Nonlocal Cancellation in a Theta-Kernel Decomposition of the Riemann \( \Xi \)-Growth Derivative: An Obstruction to Phase-Aligned Blockwise Positivity

1. Introduction

1.1. Setting

Write

$$\mathrm{\Phi }\left(u\right)=\sum _{n\ge 1}\left(2{\pi }^2{n}^4{e}^{9u/2}-3\pi n^2{e}^{5u/2}\right)e^{-\pi n^2{e}^{2u}},D\left(z\right)={\int }_0^{\infty }\mathrm{\Phi }\left(u\right)cos\left(zu\right)du,$$

(1)

and let

$$\xi \left(s\right)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta \left(s\right),\mathrm{\Xi }\left(z\right)=\xi \left(\frac{1}{2}+iz\right).$$

(2)

The kernel $\mathrm{\Phi }$ is real, even and real-analytic, and it together with all its derivatives decays doubly exponentially.

1.2. Background

The Fourier representation of $\mathrm{\Xi }$ by the theta kernel $\mathrm{\Phi }$ is the starting point of a long tradition. Pólya [1] initiated the study of RH through the zeros of Fourier transforms of such kernels; de Bruijn [2] introduced the heat deformation $H_t$ of $\mathrm{\Xi }$ and showed that $H_t$ has only real zeros for $t\ge \frac{1}{2}$; Newman [3] proved the existence of the constant $\mathrm{\Lambda }$ for which $H_t$ has only real zeros exactly when $t\ge \mathrm{\Lambda }$, RH being the assertion $\mathrm{\Lambda }\le 0$, and Rodgers and Tao [4] proved $\mathrm{\Lambda }\ge 0$. A convenient modern account, with the normalisation used here, is the Polymath treatment of the flow [5]. Positivity and hyperbolicity conditions on objects attached to $\mathrm{\Xi }$ remain an active line of attack; see for instance the work of Griffin, Ono, Rolen and Zagier on Jensen polynomials [6]; the classical Turán-inequality analysis of the same theta kernel by Csordas, Norfolk and Varga [7] belongs to the same tradition. In [8,9] the present author studied the onset of Jensen hyperbolicity, in the first paper exhibiting a structural obstruction to that route and in the second a candidate moment-positivity structure; the present obstruction is of a similar spirit to the former but arises in a different representation and by a different mechanism. A second strand concerns the growth of $\left|\xi \right|$ away from the critical line. Hinkkanen [10] and Lagarias [11] showed that RH is equivalent to $Re({\xi }^{\prime }/\xi )\left(s\right)>0$ on $Res>\frac{1}{2}$; Sondow and Dumitrescu [12] reformulated this as horizontal monotonicity of $\left|\xi \right|$ in zero-free half-planes; Goldštein and Grigutis [13] study the positivity of $Re({\xi }^{\prime }/\xi )$ in the right half of the critical strip, including the behaviour in the hypothetical presence of off-line zeros. These criteria situate the present work: our object, ${\partial }_y{\left|D(x+iy)\right|}^2$, is exactly the growth derivative appearing there (Theorem 2).

1.3. The Question, and What We Prove

Given the exact decomposition of ${\partial }_y{\left|D\right|}^2$ into phase-aligned blocks constructed in §§3–5, it is natural to hope that positivity can be established block by block. We show that it cannot.

The main result (Theorem 5) is that the longitudinal and transverse sectors cancel to all algebraic orders: each is of size $x^{-5}$, their sum is exponentially small, and $E/S\to -1$. Its sharp local consequence (Theorem 6) is that some aligned blocks are necessarily negative.

Novelty.

This paper does not propose another sufficient positivity condition for RH. It proves that a natural localisation of the growth derivative into independently controlled aligned blocks is structurally incompatible with the nonlocal cancellation carried by the exact theta-kernel representation. Both theorems are unconditional: they are statements about the internal sign structure of one particular exact decomposition, and they hold whether or not RH is true.

Scope.

We prove that the individual phase-aligned blocks cannot all be nonnegative, and that estimating the two sectors independently destroys the quantity whose sign is at issue. We do not claim that no use of the partition can succeed: grouping blocks, retaining correlations between the sectors, or resummation identities are not excluded. What is excluded is the strategy of blockwise positivity or independent blockwise domination.

Organisation.

Section 2 proves the two elementary properties of the theta kernel used throughout — positivity, strict monotonicity, and ${\mathrm{\Phi }}^{\prime \prime }\left(0\right)<0$ — so that the paper is unconditional from the outset. Section 3 derives the exact two-term representation of ${\partial }_y{\left|D\right|}^2$, fixes the normalisation $\mathrm{\Xi }=4D$, reduces to $x>0$, and records the growth criterion linking the object to RH. Section 4 establishes the master identity and identifies the longitudinal envelope $Q_y$ and the transverse residual ${\epsilon }_x$. Section 5 proves the phase-aligned moment lemma — the structural fact that a sine-block annihilates one Taylor order and a cosine-block two — and deduces the oscillatory hierarchy, with explicit leading coefficients for $C_m^{\left(Q\right)}$ and $C_m^{\left(\epsilon \right)}$. Section 6 contains the main result, the nonlocal cancellation theorem: the two sectors cancel to all algebraic orders, so that $E/S\to -1$ while each is of size $x^{-5}$. Section 7 draws the sharp local consequence, the no-go theorem for blockwise positivity. Section 8 collects the numerical illustrations, and Section 9 concludes.

2. The Kernel: Positivity and Monotonicity

All results used in the main theorems of §§6–7 are unconditional; the numerical material of §8 illustrates them but is never used in their proofs. We therefore begin by proving the two elementary properties of $\mathrm{\Phi }$ that are used throughout. Set

$$q_n=\pi n^2,x=e^{2u}.$$

(3)

Since $2{\pi }^2{n}^4=2q_n^2$ and $3\pi n^2=3q_n$,

$$\mathrm{\Phi }\left(u\right)=x^{5/4}\sum _{n\ge 1}{q}_n\left(2q_{n}x-3\right)e^{-q_{n}x}.$$

(4)

Lemma 1

(Positivity and strict decrease). $\mathrm{\Phi }>0$ on $\mathbb{R}$, and ${\mathrm{\Phi }}^{\prime }\left(u\right)<0$ for every $u>0$. Consequently

$$p:=-{\mathrm{\Phi }}^{\prime }/\mathrm{\Phi }>0\mathrm{on}(0,\infty ),p\left(0\right)=0.$$

(5)

Proof. 

Positivity. For $u\ge 0$ we have $x\ge 1$, so $2q_{n}x-3\ge 2\pi -3>0$ for every $n\ge 1$; every summand in (4) is positive. Evenness gives $\mathrm{\Phi }>0$ on $\mathbb{R}$.

Monotonicity. Differentiating (4) with respect to x and using ${\mathrm{\Phi }}^{\prime }\left(u\right)=2x\frac{d\mathrm{\Phi }}{dx}$, one finds after dividing by $x^{5/4}>0$ that ${\mathrm{\Phi }}^{\prime }\left(u\right)<0$ is equivalent to

$$\sum _{n\ge 1}{q}_n\left(4t_n^2-15t_n+\frac{15}{2}\right)e^{-q_{n}x}>0,t_n:=q_{n}x.$$

(6)

The quadratic $4t^2-15t+\frac{15}{2}$ is positive for $t>\frac{15+\sqrt{105}}{8}=3.155869\dots$. Since $t_n\ge t_1=\pi x$, for $u\ge \frac{1}{100}$ we get $t_n\ge \pi e^{1/50}=3.205057\dots >3.155869$, so every summand in (6) is positive and ${\mathrm{\Phi }}^{\prime }\left(u\right)<0$.

For $0<u<\frac{1}{100}$ we argue instead from ${\mathrm{\Phi }}^{\prime }\left(0\right)=0$ and ${\mathrm{\Phi }}^{\prime }\left(u\right)={\int }_0^u{\mathrm{\Phi }}^{\prime \prime }\left(v\right)dv$, using Lemma 3 below: ${\mathrm{\Phi }}^{\prime \prime \prime }\left(0\right)=0$ (as $\mathrm{\Phi }$ is even) and $0<{\mathrm{\Phi }}^{\left(4\right)}<444$ on $[0,\frac{1}{100}]$, so $|{\mathrm{\Phi }}^{\prime \prime \prime }\left(v\right)|\le 444v$ there, and hence by Lemma 2

$${\mathrm{\Phi }}^{\prime \prime }\left(v\right)\le {\mathrm{\Phi }}^{\prime \prime }\left(0\right)+{\int }_0^{v}444wdw={\mathrm{\Phi }}^{\prime \prime }\left(0\right)+222v^2<-8.02+0.023<-7.99<0.$$

(7)

Therefore ${\mathrm{\Phi }}^{\prime }\left(u\right)={\int }_0^u{\mathrm{\Phi }}^{\prime \prime }\left(v\right)dv<0$.    □

Lemma 2

(${\mathrm{\Phi }}^{\prime \prime }\left(0\right)<0$). With $q_n=\pi n^2$,

$${\mathrm{\Phi }}^{\prime \prime }\left(0\right)={\displaystyle \frac{1}{4}}\sum _{n\ge 1}{q}_n\left(32q_n^3-224q_n^2+330q_n-75\right)e^{-q_n},$$

(8)

and ${\mathrm{\Phi }}^{\prime \prime }\left(0\right)<-8.02<0$. Consequently $p^{\prime }\left(0\right)=-{\mathrm{\Phi }}^{\prime \prime }\left(0\right)/\mathrm{\Phi }\left(0\right)>0$.

Proof. 

Differentiating the n-th summand $\left(2q_n^2{e}^{9u/2}-3q_n{e}^{5u/2}\right)e^{-q_n{e}^{2u}}$ twice at $u=0$ and collecting terms gives (8). Throughout this proof $T_n$ denotes the n-th summand of (8) including the prefactor $\frac{1}{4}$, that is

$$T_n:=\frac{1}{4}{q}_n\left(32q_n^3-224q_n^2+330q_n-75\right)e^{-q_n},\mathrm{so}\mathrm{that}{\mathrm{\Phi }}^{\prime \prime }\left(0\right)=\sum _{n\ge 1}{T}_n.$$

(9)

The $n=1$ term is an explicit cubic in $\pi$, and $-8.719<T_1<-8.718$.

For $n\ge 2$ we have $q_n\ge 4\pi$. The cubic $P_3\left(q\right):=32q^3-224q^2+330q-75$ satisfies $P_3\left(4\pi \right)>32200>0$, while $P_3^{\prime }\left(q\right)=96q^2-448q+330>0$ for $q>3.751$ (the larger root of $P_3^{\prime }$); hence $P_3>0$ on $[4\pi ,\infty )$, so $T_n>0$ there. Also $224q^2-330q+75>0$ for $q>1.193$, so $P_3\left(q\right)<32q^3$ on $[4\pi ,\infty )$ and therefore $0<T_n<8q_n^4{e}^{-q_n}=8{\pi }^4{t}_n$ with $t_n:=n^8{e}^{-\pi n^2}$. For $n\ge 2$,

$${\displaystyle \frac{t_{n+1}}{t_n}}={\left(1+\frac{1}{n}\right)}^8{e}^{-\pi (2n+1)}\le {\left(\frac{3}{2}\right)}^8{e}^{-5\pi }<3.87\times {10}^{-6}=:r,$$

(10)

so ${\sum }_{n\ge 2}{t}_n\le t_2/(1-r)=2^8{e}^{-4\pi }/(1-r)$ and

$$0<\sum _{n\ge 2}{T}_n<8{\pi }^4·{\displaystyle \frac{2^8{e}^{-4\pi }}{1-r}}<0.696.$$

(11)

Therefore ${\mathrm{\Phi }}^{\prime \prime }\left(0\right)<-8.718+0.696<-8.02<0$. (For reference, $\mathrm{\Phi }\left(0\right)=0.4466969005$, ${\mathrm{\Phi }}^{\prime \prime }\left(0\right)=-8.36525039$, $p^{\prime }\left(0\right)=18.7269049$.)    □

Lemma 3

(A majorant for ${\mathrm{\Phi }}^{\left(4\right)}$ near the origin). With $x=e^{2u}$ and $q_n=\pi n^2$,

$$\begin{array}{cc}\hfill {\mathrm{\Phi }}^{\left(4\right)}\left(u\right)& ={\displaystyle \frac{x^{5/4}}{16}}\sum _{n\ge 1}{q}_n{e}^{-q_{n}x}{P}_5\left(q_{n}x\right),\hfill \\ \hfill P_5\left(t\right)& =512t^5-8448t^4+41408t^3\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & -68096t^2+30930t-1875.\hfill \end{array}$$

(13)

For $0\le u\le \frac{1}{100}$ one has $0<{\mathrm{\Phi }}^{\left(4\right)}\left(u\right)<444$.

Proof. 

Differentiating the n-th summand of (4) four times in u and factoring out $x^{5/4}{q}_n{e}^{-q_{n}x}$ gives (12).

Let $0\le u\le \frac{1}{100}$, so $x\in [1,e^{1/50}]$ with $e^{1/50}<1.02021$.

Sign facts.$P_5^{\prime \prime \prime }\left(t\right)=30720t^2-202752t+248448$ has largest root $\frac{33}{10}+\frac{\sqrt{1121}}{20}<5$, so $P_5^{\prime \prime \prime }>0$ on $[4\pi ,\infty )$; together with $P_5^{\prime \prime }\left(4\pi \right)>7.29\times {10}^6$, $P_5^{\prime }\left(4\pi \right)>1.47\times {10}^7$ and $P_5\left(4\pi \right)>2.15\times {10}^7$, successive integration gives $P_5>0$ on $[4\pi ,\infty )$. Moreover

$$512t^5-P_5\left(t\right)=t^3\left(8448t-41408\right)+\left(68096t^2-30930t+1875\right)>0(t\ge 4\pi ),$$

(14)

since $8448·4\pi >106160>41408$ and the quadratic has both roots below $0.383$. Hence $0<P_5\left(t\right)<512t^5$ for $t\ge 4\pi$.

Term $n=1$. Here $t=\pi x\in [3.14159,3.20506]$. On this interval $P_5^{\prime \prime }<0$ (its values at the endpoints are $-38706$ and $-44139$, and $P_5^{\prime \prime }$ is monotone there), so $P_5^{\prime }$ is decreasing; since $P_5^{\prime }\left(3.20506\right)>28086>0$ we get $P_5^{\prime }>0$, hence $P_5$ is increasing and $0<P_5\left(t\right)\le P_5\left(3.20506\right)<42760$. As $e^{-\pi x}\le e^{-\pi }$, the term is at most $\pi e^{-\pi }·42760<5806$.

Term $n=2$. Here $t=4\pi x\le 12.8203$, and $P_5$ is increasing on $[4\pi ,\infty )$, so $P_5\left(t\right)\le P_5\left(12.8203\right)<2.5559\times {10}^7$; with $e^{-4\pi x}\le e^{-4\pi }$ the term is at most $4\pi e^{-4\pi }·2.5559\times {10}^7<1121$.

Terms $n\ge 3$. Using $0<P_5\left(t\right)<512t^5$ for $t\ge 4\pi$, these terms are together at most $512e^{5/40}{\sum }_{n\ge 3}{q}_n^6{e}^{-q_n}<0.16$.

Summing and multiplying by $x^{5/4}/16\le e^{1/40}/16$ gives ${\mathrm{\Phi }}^{\left(4\right)}\left(u\right)<444$; positivity of every summand gives ${\mathrm{\Phi }}^{\left(4\right)}\left(u\right)>0$. (The true maximum on $[0,\frac{1}{100}]$ is ${\mathrm{\Phi }}^{\left(4\right)}\left(0\right)=406.081\dots$.)    □

3. The Exact Two-Variable Representation

Definition 1.

$M(a,b)=\mathrm{\Phi }(a+b)\mathrm{\Phi }(a-b)$ and $A(a,b)=-{\partial }_{b}M(a,b)$.

M is even in b and A odd in b; since ${\mathrm{\Phi }}^{\prime }\left(0\right)=0$,

$$A(a,0)={\partial }_b^{2}A(a,0)=0,A(a,a)=-\mathrm{\Phi }\left(0\right){\mathrm{\Phi }}^{\prime }\left(2a\right).$$

(15)

Proposition 1

(Normalisation). $\mathrm{\Xi }\left(z\right)=4D\left(z\right)$.

Proof. 

In the normalisation of [5], $H_0\left(w\right)={\int }_0^{\infty }{\mathrm{\Phi }}_0\left(v\right)cos\left(wv\right)dv=\frac{1}{8}\xi \left(\frac{1}{2}+\frac{iw}{2}\right)$ with ${\mathrm{\Phi }}_0\left(v\right)={\sum }_{n\ge 1}(2{\pi }^2{n}^4{e}^{9v}-3\pi n^2{e}^{5v})e^{-\pi n^2{e}^{4v}}$. Our kernel satisfies $\mathrm{\Phi }\left(u\right)={\mathrm{\Phi }}_0(u/2)$. Substituting $u=2v$,

$$D\left(z\right)={\int }_0^{\infty }{\mathrm{\Phi }}_0(u/2)cos\left(zu\right)du=2{\int }_0^{\infty }{\mathrm{\Phi }}_0\left(v\right)cos\left(2zv\right)dv=2H_0\left(2z\right)=\frac{1}{4}\xi \left(\frac{1}{2}+iz\right)=\frac{1}{4}\mathrm{\Xi }\left(z\right).$$

(16)

   □

Theorem 1

(Exact two-term identity). For $x\in \mathbb{R}$, $y>0$,

$${\partial }_y{\left|D(x+iy)\right|}^2=4{\int }_0^{\infty }{\int }_0^{a}M(a,b)\left[asinh\left(2ya\right)cos\left(2xb\right)+bsinh\left(2yb\right)cos\left(2xa\right)\right]dbda.$$

(17)

Proof. 

D is entire and ${\partial }_{y}D=i{D}^{\prime }$ with $D^{\prime }\left(z\right)=-{\int }_0^{\infty }\mathrm{\Phi }\left(u\right)usin\left(zu\right)du$, so

$${\partial }_y{\left|D\right|}^2=2Re\left(i\overline{D}{D}^{\prime }\right)=2{\int }_0^{\infty }{\int }_0^{\infty }\mathrm{\Phi }\left(u\right)\mathrm{\Phi }\left(v\right)Im\left[usin\left(zu\right)\overline{cos\left(zv\right)}\right]dudv,$$

(18)

and, with $z=x+iy$,

$$Im\left[usin\left(zu\right)\overline{cos\left(zv\right)}\right]=u\left[sin\left(xu\right)sin\left(xv\right)cosh\left(yu\right)sinh\left(yv\right)+cos\left(xu\right)cos\left(xv\right)sinh\left(yu\right)cosh\left(yv\right)\right].$$

(19)

The measure is symmetric in $(u,v)$, so we may symmetrise. Put $u=a+b$, $v=a-b$ and set $P=cosh\left(yu\right)sinh\left(yv\right)$, $Q=sinh\left(yu\right)cosh\left(yv\right)$; then $P+Q=sinh\left(2ya\right)$, $Q-P=sinh\left(2yb\right)$, and

$$uP+vQ=asinh\left(2ya\right)-bsinh\left(2yb\right),uQ+vP=asinh\left(2ya\right)+bsinh\left(2yb\right).$$

(20)

Combined with $sin\left(xu\right)sin\left(xv\right)=\frac{1}{2}[cos\left(2xb\right)-cos\left(2xa\right)]$ and $cos\left(xu\right)cos\left(xv\right)=\frac{1}{2}[cos\left(2xb\right)+cos\left(2xa\right)]$, the symmetrised integrand collapses to $\frac{1}{2}[asinh\left(2ya\right)cos\left(2xb\right)+bsinh\left(2yb\right)cos\left(2xa\right)]$. Finally $dudv=2dadb$, the region $\{u,v>0\}$ is $\{a>|b\left|\right\}$, and the integrand is even in b; the factors $2·\frac{1}{2}·2·2=4$ give (17).    □

Proposition 2

(Reduction to $x>0$). ${\left|D(x+iy)\right|}^2$ and its y-derivative are even in x; and for $y>0$, ${\partial }_y{\left|D\left(iy\right)\right|}^2=2D\left(iy\right){\int }_0^{\infty }\mathrm{\Phi }\left(u\right)usinh\left(yu\right)du>0$.

Proof. 

Since $\mathrm{\Phi }$ is real and cos is even, $cos\left((-x+iy)u\right)=cos\left((x-iy)u\right)=\overline{cos\left((x+iy)u\right)}$ for real u, so

$$D(-x+iy)=\overline{D(x+iy)}=D(x-iy),$$

(21)

whence $\left|D\right(-x+iy\left)\right|=\left|D\right(x+iy\left)\right|$ and ${\partial }_y{\left|D\right|}^2$ is even in x. The case $x=0$ is immediate from $D\left(iy\right)={\int }_0^{\infty }\mathrm{\Phi }\left(u\right)cosh\left(yu\right)du>0$.    □

Henceforth $x>0$.

3.1. The Growth Criterion

Theorem 2

(Bridge to RH). For $y>0$, $x\in \mathbb{R}$, put $s=\frac{1}{2}+y+ix$. Then

$${\partial }_y{\left|D(x+iy)\right|}^2=\frac{1}{8}Re\left({\xi }^{\prime }\left(s\right)\overline{\xi \left(s\right)}\right),$$

(22)

an identity between entire quantities, valid also at zeros of ξ. Moreover ${\partial }_y{\left|D(x+iy)\right|}^2>0$ for all $x\in \mathbb{R}$ and all $y>0$ if and only if the Riemann hypothesis holds.

Proof. 

Let $h(\sigma ,t)={\left|\xi (\sigma +it)\right|}^2$. From $\xi \left(s\right)=\xi (1-s)$ and $\xi \left(\overline{s}\right)=\overline{\xi \left(s\right)}$ we get $h(\sigma ,t)=h(1-\sigma ,t)$, so $\left({\partial }_{\sigma }h\right)(\frac{1}{2}-y,t)=-\left({\partial }_{\sigma }h\right)(\frac{1}{2}+y,t)$. Since $\frac{1}{2}+i(x+iy)=(\frac{1}{2}-y)+ix$ we have ${\left|\mathrm{\Xi }(x+iy)\right|}^2=h(\frac{1}{2}-y,x)$, and therefore

$${\partial }_y{\left|\mathrm{\Xi }(x+iy)\right|}^2=-\left({\partial }_{\sigma }h\right)(\frac{1}{2}-y,x)=\left({\partial }_{\sigma }h\right)(\frac{1}{2}+y,x)=2Re\left({\xi }^{\prime }\left(s\right)\overline{\xi \left(s\right)}\right).$$

(23)

Now ${\left|D\right|}^2=\frac{1}{16}{\left|\mathrm{\Xi }\right|}^2$ by Proposition 1, giving (22).

The map $(x,y)\mapsto s=\frac{1}{2}+y+ix$ is a bijection from $\mathbb{R}\times (0,\infty )$ onto the half-plane $\{Res>\frac{1}{2}\}$. Suppose RH holds. Then $\xi$ has no zero with $Res>\frac{1}{2}$, and the Hadamard logarithmic derivative, summed in the standard symmetric order, gives

$$Re{\displaystyle \frac{{\xi }^{\prime }}{\xi }}\left(s\right)=\underset{T\to \infty }{lim}\sum _{\begin{array}{c}\rho ={\displaystyle \frac{1}{2}}+i\gamma \\ \left|\gamma \right|\le T\end{array}}Re{\displaystyle \frac{1}{s-\rho }}=\underset{T\to \infty }{lim}\sum _{\begin{array}{c}\rho ={\displaystyle \frac{1}{2}}+i\gamma \\ \left|\gamma \right|\le T\end{array}}{\displaystyle \frac{\sigma -\frac{1}{2}}{{|s-\rho |}^2}}>0(\sigma >\frac{1}{2}),$$

(24)

every term being positive; see [10,11]. Hence $Re\left({\xi }^{\prime }\overline{\xi }\right)={\left|\xi \right|}^{2}Re({\xi }^{\prime }/\xi )>0$. Conversely, if RH fails then by the functional equation there is a zero ${\rho }_0$ with $Re{\rho }_0>\frac{1}{2}$; at $s={\rho }_0$ we have $\xi \left(s\right)=0$ and hence $Re\left({\xi }^{\prime }\left(s\right)\overline{\xi \left(s\right)}\right)=0$, so positivity fails.    □

Remark 1.

Away from zeros, (22) is the criterion of Hinkkanen [10] and Lagarias [11]; see also [12,13].

Two s-variables occur here and should be distinguished. The substitution $\mathrm{\Xi }\left(z\right)=\xi (\frac{1}{2}+iz)$ attaches to $z=x+iy$ the point

$$s_{-}:=\frac{1}{2}+iz=\frac{1}{2}-y+ix,$$

(25)

which for $y>0$ lies in thelefthalf-plane $\{\sigma <\frac{1}{2}\}$; it is the functional equation, in the form $h(\sigma ,t)=h(1-\sigma ,t)$, that transfers the computation to

$$s:=1-\overline{s_{-}}=\frac{1}{2}+y+ix,$$

(26)

where the real-part criterion lives. In the z-variable itself one has, away from zeros, theimaginary-part condition

$${\partial }_y{\left|D\right|}^2=-2{\left|D\right|}^{2}Im{\displaystyle \frac{D^{\prime }}{D}}.$$

(27)

Remark 2

(What the classical criteria do and do not say). The criteria of [10,11,12] concern the sign of the complete logarithmic derivative or, equivalently, the horizontal monotonicity of $\left|\xi \right|$ in zero-free half-planes. They determine the sign of thetotalquantity ${\partial }_y{\left|D\right|}^2$ under RH; they say nothing about the signs of the phase-aligned block integrals $C_m^{\mathrm{tot}}$ introduced in §5, nor do they introduce the objects S, E, $Q_y$ and ${\epsilon }_x$ of the decomposition below. These are logically independent questions: an integral may be positive while the integrals of its restrictions to a partition of the domain have any prescribed pattern of signs, and conversely all those restricted integrals may be positive even when the integrand oscillates. The results of §§6–7 are unconditional statements about the internal sign structure of one particular exact decomposition, and are not deduced from — and do not imply — the classical criteria.

4. The Longitudinal–Transverse Decomposition

Definition 2.

$$\begin{array}{cc}\hfill Q_y\left(a\right)& =-{\int }_0^{a}bsinh\left(2yb\right){\partial }_{a}M(a,b)db,\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill {\epsilon }_x(a;y)& =asinh\left(2ya\right){\int }_0^{a}A(a,b)sin\left(2xb\right)db.\hfill \end{array}$$

(29)

Proposition 3.

${\partial }_{a}M(a,b)=-M(a,b)\left[p(a+b)+p(a-b)\right]$, and consequently

$$Q_y\left(a\right)={\int }_0^{a}bsinh\left(2yb\right)M(a,b)\left[p(a+b)+p(a-b)\right]db>0(a>0,y>0).$$

(30)

Proof. 

The identity is immediate from ${\mathrm{\Phi }}^{\prime }=-p\mathrm{\Phi }$. For $0\le b\le a$ we have $a\pm b\ge 0$, so $p(a\pm b)\ge 0$ by Lemma 1, with strict inequality on a set of positive measure; and $M>0$, $bsinh\left(2yb\right)\ge 0$.    □

Theorem 3

(Master identity). For $x>0$, $y>0$,

$${\partial }_y{\left|D(x+iy)\right|}^2={\displaystyle \frac{2}{x}}{\int }_0^{\infty }\left[Q_y\left(a\right)sin\left(2xa\right)+{\epsilon }_x(a;y)\right]da=:{\displaystyle \frac{2}{x}}I(x,y).$$

(31)

Proof. 

Write (17) as $4(T_1+T_2)$, where

$$T_1={\int }_0^{\infty }asinh\left(2ya\right)\left[{\int }_0^{a}Mcos\left(2xb\right)db\right]da,T_2={\int }_0^{\infty }cos\left(2xa\right)G\left(a\right)da,$$

(32)

with $G\left(a\right)={\int }_0^{a}M(a,b)bsinh\left(2yb\right)db$ and $G\left(0\right)=0$.

Integrating by parts in b, using ${\partial }_{b}M=-A$ and $M(a,a)=\mathrm{\Phi }\left(0\right)\mathrm{\Phi }\left(2a\right)$,

$${\int }_0^{a}Mcos\left(2xb\right)db={\displaystyle \frac{\mathrm{\Phi }\left(0\right)\mathrm{\Phi }\left(2a\right)sin\left(2xa\right)}{2x}}+{\displaystyle \frac{1}{2x}}{\int }_0^{a}A(a,b)sin\left(2xb\right)db,$$

(33)

whence

$$T_1={\displaystyle \frac{\mathrm{\Phi }\left(0\right)}{2x}}{\int }_0^{\infty }asinh\left(2ya\right)\mathrm{\Phi }\left(2a\right)sin\left(2xa\right)da+{\displaystyle \frac{1}{2x}}{\int }_0^{\infty }{\epsilon }_x(a;y)da.$$

(34)

Integrating by parts in a (boundary terms vanish by the decay of G) gives

$$T_2=-{\displaystyle \frac{1}{2x}}{\int }_0^{\infty }{G}^{\prime }\left(a\right)sin\left(2xa\right)da,$$

(35)

and by the Leibniz rule

$$G^{\prime }\left(a\right)=\mathrm{\Phi }\left(0\right)\mathrm{\Phi }\left(2a\right)asinh\left(2ya\right)+{\int }_0^a{\partial }_{a}M(a,b)bsinh\left(2yb\right)db=\mathrm{\Phi }\left(0\right)\mathrm{\Phi }\left(2a\right)asinh\left(2ya\right)-Q_y\left(a\right),$$

(36)

so that

$$T_2=-{\displaystyle \frac{\mathrm{\Phi }\left(0\right)}{2x}}{\int }_0^{\infty }asinh\left(2ya\right)\mathrm{\Phi }\left(2a\right)sin\left(2xa\right)da+{\displaystyle \frac{1}{2x}}{\int }_0^{\infty }{Q}_y\left(a\right)sin\left(2xa\right)da.$$

(37)

The diagonal terms of (34) and (37) cancel, giving $T_1+T_2=\frac{1}{2x}I(x,y)$.    □

Remark 3.

The cancellation is between thetransversesector (integration by parts in b) and thelongitudinalsector (integration by parts in a). It is not visible from a one-term starting point, which is why the two-term identity (17) is essential.

5. The Phase-Aligned Oscillatory Hierarchy

Fix $x>0$ and set

$$L={\displaystyle \frac{\pi }{2x}},{\alpha }_m={\displaystyle \frac{m\pi }{x}}=2mL,J_m=[{\alpha }_m,{\alpha }_m+2L],$$

(38)

so that $sin\left(2x({\alpha }_m+s)\right)=sin(\pi s/L)$ and $cos\left(2x({\alpha }_m+s)\right)=cos(\pi s/L)$. Put

$$C_m^{\left(Q\right)}={\int }_{J_m}{Q}_y\left(a\right)sin\left(2xa\right)da,C_m^{\left(\epsilon \right)}={\int }_{J_m}{\epsilon }_x(a;y)da,C_m^{\mathrm{tot}}=C_m^{\left(Q\right)}+C_m^{\left(\epsilon \right)},$$

(39)

so that $I(x,y)={\sum }_{m\ge 0}{C}_m^{\mathrm{tot}}$.

Lemma 4

(Moments over a full aligned period). With $k=\pi /L=2x$ and $T=2L$ (so $kT=2\pi$),

$$\begin{array}{cccccc}& {\int }_0^{T}sin\left(ks\right)ds=0,\hfill & \hfill & {\int }_0^{T}ssin\left(ks\right)ds=-{\displaystyle \frac{2L^2}{\pi }},\hfill & \hfill & {\int }_0^T{s}^{2}sin\left(ks\right)ds=-{\displaystyle \frac{4L^3}{\pi }},\hfill \end{array}$$

(40)

$$\begin{array}{cccccc}& {\int }_0^{T}cos\left(ks\right)ds=0,\hfill & \hfill & {\int }_0^{T}scos\left(ks\right)ds=0,\hfill & \hfill & {\int }_0^T{s}^{2}cos\left(ks\right)ds={\displaystyle \frac{4L^3}{{\pi }^2}}.\hfill \end{array}$$

(41)

Proposition 4

(Sine- and cosine-block orders). Let $F\in C^3$ on a neighbourhood of $J_m$, and $\alpha ={\alpha }_m$. Then

$$\begin{array}{cc}\hfill {\int }_{J_m}F\left(a\right)sin\left(2xa\right)da& =-{\displaystyle \frac{\pi }{2x^2}}{F}^{\prime }\left(\alpha \right)-{\displaystyle \frac{{\pi }^2}{4x^3}}{F}^{\prime \prime }\left(\alpha \right)+O\left(x^{-4}{\parallel F^{\prime \prime \prime }\parallel }_{\infty ,J_m}\right),\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill {\int }_{J_m}F\left(a\right)cos\left(2xa\right)da& =\phantom{-}{\displaystyle \frac{\pi }{4x^3}}{F}^{\prime \prime }\left(\alpha \right)+O\left(x^{-4}{\parallel F^{\prime \prime \prime }\parallel }_{\infty ,J_m}\right).\hfill \end{array}$$

(43)

Proof. 

Set $a=\alpha +s$, Taylor-expand F to second order with remainder, and apply Lemma 4. Against a sine the constant term dies; against a cosine both the constant and the linear term die. The remainder is at most $\frac{1}{6}\parallel F^{\prime \prime \prime }{\parallel }_{\infty }{\int }_0^{2L}{s}^{3}ds=\frac{2}{3}{L}^4{\parallel F^{\prime \prime \prime }\parallel }_{\infty }$.    □

This asymmetry — one annihilated Taylor order against a sine, two against a cosine — drives the entire hierarchy.

Lemma 5

(Transverse expansion). For $a,x>0$,

$$\begin{array}{c}{\int }_0^{a}A(a,b)sin\left(2xb\right)db=-{\displaystyle \frac{A(a,a)}{2x}}cos\left(2xa\right)+{\displaystyle \frac{{\partial }_{b}A(a,a)}{{\left(2x\right)}^2}}sin\left(2xa\right)\hfill \\ \hfill +{\displaystyle \frac{{\partial }_b^{2}A(a,a)}{{\left(2x\right)}^3}}cos\left(2xa\right)-{\displaystyle \frac{{\partial }_b^{3}A(a,a)}{{\left(2x\right)}^4}}sin\left(2xa\right)+{\displaystyle \frac{1}{{\left(2x\right)}^4}}{\int }_0^a{\partial }_b^{4}A(a,b)sin\left(2xb\right)db.\end{array}$$

(44)

Proof. 

Four integrations by parts in b; all endpoint contributions at $b=0$ vanish by (15).    □

Definition 3.

Put

$$g_1\left(a\right):=-asinh\left(2ya\right)A(a,a)=asinh\left(2ya\right)\mathrm{\Phi }\left(0\right){\mathrm{\Phi }}^{\prime }\left(2a\right),g_2\left(a\right):=asinh\left(2ya\right){\partial }_{b}A(a,a),$$

(45)

where ${\partial }_{b}A(a,a)=-\mathrm{\Phi }\left(0\right){\mathrm{\Phi }}^{\prime \prime }\left(2a\right)-{\mathrm{\Phi }}^{\prime \prime }\left(0\right)\mathrm{\Phi }\left(2a\right)$.

Multiplying Lemma 5 by $asinh\left(2ya\right)$ gives, uniformly on compact a-sets,

$${\epsilon }_x(a;y)={\displaystyle \frac{g_1\left(a\right)}{2x}}cos\left(2xa\right)+{\displaystyle \frac{g_2\left(a\right)}{4x^2}}sin\left(2xa\right)+O\left(x^{-3}\right).$$

(46)

Theorem 4

(Oscillatory hierarchy). Let $y>0$ and let $K⋐K_1⋐(0,\infty )$ be compact. For all x large enough that every block with ${\alpha }_m\in K$ satisfies $J_m\subset K_1$, one has, uniformly over such blocks,

$$\begin{array}{cc}\hfill C_m^{\left(Q\right)}& ={\displaystyle \frac{\pi }{2x^2}}\left(-Q_y^{\prime }\left({\alpha }_m\right)\right)-{\displaystyle \frac{{\pi }^2}{4x^3}}{Q}_y^{\prime \prime }\left({\alpha }_m\right)+O\left(x^{-4}\right),\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill C_m^{\left(\epsilon \right)}& ={\displaystyle \frac{\pi }{8x^4}}\left[g_1^{\prime \prime }\left({\alpha }_m\right)-g_2^{\prime }\left({\alpha }_m\right)\right]+O\left(x^{-5}\right).\hfill \end{array}$$

(48)

The implied constants depend only on y and on the quantities

$$\parallel Q_y^{\prime \prime \prime }{\parallel }_{\infty ,K_1},\parallel g_1^{\prime \prime \prime }{\parallel }_{\infty ,K_1},{\parallel g_2^{\prime \prime }\parallel }_{\infty ,K_1},\underset{a\in K_1,0\le b\le a}{sup}\left|{\partial }_b^{4}A(a,b)\right|.$$

(49)

Proof. 

Since $|J_m|=\pi /x\to 0$, for x large every block with ${\alpha }_m\in K$ lies in $K_1$, and Proposition 4 may be applied with derivative norms taken over $K_1$. Formula (47) is that proposition applied to $F=Q_y$. For (48), insert (46): the $g_1$ term is a cosine-block, contributing $\frac{1}{2x}·\frac{\pi }{4x^3}{g}_1^{\prime \prime }\left({\alpha }_m\right)+O\left(x^{-5}\right)$; the $g_2$ term is a sine-block, contributing $\frac{1}{4x^2}·(-\frac{\pi }{2x^2})g_2^{\prime }\left({\alpha }_m\right)+O\left(x^{-5}\right)$. The terms of order $x^{-3}$ and $x^{-4}$ in Lemma 5 are again cosine- and sine-blocks, contributing $O\left(x^{-6}\right)$; the final integral remainder is $O\left(x^{-4}\right)$ pointwise and is integrated over a block of length $\pi /x$.    □

Corollary 1

(Residual ratio). If in addition ${inf}_{a\in K}\left|Q_y^{\prime }\left(a\right)\right|>0$, then uniformly for ${\alpha }_m\in K$,

$${\Theta }_m:={\displaystyle \frac{C_m^{\left(\epsilon \right)}}{C_m^{\left(Q\right)}}}={\displaystyle \frac{g_1^{\prime \prime }\left({\alpha }_m\right)-g_2^{\prime }\left({\alpha }_m\right)}{4x^2\left(-Q_y^{\prime }\left({\alpha }_m\right)\right)}}+O\left(x^{-3}\right)=O\left(x^{-2}\right).$$

(50)

Remark 4

(The uniformity is essential). Corollary 1 holds only where $Q_y^{\prime }$ is bounded away from zero. It fails near $a=0$, near a zero of $Q_y^{\prime }$, and in the far tail. In particular, forfixedm we have ${\alpha }_m=m\pi /x\to 0$ as $x\to \infty$; since $Q_y^{\prime }\left(a\right)\asymp a^3$ near 0 (Lemma 6), the block $C_m^{\left(Q\right)}$ is then itself of size $x^{-5}$, not $x^{-2}$, and ${\Theta }_m$ does not tend to 0. The correct summary is:on compact phase regions where $Q_y^{\prime }$ is bounded away from zero, each transverse residual block is two powers of x smaller than its longitudinal counterpart.Theorem 5 shows that this local smallness does not survive global summation.

6. Nonlocal Cancellation

Lemma 6

(Small-a expansion). As $a\to 0^{+}$, uniformly for y in compacts,

$$Q_y\left(a\right)=c_y{a}^4+O\left(a^6\right),c_y:=\frac{4}{3}y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2>0,$$

(51)

so $Q_y^{\prime }\left(a\right)=4c_y{a}^3+O\left(a^5\right)>0$ for all sufficiently small $a>0$.

Proof. 

$\mathrm{\Phi }$ is even, so $p\left(0\right)=0$ and $p\left(u\right)=p^{\prime }\left(0\right)u+O\left(u^3\right)$, giving $p(a+b)+p(a-b)=2p^{\prime }\left(0\right)a+O\left(a^3\right)$ for $0\le b\le a$. Also $sinh\left(2yb\right)=2yb+O\left(b^3\right)$ and $M(a,b)=\mathrm{\Phi }{\left(0\right)}^2+O\left(a^2\right)$. Hence the integrand in (30) equals $4y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^{2}a{b}^2(1+O\left(a^2\right))$, and ${\int }_0^a{b}^{2}db=a^3/3$. Positivity of $c_y$ is Lemma 2.    □

The next lemma is a standard estimate; we record it only because we shall need it, and we claim no novelty for it. It quantifies the well-known exponential decay of $\xi$ in vertical strips, and follows from Stirling’s formula together with a Cauchy estimate. Its role here is purely comparative: it supplies the right-hand side against which the algebraic expansions of the two sectors are measured in Theorem 5.

Lemma 7

(Exponential smallness of I). Fix $y>0$. There are $A=A\left(y\right)$ and $c=c\left(y\right)$ such that

$$\left|I(x,y)\right|\le c{x}^A{e}^{-\pi x/2}(x\ge 1).$$

(52)

Proof. 

By Theorems 3 and 2, $I=\frac{x}{2}{\partial }_y{\left|D\right|}^2=\frac{x}{16}Re\left({\xi }^{\prime }\left(s\right)\overline{\xi \left(s\right)}\right)$ with $s=\frac{1}{2}+y+ix$; this is an identity between entire functions, so

$$\left|I(x,y)\right|\le \frac{x}{16}|{\xi }^{\prime }\left(s\right)\left|\right|\xi \left(s\right)|$$

(53)

holds unconditionally, with no hypothesis on the zeros of $\xi$.

Let $r_y:=min\{\frac{1}{2},\frac{y}{2}\}$, so that the closed disc $|w-s|\le r_y$ lies in the closed strip $Rew\ge \frac{1}{2}+\frac{y}{2}$, strictly to the right of the critical line. On that strip Stirling gives, uniformly,

$$\left|\Gamma (w/2)\right|{\ll }_y{|Imw|}^{(Rew-1)/2}{e}^{-\pi |Imw|/4},$$

(54)

while $\zeta$ is polynomially bounded on vertical lines with $Rew\ge \frac{1}{2}$ [14], and $\left|w(w-1)\right|{\ll }_y{|Imw|}^2$. Hence $\left|\xi \left(w\right)\right|{\ll }_y{x}^{A_1}{e}^{-\pi x/4}$ for $|w-s|\le r_y$. Cauchy’s estimate then gives $|{\xi }^{\prime }\left(s\right)|\le r_y^{-1}{max}_{|w-s|=r_y}\left|\xi \left(w\right)\right|{\ll }_y{x}^{A_1}{e}^{-\pi x/4}$. Multiplying yields the claim with $A=2A_1+1$.    □

Theorem 5

(Nonlocal cancellation). Fix $y>0$ and set

$$S(x,y)=\sum _{m\ge 0}{C}_m^{\left(Q\right)}={\int }_0^{\infty }{Q}_y\left(a\right)sin\left(2xa\right)da,E(x,y)=\sum _{m\ge 0}{C}_m^{\left(\epsilon \right)}={\int }_0^{\infty }{\epsilon }_x(a;y)da,$$

(55)

so that $S+E=I$. Then:

(i)

S has the complete asymptotic expansion

$$S(x,y)\sim \sum _{k\ge 0}{(-1)}^k{\displaystyle \frac{Q_y^{\left(2k\right)}\left(0\right)}{{\left(2x\right)}^{2k+1}}}(x\to \infty ),$$

(56)

whose first nonvanishing term gives

$$S(x,y)={\displaystyle \frac{y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2}{x^5}}+O_y\left(x^{-7}\right).$$

(57)

(ii)

(Cancellation to all algebraic orders.)For every N, $S(x,y)+E(x,y)=O_{y,N}\left(x^{-N}\right)$. Hence E has an asymptotic expansion which istermwise the negativeof (56); in particular

$$E(x,y)=-{\displaystyle \frac{y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2}{x^5}}+O_y\left(x^{-7}\right).$$

(58)

(iii)

${\displaystyle \frac{E(x,y)}{S(x,y)}⟶-1}$ as $x\to \infty$.

Proof. (i) $Q_y\in C^{\infty }\left([0,\infty )\right)$, and $Q_y$ together with all its derivatives decays doubly exponentially, so repeated integration by parts gives, with $\lambda =2x$,

$${\int }_0^{\infty }f\left(a\right)sin\left(\lambda a\right)da={\displaystyle \frac{f\left(0\right)}{\lambda }}-{\displaystyle \frac{f^{\prime \prime }\left(0\right)}{{\lambda }^3}}+{\displaystyle \frac{f^{\left(4\right)}\left(0\right)}{{\lambda }^5}}-\dots$$

(59)

By Lemma 6, $Q_y\left(0\right)=Q_y^{\prime }\left(0\right)=Q_y^{\prime \prime }\left(0\right)=Q_y^{\prime \prime \prime }\left(0\right)=0$ and $Q_y^{\left(4\right)}\left(0\right)=24c_y=32y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2$, so the terms $k=0,1$ vanish and the term $k=2$ is $Q_y^{\left(4\right)}\left(0\right)/{\left(2x\right)}^5=y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2/x^5$.

(ii) $S+E=I$, and $I=O_{y,N}\left(x^{-N}\right)$ for every N by Lemma 7. Subtracting (56) gives the expansion of E.

(iii) By Lemma 2, $y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2\ne 0$, so $S\ne 0$ for large x; now divide.    □

Remark 5

(Interpretation). Block by block, on compact phase regions where $Q_y^{\prime }$ is bounded away from zero, the transverse residual is two powers of x smaller than the longitudinal term (Corollary 1). Globally it is not: the two sectors are each of size $x^{-5}$ and cancel to all algebraic orders, the sign-carrying quantity surviving only in the exponentially small remainder. There is no contradiction: the $C_m^{\left(Q\right)}$ have mixed signs and cancel among themselves, so $\left|S\right|\ll {\sum }_m\left|C_m^{\left(Q\right)}\right|$, and the residuals survive at exactly the order of what is left. Numerically (§Section 8), at $x=30$, $y=1$ the individual blocks are of size ${10}^{-5}$ while $I\approx 3.7\times {10}^{-15}$.

Consequentlyanyargument that estimates the two sectors independently and then adds them discards, at every step, quantities enormously larger than the answer. This is intrinsic to the decomposition, not a deficiency of the estimates.

It should be emphasised what is, and what is not, new here. The exponential estimate for the complete growth derivative (Lemma 7) is standard. The point of Theorem 5 is its comparison with the algebraic endpoint expansions of the two sectors created by the exact decomposition: S and E are individually of size $x^{-5}$, with complete asymptotic expansions in $1/x$ that are termwise negatives of one another. Neither S nor E exists outside this decomposition, so no such statement is available from the classical global criteria.

7. A Representation-Specific No-Go Theorem for Blockwise Positivity

Theorem 6

(No-go). Fix $y>0$. By Lemma 6 choose ${\eta }_y>0$ with $Q_y^{\prime }>0$ on $(0,{\eta }_y]$, and fix a nondegenerate interval $K=[a_{-},a_{+}]\subset (0,{\eta }_y)$; put $\delta :={min}_{a\in K}{Q}_y^{\prime }\left(a\right)>0$. Then there is $x_1=x_1(y,K)$ such that for all $x\ge x_1$ and every m with ${\alpha }_m\in K$,

$$C_m^{\mathrm{tot}}(x,y)\le -{\displaystyle \frac{\pi \delta }{4x^2}}<0.$$

(60)

Moreover, as soon as $x>max\{x_1,\pi /(a_{+}-a_{-})\}$, at least one ${\alpha }_m=m\pi /x$ lies in K. Hence for every $y>0$ and all sufficiently large x there exists a negative aligned block, and the blocks are not all nonnegative.

Proof. 

K is a compact subset of $(0,\infty )$ on which $Q_y^{\prime }\ge \delta >0$, so Theorem 4 applies uniformly (choose any compact $K_1$ with $K⋐K_1⋐(0,\infty )$): we have $C_m^{\left(Q\right)}\le -\frac{\pi \delta }{2x^2}+O\left(x^{-3}\right)$ and $C_m^{\left(\epsilon \right)}=O\left(x^{-4}\right)$. Choose $x_1$ so that the two remainders together are bounded by $\frac{\pi \delta }{4x^2}$. For the last statement, if the grid spacing $\pi /x$ is smaller than the length $a_{+}-a_{-}$ then some multiple of $\pi /x$ lies in $[a_{-},a_{+}]$.    □

Corollary 2.

The strategy of deducing $I(x,y)>0$ from the nonnegativity of every aligned block, or from independent blockwise domination of $C_m^{\left(\epsilon \right)}$ by $C_m^{\left(Q\right)}$, cannot succeed.

Remark 6.

Theorem 6 is fully analytic and unconditional; it uses only Lemmas 2, 6 and Theorem 4, and requires no information about the global shape of $Q_y$ — in particular no uniqueness hypothesis for the zeros of $Q_y^{\prime }$. It also makes no use of RH, or of the sign of the complete integral $I(x,y)$: it remains true whether RH holds or fails.

Remark 7

(Block signs are not determined by global positivity). It is worth stressing that Theorem 6 is not a consequence of the positivity of $I(x,y)$ together with the oscillation of the integrand. Neither oscillation of an integrand, nor positivity of its total integral, constrains the signs of the integrals over complete phase-aligned blocks. Indeed let $f\left(a\right)=q\left(a\right)sin\left(2xa\right)$ with $q>0$ decreasing, and put $h=\pi /\left(2x\right)$. Substituting $a={\alpha }_m+t$ and splitting $J_m$ at its midpoint gives

$${\int }_{J_m}q\left(a\right)sin\left(2xa\right)da={\int }_0^h\left[q({\alpha }_m+t)-q({\alpha }_m+h+t)\right]sin\left(2xt\right)dt>0$$

(61)

for every m, since the bracket is positive. Thus every block integral can be strictly positive although the integrand changes sign inside each block. What drives Theorem 6 is instead thelocal geometryof $Q_y$ near the origin — the strict increase $Q_y^{\prime }>0$ supplied by Lemma 6 — together with the uniform two-power separation of Theorem 4.

Remark 8

(Why the residual cannot compensate). One might hope that $C_m^{\left(\epsilon \right)}$, whose sign is not fixed, could be positive and large enough to offset a negative $C_m^{\left(Q\right)}$. Theorem 4 forecloses this: on the relevant compact region $C_m^{\left(\epsilon \right)}$ is smaller than $|C_m^{\left(Q\right)}|$ by at least two powers of x, uniformly (${\Theta }_m=O\left(x^{-2}\right)$). The failure occurs at leading order, and is therefore not a matter of sharper estimates.

8. Numerical Illustrations

The tables and figure of this section illustrate, but do not prove, the analytic results of §§Section 6Section 7; no statement proved above depends on them. Computations use the series for $\mathrm{\Phi }$ truncated at $n\le 12$ (tail $<{10}^{-30}$ on the relevant range) and Gauss–Legendre quadrature, with $Q_y$ evaluated in log-space. A script producing every table and both panels of Figure 1 below, together with quadrature orders, interval transformations, precision settings and explicit theta-series truncation bounds, is provided as Supplementary Material.

Independence of the evaluation of I.

In the tables below $I(x,y)$ is never obtained by subtracting E from $-S$. It is computed independently as $I=\frac{x}{2}·2Re\left(\overline{D}{\partial }_{y}D\right)$ with D and ${\partial }_{y}D$ each evaluated by direct quadrature. This matters: at $x=30$ the quantity I lies several orders of magnitude below $\left|S\right|$ and $\left|E\right|$, so an ordinary-precision subtraction would return noise. The agreement between the independent value of I and $S+E$ is therefore a genuine test of Theorem 3, not a tautology.

The master identity.

Relative errors between ${\partial }_y{\left|D(x+iy)\right|}^2$ computed directly and via (31): $2.8\times {10}^{-13}$, $1.6\times {10}^{-13}$, $1.6\times {10}^{-12}$, $5.6\times {10}^{-11}$, $3.1\times {10}^{-9}$ at $(x,y)=(3,0.7)$, $(5,1)$, $(7,1.3)$, $(12,0.4)$, $(20,2)$.

Nonlocal cancellation (Theorem 5).

With $y=1$ and $y{p}^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2=3.73673$:

x	$S(x,1)$	$x^{5}S(x,1)$
40	$4.1355\times {10}^{-8}$	$4.2347$
80	$1.1759\times {10}^{-9}$	$3.8532$
160	$3.5909\times {10}^{-11}$	$3.7654$
240	$4.7088\times {10}^{-12}$	$3.7494$

x	y	$I=S+E$	$E/S$
10	1	$1.0292\times {10}^{-4}$	$-1.647$
20	1	$2.2426\times {10}^{-9}$	$-0.9990$
30	1	$3.7368\times {10}^{-15}$	$-1.0000$
30	5	$2.5060\times {10}^{-12}$	$-1.0000$

Negative blocks (Theorem 6), $y=1$.

Blocks on which $Q_y^{\prime }$ changes sign are omitted; the rows listed are blocks lying either strictly inside or strictly outside the region where $Q_y^{\prime }>0$.

x	m	$C_m^{\left(Q\right)}$	$C_m^{\left(\epsilon \right)}$	$C_m^{\mathrm{tot}}$	sign
20	0	$-3.3737\times {10}^{-5}$	$-6.889\times {10}^{-6}$	$-4.0626\times {10}^{-5}$	−
20	1	$-2.4972\times {10}^{-5}$	$\phantom{-}8.261\times {10}^{-6}$	$-1.6712\times {10}^{-5}$	−
20	2	$\phantom{-}5.7119\times {10}^{-5}$	$-3.646\times {10}^{-6}$	$\phantom{-}5.3472\times {10}^{-5}$	+
30	0	$-5.7146\times {10}^{-6}$	$-1.945\times {10}^{-6}$	$-7.6597\times {10}^{-6}$	−
30	1	$-4.1835\times {10}^{-5}$	$\phantom{-}2.889\times {10}^{-6}$	$-3.8947\times {10}^{-5}$	−
30	3	$\phantom{-}3.1579\times {10}^{-5}$	$-1.213\times {10}^{-6}$	$\phantom{-}3.0366\times {10}^{-5}$	+

Graphical summary.

Figure 1 displays the two proved core phenomena. Panel (a) shows the convergence of $x^{5}S(x,1)$ to the level $p^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2=3.73673$ predicted by Theorem 5(i). Panel (b) shows the block structure at $(x,y)=(30,1)$: the pre-transition blocks $C_m^{\mathrm{tot}}$ are negative (Theorem 6), the transverse contributions $C_m^{\left(\epsilon \right)}$ are visibly smaller than the longitudinal ones $C_m^{\left(Q\right)}$ (Theorem 4), and the sign reversal occurs once ${\alpha }_m$ passes beyond the region where $Q_y^{\prime }>0$.

Figure 1. The two proved core results. (a) $x^{5}S(x,1)$ against x, with the limiting level $p^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2=3.73673$ of Theorem 5(i) shown dashed. (b) the longitudinal blocks $C_m^{\left(Q\right)}$, the transverse blocks $C_m^{\left(\epsilon \right)}$ and their sum $C_m^{\mathrm{tot}}$ at $(x,y)=(30,1)$, in units of ${10}^{-5}$, plotted against the left endpoint ${\alpha }_m$ of the block. The blocks lying strictly inside the region where $Q_y^{\prime }>0$ are negative, the transverse contribution is uniformly the smaller of the two, and the sign reverses further out. Both panels illustrate the analytic statements of Theorems 5 and 6; they do not prove them.

Figure 1. The two proved core results. (a) $x^{5}S(x,1)$ against x, with the limiting level $p^{\prime }\left(0\right)\mathrm{\Phi }{\left(0\right)}^2=3.73673$ of Theorem 5(i) shown dashed. (b) the longitudinal blocks $C_m^{\left(Q\right)}$, the transverse blocks $C_m^{\left(\epsilon \right)}$ and their sum $C_m^{\mathrm{tot}}$ at $(x,y)=(30,1)$, in units of ${10}^{-5}$, plotted against the left endpoint ${\alpha }_m$ of the block. The blocks lying strictly inside the region where $Q_y^{\prime }>0$ are negative, the transverse contribution is uniformly the smaller of the two, and the sign reverses further out. Both panels illustrate the analytic statements of Theorems 5 and 6; they do not prove them.

9. Conclusions

The theta-kernel decomposition of ${\partial }_y{\left|D(x+iy)\right|}^2$ into phase-aligned blocks is exact, and admits a complete oscillatory hierarchy; the dependencies of the implied constants are specified in Theorem 4. Nevertheless it cannot deliver the positivity it was designed to capture. The reason is not the sharpness of any estimate but the architecture of the decomposition: the longitudinal and transverse sectors are each of size $x^{-5}$ and cancel to all algebraic orders, so that the sign-carrying quantity lives entirely in an exponentially small remainder (Theorem 5). The concrete symptom is that some aligned blocks are necessarily negative (Theorem 6).

Within this exact phase-aligned decomposition, positivity is intrinsically nonlocal. Any successful approach through this representation must retain the correlation between the two sectors — for instance by grouping blocks, or by a resummation identity that keeps track of the cancellation — rather than estimating them independently. We emphasise once more the limits of the claim. Theorems 5 and 6 are unconditional statements about one particular decomposition; they are not statements about every conceivable local approach to the growth criterion, and they neither assume nor bear on the truth of the Riemann hypothesis itself.