Start with the most innocent pole canceller:
$$ F_q(s)=(1-q^{1-s})\zeta(s). $$
Its Dirichlet coefficients are \(1-q1_{q\mid n}\), so their partial sums are not asymptotic to anything. They are exactly
$$ \boxed{A_q(N)=N\bmod q.} $$
That tiny sawtooth turns the critical-line Hardy norm into
$$ \boxed{ \frac1{2\pi}\int_{\mathbb R} \frac{|F_q(1/2+it)|^2}{1/4+t^2}\,dt = \sum_{n\ge1}\frac{(n\bmod q)^2}{n(n+1)}. } $$
I like this identity an unreasonable amount. The Fourier transform of the Cauchy weight produces \(1/\max(m,n)\), and layering that kernel by its maximum turns the whole double sum into squares of coefficient prefixes.
But one geometric scale leaves a positive Hardy slack even after every known vertical zero of the factor is removed. Varying \(q\) cannot prove RH.
The multiscale version is literally Nyman#
Now take any finite Dirichlet polynomial
$$ A_c(s)=\sum_d c_d d^{-s}, \qquad \sum_d\frac{c_d}{d}=0, $$
and put \(F_c=A_c\zeta\). Its coefficient prefixes are
$$ B_c(y)=\sum_d c_d\left\lfloor\frac yd\right\rfloor. $$
The same max-kernel calculation gives
$$ \|F_c\|_{\mathcal H_C}^2 = \int_1^\infty\frac{|B_c(y)|^2}{y^2}\,dy. $$
Under \(x=1/y\), pole cancellation turns the prefix into the negative fractional-part function
$$ f_c(x)=\sum_d c_d\left\{\frac1{dx}\right\}. $$
So there is no analogy to negotiate:
$$ \boxed{ \|F_c\|_{\mathcal H_C} = \|f_c\|_{L^2(0,1)}. } $$
After normalizing at \(s=1\),
$$ \boxed{ \frac{\|F_c\|_{\mathcal H_C}^2}{|F_c(1)|^2}-1 = \left\|1+\frac{f_c}{F_c(1)}\right\|_2^2. } $$
The proposed multiscale BSY optimization and the classical Nyman-Beurling approximation problem are the same quadratic problem in two coordinate systems.
Can numerator zeros eat the slack?#
There is one apparent escape. Deliberately give \(A_c\) zeros in \(\Re s>1/2\), then subtract their known Blaschke mass from the Hardy budget.
For supports in one geometric progression, write
$$ A(s)=P(q^{-s}), \qquad r=q^{-1/2}. $$
The critical line is the circle \(|w|=r\) in the variable \(w=q^{-s}\). Every numerator root \(\alpha\) inside that circle is removed by the exact reflection
$$ \boxed{ \alpha\longmapsto\frac{r^2}{\overline\alpha}. } $$
The reflected factor has the same modulus on \(|w|=r\), and its center gain is exactly the Blaschke mass that was subtracted. After reflecting every interior root, the remaining multiplier is periodic on the critical line.
Even if polynomial factors are relaxed to arbitrary measurable periodic multipliers, fiberwise least squares leaves a positive residual \(\delta_q\), uniformly over every base and every degree. One-frequency numerator engineering cannot close the gap.
The completely corrected infimum#
For an arbitrary finite multiplier, let \(d_c\) be its exact Nyman error and \(\mathcal B_A(c)\) the complete Blaschke mass of its additional numerator zeros. Define
$$ \boxed{ \mathcal R(c) = \frac12\log(1+d_c^2)-\mathcal B_A(c). } $$
Poisson-Jensen gives
$$ \mathcal R(c)\ge\mathcal B_\zeta, $$
where \(\mathcal B_\zeta\) is the off-line zeta-zero mass. Conversely, under RH a Nyman approximating sequence makes \(d_c\to0\), hence \(\mathcal R(c)\to0\). Therefore
$$ \boxed{ \mathrm{RH} \quad\Longleftrightarrow\quad \inf_c\mathcal R(c)=0. } $$
That is the exact stopping point. Raw max-kernel optimization is Nyman. One-frequency inner engineering has a uniform periodic floor. Fully unrestricted inner-corrected optimization is not a softer intermediate problem: it is RH itself.
What survives is very specific and very hard. A genuinely incommensurate finite support would have to control its Nyman error and its numerator Blaschke mass together, in a way not already swallowed by the exact criterion.
Notebook references: K-0166, O-0311, K-0167, O-0312, K-0168, O-0313, C-0233