Erdős Problem #1153

Erdős Problem 1153: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and scope

For distinct interpolation nodes $x_1,\ldots,x_n\in[-1,1]$, let $l_j$ be their Lagrange basis polynomials and $\lambda(x)=\sum_j|l_j(x)|$. The question asks whether every fixed nondegenerate interval $I\subset[-1,1]$ satisfies \[ \max_{x\in I}\lambda(x)\ge(2/\pi-o(1))\log n. \] Tao has proved the stronger lower bound $(2/\pi)\log n-O_I(1)$ for arbitrary nodes; see Local Bernstein theory, and lower bounds for Lebesgue constants, https://arxiv.org/abs/2603.21453, and the earlier proof note https://terrytao.wordpress.com/wp-content/uploads/2026/02/erdos-1153-1.pdf. The present attempt does not reconstruct the general theorem. It proves the sharp local lower bound for Chebyshev nodes and a necessary potential-flatness condition for any hypothetical small-Lebesgue-function configuration. The arbitrary-node problem remains unresolved within this attempt.

A complete sharp calculation for Chebyshev nodes

Take $x_j=\cos\theta_j$, where $\theta_j=(j-\tfrac12)\pi/n$, the roots of $T_n$. Since $T_n(\cos\theta)=\cos(n\theta)$ and $|T_n'(x_j)|=n/\sin\theta_j$, \[ |l_j(\cos\theta)|=\frac{|\cos(n\theta)|\sin\theta_j} {n|\cos\theta-\cos\theta_j|}.\tag{1} \] Choose a fixed $\theta_0\in(0,\pi)$ with $\cos\theta_0$ in the interior of $I$. For every large $n$, choose an integer $i$ with $\theta=i\pi/n$ close to $\theta_0$. Then $\cos\theta\in I$ and $|\cos(n\theta)|=1$. Fix a small constant $c>0$ so that $[\theta-c,\theta+c]$ stays uniformly inside $(0,\pi)$.

Uniformly for $0<|t-\theta|\le c$, \[ \frac{\sin t}{|\cos\theta-\cos t|} =\frac1{|t-\theta|}+O_I(1).\tag{2} \] Indeed the denominator divided by $|t-\theta|$ is the average of $\sin u$ over the interval between $t$ and $\theta$. This average is bounded below, and differs from $\sin t$ by $O(|t-\theta|)$, proving $2$. On each side of $\theta$, the node distances are $(r+\tfrac12)\pi/n$, for $r=0,\ldots,L_\pm-1$, with $L_\pm=cn/\pi+O(1)$. Keeping only these positive terms in $1$ gives \[ \begin{aligned} \lambda(\cos\theta) &\ge\frac1\pi\left( \sum_{r=0}^{L_--1}\frac1{r+1/2}+ \sum_{r=0}^{L_+-1}\frac1{r+1/2}\right)-O_I(1)\\ &=\frac2\pi\log n-O_I(1). \end{aligned}\tag{3} \] The last equality follows by comparing each decreasing harmonic sum with its integral. Thus the requested sharp coefficient holds locally on every fixed interval for this important node family; it is not merely a bound on the full interval $[-1,1]$.

A quantitative obstruction for arbitrary nodes

Let \[ P(x)=\prod_{j=1}^n(x-x_j),\quad A=\left(\sum_{j=1}^n\frac1{|P'(x_j)|}\right)^{-1},\quad d(x)=\min_j|x-x_j|. \] For $x$ not a node, the exact basis formula gives \[ \lambda(x)=|P(x)|\sum_{j=1}^n\frac1{|P'(x_j)|\,|x-x_j|}. \] Since all the distances are at most $2$ and at least $d(x)$, and since $\sum_jl_j(x)=1$, we obtain \[ \frac{|P(x)|}{2A}\le\lambda(x)\le\frac{|P(x)|}{A d(x)}, \qquad\lambda(x)\ge1.\tag{4} \] The identity $\sum l_j=1$ holds because both sides are polynomials of degree at most $n-1$ agreeing at all $n$ nodes.

Suppose that $\lambda(x)\le C\log n$ throughout $I$, where $C$ is fixed. Equations $4$ imply \[ A d(x)\le|P(x)|\le2AC\log n.\tag{5} \] Define the logarithmic potential of the empirical node measure by \[ U(x)=\frac1n\log\frac1{|P(x)|},\qquad a_0=-\frac1n\log A. \] Remove from $I$ the union of the $n$ intervals of radius $n^{-10}$ about the nodes. Their total length is at most $2n^{-9}$. At every remaining point, $5$ gives \[ a_0-\frac{\log(2C\log n)}n \le U(x)\le a_0+\frac{10\log n}{n}.\tag{6} \] The lower bound holds at all nonnodal points of $I$, and at nodes in the extended sense $U=+\infty$. Thus a small Lebesgue function forces the empirical logarithmic potential to be nearly constant outside a quantitatively tiny exceptional set.

Remaining gap

Neither $6$ nor the Chebyshev calculation makes arbitrary node sets locally uniform. A general proof must turn potential control into the local derivative and zero-distribution estimates needed for the sharp harmonic-sum coefficient, including configurations with badly clustered nodes. Those steps in Tao's argument are not proved here. Therefore $3$ and $6$ are proved progress, not a claimed new solution of the full question.

Completion rate estimate: 45%.