Start with the very friendly-looking quotient
$$ E_h(s)=\frac{\zeta(s-h)\zeta(s+h)}{\zeta(s)^2}. $$
Its logarithm has Euler coefficients
$$ \frac{p^{kh}+p^{-kh}-2}{k} = \frac{(p^{kh/2}-p^{-kh/2})^2}{k} \ge0. $$
So exponentiating gives an ordinary Dirichlet series with nonnegative coefficients. This is exactly the sort of object one would like to point at the nontrivial zeros of \(\zeta\).
It points at a real pole first.
The leftmost shift is \(-h\), so \(E_h\) has an uncancelled pole at \(s=1+h\). Every pole supplied by a denominator zero \(\rho\) sits at \(s=\rho\), strictly to its left because \(\Re\rho<1\).
The whole obstruction chain begins with that tiny ordering fact. Then it keeps surviving every reasonable attempt to blur the shifts.
More shifts do not make the edge disappear#
For a finite product
$$ E(s)=\prod_j\zeta(s+\alpha_j)^{c_j}, \qquad c_j\in\mathbb Z, $$
let \(\alpha_0\) be the leftmost active shift. At a prime,
$$ a_p=\sum_jc_jp^{-\alpha_j}, $$
so
$$ p^{\alpha_0}a_p\longrightarrow c_0. $$
If every ordinary Dirichlet coefficient is nonnegative, then \(c_0>0\). The leftmost zeta pole therefore survives with positive order at \(1-\alpha_0\). It shields every nontrivial-zero pole coming from a later denominator shift.
Making the product countable does not help under ordinary convergence. Absolute exponent summability at one prime forces the shift set to be locally finite, hence to have an isolated minimum, and the same normalized prime-coefficient limit returns.
If shifts instead accumulate at their lower edge, their real zeta-pole divisors accumulate at the reflected edge. Under the required locally normal, nonvanishing reduced cofactors, that accumulation forbids a nonzero meromorphic continuation. Infinitely many shifts are not yet a singular operation. They are just a more crowded version of the same problem.
A continuum becomes a branch cut#
The next escape is much prettier. Replace the exponents by a finite signed measure:
$$ E_\mu(s) = \exp\left( \int\log\zeta(s+\alpha)\,d\mu(\alpha) \right). $$
Let \(a=\min\operatorname{supp}\mu\), and inspect the reflected edge \(s_0=1-a\). The logarithmic derivative contains the Cauchy transform
$$ -\frac{E_\mu'(s_0+z)}{E_\mu(s_0+z)} = \int\frac{d\nu(t)}{z+t}+H(z), $$
where \(\nu\) is the translated edge measure and \(H\) is holomorphic.
A Cauchy transform can cross the reflected support meromorphically only where its measure is atomic. Stieltjes inversion recovers continuous mass from the boundary jump; a holomorphic crossing has no jump. The residues must also be integers because this is the logarithmic derivative of a single-valued meromorphic function.
So meromorphicity forces
$$ \mu|_{[a,a+\delta)} = \sum_j c_j\delta_{\alpha_j}, \qquad c_j\in\mathbb Z\setminus\{0\}. $$
The continuum atomizes back into shifted-zeta powers. If ordinary Dirichlet coefficients are nonnegative, the leftmost integer mass is positive and the pole shield returns.
There is an explicit little witness for the alternative. Lebesgue mass on \([0,h]\) gives positive prime coefficients
$$ \int_0^h p^{-\alpha}\,d\alpha = \frac{1-p^{-h}}{\log p}>0, $$
but near \(s=1\) the logarithm contains
$$ z\log z-(z+h)\log(z+h). $$
The pole did not vanish. It melted into a logarithmic branch point. I like how literal this is.
Distributional derivatives only make the singularity worse#
Now let \(T\) be a compactly supported distribution and define
$$ E_T(s) = \exp\left\langle T_\alpha,\log\zeta(s+\alpha) \right\rangle. $$
Meromorphic continuation of its edge Cauchy transform first forces \(T\) to have finite point support. Such a distribution is a finite sum of delta derivatives, and
$$ \left\langle \delta_{t_j}^{(k)},\frac1{z+t} \right\rangle = \frac{k!}{(z+t_j)^{k+1}}. $$
But a logarithmic derivative of a meromorphic function has only simple poles. Every \(k\ge1\) is therefore forbidden. Again only integer delta masses survive, and again coefficient positivity makes the leftmost one positive.
The witness \(T=\delta_0'\) is wonderfully rude:
$$ \left\langle \delta_0',\log\zeta(s+\alpha) \right\rangle = -\frac{\zeta'}{\zeta}(s) = \sum_{n\ge2}\Lambda(n)n^{-s}. $$
This has positive prime weights, but exponentiating near \(s=1\) gives
$$ \exp\left(\frac1{s-1}+O(1)\right), $$
an essential singularity. Distributional differentiation did preserve the prime positivity. It also made the meromorphic failure spectacular.
Completion changes the game#
The gamma factors cancel the real zeta-pole edge, so the previous argument cannot simply be repeated after completion. Instead, the exact inverse Laplace carrier of \(L''=(\log\xi)''\) splits into two pieces:
$$ L''(s) = \int_0^\infty e^{-st}\kappa(t)\,dt + \sum_{n\ge2} \Lambda(n)\log n\,e^{-s\log n}, $$
with
$$ \boxed{ \kappa(t) = t\left(\frac1{e^{2t}-1}-e^t\right). } $$
The first piece is the archimedean continuum. The second is the literal prime-power comb.
A real shift packet acts on both through one multiplier
$$ M(z)=\left\langle T_\alpha,e^{-\alpha z}\right\rangle. $$
The shifted carrier is
$$ d\nu_M(t) = \kappa(t)M(t)\,dt + \sum_{n\ge2} \Lambda(n)\log n\,M(\log n)\delta_{\log n}(t). $$
Here is the part I keep staring at:
$$ \kappa(t)<0 \qquad(t\ge\log2), $$
while the atom at a prime logarithm has positive base weight \((\log p)^2\).
If \(\nu_M\) were a nonnegative measure, its continuous part would require
$$ M(t)\le0 \qquad(t\ge\log2), $$
but its prime atom would require
$$ M(\log p)\ge0. $$
Therefore
$$ \boxed{ M(\log p)=0 \quad\text{for every prime }p. } $$
The finite places and the archimedean place demand opposite signs from the same number at exactly the same locations. Completion removed the pole shield and exposed something harsher.
For compactly supported \(T\), the multiplier \(M\) is entire of exponential type, so it has only \(O(R)\) zeros in \(|z|\le R\). The prime logarithms contribute
$$ \pi(e^R)\asymp\frac{e^R}{R} $$
distinct zeros. Thus \(M\equiv0\), hence \(T=0\).
Compact completed shifts also fail a separate test. If
$$ X_T(s) = \exp\left\langle T_\alpha,\log\xi(s+\alpha) \right\rangle, $$
then the first nonzero shift moment leaves superexponential growth, a nonconstant power law, or an algebraic correction on the positive real axis. An absolutely convergent ordinary Dirichlet series instead has an exponential tail. So \(X_T\) is not such a Dirichlet series even with signed or complex coefficients.
Letting the support escape is still not enough#
Compact support was convenient, but it was not the real zero-count threshold.
If \(M\) is any nonzero entire function of finite order, Jensen's formula permits only polynomially many zeros in radius \(R\). The prime logarithms still supply \(e^R/R\). So no noncompact shift packet with a finite-order entire bilateral Laplace multiplier can make the completed logarithmic carrier nonnegative, provided the interchange is justified and produces a genuine measure. Gaussian shift densities are already too tame.
There is a different closure for one-sided noncompact measures. If \(\mu\) is a finite signed measure on \([0,\infty)\), then
$$ M_\mu(z) = \int_0^\infty e^{-\alpha z}\,d\mu(\alpha) $$
is bounded and analytic in \(\Re z>0\). Under the Cayley map
$$ \phi(z)=\frac{z-1}{z+1}, $$
a sufficiently large prime logarithm contributes
$$ 1-|\phi(\log p)| = \frac{2}{\log p+1}. $$
Zeros of a nonzero bounded analytic function must satisfy the Blaschke condition, but
$$ \sum_p\frac{2}{\log p+1} =\infty. $$
In fact the primes through \(e^R\) already contribute on the scale \(e^R/R^2\). So bounded half-plane multipliers die too.
One argument is entire-function zero density. The other is Hardy-space zero geometry. The prime logarithms are a uniqueness set in both worlds.
The surviving objects have to be genuinely singular#
This chain does not prove RH, and it does not exclude every imaginable positive final function.
Before completion, the surviving possibilities include noncompact or infinite-order functionals, infinite-variation and zeta-regularized packets, or another operation that invalidates the literal meromorphic edge factorization.
After completion, the theorem concerns positivity of the logarithmic inverse Laplace carrier. It does not rule out positivity that appears only after exponentiation while the logarithmic spectrum remains signed. It also leaves very large infinite-order entire multipliers, unbounded half-plane transforms outside the relevant Hardy/Nevanlinna classes, two-sided transforms not bounded in the right half-plane, and distributional carriers that are not genuine measures.
But "use more shifts" is no longer an explanation. Finite shifts, countable shifts, compact measures, compact distributions, finite-order entire multipliers, and bounded half-plane multipliers have each paid their exact bill.
At the end, the bill is this:
$$ \text{gamma continuum}<0, \qquad \text{prime atom}>0, \qquad \text{one shared multiplier}. $$
The only tame compromise is zero.
Notebook references: O-0334, O-0335, O-0336, O-0338, O-0339, O-0340, O-0341, C-0248, C-0250, C-0261