GPT 6 Astra Ultra attempt on Math-6183
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Erdős Problem 1129: optimal nodes and the endpoint convention
Statement.
For distinct ordered nodes $-1\leq x_1<\cdots<x_n\leq1$, let \[ l_k(x)=\prod_{j\ne k}\frac{x-x_j}{x_k-x_j}, \qquad L_{\boldsymbol x}(x)=\sum_k|l_k(x)|, \qquad\Lambda(\boldsymbol x)=\max_{[-1,1]}L_{\boldsymbol x}. \] We describe all minimizers, relative to the established de Boor--Pinkus theorem for fixed endpoints. The distinction between fixed and movable endpoints matters: the literal formulation at https://www.erdosproblems.com/1129 does not have a unique minimizer when $n\geq3$.Explicit external theorem.
For $n\geq3$, among canonical arrays $-1=t_1<\cdots<t_n=1$, there is a unique array $\boldsymbol t^*$ whose $n-1$ internal interval maxima of $L_{\boldsymbol t^*}$ all equal a constant $\Lambda_n^*$. For every other canonical array, the least internal maximum is strictly below $\Lambda_n^*$ and the greatest is strictly above it. This is the consequence of Theorems 1 and 2 of C. de Boor and A. Pinkus, Proof of the conjectures of Bernstein and Erdős concerning the optimal nodes for polynomial interpolation, J. Approximation Theory 24 $1978$, 289--303, https://doi.org/10.1016/0021-9045(78)90014-X. Their paper fixes both endpoints; its indexing uses $n+1$ nodes. This deep theorem is imported, not reproved here. Reflection and uniqueness show that $\boldsymbol t^*$ is symmetric.All minimizers with movable endpoints.
Write $L^*=L_{\boldsymbol t^*}$. Then the answer to the literal problem is precisely \[ \boxed{\quad x_i=c+h t_i^*,\quad h>0,\quad -1\leq c-h<c+h\leq1,\quad L^*((-1-c)/h)\leq\Lambda_n^*,\ L^*((1-c)/h)\leq\Lambda_n^*.\quad} \] To prove necessity, normalize an arbitrary array by \[ c=(x_1+x_n)/2,\qquad h=(x_n-x_1)/2,\qquad t_i=(x_i-c)/h. \] The Lagrange product formula gives $L_{\boldsymbol x}(c+hu)=L_{\boldsymbol t}(u)$. The global maximum is at least the largest internal interval maximum, hence at least $\Lambda_n^*$. The canonical optimal array attains that lower bound. If any other array attains it, the strict part of the external theorem forces $\boldsymbol t=\boldsymbol t^*$, and the two displayed endpoint inequalities are necessary.They are also sufficient. For $u>1$, each $|l_k^*(u)|$ is a product of positive increasing factors divided by a positive constant, so $L^*(u)$ is increasing. For $u<-1$ it increases as $u$ moves left. Thus the exterior interval maxima occur at the domain endpoints $-1,1$, while every internal maximum equals $\Lambda_n^*$.
Equivalently, let $r_n>1$ be the unique solution $L^*(r_n)=\Lambda_n^*$. It exists since $L^*(1)=1<\Lambda_n^*$, exterior growth is strict, and $L^*(u)\to\infty$. Symmetry gives the alternative constraints \[ (1+|c|)/r_n\leq h\leq1-|c|. \] This makes the nonuniqueness explicit.
A fully calculated example.
For three canonical nodes $-1,u,1$, the two internal maxima are \[ 1+\frac{(1+u)^2}{4(1-u)},\qquad 1+\frac{(1-u)^2}{4(1+u)}. \] If $u\geq0$, the first is at least $5/4$; if $u\leq0$, the second is at least $5/4$. Equality of their maximum to $5/4$ forces $u=0$. For nodes $-t,0,t$, direct evaluation outside their span gives \[ \Lambda(-t,0,t)=\max\{5/4,\ 2/t^2-1\}. \] Thus every $\sqrt{8/9}\leq t\leq1$ is a minimizer. Only the first of these has both exterior maxima equal to the internal ones. Unqualified uniqueness or equality of all $n+1$ interval maxima therefore cannot characterize the literal problem.For $n=2$, the minimum $1$ is attained only at $-1,1$, since the Lebesgue function is $1$ between the two nodes and strictly larger outside. For $n=1$, it is identically $1$, so every node is optimal.