Neutral Tents and the Square-Root Response Obstruction

Abstract

Two exact zero moments and bounded increments do not give a uniform square-root response bound.

Fix phi to be the positive golden ratio and q=phi^(-2). The kernel k is precisely BinetPositiveResponse.response(q,hq), where hq proves q>0. Thus k is the Dirichlet inverse of the literal Binet logarithmic coefficient applied to the arithmetic logarithm. It is not a free kernel or an assumed response law.

For positive integer m, use u(m)=1/(m(m+1)) and v(m)=log(m)/m-log(m+1)/(m+1). Let K(J) be the sum of k(d) over 1<=d<=J, and let T(f,N) be the sum of k(d)f(floor(N/d)) over 1<=d<=N. All sequences are real-valued and indexed by natural numbers.

Write S(f) for the set of |f(m)|/sqrt(m) at positive integer m. The predicate I(f) below requires f(0)=0, summability of both positive-index moment series, both exact zero moments, nonemptiness and upper boundedness of S(f), sup S(f)<=1, the pointwise envelope |f(m)|<=sqrt(m) for every natural m, and |f(m)-f(m-1)|<=1 for every m>=1. It has no finite-support restriction.

$I\left(f\right)⟺(f\left(0\right)=0\wedge \mathrm{Summable}\left(m\mapsto f\left(m+1\right)u\left(m+1\right)\right)\wedge \mathrm{Summable}\left(m\mapsto f\left(m+1\right)v\left(m+1\right)\right)\wedge {\sum }_{m\ge 1}f\left(m\right)u\left(m\right)=0\wedge {\sum }_{m\ge 1}f\left(m\right)v\left(m\right)=0\wedge \mathrm{Nonempty}\left(S\left(f\right)\right)\wedge \mathrm{BddAbove}\left(S\left(f\right)\right)\wedge \sup \left(S\left(f\right)\right)\le 1\wedge (\forall m\in \mathbb{N},|f\left(m\right)|\le \sqrt{m})\wedge (\forall m\ge 1,|f\left(m\right)-f\left(m-1\right)|\le 1))$

The final quantifier uses O(f), the original zero value, two zero moment identities, bounded unit supremum norm, and unit adjacent increments. This class has no finite-support restriction. Every constructed member of I also belongs to O.

$O\left(f\right)⟺(f\left(0\right)=0\wedge {\sum }_{m\ge 1}f\left(m\right)u\left(m\right)=0\wedge {\sum }_{m\ge 1}f\left(m\right)v\left(m\right)=0\wedge \mathrm{Nonempty}\left(S\left(f\right)\right)\wedge \mathrm{BddAbove}\left(S\left(f\right)\right)\wedge \sup \left(S\left(f\right)\right)\le 1\wedge (\forall m\ge 1,|f\left(m\right)-f\left(m-1\right)|\le 1))$

Theorem 1.1 (The actual response is unbounded on the entire neutral input class).

$$(\exists F:\mathbb{N}\to \mathbb{N}\to \mathbb{R},\forall J\in \mathbb{N},2\le J\Rightarrow (\mathrm{Finite}\left(\mathrm{support}\left(F\left(J\right)\right)\right)\wedge I\left(F\left(J\right)\right)\wedge \frac{K\left(J\right)}{16\sqrt{J}}\le \frac{T\left(F\left(J\right),64J^3\right)}{\sqrt{64J^3}}))\wedge \underset{J\to \infty }{\lim }\frac{K\left(J\right)}{16\sqrt{J}}=\infty \wedge \neg (\exists C\in \mathbb{R},\forall f:\mathbb{N}\to \mathbb{R},O\left(f\right)\Rightarrow \forall N\in \mathbb{N},1\le N\Rightarrow \frac{|T\left(f,N\right)|}{\sqrt{N}}\le C)$$

Proof. Machine-checked in Lean as D5/S3/Arith/FibonacciAtomic/NeutralTentResponseObstruction.neutral_tent_response_obstruction (✓ std3). ∎

Source. Repository-derived.

Commentary.

There is one family F, with f_J=F(J), whose members have finite support and satisfy every condition in I simultaneously for every J>=2, including J=2. At the same integer cutoff N_J=64J^3, the signed response divided by sqrt(N_J) is at least K(J)/(16sqrt(J)). This lower bound tends to positive infinity as J tends to infinity through the integers.

For each 2<=j<=J put m_j=floor(N_J/j) and R_j=floor(sqrt(m_j)/8). Three literal max/absolute-value tents, centered at m_j-2R_j, m_j and m_j+2R_j, are combined with coefficients (-alpha_j,1,-eta_j)/2. The ratio v(m)/u(m) is strictly increasing: its adjacent difference is (m+1)log((m+1)^2/(m(m+2)))>0. Ordered positive weighted averages therefore give the two exact compensating coefficients, with 0<alpha_j<1 and 0<eta_j<2.

Each pulse has both exact zero moments, center value R_j/2, the square-root envelope and unit increments. An adjacent pair can meet the nonzero support of at most one individual tent; the same property holds between different j blocks. This checks the zero seams as well as the interior slopes. All supports lie in the finite interval from 1 to N_J, so the finite moment cancellations are also the exact infinite sums.

The endpoint quotient identities are floor(N_J/(m_j-3R_j+1))=j and floor(N_J/(m_j+3R_j))=j-1. Consequently only d=j samples the j-th pulse, at its center; d=1, d=J+1 and all other divisor indices contribute zero to that pulse. The exact response is one half of the sum of k(j)R_j over 2<=j<=J. Since R_j>=J and k(1)=0, this is at least J K(J)/2. The identity sqrt(N_J)=8J sqrt(J) gives the original 1/16 constant.

The actual Binet contract supplies a positive constant c with k(j)>=c log(j). Thus K(J)>=(J-1)c log(2) for J>=2, and K(J)/(16sqrt(J)) is at least c log(2)sqrt(J)/32. A uniform constant for every member of O and every positive cutoff would contradict this divergence. The finite-support family is used to refute a bound on the entire original class O, rather than defining that class.

The inputs change with J. They are not the actual arithmetic partial sums H, and no fixed finite-support input is asserted to have a divergent normalized response. The obstruction proves no improvement to a Robin estimate or the Riemann hypothesis.

References

• Truth anchor: D5/S3/Arith/FibonacciAtomic/NeutralTentResponseObstruction.neutral_tent_response_obstruction
• Dependency: D5/S3/Arith/FibonacciAtomic/BinetPositiveResponse