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Source. P. Erdős, Problems and results on the theory of interpolation. II, Acta Math. Acad. Sci. Hungar. 12 (1961), 235--244 (source card): the notation and Theorem 1 on p. 235, the Chebyshev comparison on p. 236, Lemmas 1--6 on pp. 236--240, and the proof of the theorem on pp. 240--242.

Read depth. Claims checked: the notation, the statement and the strengthened form (18) were read clause by clause on the page images. The proof was read but not checked step by step, and nothing here is independently reviewed.

Statement

Notation (p. 235): −1≤x1<x2<⋯<xn≤1-1\le x_1<x_2<\cdots<x_n\le1 are nn points, ωn(x)=∏i=1n(x−xi)\omega_n(x)=\prod_{i=1}^n(x-x_i) and lk(x)=ωn(x)/(ωn′(xk)(x−xk))l_k(x)=\omega_n(x)/\bigl(\omega_n'(x_k)(x-x_k)\bigr), the fundamental polynomials of Lagrange interpolation at these points. Throughout the paper c,c1,c2,…c,c_1,c_2,\ldots denote positive absolute constants (p. 235).

Theorem 1 (p. 235, quoted). "Let −1≦x1<x2<⋯<xn≦1-1\leqq x_1<x_2<\cdots<x_n\leqq1. Then

max⁡−1≦x≦1∑k=1n∣lk(x)∣>2πlog⁡n−c1."\max_{-1\leqq x\leqq1}\sum_{k=1}^n|l_k(x)|>\frac2\pi\log n-c_1."

So the constant c1c_1 depends neither on nn nor on the nodes.

The form the proof gives (pp. 240--242). The paper proves more: if x0x_0 is the point of (−1,+1)(-1,+1) at which ∣ωn(x)∣|\omega_n(x)| attains its maximum there (p. 237), then ∑k=1n∣lk(x0)∣>2πlog⁡n−c1\sum_{k=1}^n|l_k(x_0)|>\frac2\pi\log n-c_1 for a sufficiently large absolute c1c_1 (the paper's (18), p. 240). The point x0x_0 depends on nn and on the nodes.

Context

The paper places the theorem after Faber's bound 112log⁡n\frac1{12}\log n (its (1), p. 235), Bernstein's assertion of (1−ε)2πlog⁡n(1-\varepsilon)\frac2\pi\log n for n>n0n>n_0 (its (2), p. 235), whose proof for algebraic interpolation Erdős writes he could not reconstruct, and the bound 2πlog⁡n−clog⁡log⁡n\frac2\pi\log n-c\log\log n of Erdős and Turán in the same volume (their card). On p. 236 the paper notes that the result cannot be improved much: for the roots of the nnth Chebyshev polynomial TnT_n the maximum is less than 2πlog⁡n+c2\frac2\pi\log n+c_2, and the maximum over each gap between consecutive Chebyshev roots lies between 2πlog⁡n−c2\frac2\pi\log n-c_2 and 2πlog⁡n+c2\frac2\pi\log n+c_2, which the paper calls known. So the coefficient 2π\frac2\pi is sharp and the loss is at most a constant.

Proof pointer

Pp. 236--242. The proof works at the maximum point x0=cos⁡ϑ0x_0=\cos\vartheta_0 of ∣ωn∣|\omega_n| on (−1,+1)(-1,+1). Normalizing ωn(x0)=1\omega_n(x_0)=1, Bernstein's inequality bounds ∣ωn′(xk)∣|\omega_n'(x_k)| (the paper's (19)), which reduces the sum at x0x_0 to a lower bound for 1n∑k∣(1−xk2)1/2/(x0−xk)∣\frac1n\sum_k|(1-x_k^2)^{1/2}/(x_0-x_k)| (its (20), p. 241). Lemma 1 (p. 236; its proof is left to the reader, p. 237) gives the required size of the corresponding sum over Chebyshev roots. The nodes are then counted in the intervals ItI_t, the images under x=cos⁡ϑx=\cos\vartheta of intervals of length tπ/nt\pi/n with one endpoint at ϑ0\vartheta_0 (p. 237). If for every large tt every ItI_t holds more than t(1−(log⁡t)−2)t\bigl(1-(\log t)^{-2}\bigr) nodes, Lemma 2 (p. 237) gives the bound; if some ItI_t holds more than t3t^3 nodes, Lemma 3 (p. 237) does. Otherwise Lemma 6 (pp. 238--240, which the paper calls the most difficult part), built on a known polynomial estimate (Lemma 4, p. 238) and M. Riesz's theorem on the distance from a maximum point to a root (Lemma 5, p. 238), produces one fundamental polynomial large enough at x0x_0 to make up the loss (pp. 241--242).

Bears on

  • Problem 1153: the theorem is the case a=−1a=-1, b=1b=1 of the question, with the loss o(1)log⁡no(1)\log n in the stronger form of an absolute constant; it says nothing about a shorter fixed interval [a,b][a,b]. The problem's claim page for this paper, Erdős 1961, records it as an accepted partial claim.
  • Problem 1129: the theorem bounds the minimal Lebesgue constant below by 2πlog⁡n−c1\frac2\pi\log n-c_1, and with the Chebyshev bound of p. 236 fixes its size up to an additive constant. It does not describe the minimizing nodes, which the problem asks for.
  • Problem 1132: the form (18) gives, for each nn, a point x0x_0 of (−1,1)(-1,1) with Ln(x0)>2πlog⁡n−c1L_n(x_0)>\frac2\pi\log n-c_1, but x0x_0 moves with nn; the problem's first question asks for one point that works for infinitely many nn, and the theorem answers neither question.