Math-6205 · mathematics · Mathematics

Math-6205

Erdős Problem #1151

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Erdős problems openaigpt 6 astra ultra 5% cataloguedOpenOpen

Statement

Let $L_nf$ be the polynomial of degree at most $n-1$ interpolating a continuous real function $f$ at the $n$ Chebyshev nodes in $[-1,1]$. For a prescribed closed set $A\subseteq[-1,1]$, can one choose $f$ so that the set of subsequential limits of the values $L_nf(x)$ is precisely $A$?

Restatement from Math-6205-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6205-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6205

Erdős Problem 1151 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Let $L_nf$ be the polynomial of degree at most $n-1$ interpolating a continuous real function $f$ at the $n$ Chebyshev nodes in $[-1,1]$. For a prescribed closed set $A\subseteq[-1,1]$, can one choose $f$ so that the set of subsequential limits of the values $L_nf(x)$ is precisely $A$?

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.

The official statement does not specify the quantifier on x. Its commentary expressly asks whether x is fixed and arbitrary and discusses special x=cos(pi*p/q) with odd p,q. No choice of that missing quantifier is asserted here; it must be settled before a proof target is fixed.

ATTACK PLAN AND WORK

Every singleton target is realizable: the constant function $f\equiv a$ has $L_nf(x)=a$ for every $n$ and every $x$, so its cluster set is exactly $\{a\}$. For the standard roots-of-$T_n$ Chebyshev nodes, zero is an interpolation node for every odd $n$. At $x=0$ this gives $L_{2j+1}f(0)=f(0)$, so the cluster set must contain $f(0)$. This is a necessary condition at that point, not a disproof of realizing an arbitrary nonempty closed target, since $f(0)$ can be chosen.

A useful operator obstruction is more general. Put $\Lambda_n(x)=\sum_i|\ell_{i,n}(x)|$. If $\sup_n\Lambda_n(x)=M<\infty$, then interpolation converges to $f(x)$ at this $x$ for every continuous $f$. To see this, approximate $f$ uniformly by a polynomial $q$ within $\delta$. For $n>\deg q$, $L_nq=q$, and \[|L_nf(x)-f(x)|\le(M+1)\delta.\] Letting $\delta\downarrow0$ proves the assertion, using polynomial approximation of continuous functions on a compact interval. Thus a target with several cluster points requires unbounded evaluation norms at the chosen $x$.

The missing quantifier on $x$ is preserved. Neither this argument nor the zero-node observation silently identifies the special evaluation points intended by the source.

VERIFICATION AND REMAINING GAP

For a specified evaluation point and node convention, construct a continuous function with exactly the prescribed nonsingleton cluster set. Unbounded norms alone do not prescribe that cluster set or exclude unwanted subsequential limits.

SOURCES

https://www.erdosproblems.com/1151

https://www.erdosproblems.com/forum/thread/1151?order=newest

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.

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