Erdos 1941 divergence properties lagrange interpolation parabolas

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remark_p315: Records Erdős's closing remark that at every point of (-1,1) some continuous function has Chebyshev-node interpolation polynomials whose arithmetic means tend to infinity.

theorem_1: Erdős's theorem that at a point x_0 = cos(p pi/q) with p and q odd some continuous function has Lagrange interpolation polynomials at the Chebyshev nodes tending to infinity at x_0.

theorem_2: Erdős's Theorem 2 that at points other than cos(p pi/q) with p and q odd the Chebyshev-node interpolation polynomials of every continuous function converge along a subsequence; as printed it fails at -1/2.

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P. Erdős: On divergence properties of the Lagrange interpolation parabolas, Ann. of Math. (2) 42 (1941), 309--315 MR 2,283d; Zentralblatt 24,307. The copy read for this card is the Rényi Institute's Erdős archive scan, which prints the journal header and no copyright or license line; the article's Crossref record (DOI 10.2307/1968999, read 2026-10-02) names no license and its JSTOR page was not read, and the journal's site shows the footer "Copyright © 2026 Annals of Mathematics" and names no license (https://annals.math.princeton.edu/, read 2026-10-02), every other right reserved.

For Lagrange interpolation at the roots of the n-th Chebyshev polynomial T_n, Erdős proves (Theorem 1) that at any point x_0 = cos(ppi/q) with p == q == 1 (mod 2) (the introduction adds a coprimality condition, printed as (p_1,q) = 1) there is a continuous f whose interpolation polynomials satisfy L_n(f(x_0)) -> infinity, and remarks that f can instead be made to force convergence to any prescribed value. Theorem 2 is stated as the complementary statement: if x_0 is not of that form, then for every continuous f some subsequence n_1 < n_2 < ... has L_{n_k}(f(x_0)) -> f(x_0). As printed it is false: the nodes are symmetric about 0, so Theorem 1 applied to f(-x) gives divergence at -cos(ppi/q) = cos((q-p)pi/q), whose numerator is even; for instance |T_n(-1/2)| >= 1/2 for every n, and x_0 = -1/2 = cos(2pi/3), the reflection of cos(pi/3), admits a continuous f with L_n(f(x_0)) -> infinity. By the same reflection every cos(ppi/q) with q odd inside (-1,1) belongs to the exceptional set. The introduction (p. 309) attributes to Erdős and Turán the statement that divergence to infinity "does not hold for any other point", stated with a misprint in Ann. of Math. 38 (1937), p. 155. The proofs rest on six lemmas: Lemma 1 separates distinct Chebyshev nodes of different orders, |x_i^{(m)} - x_j^{(n)}| > 1/m^3 for m >= n (the print states no distinctness hypothesis, but nodes of orders n and 3n can coincide); Lemma 2 bounds |T_n(x_0)| and the distance from x_0 to the nodes below at the exceptional points inside (-1,1); Lemma 3 bounds a sum of the fundamental polynomials |l_k^{(n)}(x_0)| above, Lemma 4 bounds single terms below and Lemma 5 bounds a sum below; and Lemma 6, a Diophantine approximation of x_0 by fractions (2r-1)/(2n_k), produces subsequences along which |T_{n_k}(x_0)| < c_13/n_k. A closing remark states that at every x in (-1,1) there is a continuous f whose arithmetic means (1/n) sum_{m <= n} L_m(f(x_0)) tend to infinity, proved very similarly to Theorem 1. Problem 1151 asks for a proof that at Chebyshev nodes every closed A in [-1,1] is the set of limit points of L^n f(x) for some continuous f; Theorem 1 gives the case of divergence to infinity at the points cos(ppi/q) with p, q odd inside (-1,1), and its remark, stated without proof, the case of convergence to a prescribed value.

Source: https://users.renyi.hu/~p_erdos/1941-02.pdf.

Read status: claims checked for Theorems 1 and 2 and the closing remark (pp. 311--315) against the print; the proofs were followed but not checked line by line. Result pages: Theorem 1 (p. 311), Theorem 2 (p. 313), Remark (p. 315).

Bears on.

• #1151: the problem page reads its Statement at a fixed point cos(pi p/q) with p, q odd, the empty set meaning divergence to infinity. Theorem 1 gives a continuous f with L_n(f(x_0)) -> infinity at every such point inside (-1,1), the case of the empty set (its proof does not cover x_0 = -1, where p/q is an odd integer); the remark on p. 313, stated without proof, concerns convergence to a single given value. The paper treats no other closed set. Theorem 2 makes no assertion at these points.

Results to transcribe.

• Theorem 1 (p. 311): For x_0 = cos(p*pi/q) with p == q == 1 (mod 2) there is a continuous f with L_n(f(x_0)) -> infinity; by a remark stated without proof (p. 313), f can also be arranged so that L_n(f(x_0)) converges to any prescribed value. The proof treats x_0 inside (-1,1); at x_0 = -1 (p/q an odd integer) the first bound of Lemma 2 fails.
• Theorem 2 (p. 313): As printed, if x_0 is not of the form cos(ppi/q) with p == q == 1 (mod 2), then for every continuous f there is a subsequence n_1 < n_2 < ... with L_{n_k}(f(x_0)) -> f(x_0); the statement fails at x_0 = -1/2 = cos(2pi/3), the reflection of cos(pi/3), and by the same reflection every cos(ppi/q) with q odd inside (-1,1) belongs to the exceptional set.
• Lemma 1 (p. 309): As printed, x_i^{(m)} - x_j^{(n)} > 1/m^3 for m >= n; the proof bounds |x_i^{(m)} - x_j^{(n)}| and needs the two nodes to be distinct, since for instance x_j^{(n)} is also a node of order 3n.
• Lemma 6 (p. 313): As printed, if x_0 is not p/q with p == q == 1 (mod 2), the inequality |x_0 - (2r-1)/(2n_k)| < c_14/n_k^2 has infinitely many solutions; it gives integers n_k with |T_{n_k}(x_0)| < c_13/n_k.
• Closing remark (p. 315): For every x_0 in (-1,1) there is a continuous f with lim_{n -> infinity} (1/n) sum_{m <= n} L_m(f(x_0)) = infinity.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.