Route-D v16, the float rehearsal: what the reduced space fixed, and what it did not

Phase-2 P2, Route-D leg 16. Figure: writeup/figures/fig33_route_d_v16_rehearsal.png. Data: writeup/data/p2_route_d_v16_rehearsal.json (rebuild the figure with writeup/4_p2_lottery/p2_route_d_v16_evidence.py; regenerate with experiments/p2_route_d_v16_rehearsal.py, ~5 min, deterministic). Code: solver/reduced_certificate.py + test_reduced_certificate.py (16/16; suite 22 files).

Rigor level: 1. This leg's headline is a NEGATIVE with a named repair. Nothing here is interval-enclosed, nothing is a certificate, and, per v15, even a successful certificate in this family would be a capability demonstration rather than a novel result.

0. What this leg is and why it is small

v15 re-priced the whole lane: computer-assisted certification of 1D toy-model profiles is routine for the groups working this family, and the two-scale traveling wave's existence appears already proved. So the certificate was demoted to a one-chunk capability build: do the float rehearsal, find out whether the pipeline closes, stop. That is exactly what this is, and the answer is no, and here is precisely why, which is worth more than a yes would have been.

The rule being honoured is banked lesson 1: do the cheap float rehearsal before hardening. solver/interval.py has existed since v1 and has still never been pointed at anything, correctly.

1. Y₀ reaches machine precision: the first genuinely good number in the ledger

The defect that matters is the residual of the interpolant as a function, not the nodal vector Newton zeroed. That distinction is v12's, and it survives the change of formulation: Newton enforces K conditions, the certificate asks about F on all of [0,1].

Fitted above the float floor: K^−13.0 (a=0.3) and K^−16.3 (a=0.4), both steeper than the K^−(2/a+1) the (1−v)^{1/a} endpoint branch predicts (−7.7 and −6.0), the fits are taken only over points above 1e−11, but they are still contaminated by proximity to the floor, so the honest statement is "faster than algebraic and it reaches machine precision", not a rate. The nodal row is there as the control: it is flat at 1e−14 by construction, and the off-node value is nine orders larger at K=32, which is what tells you the measurement is about the function rather than the grid.

For scale: the GA carried a defect floor of ~1e−2 for five legs, v11's Newton took the whole-line nodal residual to 1e−14 but the interpolant defect stayed at 1e−4–ish, and v12's budget, computed at the anchor and therefore wrong anyway, was 2.4e−4. 1.5e−12 is the first time the number a certificate actually needs has been at machine level.

Z₀ = ‖I − A·DF‖ is roundoff (1.6e−11), as it must be when A is the numerical inverse of the same matrix. Reported for ledger completeness, not because it is informative.

2. Z₂ does not exist in the sup setting, and the far field is not why

This is the leg's result. Two independent obstructions.

2a. The finite Hilbert transform is unbounded on the sup norm, on a bounded interval

Removing the far field killed the decay grading that v3–v11 needed. It did nothing whatever to the smoothness one. H unbounded on L^∞ is v4's W3, and it is a local statement about a jump; a bounded domain does not repair it.

Sampling cannot establish this and cannot refute it (banked lesson 9): the adversary has to be built. Inside the actual perturbation space δe = (1−v²)δs, take Chebyshev partial sums of a step:

Linear in log K at +0.499 per e-fold (max fit residual 0.068). The classical constant for a clean jump is 2/π = 0.637; the deficit is the (1−v²) factor damping the jump region.

The instrument check that makes this trustworthy. The naive probe, a single high Chebyshev mode, also reports a divergence: 0.998 → 2.935 → 6.436 at K = 64/128/256. It is entirely the quadrature. Refine the rule and it collapses:

One row moves under a 4× refinement of the instrument and one does not. That is banked lessons 9 and 14 in a single table, which is why both rows are kept in the module and in the figure rather than only the one that supports the conclusion.

2b. The nonlinearity's second derivative is bounded exactly for a ≤ 1/2

N(e) = e^{1/a} has N''(e) = p(p−1)e^{p−2} with p = 1/a, and e vanishes linearly at the support edge. So sup|N''| is finite iff p ≥ 2 iff a ≤ 1/2, which is exactly where Ω = −e^{1/a} loses C².

Displayed as growth under an edge cutoff tightened over eight decades (1e−2 → 1e−10), because quoting whatever value a grid happened to reach would hide the divergence:

Flat to every displayed digit for a ≤ 1/2; growing without bound above. At a = 1/2 the value is p(p−1) = 2 exactly, checkable by hand, and the code reproduces 2.000000000000.

Three things this is not. - It is not a statement about the equation. v14 solves the profile cleanly and grid-converged to a = 1.2. The traveling wave exists; it is the certificate's norm that fails. - It is norm-dependent. A domain weight vanishing like (1−v)^{(2−1/a)/2} restores finiteness, at the price of requiring perturbations to vanish at the edge. That is the v5-U3c trade in a new place: check which side of the inequality a marginality lives on before pricing it (lesson 13). - It is not an explanation of a*. The coincidence with the independently measured survival boundary a* ≈ 0.5–0.55 (measured on a > 0, for the a > 0 continuation of the a = 0 two-scale traveling wave, not for the published two-scale scenario that arXiv:2603.25104 scopes to a ≤ 0) is striking and is recorded for exactly that reason: recording it is how the next person gets to disprove it. v12's a = 1/3 control is the precedent: a tantalising arithmetic coincidence at a* that a control killed. No control has been run here, so it stays an observation.

3. Z₁ is not computed at all

The infinite-dimensional tail (the part of the operator outside the K-mode subspace) is the entire content of a real computer-assisted proof, and nothing in this leg bounds it. It is reported as None in the assembled rehearsal rather than as zero, and rehearsal() deliberately refuses to return a radii polynomial: assembling one from a ledger with an uncomputed entry, or with an infinity in it, is the exact failure mode banked as lesson 15.

4. The verdict, and the next brick

The sup-to-sup rehearsal does not close. Not because of the far field, v14 genuinely removed that, but because the smoothness half of the requirement was never about the far field in the first place.

The repair is v5's, and it is measured here rather than assumed. Re-running the same adversary against ‖δe‖_γ = sup|δe| + [δe]_γ:

The divergence stops at γ ≳ 0.35, and γ = 0.15 still creeps, which matters: it shows the threshold is a real threshold and not an artefact of dividing by any seminorm at all.

v5 U1 found the same γ ≳ 0.35 on the whole line. That is a genuine independent check rather than a restatement, because v5's norm also carried a decay grading and this one has none, so the threshold belongs to the smoothness half, which is what one would want to be true and had not been separated before.

So the next brick is the Hölder version of the reduced space, and the good news is that it is a smaller job than the whole-line one was: v5's holder_norms, v7's derivative-gain closure and v8's weighted-Hölder bound on H all adapt, and on a compact interval there is no decay grading, no resonance, no matching radius X₀, and no tail bound to price. Roughly: v7–v9 without the half that was expensive.

5. Gate-check against the Clay chain

(a) Which link? L1, and only in the sense of making the pipeline honest. v15 already established that L1 is occupied territory, so this is capability work by design.

(b) Is another L1 leg the best use of the next chunk? No. One more would be defensible, the Hölder version is well-specified and cheap, but v15's ranking stands: the DSS lane is the swing, and verifying the two load-bearing readings from the literature (blocked on PDF access in this container) beats both. If a session does do the Hölder leg, it should be framed as finishing the capability, not as chasing a result.

(c) Cheaper experiment that kills the route? Still "read one paper", still blocked on access. That has not changed and should keep being said.

6. What is NOT claimed

• No certificate. Z₁ uncomputed, Z₂ infinite in the setting tested, no radii polynomial assembled, nothing interval-enclosed.
• Y₀ = 1.5e−12 is not a budget. It is one input, and the budget it would feed does not exist yet. A number under a budget is not a result until the budget was computed for the same object (lesson 30), and here there is no budget at all.
• The a ≤ 1/2 threshold is about the norm, not the equation, and is not offered as an explanation of a*.
• The K^−13 fit is not a rate. It is contaminated by the float floor; the defensible claim is "faster than the algebraic prediction, and it reaches machine precision".
• v14's "cold start converges at every a" is slightly overstated and is corrected here: at a = 0.7 the cold start misses the basin and continuation from the neighbouring a is needed. Every other value in 0.2…1.2 converges cold. Small, but it was a claim.