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Source. Theorem 2, p. 1171, and the remark after it, p. 1172, of P. Erdős, "Some remarks on polynomials," Bull. Amer. Math. Soc. 53 (1947), 1169-1176. Pages are the journal's own, as on the source card.

Setting

Page 1171. For nodes $x_1,\ldots,x_n$ in $[-1,1]$ the paper puts $\omega(x)=\prod_{i=1}^n(x-x_i)$ and $l_k(x)=\omega(x)/\bigl(\omega'(x_k)(x-x_k)\bigr)$, the fundamental functions of Lagrange interpolation (the print writes $\omega(x_k)$ in the denominator, where the derivative is meant). It writes the nodes there as $-1=x_0<x_1\le\cdots\le x_n=x_{n+1}=1$; Theorem 2 itself uses the ordering given below.

The paper records, as context (p. 1171), that the problem of finding the nodes for which $\max_{-1\le x\le1}\sum_{k=1}^n\lvert l_k(x)\rvert$ is minimal is unsolved, and that it has been conjectured, but never proved, that the minimum is attained when the $n+1$ sums

$$\max_{x_i\le x\le x_{i+1}}\sum_{k=1}^n\lvert l_k(x)\rvert, \qquad i=0,1,\ldots,n, \tag{4}$$

are all equal. For the roots of the Chebyshev polynomial $T_n$ each sum (4) equals $\tfrac{2}{\pi}\log n+O(1)$. It cites S. Bernstein's lower bound $(1+o(1))\tfrac{2}{\pi}\log n$ for the maximum over $[-1,1]$, for any nodes, and states the author's own unpublished bound $\tfrac{2}{\pi}\log n-c$ with $c$ an absolute constant.

Statement

Theorem 2 (p. 1171). Let $-1=x_0\le x_1\le\cdots\le x_n\le x_{n+1}=1$. Then for some $i$

$$\max_{x_i<x<x_{i+1}}\sum_{k=1}^n\lvert l_k(x)\rvert<n^{1/2}. \tag{5}$$

Remark (p. 1172). The paper says that $n^{1/2}$ in (5) can very likely be improved to $c\log n$, and that it is likely that

$$\min_{0\le i\le n}\ \max_{x_i\le x\le x_{i+1}}\sum_{k=1}^n\lvert l_k(x)\rvert$$

assumes its maximum, over the choice of nodes, when all the sums (4) are equal. (The print writes the range of the minimum as $0\le x_i\le n$.)

Read depth. Claims checked: the statement, the remark and the context above were read clause by clause on the print, and the short proof on p. 1172 was read.

Proof pointer

Page 1172. If two consecutive points coincide, (5) is immediate. Otherwise the equation $\sum_{k=1}^n l_k^2(x)=1$ has at most $2n-2$ solutions, and the $n$ nodes $x_1,\ldots,x_n$ are among them, so for some $i$ with $1\le i\le n-1$ the sum of squares is below $1$ throughout $x_i<x<x_{i+1}$. The Cauchy-Schwarz inequality then gives $\sum_k\lvert l_k(x)\rvert<n^{1/2}$ there.

Bears on

• Problem 1130: the source of the question. The remark after Theorem 2 (p. 1172) conjectures the bound $c\log n$ for the least of the interval maxima and that the equal-sums nodes maximize that least value, which are the problem's two questions. Theorem 2 proves only the bound $n^{1/2}$. The paper does not settle the problem.
• Problem 1129: background. The paper states (p. 1171) that the problem of the nodes minimizing the maximum of $\sum_k\lvert l_k(x)\rvert$ over $[-1,1]$ is unsolved, records the conjecture that the equal-sums nodes are the minimizers, and quotes the logarithmic bounds above. It does not settle the problem.