Put
$$ \phi(t)=\frac{\xi(1/2)}{\xi(1/2+t)}. $$
This is unconditionally the characteristic function of a symmetric self-decomposable law.
I did not expect that sentence to be true without RH.
The actual xi law#
Write an upper centered zeta zero as \(\alpha=\delta+i\gamma\), with \(-1/2<\delta<1/2\) and \(\gamma>0\). After pairing the functional-equation symmetries, the exact Levy-Khintchine exponent is
$$ \boxed{ \log\frac{\xi(1/2+t)}{\xi(1/2)} = 2\int_0^\infty (1-\cos(tx))k(x)\frac{dx}{x}, } $$
where
$$ k(x) = \sum_{\delta=0}e^{-\gamma x} + \sum_{0<\delta<1/2} 2e^{-\gamma x}\cos(\delta x). $$
The off-axis packets look dangerous because of the cosine. They are still positive and decreasing for \(0<x\le\pi\): the critical strip gives \(0\le\delta x<\pi/2\), and
$$ -\frac d{dx} \left(2e^{-\gamma x}\cos(\delta x)\right) = 2e^{-\gamma x} \left(\gamma\cos(\delta x)+\delta\sin(\delta x)\right)>0. $$
The rest of the half-line is where the arithmetic enters. Every zero through
$$ T_0=3000175332800 $$
is verified on the critical line. Beyond \(T_0\), a deliberately coarse explicit count \(N(T)\le T^2\) makes the entire possible unknown tail smaller than the first critical-line packet. For \(x\ge\pi\),
$$ 2T_0^2e^{-xT_0}<e^{-\gamma_1x}, \qquad 2T_0^3e^{-xT_0}<\gamma_1e^{-\gamma_1x}. $$
The first inequality protects \(k\); the second protects \(-k'\). So \(k(x)>0\) and \(k'(x)<0\) for every \(x>0\).
That makes
$$ \nu(dx)=\frac{k(|x|)}{|x|}\,dx $$
a symmetric Levy measure. More specifically, for every \(0<c<1\),
$$ \frac{\phi(t)}{\phi(ct)} = \exp\left( -2\int_0^\infty (1-\cos(tx))[k(x)-k(x/c)]\frac{dx}{x} \right) $$
is again infinitely divisible. This is exactly the class-\(L\) factorization: reciprocal centered \(\xi\) is self-decomposable.
That theorem is real. It is also not RH.
The counterfeit#
Start with
$$ \boxed{ Y(z)= \frac14(1+z^2)^3(z^4+4) \cosh\left(\frac{\pi z}{2}\right). } $$
This is an even real entire function of order one. It satisfies \(Y(0)=1\) and is strictly positive on the real axis.
It also has the four zeros
$$ \boxed{\pm1\pm i.} $$
The quartic is not being subtle:
$$ z^4+4 = (z-(1+i))(z-(1-i))(z-(-1+i))(z-(-1-i)). $$
So this function definitely does not have all its zeros on the imaginary axis. The rude part is how much probabilistic structure its reciprocal still has.
Its reciprocal is self-decomposable too#
For \(x>0\), define
$$ k(x) = e^{-x}(3+2\cos x) +\frac1{2\sinh x}. $$
Two Frullani identities and Euler's product for the hyperbolic cosine give the exact Levy exponent
$$ \boxed{ \log Y(t) = 2\int_0^\infty (1-\cos(tx))k(x)\frac{dx}{x}. } $$
Equivalently,
$$ \frac1{Y(t)} = \exp\left( -\int_{\mathbb R}(1-\cos(tx)) \frac{k(|x|)}{|x|}\,dx \right) $$
is the characteristic function of a symmetric infinitely divisible law.
But \(k\) is not merely positive. It is strictly decreasing. The oscillatory piece obeys
$$ \frac d{dx}\left[e^{-x}(3+2\cos x)\right] = -e^{-x}(3+2\cos x+2\sin x) \le -e^{-x}(3-2\sqrt2)<0, $$
and \(1/(2\sinh x)\) is decreasing too.
That is exactly the one-dimensional class-\(L\) condition. For every \(0<c<1\), the quotient
$$ \frac{Y(ct)}{Y(t)} $$
is another infinitely divisible characteristic function, so \(1/Y\) is self-decomposable.
This is a premise-matched counterfeit for the theorem above: even, real, order one, normalized, positive on the real line, infinitely divisible, and self-decomposable. Four zeros still wandered off the axis.
The density is positive everywhere#
There is one more pleasant trap. The reciprocal factors as
$$ \frac1{Y(t)} = \operatorname{sech}\left(\frac{\pi t}{2}\right) \left[ \frac14(1+t^2)^3(t^4+4) \right]^{-1}. $$
The first factor has inverse Fourier density
$$ \frac{\operatorname{sech}x}{\pi}, $$
which is strictly positive on the whole real line. Convolving it with the probability law supplied by the second factor makes the full reciprocal density strictly positive everywhere.
So positive reciprocal Fourier density does not rescue the argument either. Neither does the entire-function symmetry. All of those properties coexist with the explicit quartet \(\pm1\pm i\).
Where the counterfeit finally breaks#
The decreasing Levy numerator is not completely monotone. As \(x\to\infty\),
$$ k(x) = e^{-x}(4+2\cos x)+O(e^{-3x}), $$
and for fixed \(n\),
$$ (-1)^nk^{(n)}(x) = e^{-x} \left[ 4+2(\sqrt2)^n \cos\left(x-\frac{n\pi}{4}\right) \right] +O(3^ne^{-3x}). $$
Choose arbitrarily large
$$ x\equiv\pi+\frac{n\pi}{4}\pmod{2\pi}. $$
The leading bracket becomes \(4-2(\sqrt2)^n\), which is negative for every \(n\ge3\). The tiny cosine modulation that allowed the off-axis quartet eventually appears in the derivative signs. I keep staring at that seam: monotone decay misses it, complete monotonicity sees it.
The surviving condition is already RH-sized#
Suppose instead that an even entire \(Y(z)=E(z^2)\) had a completely monotone Levy numerator
$$ k(x)=\int_{(0,\infty)}e^{-ax}\,d\mu(a). $$
Differentiating the Levy exponent then gives
$$ \boxed{ \frac{E'(w)}{E(w)} = \int_{(0,\infty)} \frac{d\mu(a)}{w+a^2}. } $$
The logarithmic derivative is Stieltjes. It is analytic away from the negative real axis, and the differential equation \(E'=HE\) forbids a zero anywhere in that slit plane. Therefore every zero of \(Y\) is imaginary.
For the centered Riemann \(\xi\)-function, this is precisely the existing Stieltjes/Loewner formulation equivalent to RH. Complete monotonicity survives the counterfeit, but it does not produce a cheaper theorem. It returns the full zero-location problem in exact integral form.
That leaves a clean probabilistic ladder for reciprocal entire functions:
$$ \text{positive density} \;<\; \text{infinite divisibility} \;<\; \text{self-decomposability} \;<\; \text{complete monotonicity}. $$
The actual reciprocal centered \(\xi\) now reaches the third rung unconditionally. The first three are still compatible with four explicit off-axis zeros. The fourth would force the axis, but proving it for reciprocal \(\xi\) is already RH-strength. The oscillation had to be excluded somewhere. It was hiding in the third derivatives.
Notebook references: C-0237, O-0321, O-0050