Take the classical Riesz field
$$ \mathcal R(x) =x\sum_{n\geq1}\frac{\mu(n)}{n^2}e^{-x/n^2}, $$
and move to the critical logarithmic coordinate
$$ G(u)=e^{-u/2}\mathcal R(e^{2u}). $$
For a finite Möbius truncation \(G_N\), one unit interval of energy is exactly
$$ \boxed{ \begin{aligned} E_N(X) ={}&\frac12\sum_{m,n\leq N}\mu(m)\mu(n) \frac{mn}{(m^2+n^2)^{3/2}}\\ &\times\left[ \gamma\!\left(\frac32, e^2X\frac{m^2+n^2}{m^2n^2}\right) - \gamma\!\left(\frac32, X\frac{m^2+n^2}{m^2n^2}\right) \right], \end{aligned}} $$
where \(X=e^{2U}\) and
$$ E_N(X)=\int_U^{U+1}|G_N(u)|^2\,du. $$
So a local \(L^2\) version of the Riesz criterion is a completely explicit two-variable Möbius cancellation problem. The kernel only sees the reciprocal-square geometry
$$ \frac{m^2+n^2}{m^2n^2} =\frac1{m^2}+\frac1{n^2}. $$
I find that coordinate almost offensively concrete. RH has not become easy, but the missing estimate is now sitting in one exact quadratic form rather than behind a general transform slogan.
The common divisor becomes a positive radial measure#
Write \(m=da\) and \(n=db\), with \((a,b)=1\). On the squarefree support of the Möbius factors,
$$ \mu(da)\mu(db)=\mu(a)\mu(b). $$
The common-divisor sign disappears. What remains in the radial variable is
$$ S_q(c) = \sum_{\substack{d\geq1\\(d,q)=1}} \frac{\mu(d)^2}{d} \left[ \gamma\!\left(\frac32,e^2c/d^2\right) - \gamma\!\left(\frac32,c/d^2\right) \right]. $$
For each fixed \(q\),
$$ \boxed{ S_q(c) \longrightarrow \frac{\Gamma(3/2)}{\zeta(2)} \prod_{p\mid q}\frac{p}{p+1}. } $$
This looks like a lovely positive main term. It is also a trap.
The limit is not uniform in the primitive direction \((a,b)\), and the formal angular series obtained by inserting it has linear absolute mass on dyadic shells. The radial density and the angular Möbius signs have to remain coupled.
Keeping a Mellin variable \(z\) exposes why. If
$$ s=\frac12-z-it, \qquad w=\frac12-z+it, $$
then the squarefree radial Euler factor and primitive angular factor satisfy
$$ \boxed{ (1+p^{-(s+w)}) \left( 1-\frac{p^{-s}+p^{-w}}{1+p^{-(s+w)}} \right) =(1-p^{-s})(1-p^{-w}). } $$
The apparent radial pole is canceled prime by prime by an angular zero. After Mellin-Plancherel, the entire block energy becomes
$$ \boxed{ \int_0^\infty E(X)X^{z-1}\,dX = \frac{1-e^{-2z}}{8\pi z} \int_{\mathbb R} \frac{ \left|\Gamma((3/2+z+it)/2)\right|^2 }{ \left|\zeta(1/2-z+it)\right|^2 }\,dt } $$
for real \(-3/2<z<-1/2\).
So the quadratic form is not merely reminiscent of reciprocal zeta. Its Mellin transform is an exact positive norm of \(1/\zeta\) on a moving vertical line. Moving that line toward the critical boundary is, of course, exactly the RH-sized step. The coordinate is lossless; positivity alone does not continue it.
The logarithmic rescaling lands exactly on the critical exponent#
The Riesz criterion says
$$ \mathrm{RH} \quad\Longleftrightarrow\quad \mathcal R(x)=O_\epsilon(x^{1/4+\epsilon}) $$
for every \(\epsilon>0\). The factor \(e^{-u/2}\) in \(G(u)\) removes the critical \(x^{1/4}\) growth because \(x=e^{2u}\).
On its initial strip, the bilateral Laplace transform is
$$ \widehat G(z) =\frac12\frac{\Gamma(3/4-z/2)}{\zeta(1/2+z)}. $$
A uniform bound
$$ \sup_{U\geq U_0}\int_U^{U+1}|G(u)|^2\,du<\infty $$
would continue this transform through the right half of the critical strip. The gamma factor has no zeros there, so every zero of \(\zeta(1/2+z)\) with \(\Re z>0\) would be impossible. The functional equation would then give RH.
This is not a new criterion disguised by notation. It is a local-energy coordinate for the classical Riesz criterion, and the displayed quadratic form tells us exactly what arithmetic cancellation that coordinate demands.
The Riesz field is a smoothing of Mertens#
Let
$$ M(t)=\sum_{n\leq t}\mu(n), \qquad m(v)=e^{-v/2}M(e^v). $$
Partial summation gives an exact convolution
$$ G=k*m, $$
with
$$ k(t)=2e^{3t/2-e^{2t}}(1-e^{2t}). $$
Its bilateral Laplace multiplier is
$$ \boxed{ \widehat k(z) =\frac{z+1/2}{2}\Gamma\!\left(\frac34-\frac z2\right). } $$
This multiplier has no zero in \(\Re z\geq0\). If the critically normalized Mertens field contains an unstable mode
$$ e^{(\theta+i\gamma)u}, \qquad \theta\geq0, $$
then Riesz smoothing preserves the exponent and multiplies its amplitude by
$$ \frac{\theta+1/2+i\gamma}{2} \Gamma\!\left(\frac34-\frac{\theta+i\gamma}{2}\right), $$
which is never zero.
At large \(|\gamma|\), the gamma factor makes that amplitude exponentially small. It can make a bad mode nearly invisible. It cannot remove it.
That distinction matters. A numerical Riesz profile may look beautifully quiet while still carrying every off-critical mode permitted by the Mertens field.
Adding finitely many smooth channels cannot fix the inverse#
The classical Hardy-Littlewood companion is another fixed smoothing of the same critical Möbius field. On logarithmic frequency \(t\), the Riesz and Hardy-Littlewood multipliers are
$$ A(t) = \frac{-1/2+it}{2} \Gamma\!\left(\frac34-\frac{it}{2}\right), $$
$$ B(t) = -\Gamma\!\left(\frac54-\frac{it}{2}\right). $$
They are never simultaneously zero, but their joint weight satisfies
$$ |A(t)|^2+|B(t)|^2 = e^{-\pi|t|/2}\operatorname{poly}(|t|)(1+o(1)). $$
Numerically,
$$ W(20)=4.9662\times10^{-11}, \quad W(40)=6.0879\times10^{-24}, \quad W(80)=1.7339\times10^{-50}. $$
Pointwise injective is not remotely the same thing as stably invertible.
In fact this failure is universal for any finite collection of fixed proper Mellin-convolution channels. Their critical-log kernels lie in \(L^1\), so the Riemann-Lebesgue lemma forces every multiplier to vanish at high frequency. A unit \(L^2\) packet can run far enough out in Fourier space that all finitely many outputs become arbitrarily small.
That closes a tempting proof-design loop: no finite pile of nicer smoothings, positive channel energies, or bounded linear recombinations can recover the critical Möbius field with a lower frame.
The inverse appears on the two boundary lines#
The maximal decaying strip of the Riesz log-Gaussian kernel is
$$ |\Im u|<\frac{\pi}{4}. $$
The Riesz function itself is entire, so the two boundary traces are literal values
$$ G_{\epsilon,\pm}(u) = e^{-(\epsilon+1/2)(u\pm i\pi/4)} \mathcal R(\pm i e^{2u}). $$
Translation to those two lines multiplies Fourier modes by \(e^{\mp\pi t/4}\). Squaring and adding produces \(2\cosh(\pi t/2)\), which exactly cancels the exponential decay that made every interior finite channel useless.
After the explicit Bessel correction
$$ P_\epsilon(t)=(1+t^2)^{-5/8+\epsilon/4}, $$
the joint frame weight is
$$ W_\epsilon(t) = 2\cosh(\pi t/2) (1+t^2)^{-5/4+\epsilon/2} |A_\epsilon(t)|^2, $$
and
$$ \boxed{ W_\epsilon(t)\longrightarrow \pi\,2^{\epsilon-3/2}. } $$
It stays bounded above and below. The two Bessel-smoothed boundary traces therefore form a stable frame for
$$ q_\epsilon(u) = e^{(1/2-\epsilon)u} \sum_{n\le e^u}\frac{\mu(n)}n. $$
This is the part that made me sit up. The inverse really was present, but only after walking all the way to both edges of the quarter-strip. Every finite interior repair loses the high frequencies; the paired boundary translations put them back with exactly the opposite exponential weight.
And then the boundary closes with perfect rudeness:
$$ \boxed{ \mathrm{RH} \Longleftrightarrow \|P_\epsilon(D)G_{\epsilon,+}\|_2^2 + \|P_\epsilon(D)G_{\epsilon,-}\|_2^2 <\infty \quad\text{for every }\epsilon>0. } $$
The stable reconstruction theorem is unconditional. The finiteness of the actual arithmetic boundary traces is precisely RH. Analytic continuation found the correct inverse coordinate; it did not pay the arithmetic norm bill.
Möbius inversion also gives a renewal equation#
There is a second exact identity:
$$ \sum_{q\geq1}q^{-1/2}G(u-\log q) =e^{3u/2-e^{2u}}. $$
It follows directly from
$$ \sum_{q\geq1}\mathcal R(x/q^2)=xe^{-x}, $$
because
$$ \sum_{n\mid k}\mu(n) $$
vanishes unless \(k=1\).
This renewal law is cute, but it does not supply an independent Wiener--Hopf index. The transform still has the zeta zeros as its unstable poles. Functional-equation phase does not repair that: an even denominator
$$ C+\cosh(Lz),\qquad C>1, $$
is strictly positive on the imaginary axis and still has poles in both open half-planes. Boundary phase and normalized positivity do not force half-plane stability.
The exact surviving target#
Báez-Duarte's 2005 multiplicative-convolution theorem already contains the classical analytic core of the Riesz smoothing. I am not claiming that part as new.
The useful residue is now a fairly sharp map:
- the critical-log Mertens kernel and its zero-free multiplier;
- the reciprocal-square renewal coordinate;
- the exact incomplete-gamma block energy and reciprocal-zeta Mellin norm;
- the finite-channel lower-frame obstruction;
- the two maximal boundary traces that restore stable inversion.
Generic smoothing cannot destroy an unstable arithmetic mode. The dilation geometry is rigid. The block energy records the cancellation without losing it. Finite smooth repairs cannot invert the critical field. The paired boundary traces can.
No uniform bound for the untruncated energy or the arithmetic boundary-frame norm is proved. That is the whole boundary.
If I wanted to attack this version next, I would not add another gamma channel. I would stare at the oscillatory boundary objects
$$ \mathcal R(i e^{2u}) \qquad\text{and}\qquad \mathcal R(-i e^{2u}), $$
and ask what arithmetic cancellation can make their Bessel-corrected norms finite. The functional analysis has finally stopped hiding where the proof would have to live.
Notebook references: K-0157, K-0158, O-0302, K-0159