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Explicit one-sided arithmetic reduction

2026-09-30. This is a direct calculation in the unchanged scalar quotient specified in ../../one_prime/notes/arithmetic_space.md. RH is OPEN. No new topology, spectral encoding, or enlargement of the relation space is used. The inverse-Euler-operator/cutoff architecture already occurs in Meyer, math/0311468v1, §5.3, Lemmas 5.3–5.4. No novelty claim is made for the construction or estimates below.

1. Fixed data and the actual representative

Write S f(x)=2 sum_{m>=1} f(mx) for even Schwartz functions on R. If f(0)=integral_R f=0, then J S f(t)=exp(t/2) S f(exp(t)) belongs to the actual arithmetic range V, hence W. Its finite adelic components are exactly product_p 1_{Z_p}. No Euler factor is removed or altered.

Fix once and for all smooth real functions chi and psi on the positive axis:

chi=0 on x<=1/2, chi=1 on x>=1, 0<=chi<=1;
psi is supported in (1/2,1), and integral_0^infinity psi(x) dx=1.

For g in E and a=2^n, n>=0, define

H(x)=x^{-1/2}g(\log x),\quad H_a(x)=a^{-1/2}H(x/a), \quad F_a(x)=\frac12\sum_{m\ge1}\mu(m)H_a(mx). \tag{1}

The series and each derivative converge locally absolutely on x>0. Define

c_a=\int_0^\infty\chi(x)F_a(x)\,dx,\qquad f_a(x)=\chi(x)F_a(x)-c_a\psi(x)\quad(x>0), \tag{2}

and extend f_a evenly to R. This is a Schwartz function, vanishing near 0, with integral_R f_a=0. In particular, the correction in (2) enforces the existing arithmetic moment condition. It is not a new counterterm in the Weil form. Choices of chi and psi are not canonical; the proof and rate are valid for any one fixed such choice, independent of a, g and zeros.

Set

w_n(g)=-J S f_a\in V\subseteq W,\qquad R_n(g)(t)=\sqrt{x}\bigl(H_a(x)-S f_a(x)\bigr),\quad x=e^t. \tag{3}

These are explicit linear constructions from g using the usual integer Möbius coefficients, one cutoff, one integral, and arithmetic summation. They satisfy

\boxed{T_2^n g+w_n(g)=R_n(g),\qquad \operatorname{supp}R_n(g)\subseteq(-\infty,0].} \tag{4}

Indeed, for x>=1, chi(mx)=1 and psi(mx)=0 for every m>=1. Absolute convergence permits the divisor regrouping

2\sum_{m\ge1}F_a(mx) =\sum_{r\ge1}H_a(rx)\sum_{d\mid r}\mu(d)=H_a(x). \tag{5}

Both sides of (3) lie in E, by the pole-free Poisson formula for f_a. Thus (4) is an E identity, not merely a pointwise equality. R_n is flat at t=0. It generally has a noncompact left tail. It does not give a representative in one compact dyadic fundamental interval.

2. Unconditional bound: exponent 1 becomes 1/2

Put P=p_4(g). Differentiating H gives, for 0<=j<=4,

|x^j H^{(j)}(x)| \le C_j P\min(x^{7/2},x^{-9/2}). \tag{6}

The elementary lattice estimate

\sum_{m\ge1}\min((mh)^{7/2},(mh)^{-9/2}) \le C h^{-1}(1+h)^{-7/2}

and |mu(m)|<=1 yield, for j<=3,

|F_a^{(j)}(x)|\le C P\sqrt a\,x^{-j-1}(1+x/a)^{-7/2}. \tag{7}

Consequently

|c_a|\le CP\sqrt a(1+\log a), \quad \|f_a''\|_{L^1(\mathbb R)}+ \|x f_a'''\|_{L^1(\mathbb R)} \le CP\sqrt a(1+\log a). \tag{8}

For the first inequality, split the integral at a: the interval [1/2,a] costs at most a logarithm and the tail is integrable by (7). For the second, all derivatives of chi are in [1/2,1]; away from that interval integrate x^{-3} in the f_a'' and x f_a''' terms. The fixed psi contribution is bounded by |c_a|. All constants depend only on the fixed cutoffs and the seminorm convention, not on a or g.

With Fourier convention exp(-2 pi i x xi), Poisson for pole-free even f is

S f(x)=x^{-1}S\widehat f(x^{-1}),\qquad |S f(x)|\le\frac{x}{12}\|f''\|_1. \tag{9}

Here 2 zeta(2)/(2 pi)^2=1/12. Let D=x d/dx. Df is also even and pole-free, since integral Df=-integral f=0, and

(Df)''=2f''+xf''',\quad \partial_t(J S f)=J S\bigl(\tfrac12 f+Df\bigr). \tag{10}

Applying (9) to f and (1/2)f+Df gives, for 0<x<=1,

|J S f|+|\partial_t J S f| \le Cx^{3/2}(\|f''\|_1+\|xf'''\|_1). \tag{11}

The p_1 weight on this half-axis is x^{-1}(1-log x), and sup_{0<x<=1} x^{1/2}(1-log x)<infinity. The unreduced term obeys the one-sided bound p_1(T_2^n g; t<=0)<=C a^{-4}p_4(g). Together with the exact vanishing for t>=0, this proves

\boxed{p_1(R_n(g))\le C\,2^{n/2}(1+n)\,p_4(g).} \tag{12}

Taking the infimum over all representatives g of a class, after applying this explicit construction to each representative, gives

q_1(T_2^n[g])\le C\,2^{n/2}(1+n)q_4([g]). \tag{13}

No assumption that R_n descends as an E-valued map on the quotient is needed: (4) proves the class identity for every input representative. This is an inter-seminorm estimate, not a bound on the q_1 Banach operator norm and not a proof that its full spectral radius is sqrt(2). The exponent 1/2 is an achieved upper bound, not a proved optimal rate.

This reaches the task's Stage B / Level 3 literally. Spectrally it only implies Re rho<=1, which is already known. It is not new zero-location information or progress to RH. The full integer Möbius function, hence all primes, enters (1); only the return action is dyadic.

3. Exactly what stronger arithmetic cancellation would buy

Assume, conditionally, for a fixed 0<theta<1,

|M(X)|=\left|\sum_{m\le X}\mu(m)\right|\le K_\theta X^\theta \quad(X\ge1). \tag{14}

Summation by parts, applied also to K_j(y)=y^j H^{(j)}(y), gives

|F_a^{(j)}(x)| \le C_\theta K_\theta p_4(g) a^{\theta-1/2}x^{-j-\theta}\quad(0\le j\le3). \tag{15}

For example, the relevant integral is integral_0^infinity y^theta |K_j'(y)|dy, finite by (6) and p_4. F_a is integrable at 0 by theta<1 and rapidly decreasing at infinity. Its Mellin transform is, initially for Re s>1,

\mathcal M F_a(s)=\frac{\mathcal M H_a(s)}{2\zeta(s)}. \tag{16}

Taking real s down to 1 is justified by dominated convergence on the left; on the right, zeta has its known simple pole. Thus integral F_a=0. This step does not presuppose a zero-free half-plane. Therefore

c_a=-\int_0^\infty(1-\chi)F_a, \quad |c_a|\le C_\theta K_\theta p_4(g)a^{\theta-1/2}/(1-\theta). \tag{17}

Equations (15), (17) and the same Poisson estimates prove for the same explicit representatives (1)–(3)

\boxed{p_1(R_n(g))\le C_\theta K_\theta,2^{n(\theta-1/2)}p_4(g).} \tag{18}

The mean cancellation in (17) is essential. Estimating integral|F_a| on [1/2,infinity) would discard it and could retain an a^{1/2} cost. Moment cancellation cannot simply be inferred from local estimates.

The classical condition M(X)=O_epsilon(X^{1/2+epsilon}) for every epsilon>0 is equivalent to RH; see Báez-Duarte math/0202141v2, §2.1 and Soundararajan 0705.0723v2, introduction Eq.(1). In particular RH implies (14) with every theta in (1/2,1), which would prove the desired subexponential estimate by (18). Conversely that estimate for these representatives implies RH by the nonzero zero evaluations and reflection. Consequently the following endpoint assertion about this fixed, zero-independent algorithm is RH-equivalent:

\forall\epsilon>0\ \exists C_\epsilon: p_1(R_n(g))\le C_\epsilon2^{\epsilon n}p_4(g) \quad(g\in E,n\ge0). \tag{19}

For epsilon>=1/2, (19) under RH follows by choosing any smaller positive epsilon first. Equation (19) is rejected as an independent bridge. We have not proved that every possible reduction algorithm must pass through Mertens estimates, or that arithmetic relations cannot give a better estimate by a different method.

4. Arithmetic fidelity

Every w_n is produced by the genuine test function eta_n=f_a tensor product_p 1_{Z_p}. The value and integral constraints hold exactly. Compact-unit averaging fixes this eta_n already. The archimedean Schwartz function, its Fourier transform, and both pole conditions are in the construction from the start. This preserves the actual W and all its Mellin zero evaluations/jets. No zero ordinate or real part is used in chi, psi, f_a, c_a or w_n. No topology is weakened. The archimedean local factor is not replaced by an artificial Gamma factor.

Decision: retain the unconditional identity and rate (12) as a constructive bound in the fixed quotient. Stop the subexponential endpoint route after identifying (19) as RH-equivalent; it is not an independently proved arithmetic cancellation principle.