Let \(\lambda_n\) be the Li coefficients of the Riemann xi function. Li's criterion says
$$ \mathrm{RH} \iff \lambda_n\geq 0\quad\text{for every }n. $$
I expected a false RH to make the sequence oscillate violently. What I did not expect was that the negative terms would be forced to occur with positive lower density.
Not merely infinitely often. Not on an unknown thin subsequence. A positive fraction of all sufficiently large indices.
The nearest pole shell does it#
The literal generating function is
$$ F(z) = \frac{d}{dz}\log\Xi\left(\frac1{1-z}\right) = \sum_{n\geq1}\lambda_n z^{n-1}. $$
A zeta zero \(\rho\) becomes a pole at
$$ z_\rho=1-\frac1\rho. $$
The critical line maps to the unit circle. A zero with \(\Re\rho>1/2\) maps strictly inside it because
$$ |z_\rho|^2 = 1+\frac{1-2\Re\rho}{|\rho|^2} <1. $$
Assume RH is false and let \(R<1\) be the smallest modulus of any such interior pole. Only finitely many poles lie on this nearest shell. Darboux's method then gives
$$ \lambda_n = -\sum_{\rho\in\mathcal Z_R}m_\rho z_\rho^{-n} +O(q^{-n}), \qquad R<q<1, $$
where \(\mathcal Z_R\) is the nearest shell.
Writing \(z_\rho=Re^{i\theta_\rho}\), the dominant term is
$$ -R^{-n}P_n, \qquad P_n=\sum_{\rho\in\mathcal Z_R}m_\rho e^{-in\theta_\rho}. $$
The sequence \(P_n\) is a real trigonometric wave on the closure of a cyclic orbit in a finite torus. Its Haar mean is zero, it is not identically zero, and no zero-frequency character is present. Therefore \(P_n\) exceeds some fixed \(\delta>0\) on a set of positive natural density.
The exponentially smaller remainder cannot change those signs forever. Hence
$$ \boxed{ \liminf_{N\to\infty} \frac1N\#\{n\leq N:\lambda_n<0\}>0 } $$
whenever RH is false. Equivalently,
$$ \boxed{ \mathrm{RH} \iff \#\{n\leq N:\lambda_n<0\}=o(N). } $$
I like this criterion because it is weaker than coefficientwise positivity but still exact. A false RH is not allowed to hide in a sparse exceptional set.
The prime side becomes one quadratic kernel#
Write the arithmetic decomposition as
$$ \lambda_n=A_n-S_f(n), \qquad A_n=\frac12n\log n+O(n). $$
The finite-prime term has the exact regularized Laguerre form
$$ S_f(n) = \lim_{X\to\infty} \int L_{n-1}^{1}(t)\,d\nu_X(t), $$
where
$$ d\nu_X(t) = \sum_{m\leq X}\frac{\Lambda(m)}m\delta_{\log m}(t)-dt. $$
Now sum the square with weight \(1/n\). The entire family collapses to
$$ \sum_{n=1}^{M}\frac{S_f(n)^2}{n} = \lim_{X\to\infty} \iint \mathcal K_M(x,y)\,d\nu_X(x)d\nu_X(y), $$
with
$$ \mathcal K_M(x,y) = \sum_{n=1}^{M} \frac1nL_{n-1}^{1}(x)L_{n-1}^{1}(y). $$
The cute part is that this is not a long spectral sum after all. The Christoffel-Darboux identity reduces it to two adjacent polynomials:
$$ \boxed{ \mathcal K_M(x,y) = \frac{ L_{M-1}^{1}(x)L_M^{1}(y) - L_M^{1}(x)L_{M-1}^{1}(y)} {x-y}. } $$
If one could prove the dyadic estimate
$$ \sum_{N<n\leq2N}\frac{S_f(n)^2}{n} = o(N^2\log^2N), $$
then negative Li coefficients would have density zero, and the shell theorem would force RH.
Why the usual PNT envelope is useless here#
There is an exact growing-order Laguerre transform
$$ S_f(n) = -\gamma n + \int_0^\infty L_{n-2}^{2}(t)B(t)\,dt, $$
where classical zero-free regions give a pointwise envelope resembling
$$ |B(t)|\leq \operatorname{poly}(t)e^{-c\sqrt t}. $$
That looks strong until the polynomial order grows with \(n\).
Put \(d=n-2\) and \(t_d=4d(d+2)\). At that quadratic spatial scale, the leading monomial of \(L_d^2\) dominates all lower terms. A compact bump with the matching sign,
$$ B_d(t) = (-1)^d e^{-c\sqrt t}\mathbf 1_{[t_d,t_d+1]}(t), $$
obeys the pointwise PNT-shaped envelope but produces
$$ \left| \int L_d^2(t)B_d(t)\,dt \right| \geq \exp(d\log d-O_c(d)). $$
That is wonderfully rude. Any proof using only absolute values and pointwise PNT decay can permit a supertempered Li transform.
The surviving target is therefore genuinely quadratic and arithmetic: cancellation of the centered von Mangoldt measure against the dyadic Christoffel-Darboux kernel. The exact reduction is finished. The estimate is not.
Notebook references: C-0235, K-0171, O-0316