GPT 6 Astra Ultra attempt on Math-6206

Erdős Problem 1152 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

For each $n\geq1$, choose $n$ distinct interpolation nodes $x_{1n},\ldots,x_{nn}$ in $[-1,1]$, and let $\epsilon(n)\to0$. Must there be a continuous $f:[-1,1]\to\mathbb R$ such that every polynomial sequence satisfying \[ \deg p_n<(1+\epsilon(n))n,\qquad p_n(x_{kn})=f(x_{kn})\quad(1\leq k\leq n) \] fails to converge to $f(x)$ for almost every $x\in[-1,1]$?

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

ATTACK PLAN AND WORK

There is an exact algebraic description of the extra freedom. Set $w_n(x)=\prod_{j=1}^n(x-x_{jn})$ and let $L_nf$ be the degree-at-most-$n-1$ interpolant. Every interpolating polynomial has the unique form \[p_n=L_nf+w_nq_n.\tag{1}\] Indeed its difference from $L_nf$ vanishes at $n$ different points and is therefore divisible by $w_n$; polynomial division gives uniqueness. For $q_n\ne0$, the two summands in $1$ have different degree ranges, so there is no leading-term cancellation and \[\deg p_n=n+\deg q_n<(1+\epsilon(n))n \quad\Longleftrightarrow\quad \deg q_n<\epsilon(n)n.\tag{2}\] For nonnegative $\epsilon(n)$ this leaves $\lceil\epsilon(n)n\rceil=o(n)$ free coefficients, with zero free coefficients when $\epsilon(n)=0$. If a negative degree allowance excludes $L_nf$ itself, existence of admissible interpolants must first be checked; a vacuous universal statement proves no interpolation phenomenon.

The desired result therefore requires one $f$ whose interpolation errors cannot be repaired almost everywhere by any of the low-degree multipliers in $1$. Simply showing divergence of the unique degree-$n-1$ interpolants only handles $q_n=0$, and does not address the quantifier over all correction sequences.

VERIFICATION AND REMAINING GAP

Construct one continuous function defeating every permitted multiplier sequence. The dimension count $o(n)$ is not a measure-theoretic obstruction by itself and supplies no almost-everywhere divergence conclusion.

SOURCES

https://www.erdosproblems.com/1152

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$5\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.