Phase-2 P2 (Route D v3: which space pair can carry the certificate

Status: Level-1 tooling + a no-go theorem for an entire category of function spaces, with its numerical face, its control, and the replacement it points to) NOT a certificate. Clay odds unchanged (~0.05%).

Route-D v2 ran the float dress rehearsal of the Newton–Kantorovich bounds at the exact a = 0 anchor and got an honest structural negative: the ball closes at no truncation, because the transport term c(1 + cos θ)∂_θ degenerates at θ = ±π (i.e. at X = ∞). It ended with a constructive half (the linearized inverse loses exactly one power of decay, so grade the codomain by one mode power and ‖A‖ = 3.000 goes flat in N) and called that "the asymmetric space pair a working certificate must use."

That proposal came with a condition attached, and this leg discharges it:

the quadratic term D²F[h,h] = 2 h H(h) must also land in the graded codomain; do this on paper first, and if no consistent pair exists, that is itself the answer and the leg stops cheaply.

No consistent pair exists. Not for the grading v2 proposed, and not for any pair of diagonally weighted ℓ¹ spaces: the two Newton–Kantorovich requirements are separated by exactly one grading power, and the separation is conserved; you can move it between the two bounds but you cannot remove it. That is the no-go. The same measurements then identify what does work, and it is not a weighted-ℓ¹ pair at all: decay-graded spaces, in which both requirements hold simultaneously, at a quantified price with an interior optimum.

Rebuild the figure from committed data (no re-derivation): python writeup/4_p2_lottery/p2_route_d_v3_evidence.py → writeup/figures/fig21_p2_route_d_v3_spaces.png (reads writeup/data/p2_route_d_v3_spaces.json; regenerate the data (deterministic, about a minute) with python experiments/p2_route_d_v3_spaces.py).

Code: solver/decay_grading.py + test_decay_grading.py (7/7). Full suite now 10 files green.

0. Why this was worth doing before writing any more solver

The v2 note ranked the v3 rebuild as a "real, multi-brick build": rederive the bounds in the graded pair, handle the far field with the exact ODE inverse rather than a diagonal model, and only then bring in interval arithmetic. All of that is downstream of one question that costs an afternoon of algebra:

is there a pair of spaces in which every Newton–Kantorovich bound is finite?

The radii polynomial needs four quantities, and they pull on the space pair in opposite directions:

A bounded pulls Y to be stronger than X (the inverse of a degenerate transport loses decay, so its input must supply the deficit). The quadratic pulls Y to be no stronger than X (products do not gain decay). v2 measured the first pull and proposed a Y one power stronger. Nobody had measured the second.

This is the discipline lesson from v2 §11 applied one level up: do the cheap rehearsal before hardening. v2's rehearsal saved interval-enclosing bounds that could never close. This leg's rehearsal saves building a whole two-region solver in a space pair that could never have been consistent.

1. The quadratic term is a pure convolution

The starting point is an identity worth banking on its own.

With H(cos kθ) = sin kθ for every k ≥ 0 and H(1) = 0 (unconditional; the Hardy/Cayley argument of v2 §1), write h = Σ_{k≥0} h_k cos kθ. Then

Q(h) := h H(h) = Σ_{j≥0, k≥1} h_j h_k cos(jθ) sin(kθ)
              = ½ Σ_{j≥0, k≥1} h_j h_k [ sin((j+k)θ) + sin((k−j)θ) ].

The difference part is antisymmetric in (j,k) once the missing k = 0 row is restored, and the same k = 0 row is exactly what the sum part is missing. The two corrections cancel identically, leaving

(Q) Q(h) = ½ Σ_{m≥0} ( Σ_{j+k=m} h_j h_k ) sin(mθ): a pure convolution. No difference frequencies at all.

Structurally this is Hardy space: h + iH(h) is a boundary value of a function holomorphic in the upper half plane, 2 h H(h) is the imaginary part of its square, and squaring a holomorphic function cannot produce difference frequencies.

solver/decay_grading.quadratic_coeffs implements (Q) as a single np.convolve. It is a deliberately independent second build of the quadratic part of solver/nk_fourier.residual (a double loop over sum and difference frequencies): the "build the same object twice" rule that caught the k = 0 fold bug in Route-D v1. They agree to 3.2e-17 on random inputs, and exactly (0.0) at the anchor. (Gate 4.)

Two immediate consequences.

1. A sharp constant. For X = ℓ¹_u (cosine side) and Y = ℓ¹_v (sine side), (Q) gives termwise ‖Q(h)‖_Y ≤ ½ S ‖h‖²_X with

S := sup_{j,k ≥ 0} v_{j+k} / (u_j u_k) .

And it is necessary: take h = ξ e_j + η e_k, whose image contains q_{j+k} = ξη, and optimize ξ, η > 0 at fixed ‖h‖_X = ξu_j + ηu_k; the minimum of (ξu_j + ηu_k)²/(ξη) is 4u_j u_k, so any valid constant M obeys v_{j+k} ≤ 4M u_j u_k. Hence

‖Q(h)‖_Y ≤ M‖h‖²_X for all h ⟺ S < ∞, with S/4 ≤ M ≤ S/2.

Gate 5 checks this two-sidedly on four weight pairs by direct maximization over the two-mode optimizer plus random probes.

2. A free 2× improvement on v2. In the unweighted case S = 1, so M = ½: half the Wiener-algebra constant ‖hH(h)‖ ≤ ‖h‖‖H(h)‖ ≤ ‖h‖² that v2 used. It changes no v2 conclusion (its certification budget was identically zero), but it is the sharp constant and it is what the v3 accounting uses.

2. The no-go, on paper

Take positive weights u = (u_k)_{k≥0} on the domain, v = (v_m)_{m≥1} on the codomain.

Requirement II (quadratic). By §1, S = sup_{j,k} v_{j+k}/(u_j u_k) < ∞. Setting k = 0 already gives

v_m ≤ S u_0 u_m        for every m,   i.e.   v_m / u_m ≤ S u_0   UNIFORMLY.

If the constant mode is gauged out of the domain (as the a0 normalization of v2 D3 does, so that only j,k ≥ 1 are available), take k = 1 instead: v_m ≤ S u_1 u_{m−1}, and for any weight that is non-decreasing (higher modes controlled at least as strongly, the natural choice, and the one every implementation makes) this again gives v_m/u_m ≤ S u_1, bounded.

Requirement I (bounded inverse). v2 D6 measured ‖A e_m‖_{ℓ¹} ≈ 1.97 m and derived it from the far-field ODE. In weighted norms the smallest admissible codomain weight is v_m^min(u) = ‖A e_m‖_{ℓ¹_u}, so requirement I is v_m ≳ ‖A e_m‖_{ℓ¹_u}. §3 measures this for weighted u as well, and the answer is the same: v_m ≳ κ · m · u_m with κ ≈ 1.3–3.2 over s ∈ [0,2].

They are incompatible. v_m/u_m cannot be both ≤ S u_0 (bounded) and ≥ κ m (linearly growing). The failure margin grows linearly in the mode number, so no clever tuning at finite m helps. ∎

The proof is almost embarrassingly cheap once (Q) is in hand, and it is the whole point of doing the algebra before the code. The mechanism is worth stating in words: the constant mode (or any fixed low mode) multiplies everything at full strength. e_0 · H(h) = H(h), no decay gained. A codomain strictly stronger than the domain therefore cannot receive the quadratic, however the weights are chosen.

3. The no-go, measured

experiments/p2_route_d_v3_spaces.py runs the theorem as an experiment, over the two-parameter family

u_k = (1+k)^s   (domain),        v_m = (1+m)^t   (codomain),

with s ∈ [0,2], t ∈ [−1.5,3] on a 0.25 grid, the gauge row carrying weight 1 in every pairing (as in v2 D6).

S2: what the inverse costs. The ratio v_m^min/(m u_m), which the theorem needs to be bounded below:

All O(1): the price really is exactly one mode power, for weighted domains as well as flat ones. (The s = 0 column reproduces v2's 1.97 m.)

S3: the map. Both requirements turn out to depend only on the gap g = t − s, and their growth exponents are exact complements:

So ‖A_N‖ ~ N^{1−g} and S_K ~ K^{g}:

The conservation law. A certificate needs both exponents to be zero. Their sum is ≥ 1 at every g, and exactly 1 on 0 ≤ g ≤ 1. The measured minimum over the entire family is 0.98. The one power the far field loses has to be paid by one bound or the other; the choice of weights only decides which.

In (s,t) coordinates: a bounded inverse needs t ≥ s+1, a bounded quadratic needs t ≤ s, and the unit strip between them is empty. Both boundaries are pinned grid-independently by evaluating the exponents on the candidate lines (t = s−1: 1.96; t = s: 0.98; t = s+1: 0.00–0.06; algebra at t = s: 0.00, at t = s+0.25: 0.25, all essentially constant in s).

S4: the control. The v2 D4 ablation, re-run on the space-pair question: replace the transport factor (1 + cos θ) → 1, change nothing else. The inverse boundary moves from t ≥ s+1 to t ≥ s−1, down by exactly two powers. That is the right number: a non-degenerate first-order transport gains one power when inverted, the true one loses one, and 1 + cos θ vanishes to order two at θ = ±π. The regions then overlap on a full unit strip (s−1 ≤ t ≤ s), the minimum exponent sum drops from 0.98 to 0.00, and the overlap is non-empty at 9/9 values of s.

The obstruction is the far-field degeneracy of this specific operator, not the Newton–Kantorovich method and not the ℓ¹ framing. Remove the degeneracy and the same machinery has an admissible space pair immediately.

This is the ablation discipline of v2 D4 doing its job again: a divergence is a suspicion, a divergence that disappears when exactly one feature is removed is an attribution.

4. Why weights were never going to work: a category error

The no-go is sharp, but it should not be read as "the far field is un-invertible". It says something narrower and more useful.

A diagonal weight on Fourier coefficients measures smoothness, not decay. The single mode cos kθ equals (−1)^k at θ = π: it does not decay at X = ∞ at all, for any k, and no weight u_k can see that. Decay in X is a statement about the oscillatory-in-k structure of the coefficient sequence (alternating coefficients concentrate mass at θ = π), i.e. about finitely many linear moment conditions, not about a diagonal weight. Asking a weighted-ℓ¹ pair to express "loses one power of decay" was a category error, and the conservation law of §3 is what the category error looks like when you measure it.

solver/decay_grading.cos_power_coeffs makes this concrete. The canonical element of the class "decays like X^{-α}" is

f_α(X) = (1 + X²)^{-α/2} = |cos(θ/2)|^α ,

whose exact cosine coefficients follow from the generalized binomial series via a stable two-term recursion (a_{k+1} = a_k (α/2 − k)/(α/2 + k + 1); no Γ of a negative argument is ever formed). Gate 1 pins it against α = 0, 2, 4, where the answer is a finite trig polynomial, in particular f_2 = (1 + cos θ)/2 is the anchor's own shape. For non-even α the ratio tends to −1: the coefficients alternate and decay like k^{-α-1}. Alternation is the decay. Combined with H(cos kθ) = sin kθ, this makes H f_α computable to machine precision with no quadrature at all, which is what lets the far field be measured rather than assumed. (Gate 2 checks it against the closed-form pairs (1+X²)^{-1} ↦ X/(1+X²) and (1+X²)^{-2} ↦ X(3+X²)/2(1+X²)², to 1e-16.)

5. The far field is resonant at the anchor's own decay rate

v2 D6 derived the far-field model operator and its "one lost power" from

L h = −c h_X − h/X ,      (X^{1/c} h)' = −(1/c) X^{1/c} g .

Between decay-graded sup norms ‖h‖ = sup X^α|h| and ‖g‖ = sup X^{α+1}|g| that solution operator has norm

‖L^{-1}‖ = 2 / |α − 2|.

solver/decay_grading.farfield_inverse_norm measures this independently: a second-order (trapezoidal, in τ = log X) discretization on [1, 10¹²], with the integration direction chosen by which solution the space admits, and the induced norm obtained in one pass because the discretization is an M-matrix. Agreement with the exact finite-domain value is ≤ 0.008 % across 15 values of α on both sides of the pole. (Gate 7. Comparing instead against the infinite-domain 2/|α−2| shows a 6 % gap at α = 1.9, that is the honest finite-domain effect (X₀/X_max)^{|2−α|}, not a discretization error.)

The pole at α = 2 is not an artifact:

• X^{-1/c} = X^{-2} at c = ½ is the homogeneous far-field solution, and
• X^{-2} is the decay of the anchor Ω₂ = −1/(1+X²).

The space boundary sits exactly on the kernel. And the same resonance is visible from the operator side without any ODE model at all: applying the full linearization to f_α and using Ω₂ ~ −X^{-2}, H(Ω₂) = −X/(1+X²) ~ −1/X,

DF f_α = f_α H(Ω₂) + Ω₂ H(f_α) − c (f_α)_X  ~  ( c α − 1 ) X^{-α-1} + O(X^{-3}),

so lim X^{α+1} DF[f_α] = cα − 1 = (α−2)/2, which vanishes at α = 2. Measured from the exact Fourier coefficients (no quadrature) at X = 3·10³, 10⁴, to ≤ 0.3 %. Two independent routes, same resonance.

v2's "the inverse loses exactly one power" is this resonance seen at integer grading. Generic α loses nothing; α = 2, the anchor's own rate, loses a logarithm, and integer-graded weights round that up to a full power.

Note also that the expansion above requires α < 2: for α > 2 the coupling term Ω₂ H(f_α) ~ X^{-3} dominates X^{-α-1}, so perturbations decaying faster than the anchor leave the natural codomain class. The admissible window is 1 < α < 2: α > 1 for integrability (which the far-field law H(f)(X) ~ (∫f)/(πX) needs), α < 2 from the coupling.

6. The pair that works, and what it costs

In the decay-graded pair

X = { h : |h| ≲ X^{-α} },        Y = { g : |g| ≲ X^{-α-1} },        1 < α < 2,

every term lands where it should:

The quadratic line is the one the whole leg turns on. H(h)(X) → (∫h)/(πX) for any integrable h, so h·H(h) decays one power faster than h: precisely the gain the far-field inverse needs and precisely the gain no diagonal weight can express. Measured directly on f_α:

X^{α+1} f_α H(f_α) → (∫f_α)/π, with ∫f_α = √π Γ((α−1)/2)/Γ(α/2),

verified against that closed form at α = 1.2, 1.5, 1.8 (gate 6; the approach to the limit is O(X^{1−α}), the known second term of the multipole expansion, which is fitted before comparing, and the extrapolated limit matches to < 0.1 %).

The price. Detuning off the resonance to α = 2 − ε buys a bounded far-field inverse at cost 2/ε, while the quadratic constant C_Q(α) = (∫f_α)/π blows up like 2/(π(α−1)) as α → 1. Combining them in the same Z₂ = 2‖A‖M convention v2 used:

There is an interior optimum at α* ≈ 1.44 ((2−α)/∫f_α maximized; α = 3/2 is within 1 % of it and is the natural round choice). The optimum is broad: the budget is within 1 % over α ∈ [1.35, 1.55].

So a Route-D v3 build has a design parameter it did not know it had: certify profiles decaying like X^{-3/2}, with residuals measured in X^{-5/2}. Not X^{-2}, which is the anchor's own rate and the one every naive choice would pick, and which is exactly the rate at which the far-field inverse is unbounded.

This is a scoping estimate and nothing more. It uses leading-order far-field constants only; it contains no compact-core contribution to Z₁ or Z₂; it is plain float64; nothing in it is interval-enclosed. It sizes the next brick. It certifies nothing.

7. What this changes for Route D

Retired. v2 D6's literal recipe: "grade the codomain by one mode power, ‖A‖ = 3.000 flat, that is the pair to use" — is retired. The measurement was right, the inference was not: in that pairZ₂diverges exactly as fast as‖A‖` converges.

Established. The whole diagonal-weight category is closed, with a proof, a 2-parameter numerical sweep, a grid-independent boundary check, and a control that attributes the obstruction causally.

Specified. The replacement is a decay-graded (two-region) pair with an identified optimum α ≈ 3/2 and an estimated Z₂ ≈ 13. That is a much sharper starting point than v2 handed over, but the honest reading of the number is sobering: a Z₂ of order 10 with a Z₁ yet to be paid for leaves a certification budget of order 10^{-2} at best, and the a ≠ 0 residual floor is ~10^{-2}. The margin, if it exists at all, is thin.

Still open (unchanged by this leg). The compact-core block; a rigorous enclosure of the far-field inverse rather than an asymptotic one; the matching between core and far field; and, downstream of all of it, the a ≠ 0 case where no exact anchor exists.

8. Honest ceiling

Everything here is plain float64. Nothing is interval-enclosed and nothing is rigorous. This leg does not climb the rigor ladder: it closes one more route with a reason, retires an inference that looked constructive, and gives the replacement an address and a design parameter. Even the eventual success it scouts would be a computer-assisted toy-model certification in the Chen–Hou / Gómez-Serrano genre, of a profile that at a = 0 is already known in closed form. 1D HL is a toy model of the boundary behaviour of Hou–Luo / 3D axisymmetric Euler. Overall Clay odds remain ~0.05%.

Reproduce

python test_decay_grading.py                                  # 7/7 gates
python experiments/p2_route_d_v3_spaces.py                    # ~1 min, deterministic
python writeup/4_p2_lottery/p2_route_d_v3_evidence.py         # fig21 from committed data