s > 3: radialization fails

Hundertmark–Pattakos–Schulz, arXiv:2504.18487v1, Lemma 4.3 and Proposition 4.5. Opened this session: the abs and the HTML, §4 and Lemma 5.9. Replay: python3 s_gt_3.py or ./run_s_gt_3.sh (writes certs/s_gt_3.json).

Result

I_s(ν) goes negative for every tested s > 3 on an explicit family of measures orthogonal to radial functions. There is no certified path to s > 3 along the HPS argument. The function b(4) from (2.7) is

b(4) = 1.08302491750762443520… ∈ (1.08302, 1.08303).

That number is smaller than the printed leading coefficient 1.1185, but it cannot be used in Theorem 2.2: Lemma 4.3 does not extend, so Proposition 4.5 does not give β_4 ≥ b(4)^{-1}.

Status: residue. Not a dent of 1.1185.

What HPS need

Lemma 4.3: I_s(ρ) ≥ I_s(ρ̄) for s ∈ (1, 3], where

I_s(μ) = ∬ (|x|^s + |y|^s) / (2 |x−y|) dμ(x) dμ(y).

The cross term against the radial part vanishes by Newton. The claim reduces to I_s(ν) ≥ 0 for ν = ρ − ρ̄, which is orthogonal to every radial function. IMS plus the improved Hardy constant c_H = d²/4 = 9/4 (Ekholm–Frank 2006, Lemma 2.4; HPS (4.15)) gives

I_s(ν) ≥ (9/4 − s²/4) × (a nonnegative remainder).

The prefactor is nonnegative only for s ≤ 3. Remark 2.3 conjectures that Theorem 2.2 still holds for all s ≥ 1 with the same b(s). Lemma 5.9 already compares α_{N,s} to β_s for 2 ≤ s ≤ 4. The missing piece on (3, 4] is radialization.

Proposition 4.5 itself, if one already has I_s(ρ) ≥ I_s(ρ̄), is not restricted to s ≤ 3: Newton and the elementary minimum of (1 + t^s)/(1 + t^{s−1}) work for every s > 1. The obstruction is only Lemma 4.3.

Two-shell spherical harmonics

Put surface densities α Y_{ℓm} and β Y_{ℓm} on spheres of radii t ∈ (0, 1] and 1. This is orthogonal to every radial test function for ℓ ≥ 1, and it lies in H^{-1}. The Coulomb potential is elementary. With Y_{ℓm} L²-normalised on S²,

I_s(ν) = [4π / (2ℓ+1)] Q_s(α, β; t, ℓ),
Q_s    = α² t^{s−1} + β² + αβ (t^{s+ℓ} + t^ℓ).

Q has a negative eigenvalue if and only if (t^{s+ℓ} + t^ℓ)/2 > t^{(s−1)/2}. For small t this is the comparison of t^ℓ with t^{(s−1)/2}, i.e. s ≷ 2ℓ+1. That is exactly the Hardy threshold for angular momentum ℓ: ∫ |∇u|² ≥ (ℓ(ℓ+1) + 1/4) ∫ u²/|x|², so s ≤ 2√(ℓ(ℓ+1)+1/4) = 2ℓ+1.

Dipole ℓ = 1: threshold s = 3. Quadrupole ℓ = 2: threshold s = 5, so two-shell quadrupoles stay nonnegative on (3, 4].

At s = 3, ℓ = 1 the discriminant is an identity,

(t⁴ + t)² − 4 t² = t² (t³ − 1)(t³ + 3) ≤ 0    for t ∈ (0, 1],

so Q ≥ 0 for every real (α, β). Interval arithmetic on the examples in the certificate agrees.

Certified negatives

Opposite-sign, inner-heavy dipoles:

The s = 4 line is a dyadic rational:

16² · (1/8)³ + 1 − 16 · ((1/8)⁵ + 1/8)
  = 1/2 + 1 − 2 − 1/2048
  = −1025/2048 < 0.

The s = 7/2 line is negative because 1/√2 − 1 < 0. Both sit in certs/s_gt_3.json as interval enclosures and as these closed forms. A Fibonacci-sphere pairwise sum at s = 4, omitting the same-shell diagonal (a positive self-energy), is already negative, so the sign does not come from a multipole-algebra slip.

Thin radial Gaussians around the same two radii, same weights, give the same signs as volume densities: I_4 < 0, I_{3.5} < 0, I_3 > 0. So the surface-measure idealisation is not the source of the negativity.

The listed one-scale examples stay positive

These are the measures named in the brief. All are orthogonal to radial functions. None of them makes I_s negative for s ∈ {3, 3.1, 3.5, 4}.

• Pure dipole, R(r) = r e^{-r} times Y_{10} (one-sign radial profile). I_s > 0.
• Pure quadrupole, R(r) = r² e^{-r} times Y_{20}. I_s > 0.
• One-shell Y_{ℓm}, any s. Here |x|^s is constant, so I_s = R^s I_0 and Coulomb is positive definite.
• Two opposite Gaussians at ±μ e_z (regularised opposite Diracs, same radius). I_s > 0 on the three widths that were integrated.
• Same-sign two-shell dipole. The quadratic form has positive coefficients on the diagonal and the cross term; Q > 0.
• Two-shell quadrupole, opposite weights, s ≤ 4 < 5. Q > 0.

A generic sign-changing radial profile is not enough. The negative mode wants a large inner dipole and a small opposite outer dipole, with radius ratio past t_* ∼ (1/4)^{1/(s−3)}. For s = 3.1 that is t ≲ 10^{-6}. For s = 4 it is t < 1/4. The two-exponential e^{-r} − e^{-r/8} does not put enough weight on the inner scale and stays positive.

Hardy for Coulomb potentials

The informal hope is that V = (−Δ)^{-1} ν is not the Hardy optimiser, so the constant in front of ∫ V²/|x|² might exceed 9/4 and buy an s = 3+ε.

Hardy on V itself is the wrong object. IMS applies Hardy to u = |x|^{s/2} V. For every two-shell dipole in the table, Q_H(V) = ∫|∇V|² / ∫ V²/|x|² sits at 4.5 or a few parts in 10^2 above it (exactly 9/2 for a single shell). If one mistakenly used c_H = 9/2 on V, the threshold would look like s ≤ √18 ≈ 4.24. That is not a proof: the identity is not Hardy on V, and the closed form already gives I_{3.1} < 0.

The unregularized remainder ∫ |x|^{s−2} V² diverges at infinity for a dipole tail V ∼ 1/r² as soon as s ≥ 3. HPS know this — they replace |x|^s by |x|^s/(1+ε|x|^s) before IMS. After that cutoff the only uniform constant they have is 9/4, and the examples show it is sharp in the dipole sector: for every s > 3 there is a ν with I_s(ν) < 0. There is no uniform s = 3+ε.

A quadrupole-only restriction would allow s ≤ 5, but a general ν ⊥ radial has an ℓ = 1 piece.

β_s without radialization

A lower bound on β_s is what Theorem 2.2 consumes. A search is an upper bound.

The pointwise minorant (|x|^s+|y|^s)/(2|x−y|(|x|^{s−1}+|y|^{s−1})) equals (1+t^s)/(1+t^{s−1}) · max(|x|,|y|) / (2|x−y|). The geometry factor is at least 1/2, so

β_s ≥ (1/2) b(s)^{-1}.

For s = 4 that is β_4 ≥ 0.46166…, hence a leading coefficient at most 2.17, worse than Lieb. The |x−y| in the denominator is why a pointwise bound does not beat 1.1185.

Trials (upper bounds only, certs/s_gt_3.json):

• Two shells with angular modulation 1 ± ε cos θ, |ε| ≤ 1 so the measure stays a probability: β_{3.5} ≤ 0.969, β_4 ≤ 0.972. Both sit above b(s)^{-1}.
• Nam-like radial r^{-p} on [1, n], Newton: β_{3.5}^{rad} ≤ 0.937, β_4^{rad} ≤ 0.950. Still above b(s)^{-1}, as they must (β_s^{rad} ≥ b(s)^{-1} for every s, no radialization required).

None of these is a lower bound. None of them shows β_s < b(s)^{-1}. The negative I_s(ν) does show that a probability ρ = ρ̄ + εν with a small opposite-dipole modulation has I_s(ρ) < I_s(ρ̄), so the radial infimum is not the full infimum. Whether the drop crosses b(s)^{-1} is open. Crossing it would make the Remark 2.3 coefficient strictly too small for this proof; it would not, by itself, produce a new ionization upper bound.

b(s)

b(s) = max_{t∈[0,1]} (1 + t^{s−1}) / (1 + t^s) = (s−1) / (s t_0),
t_0^s + s t_0 + 1 − s = 0.

b(3) matches the q1 replay of (2.9). b(4) matches a 20 000-point grid maximum of (1+t³)/(1+t⁴) to 10^{-8}, and the root of t⁴ + 4t − 3 = 0 is isolated in a tiny interval before the quotient is evaluated.

What is proved vs still open

Proved here: Lemma 4.3 does not extend past s = 3. Explicit ν ⊥ radial have I_s(ν) < 0 for every s ∈ {3.1, 3.5, 4}, with a rational certificate at s = 4. The improved-Hardy gap cannot be opened uniformly for Coulomb potentials of such ν.

Not proved: a lower bound on β_s for s > 3 other than the useless (1/2) b(s)^{-1}. Not proved: β_s < b(s)^{-1}. Not a dent of the printed 1.1185. Remark 2.3 remains a conjecture. Lemma 5.9 on (3, 4] has nothing to feed into if β_s is not bounded from below by b(s)^{-1}.