,\dd t,
                                  &&T>2v,\label{local:primitive-abel}\\
 \left|\int_0^T h(t)\,\dd t\right|
 &\le A_gK_f\log(T/(2v)).\label{local:primitive-bound}
\end{align}
For $T\le2v$ this integral is zero.  The last inequality uses the
exact equality
$(T-2v)/T+\int_{2v}^T(t-2v)t^{-2}\,\dd t=\log(T/(2v))$.
No primitive estimate for $g$ is needed.

\begin{lemma}[One finite strict energy step]\label{local:energy-step}
Suppose $u>2w$, $|f|,|g|\le A$, and the exact equation
$f+h=e$ holds on $[w,u]$, with $|e|\le E$. For an integer
$N\ge N_W$, assume
\begin{equation}\label{local:primitive-test}
 K_f\frac{\log u}{w}\le\frac{A}{4N^3}.
\end{equation}
Then
\begin{equation}\label{local:one-step}
 |f(u)|\le\sqrt{\tfrac12+C_W/N}\,A^3
                         +(1+A)E+A^2\sqrt{w/u}.
\end{equation}
In particular, for
\begin{equation}\label{local:numerical-tests}
 A=\frac{257}{256},\quad N=N_0,\quad
 E\le\frac1{256},\quad \sqrt{w/u}\le\frac1{256},
\end{equation}
one has $|f(u)|<3/4$.
\end{lemma}
\begin{proof}
Use Lemma~\ref{lem:s049-wirsing} with $X=u,Y=w,F=G=A^2$.
Equations \eqref{local:primitive-bound} and
\eqref{local:primitive-test} imply its primitive hypothesis, because
$A K_f\log u/w\le A^2/(4N^3)$.  Put
$\kappa_N=1/2+C_W/N$. The conclusion is
$\int_w^u h^2\le\kappa_N A^4(u-w)$.
The $L^2$ triangle inequality in the exact equation, restoring the
entire initial interval $[0,w]$, gives
\[
 \frac{\|f\|_{L^2(0,u)}}{\sqrt u}
 \le\sqrt{\kappa_N}A^2+E+A\sqrt{w/u}.
\]
Cauchy--Schwarz at $u$ gives
$|h(u)|\le A\|f\|_2/\sqrt u$; the pointwise arithmetic equation
at $u$ proves \eqref{local:one-step}, also when an endpoint is an
integer or prime power. For $N=N_0$, $\kappa_N\le209/400$ and
$\sqrt{209/400}<29/40$. The resulting rational upper bound is
\[
 \frac{29}{40}\left(\frac{257}{256}\right)^3
 +\frac{1+257/256}{256}
 +\frac{(257/256)^2}{256}
 =\frac{500158277}{671088640}<\frac34.
\]
\end{proof}

\subsection{The common growing-degree schedule}

We now verify all preceding finite conditions on exactly the family
needed later.  Fix $K>400$.  Let $X$ be the outer variable, and write
\begin{equation}\label{local:outer-schedule}
 L=\log X,\quad r=\log L,\quad Y_0=\frac{\sqrt L}{2},\quad
 \lambda=\frac{K\log r}{r},\quad \delta=\lambda/2,
 \quad T=\left\lceil\frac{24}{\lambda}\right\rceil,
 \quad\eta=\lambda/8=\delta/4.
\end{equation}
All these parameters remain fixed during this outer proof.  Eventually
$0<\delta\le1/4$.  At each root logarithm
\begin{equation}\label{local:root-family}
 Y_0\le u\le L+1,\qquad q\in[1/2,1),\qquad
 \kappa=\log(1/(1-q))\le u^{1-\delta},
\end{equation}
put
\begin{equation}\label{local:root-cuts}
 \alpha_0=1-\frac{3\delta}{4},\qquad
 \beta=\frac{1-\delta}{1-\delta/2},\qquad
 V=u^{\alpha_0},\quad v=u^\beta,\quad
 W=V/\theta,\quad w=2v/\theta,\qquad p=1-\delta/3.
\end{equation}
Here $\theta$ will be one fixed number, independent of $X,u,q$.

Let $C_a$ be the absolute cutoff-head constant in
Lemma~\ref{grow:certificate}, and let $C_r$ cover the $\theta$ terms in
\eqref{local:complete-error} when $A_f,A_g\le2$.
Choose $0<\theta\le1/32$ once, small enough that
\begin{equation}\label{local:fixed-theta}
 16C_a\theta^2\le\frac1{64},\qquad
 C_a\theta^2+C_r\theta\le\frac1{1024}.
\end{equation}
The first allowance is kept for the final exponential sum; it is not
claimed to vanish.  The second supplies room in the present finite
energy test.

\begin{proposition}[Uniform strict coefficient on the required family]
\label{local:strict-family}
For all sufficiently large $X$, with a threshold depending only on
$K$, $\theta$ and the fixed elementary constants, every interval in
\eqref{local:root-family} satisfies
\begin{equation}\label{local:three-quarters}
 |R(e^u)-R(qe^u)|\le\tfrac34(1-q)e^u.
\end{equation}
All bounds are uniform in $u\in[Y_0,L+1]$ and in $q$.
\end{proposition}
\begin{proof}
First verify the complete arithmetic certificate.  The exact
identities
\begin{equation}\label{local:exponent-gaps}
 \alpha_0(1-\eta)=1-\delta+3\delta^2/16,
 \qquad \beta-\alpha_0
 =\delta\left(\frac34-\frac1{2-\delta}\right)\ge\delta/7
\end{equation}
show that $\kappa\le V^{1-\eta}$.  Also
$\eta(T+1)>3$, $T\le25/\lambda=O_K(r/\log r)$, and
$V\ge Y_0^{13/16}$.  This exponential growth in $r$ implies
$V^{1-\eta}\ge2T$, $V\ge100(1+J)T^2$ and $V\ge12C_MT$
after one fixed threshold.  The remaining cutoff tests follow from
\[
 \frac{\theta V}{u^{1-\delta}}
       =\theta u^{\delta/4}\longrightarrow\infty,
 \qquad \log u\le\log(L+1)=O(r),
\]
uniformly, so $\kappa\le V/6$ and
$\theta V\ge\kappa+\log u$.  Also
$W/u=\theta^{-1}u^{-3\delta/4}\to0$ uniformly, and $w/W\ge2$
by \eqref{local:exponent-gaps}.  These verify every finite size and
mass condition in Lemma~\ref{grow:certificate}; the additional
$w\le u$ needed below follows from the displayed ratio $w/u$ shortly.

For all $w\le t\le u$, its degree-two conclusion gives
\begin{equation}\label{local:degree-two}
 |d_2(t,q)|\le C_a\theta^2+
      \frac{C_H(1+J)T^4}{V}+8T^3\sqrt{W/t}.
\end{equation}
The entire boundary below $W$ is included in the last term.
Because $W/w=V/(2v)$, \eqref{local:exponent-gaps} gives
\[
 T^3\sqrt{W/w}\le C T^3e^{-\delta\log u/14}.
\]
Eventually $\log Y_0\ge r/3$, whence
$\delta\log u\ge K\log r/6$.  Consequently
\begin{equation}\label{local:vanishing-boundary}
 T^3\sqrt{W/w}=O_K(r^{3-K/84})=o(1).
\end{equation}
The shared $T^4/V$ row also tends uniformly to zero.  The only
nonvanishing arithmetic term in \eqref{local:degree-two} is the
explicit $C_a\theta^2$ allowance.

Next check the raw amplitudes, without using any strict prime result.
For $v\le t\le u$,
\[
 \frac\kappa v\le
 u^{-\delta(1-\delta)/(2-\delta)}
 \le u^{-3\delta/7}=O_K(r^{-K/14}).
\]
Equation \eqref{local:raw-sieve} therefore gives $|f|\le1+o(1)$
uniformly; its logarithmic, proper-power and floor rows vanish as
well.  In the integer mesh defining $g$, every relevant lower knot
has logarithm $Z\ge v-\log2$.  Since $\beta\ge6/7$,
\[
 Z^{p-1}=Z^{-\delta/3}=O_K(r^{-K/21}).
\]
The unrounded length $\exp(Z-Z^p)$ is eventually at least
$e^{Z/2}$ uniformly, justifying the retained factor $1/2$ after taking
its floor.  The mesh upper endpoint differs
from the evaluation logarithm by at most $\log2$.
Equation \eqref{local:raw-mesh} now proves $|g|\le1+o(1)$
uniformly.  Hence both actual amplitudes are at most $A=257/256$
after one common threshold.  This uses only the elementary interval
sieve, including its proper-prime-power and integer errors.

The smoothing geometry also holds throughout $[w,u]$.  In fact
$\theta w=2v$ and
\[
 \frac wu=\frac2\theta
       e^{-\delta\log u/(2-\delta)}\longrightarrow0,
 \qquad \sqrt{w/u}=O_{K,\theta}(r^{-K/24}).
\]
Thus $u>2w$, $\sqrt{w/u}\le1/256$ and
$\kappa\le(1-\theta)w/3$ eventually.  Moreover
\begin{equation}\label{local:smoothing-gap}
 \beta p-(1-\delta)
 =\frac{(1-\delta)\delta}{6(1-\delta/2)}\ge\delta/8.
\end{equation}
It follows that
$v^p/\kappa\ge u^{\delta/8}\ge r^{K/48}$ whenever
$\kappa>0$.  Since $2v-\log4\ge v$ eventually, the last term
of \eqref{local:complete-error} is at most
\[
 u\exp\{\kappa-(2v-\log4)^p\}
 \le u\exp(-v^p/2)\le u\exp(-u^{3/4}/2)=o(1).
\]
The other nonconstant terms there are bounded by $C/w$ or by
exponentials tending to zero uniformly: $\kappa/v=o(1)$ and
$\theta t\ge2v$ handle the count/cell term; $\kappa/t=o(1)$
handles the divisor-average, tail-count and ordinary-floor terms.
Their constants have not acquired a dependence on $\delta$.
Together with \eqref{local:degree-two}--\eqref{local:fixed-theta},
this proves the complete forcing bound
\[
 \sup_{w\le t\le u}|f(t)+h(t)|
 \le C_a\theta^2+C_r\theta+o(1)\le\frac1{256}.
\]

Finally, the numerical primitive test is uniform.  Since $\beta\ge6/7$
and $u\le L+1$,
\[
 K_f\frac{\log u}{w}
 \le\frac{\theta K_f\log(L+1)}{2Y_0^{6/7}}
 \longrightarrow0.
\]
It is therefore eventually at most
$(257/256)/(4N_0^3)$, a fixed positive number, exactly as required by
\eqref{local:primitive-test}.  All the finite hypotheses of
Lemma~\ref{local:energy-step} are now verified.  Its rational
inequality proves \eqref{local:three-quarters}.
\end{proof}

The argument has a single strict energy stage.  The degree-two source,
all smoothing errors and the initial support are retained before that
stage; no already strict bound for $g$ is assumed.  The abstract
real-analysis lemma cited above is independent of the preliminary
arithmetic saving and the main theorem.


\section{Balog's identity with exponential weights}
\label{exp:balog-section}
Lu applies the finite-degree identity~\eqref{main:balog} after choosing
a growing $k$. We use its exponential generating form, so that the
same arithmetic identity incorporates every degree at once. The local
estimates already proved will control its smooth mean, residual and
infinite-degree tail. For $0<\sigma<1/2$ put
\[
 A_\sigma(n)=\sum_{d\mid n}\mu(d)d^{-\sigma}
 =\prod_{\substack{q\mid n\\q\text{ prime}}}(1-q^{-\sigma}),
 \qquad A_\sigma(1)=1.
\]
This coefficient is nonnegative, and all prime powers remain.
Multiply~\eqref{main:balog} by $\sigma^k/k!$ and sum over $k\ge0$.
For each fixed $X$, all rearrangements are absolutely convergent.
On the right this follows uniformly from~\eqref{main:balog-source-bound}:
$\sum_{k\ge0}\sigma^k|\mathcal B_k(X)|/k!
\le2C_0X/(1-2\sigma)$.
On the left there are only finitely many $n$, and their exponential
series converge absolutely. For each such $n$,
\[
 \sum_{i\ge1}\frac{\sigma^i}{i!}\Lambda_i(n)
 =\sum_{d\mid n}\mu(d)[(n/d)^\sigma-1]
 =n^\sigma A_\sigma(n)-e_0(n).
\]
Thus the $n=1$ term in this sum is zero, while the first term of
Balog's identity sums to $X^\sigma R(X)$. Dividing by $X^\sigma$
gives exactly
\begin{align}
 R(X)+\sum_{2\le n\le X}A_\sigma(n)R(X/n)
 &=\mathcal B_\sigma(X),\label{exp:identity}\\
 \mathcal B_\sigma(X)
 &=X^{-\sigma}\sum_{k\ge0}\frac{\sigma^k}{k!}\mathcal B_k(X)
   =\sum_{d\le X}\mu(d)d^{-\sigma}\Rtilde(X/d),\notag\\
 |\mathcal B_\sigma(X)|
 &\le\frac{C_0X^{1-\sigma}}{1/2-\sigma}.
 \label{exp:forcing}
\end{align}
This is the generating form of Balog's identity, with its complete
source. The factorial in~\eqref{main:balog-source-bound} was retained
until summation; neither a finite-degree truncation nor a suppressed
unit term is used here.

\subsection{The smooth mean and its constant centering}
For real $s>1$ define $\zeta(s)=\sum_{n\ge1}n^{-s}$, and put
$b_\sigma=1/\zeta(1+\sigma)$. Absolute convergence and
$\mu*\mathbf1=e_0$ give
\[
 \zeta(1+\sigma)\sum_{d\ge1}\frac{\mu(d)}{d^{1+\sigma}}=1,
 \qquad 0<b_\sigma\le\sigma,
\]
where the inequality follows from
$\zeta(1+\sigma)\ge\int_1^\infty t^{-1-\sigma}\dd t=1/\sigma$.
Define the complete smooth coefficient and exponential residual by
\begin{align}
 \beta_\sigma(t)&=\sum_{d\le t}\frac{\mu(d)}{d^{1+\sigma}}
                   -t^{-\sigma}m(t)
       =\sigma\int_1^t m(v)v^{-1-\sigma}\dd v,\label{exp:beta}\\
 u_\sigma(n)&=\sum_{i\ge1}\frac{\sigma^i}{i!}u_i(n)
       =n^\sigma[A_\sigma(n)-\beta_\sigma(n)]-e_0(n),
       \label{exp:residual}\\
 U_\sigma(t)&=\sum_{n\le t}u_\sigma(n)
             =\sum_{i\ge1}\frac{\sigma^i}{i!}U_i(t).\notag
\end{align}
Partial summation proves \eqref{exp:beta}; its two finite terms
cancel at $1$ and their jumps cancel at every integer.
Summing the exponential series in the finite divisor formulas for
$\Lambda_i,m_i$ proves \eqref{exp:residual}. These series converge
absolutely at each finite argument, and $u_\sigma(1)=0$.

The correction from $\beta_\sigma(t)$ to the constant $b_\sigma$ is
\begin{equation}\label{exp:average-correction}
 e_\sigma(t):=t^\sigma[b_\sigma-\beta_\sigma(t)]
   =\sigma\int_0^\infty e^{-\sigma v}m(te^v)\dd v.
\end{equation}
Indeed partial summation to infinity gives
$b_\sigma=\sigma\int_1^\infty m(v)v^{-1-\sigma}\dd v$.
Lemma~\ref{arith:lem-mobius} supplies a fixed $C_3,t_0$ with
$|m(t)|\le C_3(\log t)^{-3}$ for $t\ge t_0$. Consequently,
for every $\sigma>0$ and $t\ge t_0$,
\begin{equation}\label{exp:average-bound}
 |e_\sigma(t)|\le C_3\sigma\int_0^\infty(\log t+v)^{-3}\dd v
       =\frac{C_3\sigma}{2(\log t)^2}.
\end{equation}
Thus constant centering retains the exact identity
\begin{equation}\label{exp:coefficient-split}
 w_\sigma(n):=A_\sigma(n)-b_\sigma
 =e_0(n)+n^{-\sigma}[u_\sigma(n)-e_\sigma(n)].
\end{equation}
At $n=1$, $e_\sigma(1)=b_\sigma$, so the right side is $1-b_\sigma$.

\subsection{The outer parameters and the comparison envelope}
Fix $K>400$. For each sufficiently large outer $X$, put
\begin{equation}\label{exp:parameters}
 \begin{gathered}
 L=\log X,\quad r=\log L,\quad
 \lambda=K\frac{\log r}{r},\quad p=1-\lambda,\quad
 \delta=\lambda/2,\quad c=e^{\sqrt L},\quad Y_0=\sqrt L/2,\\
 D=\lceil3/\lambda\rceil,\quad T=\lceil24/\lambda\rceil,
 \quad \eta=\lambda/8,\quad r_*=1/16,\\
 \alpha=\frac{\Phi_K(L)}L,\qquad \sigma=32\alpha.
 \end{gathered}
\end{equation}
The exponent $p=1-\lambda$ here specifies the final coefficient mesh;
it is distinct from the internal smoothing exponent $1-\delta/3$
used to prove Proposition~\ref{local:strict-family}.
The cutoff $\theta$ is the single fixed small constant used in the
local proof. Every parameter in~\eqref{exp:parameters} is frozen while
performing the finite sums, local comparisons and integrals for this
outer $X$. Eventually $\lambda\le1/4$ and $0<\sigma\le1/4$.
The degree estimates below are uniform over the whole latter range,
not just the final choice of $\sigma$.

Choose a fixed $L_*>1$ so large that $\Phi_K$ is increasing on
$[L_*,\infty)$. Extend it to $[0,L_*]$ by
$\Phi_K(L)=L\Phi_K(L_*)/L_*$. The ratio $\Phi_K(L)/L$ is
nonincreasing: above $L_*$ its logarithmic derivative is
$-1/(2L)-K/(L\log L)<0$, and below $L_*$ it is constant.
Consequently, for $0\le z\le L$,
\begin{equation}\label{exp:profile-shift}
 \Phi_K(L)-\Phi_K(L-z)\le\alpha z.
\end{equation}
This uses only the monotonicity of the ratio, not a concavity theorem.
Put, for $t\ge1$,
\[
 C(t)=e^{-\Phi_K(\log t)},\qquad
 Q(t)=\frac{R(t)}{tC(t)},\qquad
 \A(t)=\max\left(1,\sup_{1\le v\le t}|Q(v)|\right).
\]
The envelope is finite on every compact interval, and
\begin{equation}\label{exp:envelope-shift}
 |R(X/n)|\le XC(X)\A(X/2)n^{-1+\alpha}
 \qquad(2\le n\le X).
\end{equation}
No bounded supremum at infinity is assumed.

Split the exact identity~\eqref{exp:identity} as
\begin{equation}\label{exp:three-parts}
 \begin{gathered}
 R(X)+I+J_b+J_w=\mathcal B_\sigma(X),\\
 I=\sum_{2\le n\le c}A_\sigma(n)R(X/n),\qquad
 J_b=b_\sigma\left[\Rtilde(X)-\sum_{n\le c}R(X/n)\right],\\
 J_w=\sum_{c<n\le X}w_\sigma(n)R(X/n).
 \end{gathered}
\end{equation}
In particular, $J_b$ retains the self term $-b_\sigma R(X)$.

\section{Growing degrees, a uniform tail and positive interpolation}
\subsection{The full infinite-degree tail}

\begin{lemma}[A factorial-retaining tail]\label{exp:uniform-tail}
For every integer $D\ge1$, $0<\sigma\le1/4$ and $h\ge4(D+1)$,
\begin{equation}\label{exp:degree-tail}
 \sum_{i>D}\frac{\sigma^i}{i!}E_i(h)
 \le\frac{15\sigma^{D+1}h^{D+2}e^{\sigma h}}{(D+1)!}.
\end{equation}
If also $h\le2Y^p\le r_*Y/2$, where $p=1-\lambda$ and
$D\ge3/\lambda$, then
\begin{equation}\label{exp:normalized-tail}
 \frac{\sum_{i>D}\sigma^iE_i(h)/i!}{\sigma e^{r_*\sigma Y}}
 \le60\,64^D Y^{-1-2\lambda}.
\end{equation}
Both constants are independent of $D,\lambda,\sigma,h,Y$.
\end{lemma}
\begin{proof}
For $u\ge0$, the positive Taylor remainder satisfies
$\sum_{i>D}u^i/i!\le u^{D+1}e^u/(D+1)!$.
Use the full integral defining $Q_i$, interchange its nonnegative
terms, and observe that
\[
 (h+v)^{D+1}\le h^{D+1}e^{v/4},\qquad
 \int_0^\infty e^{-(1-\sigma)v}(h+v)^{D+1}\dd v
 \le2h^{D+1}.
\]
The $4Q_i$ row therefore contributes at most
$8\sigma^{D+1}h^{D+1}e^{\sigma h}/(D+1)!$.
The other row contributes at most
$(3h+4)\sigma^{D+1}h^{D+1}e^{\sigma h}/(D+1)!$.
Since $h\ge1$, their sum is at most the right side of
\eqref{exp:degree-tail}. This keeps both lattice-count terms.
For $z=\sigma Y$, division by $\sigma e^{r_*\sigma Y}$ gives at most
\[
 60\,2^D Y^{2p-\lambda D}
       \frac{z^D e^{-z/32}}{(D+1)!}
 \le60\,64^D Y^{2p-\lambda D}.
\]
Here $z^D e^{-z/32}\le32^D D!$, from one term of the exponential
series, and $2p-\lambda D\le-1-2\lambda$.
The factorial was retained until this cancellation; no fixed-degree
constant has been substituted at a growing degree.
\end{proof}

\subsection{The complete coefficient sum on one block}
Start an integer mesh at $b_0=\lceil c/4\rceil$ and set
\[
 b_{j+1}=b_j+\left\lfloor b_j e^{-(\log b_j)^p}\right\rfloor.
\]
For large $X$ all increments are positive. Since each is at least one,
the knots are unbounded. We need only the finitely many blocks meeting
$(c,X]$. Let $V_\sigma$ be the chord interpolant of
$W_\sigma(t)=\sum_{n\le t}w_\sigma(n)$ on these knots, and put
$v_\sigma(n)=V_\sigma(n)-V_\sigma(n-1)$.

\begin{lemma}[A strict constant-centered slope]\label{exp:slopes}
For all sufficiently large $X$, depending only on the fixed $K$ and
the fixed elementary constants, uniformly over $0<\sigma\le1/4$,
\begin{equation}\label{exp:slope-bound}
 |v_\sigma(n)|\le\frac78\sigma n^{-(1-r_*)\sigma}
 \qquad(c<n\le X).
\end{equation}
Moreover, on each needed block $[b,b']$, with $\Delta=b'-b$,
\begin{equation}\label{exp:block-mass}
 0\le\sum_{b<n\le b'}A_\sigma(n)\le\frac{15}{8}\sigma\Delta.
\end{equation}
\end{lemma}
\begin{proof}
Fix a block and a subinterval $(s,t]\subseteq(b,b']$ of length
$\ell\ge\Delta/2$. Write $Y=\log t$ and $h=\log(t/\ell)$.
Uniformly over all these intervals, the floor in $\Delta$ gives
\[
 Y_0\le Y\le L+1,\qquad
 \ell\ge\sqrt t,\qquad h=Y^{1-\lambda}+O(1),\qquad
 h\le2Y^{1-\lambda}\le Y^{1-\delta}.
\]
For the last inequality use $\delta\log Y\ge(K/6)\log r\to\infty$.
The same observation gives $h\ge4(D+1)$ and
$2Y^{1-\lambda}\le Y/32$ eventually.
Every interval is dyadic, with $q=s/t\ge1/2$.
Thus the fixed-cutoff local construction applies at this root $Y$.
Its degree-one conclusion is
\[
 \frac{|R(t)-R(s)|}{\ell}\le\frac34.
\]
Since $u_1(n)=\omega(n)+1-m_1(n)$, the exact transfer to $U_1$
costs $O(Y^{-2})$ after normalization by $\ell$.

For $2\le i\le D$, use Lemma~\ref{grow:certificate} with
$V_Y=Y^{1-3\lambda/8}$, $W_Y=V_Y/\theta$ and the same $T,\eta$.
The uniform moment estimate~\eqref{arith:fixed-mj}, with the integer
count $\ell+1\le2\ell$, adds at most $CJi(i-1)/Y$ to its
normalized bound. Put
\[
 \rho_Y=\sqrt{W_Y/Y}=\theta^{-1/2}Y^{-3\lambda/16},
 \qquad z=\sigma Y.
\]
One term of the exponential series gives
\[
 \frac{z^{i-1}e^{-z/16}}{i!}\le\frac{16^{i-1}}i.
\]
Sum the degree-dependent powers before replacing them by any common
error. The three contributions of the finite comparison are bounded by
\begin{align}
 \text{cutoff contribution}&\le
     \frac{8C_a\theta^2}{1-16\theta}\le\frac1{64},
       \label{exp:geometric-head}\\
 \text{whole boundary contribution}&\le
     \frac{64T^3\rho_Y}{1-16\rho_Y}=o(1),
       \label{exp:geometric-boundary}\\
 \text{shared and smooth contributions}&\le
    C16^D\left(\frac{H_T}{V_Y}+\frac{JD^2}{Y}\right)=o(1).
       \label{exp:shared-rows}
\end{align}
These are normalized by $\sigma e^{r_*\sigma Y}$.
The first bound follows from $1/i\le1/2$; the second from
$1/[i^2(i-1)]\le1/4$. The fixed cutoff contribution is small,
not asserted to tend to zero.

For uniformity of the other two assertions, $\log Y\ge r/3$ gives
$\rho_Y\le C_\theta r^{-K/16}$ and
$T=O_K(r/\log r)$, so $T^3\rho_Y\to0$.
Also $\log V_Y\ge c_1r$ with a fixed $c_1>0$, whereas
$D=O_K(r/\log r)$ and $\log H_T=O_K(\log r)$.
Thus the exponential decay in $V_Y^{-1}$ and $Y^{-1}$ pays the
entire $16^D$ factor. The terminal seed remains in $H_T$.
Lemma~\ref{exp:uniform-tail} supplies the full remaining tail,
$O(64^D Y^{-1-2\lambda})=o(1)$, by the same logarithmic comparison.

The normalized degree-one contribution is at most $3/4+O(Y^{-2})$,
since its additional factor is $e^{-r_*\sigma Y}\le1$.
The absolute convergence of the exponential coefficient series follows
from its finite divisor formulas. Adding all degrees therefore gives
\begin{equation}\label{exp:long-block}
 |U_\sigma(t)-U_\sigma(s)|
 \le c_X\sigma\ell\,t^{r_*\sigma},
 \qquad c_X=\frac{49}{64}+o(1)<\frac78,
\end{equation}
uniformly in every block, long subinterval and $\sigma$.
Using the actual upper root $\log t$ avoids any change of logarithmic
normalization at a growing degree.

For $F(t)=U_\sigma(t)-U_\sigma(b)$, a short prefix is the whole
block minus a long suffix. Hence
\[
 |F(b')|\le c_X\sigma\Delta b'^{r_*\sigma},\qquad
 |F(t)|\le2c_X\sigma\Delta b'^{r_*\sigma}.
\]
The exact Abel identity is
\begin{equation}\label{exp:local-abel}
 \sum_{b<n\le b'}w_\sigma(n)
 =b'^{-\sigma}F(b')+\sigma\int_b^{b'}F(t)t^{-1-\sigma}\dd t
       -\sum_{b<n\le b'}n^{-\sigma}e_\sigma(n).
\end{equation}
The lower boundary is zero; the unit is outside these blocks.
After division by $\sigma\Delta b'^{-(1-r_*)\sigma}$, the first two
terms cost at most
\[
 c_X\bigl[1+2\{(b'/b)^\sigma-1\}\bigr]
 =\frac{49}{64}+o(1).
\]
In particular the prefix factor two multiplies only a vanishing
weight derivative. The last term in~\eqref{exp:local-abel} costs
$O(Y^{-2}(b'/b)^\sigma b'^{-r_*\sigma})=o(1)$ by
\eqref{exp:average-bound}. There are exactly $\Delta$ integers in the
block. The fixed gap between $49/64$ and $7/8$ proves the eventual
slope bound. Since $b_\sigma\le\sigma$, adding the constant mean
and using $b'^{-(1-r_*)\sigma}\le1$ proves~\eqref{exp:block-mass}.
\end{proof}

\subsection{Interpolation, with both integer endpoints}
\begin{lemma}\label{exp:interpolation}
Put $S=(\log(c/4))^{1-\lambda}$. Uniformly over $0<\sigma\le1/4$,
\begin{align}
 \mathcal E_\sigma
   &:=\sum_{c<n\le X}[w_\sigma(n)-v_\sigma(n)]R(X/n),\notag\\
 |\mathcal E_\sigma|&\le C\sigma Xe^{-S}[1+\log(X/c)],
       \label{exp:interp-bound}
\end{align}
where $C$ is an absolute constant once the earlier finite thresholds
are met.
\end{lemma}
\begin{proof}
For an integer $t$ in a mesh block $[b,b']$, write
$M_b=\sum_{b<n\le b'}A_\sigma(n)$. Constant centering cancels
exactly from the integer chord:
\begin{equation}\label{exp:positive-chord}
 W_\sigma(t)-V_\sigma(t)
 =\sum_{b<n\le t}A_\sigma(n)-\frac{t-b}{b'-b}M_b.
\end{equation}
Both terms belong to $[0,M_b]$, so their difference has absolute
value at most $M_b\le15\sigma(b'-b)/8$.
The exact cumulative interpolation error, for $c\le z\le X$, is
\begin{equation}\label{exp:prefix-floors}
 E_\sigma(z):=\sum_{c<n\le z}[w_\sigma(n)-v_\sigma(n)]
 =[W_\sigma-V_\sigma](\lfloor z\rfloor)
       -[W_\sigma-V_\sigma](\lfloor c\rfloor).
\end{equation}
Both blocks have lower knots at least $c/4$, and their lengths are
$O(ze^{-S})$ and $O(ce^{-S})$. Thus
$|E_\sigma(z)|\le C\sigma ze^{-S}$.
Set $E_\sigma(z)=0$ for $z\le c$. Expanding $R$ by its full step
function and rearranging only finite sums gives
\begin{equation}\label{exp:interp-exact}
 \mathcal E_\sigma
 =\sum_{m<X/c}\omega(m)E_\sigma(X/m)+2\gamma E_\sigma(X).
\end{equation}
The reciprocal cutoff is strict, because the original integers satisfy
$n>c$. The unit $\omega(1)=-1$ and every reciprocal jump remain.
Finally~\eqref{arith:mertens} and the harmonic estimate give
$\sum_{m\le T}|\omega(m)|/m\le C[1+\log T]$ for $T\ge1$.
Substitution proves~\eqref{exp:interp-bound}.
\end{proof}

\section{The three complete contributions and the final contraction}
\subsection{Small divisors and the constant mean}
The all-degree bound~\eqref{grow:harmonic-bound} and the exponential
coefficient formula in Section~\ref{exp:balog-section} give, for every
real $t\ge1$ and $\sigma>0$,
\[
 \begin{aligned}
 H_\sigma(t)&:=\sum_{2\le n\le t}n^{\sigma-1}A_\sigma(n)\\
 &=\sum_{i\ge1}\frac{\sigma^i}{i!}
       \sum_{n\le t}\frac{\Lambda_i(n)}n
 \le2\sum_{i\ge1}\frac{(\sigma\log t)^i}{i!}
 =2(t^\sigma-1).
 \end{aligned}
\]
The degree series converges absolutely at each finite argument;
$\Lambda_i(1)=0$ removes the unit, and no degree is truncated.
Since $H_\sigma(1)=0$, partial summation retaining the upper endpoint
gives, for every real $c\ge1$,
\begin{equation}\label{exp:small-mass}
 \begin{aligned}
 \sum_{2\le n\le c}\frac{A_\sigma(n)}n
 &=c^{-\sigma}H_\sigma(c)
       +\sigma\int_1^c H_\sigma(t)t^{-1-\sigma}\dd t\\
 &\le2(1-c^{-\sigma})
       +2\sigma\int_1^c(1-t^{-\sigma})\frac{\dd t}{t}
 =2\sigma\log c.
 \end{aligned}
\end{equation}
The endpoint term cancels the integrated subtraction exactly.
All prime powers and inclusive real cutoffs are retained; this bound
does not require $\sigma\log c$ to be small.
For the parameters~\eqref{exp:parameters},
$\sigma\log c=32/r^K$ and $\alpha\log c=r^{-K}$.
Thus~\eqref{exp:envelope-shift} implies
\begin{equation}\label{exp:I}
 \begin{aligned}
 \frac{|I|}{XC(X)}
 &\le2e^{\alpha\log c}\sigma\log c\,\A(X/2)\\
 &=\frac{64e^{r^{-K}}}{r^K}\A(X/2)
 =o(1)\A(X/2).
 \end{aligned}
\end{equation}

Writing $s=\log c=\sqrt L$, the complete mean satisfies
\begin{align}
 \frac{|J_b|}{XC(X)}\le{}&b_\sigma|Q(X)|
 +b_\sigma e^{\alpha s}(1+s)\A(X/2)
 +C_0b_\sigma e^{\Phi_K(L)-L/2}.
       \label{exp:mean-bound}
\end{align}
The first term is the retained mean self term. The second follows
from $\sum_{2\le n\le c}n^{-1+\alpha}\le e^{\alpha s}(1+s)$;
the last is the divisor-average endpoint.
Since $b_\sigma\le\sigma$, the non-self coefficient tends to zero
and the last term is $o(1)$. The self coefficient is absorbed only
at the end of the proof.

\subsection{Residual weight, source and mesh separation}
The slope and interpolation lemmas imply
\[
 |J_w|\le\frac78\sigma
       \sum_{c<n\le X}n^{-(1-r_*)\sigma}|R(X/n)|
       +|\mathcal E_\sigma|.
\]
Put $d=(1-r_*)\sigma-\alpha=29\alpha>0$.
Because $\sum_{n\ge2}n^{-1-d}\le\int_1^\infty t^{-1-d}\dd t=1/d$,
\begin{equation}\label{exp:residual-contraction}
 \frac{|J_w|}{XC(X)}
 \le\frac78\frac{\sigma}{(15/16)\sigma-\alpha}\A(X/2)
          +\frac{|\mathcal E_\sigma|}{XC(X)}
 =\frac{28}{29}\A(X/2)+\frac{|\mathcal E_\sigma|}{XC(X)}.
\end{equation}
The source in~\eqref{exp:forcing} has normalized size at most
\[
 \frac{C_0e^{\Phi_K(L)-\sigma L}}{1/2-\sigma}
 =\frac{C_0e^{-31\Phi_K(L)}}{1/2-\sigma}=o(1).
\]
For the mesh exponent, direct calculation gives
\begin{equation}\label{exp:mesh-separation}
 \log S=\frac r2-\frac K2\log r+o(1),\qquad
 \frac{S}{\Phi_K(L)}=r^{K/2}(1+o(1))\longrightarrow\infty.
\end{equation}
Both $S$ and $S-\Phi_K(L)$ dominate $r$. Consequently
\[
 \frac{|\mathcal E_\sigma|}{XC(X)}
 \le C\sigma e^{\Phi_K(L)-S}[1+\log(X/c)]=o(1).
\]
This estimate retains both floors, the strict reciprocal endpoint,
the unit, $2\gamma$ and the exponential-average correction already
used in the coefficient transfer.

\subsection{Finite-envelope closure}
Divide~\eqref{exp:three-parts} by $XC(X)$ and combine all preceding
estimates. For all sufficiently large $X$,
\begin{equation}\label{exp:closure}
 |Q(X)|\le\left(\frac{28}{29}+o(1)\right)\A(X/2)
                    +o(1)|Q(X)|+o(1).
\end{equation}
Absorb the self term now. There are fixed $q_*<1$ and $X_0$ with
$|Q(X)|\le q_*\A(X/2)+1$ for $X\ge X_0$.
Choose $M=\max(\A(X_0),1/(1-q_*))$. If
$\A(2^jX_0)\le M$, the same inequality gives
$|Q(X)|\le q_*M+1\le M$ on the next dyadic interval.
Induction proves $\A(X)\le M$ for every finite $X$.
Since
\[
 R(X)-[\psi(X)-X]=\{X\}+2\gamma
\]
is bounded, Theorem~\ref{main:theorem} follows.

\section*{Attribution and scope}
\phantomsection
\addcontentsline{toc}{section}{Attribution and scope}
The proof follows the elementary remainder method of
Wirsing~\cite{Wirsing}, Balog~\cite{Balog}, and Lu~\cite{Lu}.
The organizing identity is Balog's identity as used by Lu in
\S2, equation~(3), and \S6, Lemma~20 (printed pages~980 and~1024).
Lu's finite-degree coefficient split is in \S6, equations~(106)--(112),
and the concluding finite-envelope argument is in \S9.
Here the same identity is summed exponentially, with local estimates
quantified for a shrinking interval parameter. The finite-degree
identity is cited from Lu; its exponential form and all-degree source
bound are proved here. The standard inputs are cited
from Lu: the elementary interval sieve in \S3, equation~(17), the
finite Wirsing inequality in \S4, Lemma~3, and the fixed-power
preliminary estimates in \S6, equation~(114) and Lemma~21. Their
normalizations, endpoint corrections and all degree-uniform
consequences needed here remain explicit. The energy parameter $N_0$
is an absolute integer chosen from the cited lemma, not a parameter
that grows with $X$. Constants and thresholds are existential and
depend on the fixed $K$; numerical optimization is not attempted.

The literature comparison in Section~\ref{intro:literature} includes
Goz\'e's 2024 quantitative refinement and
Koukoulopoulos's stronger largely elementary theorem. The result is
presented as a quantitative refinement of a particular proof method.
No claim of historical priority or of the strongest elementary
remainder is made.

\section*{Document authorship and AI assistance}
\phantomsection
\addcontentsline{toc}{section}{Document authorship and AI assistance}
Rajat Bhattacharjee is the human author. He proposed and directed
the research questions, supplied source material, and determined
the scope and development of the manuscript. AI assistance through
OpenAI's Codex environment was substantial: it included exploration
of identities and estimates, proposed derivations, mathematical
drafting and revision, literature searches, and the construction
of finite computational checks. This assistance extended to the
mathematical reasoning, rather than only to language editing.

The author's assessment of mathematical correctness has relied on
GPT-6 Astra. Separate AI-agent review passes were used to challenge
derivations and check dependencies, parameter ranges, endpoint
corrections, and consistency. These were informal AI reviews;
they were not external human refereeing or formal proof verification.
Agreement among such reviews and successful finite computations do
not certify asymptotic estimates. No independent human verification
of the complete proof is claimed in this version.

AI systems are acknowledged as tools and are not listed as authors.
The author is the human contact for the manuscript and any subsequent
corrections. The theorem is presented as a research claim for
mathematical scrutiny; the authorship statement describes its
development and checking, without asserting peer-review acceptance.

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\end{thebibliography}

\end{document}