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Source. Theorem 2, p. 313, of P. Erdős, On divergence properties of the Lagrange interpolation parabolas, Ann. of Math. (2) 42 (1941), 309--315, doi:10.2307/1968999; the edition read is named on the source card.

Statement

The setting is that of Theorem 1: $L_n(f(x_0))$ is the value at $x_0$ of the Lagrange interpolation polynomial of $f$ at the roots of the Chebyshev polynomial $T_n$.

Theorem 2 (p. 313). "If $x_0\ne\cos\frac{p}{q}\pi$, $p\equiv q\equiv1$ (mod 2) then there exists for every continuous $f(x)$ a sequence of integers $n_1<n_2<\cdots$ such that $L_{n_k}(f(x_0))\to f(x_0)$."

As printed the statement is false. The nodes are symmetric about $0$. Take $x_0=\cos\frac{\pi}{3}=\frac12$, which is of the excluded form, and a continuous $f$ from Theorem 1 with $L_n(f(\frac12))\to\infty$. Then $g(x)=f(-x)$ is continuous and $L_n(g(-\frac12))=L_n(f(\frac12))\to\infty$, so no subsequence converges at $-\frac12=\cos\frac{2\pi}{3}$. That point is not of the excluded form, since $\cos\frac{p}{q}\pi=-\frac12$ forces $p/q$ to have even numerator in lowest terms. In the same way every $\cos\frac{p}{q}\pi$ with $q$ odd inside $(-1,1)$ is a point of divergence, so the exceptional set must include these points as well. The introduction (p. 309) also attributes to Erdős and Turán the statement that divergence to infinity holds at no other point than those of the excluded form, citing Ann. of Math. 38 (1937), p. 155, where the paper says it was printed with a misprint; the same reflection applies to that statement.

Proof pointer

Pp. 313--315. The paper first seeks integers $n_k$ with $|T_{n_k}(x_0)|<c_{13}/n_k$, through Lemma 6 (p. 313): as printed, if $x_0\ne p/q$ with $p\equiv q\equiv1\pmod 2$, then $\bigl|x_0-\frac{2r-1}{2n_k}\bigr|<c_{14}/n_k^2$ has infinitely many solutions. Lemma 6 is applied with $x_0$ in the role of the angle of the point divided by $\pi$, and its proof (p. 314) asserts that a rational $x_0$ has the form $\frac{2r-1}{2n_k}$, which fails for a fraction with odd denominator such as $\frac23$; this is where the argument misses the reflected points. Along such $n_k$ the fundamental polynomials other than the one at the nearest node $x_r$ have $\sum_{k\ne r}|l_k(x_0)|=o(1)$, so $l_r(x_0)=1-o(1)$ and $L_{n_k}(f(x_0))\to f(x_0)$ (pp. 314--315).

Read depth

Claims checked: the statement, Lemma 6 and the proof were read on the page images of the print. The counterexample above uses only Theorem 1 and the symmetry of the nodes.

Bears on

• Problem 1151: the problem page reads its Statement at a point $\cos(\pi p/q)$ with $p,q$ odd, where Theorem 2 makes no assertion. At other points Theorem 2 as printed would make $f(x_0)$ a limit point for every continuous $f$; it fails at the points $\cos(\pi p/q)$ inside $(-1,1)$ with $q$ odd and $p$ even.