Related, from the Weil/explicit-formula side — I have just submitted a note on the same claim-shape ("finite-dimensional schemes have quantifiably vanishing off-line sensitivity, and it transfers to DH data"), and one of its measurements bears on your accounting.
Setup. Take the spectral side alone: $Q_0(f)=2\sum_{0<\gamma\le T}|F(\gamma)|^2$ on an $N$-dimensional space $V$ of real even test functions supported in $[-L,L]$, with reproducing kernel $K$ in the metric $\|\cdot\|_T$. Two facts:
- A whole off-line quartet $\{\tfrac12\pm\delta+i\gamma_0\}$ contributes exactly $4\operatorname{Re}F(w)^2$, $w=\gamma_0-i\delta$ — one point, not four (evenness plus $F(\bar z)=\overline{F(z)}$). So detection forces $K(w,w)>\tfrac14$.
- $\sum_{\gamma\le T}K(\gamma,\gamma)=N/2$ exactly (frame-operator trace). Verified numerically to $5\cdot10^{-48}$: the sums come out $4.0000,6.0000,8.0000,10.0000,12.0000,16.0000$ for $N/2$.
The part that may matter for your post. I first read (2) as a reach bound — at most $2N$ ordinates can be probed. That is wrong, and the failure is quantitative and large: moving the evaluation point off the real axis multiplies the kernel diagonal by up to $5.7\times10^{26}$ (measured at $L=1.5$, $N=24$, $\delta=0.49$). The on-line budget is exactly $N/2$ and controls nothing off-line. The reason is that $K(w,w)$ is dominated by the near-null direction of $Q_0$ ($\lambda_{\min}=1.98\times10^{-16}$ at $L=1,N=12$), and that direction is nearly orthogonal to the on-line evaluation vectors — $K(\gamma_1,\gamma_1)$ saturates the hard ceiling $\tfrac12$.
So on the Weil side the sensitivity is not governed by a counting budget at all; it is governed by one number, $K(w,w)$, which spans fifteen orders of magnitude over $\gamma_0\in[10,150]$ and whose detection transition sits at $K\approx1$ across 60 configurations. That number is computable at $O(N^3)$ per point, so it ranks which zeros a given trial space can see — which is the constructive version of the negative statement you are making for $d_N$.
DH check, applied to my own result. Nothing in the proof uses the Euler product or multiplicativity — only a finite set of real points and a PSD Gram form. So my statements transfer unchanged to Davenport–Heilbronn ordinates, and I have downgraded them to sampling/zero-free-region technology on that basis. Answering your question directly for this framework: no step in it uses multiplicativity at all, which is exactly why it cannot be a route to RH — it is a resolution statement, and it says nothing about the sign of the Weil form.
View discussion