Erdős Problem #1132

Erdős Problem 1132 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1132

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1132.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1132.tex.

FORMAL RESTATEMENT

Let $(x_k)_{k\ge 1}$ be an infinite sequence of real numbers in $[-1,1]$ such that for each $n$ the initial segment $x_1,\dots,x_n$ consists of distinct points (otherwise the Lagrange denominators vanish). For each $n\ge 1$, define $l_k^{(n)}(x)$ (for $1\le k\le n$) to be the Lagrange fundamental polynomial associated to the nodes $x_1,\dots,x_n$: \[ l_k^{(n)}(x)=\frac{\prod_{1\le i\le n,\ i\ne k}(x-x_i)}{\prod_{1\le i\le n,\ i\ne k}(x_k-x_i)}. \] Define \[ L_n(x):=\sum_{k=1}^n \left|l_k^{(n)}(x)\right|. \] Questions:

1. Must there exist $x\in(-1,1)$ such that $L_n(x) > \tfrac{2}{\pi}\log n - O(1)$ for infinitely many $n$?
2. Is it true that for almost all $x\in(-1,1)$, \[\limsup_{n\to\infty} \frac{L_n(x)}{\log n}\ge \frac{2}{\pi}?\]

QUICK LITERATURE/CONTEXT CHECK

I only record context explicitly stated in the problem file. The file states that for each fixed set of nodes $x_1,\dots,x_n$ one has \[ \max_{x\in[-1,1]} L_n(x) > \frac{2}{\pi}\log n - O(1) \] (Erdős), and that Bernstein implies the set of $x$ with $\limsup L_n(x)/\log n\ge 2/\pi$ is everywhere dense.

ATTACK PLAN

Proof track ideas.

• Use compactness of $[-1,1]$ and continuity of $L_n$ to attempt a diagonal argument extracting a single $x$ witnessing large values infinitely often; the main obstacle is that $L_n$ changes with $n$.
• For special classes of sequences $(x_k)$, attempt to show $L_n(x)$ blows up (e.g. for clustered nodes) or has controlled growth (e.g. for well-separated nodes).

Disproof track ideas.

• Try to build a "good" infinite sequence $x_k$ such that for each fixed $x$ the growth of $L_n(x)$ is smaller than $(2/\pi)\log n$ along all large $n$.

WORK

Lemma 1132.1 (universal pointwise lower bound). For any $n\ge 1$, any distinct nodes $x_1,\dots,x_n$, and any $x\in[-1,1]$, \[ L_n(x)=\sum_{k=1}^n |l_k^{(n)}(x)|\ge 1. \]

Proof. For fixed $n$, the Lagrange polynomials satisfy the partition of unity \[ \sum_{k=1}^n l_k^{(n)}(x)=1 \] for all $x$, by the same argument as Lemma 1129.1. Applying the triangle inequality yields \[ 1=\left|\sum_{k=1}^n l_k^{(n)}(x)\right|\le \sum_{k=1}^n |l_k^{(n)}(x)|=L_n(x). \] $\square$

Lemma 1132.2 (values at nodes). For each $n\ge 1$ and each $1\le j\le n$, one has \[ L_n(x_j)=1. \]

Proof. By definition of the Lagrange basis, $l_k^{(n)}(x_j)=\delta_{kj}$. Therefore \[ L_n(x_j)=\sum_{k=1}^n |\delta_{kj}|=1. \square \]

Lemma 1132.3 (continuity and attainment of maxima). For fixed $n$ and fixed distinct nodes $x_1,\dots,x_n$, the function $x\mapsto L_n(x)$ is continuous on $[-1,1]$ and hence attains a maximum on $[-1,1]$.

Proof. Each $l_k^{(n)}(x)$ is a polynomial, hence continuous. The absolute value of a continuous function is continuous, and finite sums of continuous functions are continuous. Thus $L_n$ is continuous on the compact set $[-1,1]$, so it attains its maximum by the extreme value theorem. $\square$

FAST REALITY CHECK (a concrete infinite sequence). Take the explicit (clustered) sequence $x_k=1-2^{-k}$. For this sequence I computed the exact values of $L_n(0)$ for $2\le n\le 10$.

Sequence x_k = 1-2^{-k}
  n=2:  L_n(0) = 5
  n=3:  L_n(0) = 29
  n=4:  L_n(0) = 269
  n=5:  L_n(0) = 4589
  n=6:  L_n(0) = 151469
  n=7:  L_n(0) = 9845549
  n=8:  L_n(0) = 1270075949
  n=9:  L_n(0) = 326409519149
  n=10: L_n(0) = 167448083323949

This shows that for some sequences, $L_n(x)$ can grow extremely quickly at a fixed interior point $x=0$.

VERIFICATION

• Lemmas 1132.1--1132.2 rely only on the basic Lagrange interpolation identities and are correct whenever the nodes are distinct.
• The exact $L_n(0)$ values for $x_k=1-2^{-k}$ were computed using rational arithmetic (Fractions), so the integers listed are exact.
• The example does not address the problem's lower-bound order $(2/\pi)\log n$ in a meaningful way (growth here is much faster), but it sanity-checks that no pointwise upper bound of order $\log n$ can hold without additional assumptions on $(x_k)$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. For each $n$ and node set, $L_n(x)\ge 1$ for all $x$ and $L_n(x_j)=1$ at each node (Lemmas 1132.1--1132.2). For a concrete clustered sequence $x_k=1-2^{-k}$, one can have $L_n(0)$ extremely large even for moderate $n$ (exact values listed above).

(ii) First gap (crisp). Prove or disprove that for every infinite sequence $(x_k)$ there exists a single point $x\in(-1,1)$ such that $L_n(x)\ge \tfrac{2}{\pi}\log n - O(1)$ for infinitely many $n$.

(iii) Top 3 next moves.

• Attempt a compactness/diagonal lemma: if $\max_{x}L_n(x)$ is large for each $n$, prove existence of a point $x$ where $L_n(x)$ is frequently large by controlling the modulus of continuity of $L_n$.
• Look for an "energy" inequality relating averages of $L_n(x)$ over $x$ (or $\int L_n(x)\,dx$) to $\max_x L_n(x)$, to transfer large maxima into a large limsup set.
• Explore candidate counterexamples: sequences $(x_k)$ designed so that the locations of maxima of $L_n$ move around rapidly with $n$, potentially preventing a fixed $x$ from seeing large values infinitely often.

(iv) Minimal counterexample structure. A counterexample would be a sequence $(x_k)$ such that for every fixed $x\in(-1,1)$, $L_n(x)$ stays below $(2/\pi-\delta)\log n$ (up to $O(1)$) for all sufficiently large $n$, while still allowing $\max_x L_n(x)$ to be large for each $n$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

FORMAL RESTATEMENT

Let $(x_k)_{k\ge 1}$ be an infinite sequence of real numbers in $[-1,1]$ such that for each $n$ the initial segment $x_1,\dots,x_n$ consists of distinct points (otherwise the Lagrange denominators vanish). For each $n\ge 1$, define $l_k^{(n)}(x)$ (for $1\le k\le n$) to be the Lagrange fundamental polynomial associated to the nodes $x_1,\dots,x_n$: \[ l_k^{(n)}(x)=\frac{\prod_{1\le i\le n,\ i\ne k}(x-x_i)}{\prod_{1\le i\le n,\ i\ne k}(x_k-x_i)}. \] Define \[ L_n(x):=\sum_{k=1}^n \left|l_k^{(n)}(x)\right|. \] Questions:

1. Must there exist $x\in(-1,1)$ such that $L_n(x) > \tfrac{2}{\pi}\log n - O(1)$ for infinitely many $n$?
2. Is it true that for almost all $x\in(-1,1)$, \[\limsup_{n\to\infty} \frac{L_n(x)}{\log n}\ge \frac{2}{\pi}?\]

QUICK LITERATURE/CONTEXT CHECK

I only record context explicitly stated in the problem file. The file states that for each fixed set of nodes $x_1,\dots,x_n$ one has \[ \max_{x\in[-1,1]} L_n(x) > \frac{2}{\pi}\log n - O(1) \] (Erdős), and that Bernstein implies the set of $x$ with $\limsup L_n(x)/\log n\ge 2/\pi$ is everywhere dense.

ATTACK PLAN

Proof track ideas.

• Use compactness of $[-1,1]$ and continuity of $L_n$ to attempt a diagonal argument extracting a single $x$ witnessing large values infinitely often; the main obstacle is that $L_n$ changes with $n$.
• For special classes of sequences $(x_k)$, attempt to show $L_n(x)$ blows up (e.g. for clustered nodes) or has controlled growth (e.g. for well-separated nodes).

Disproof track ideas.

• Try to build a "good" infinite sequence $x_k$ such that for each fixed $x$ the growth of $L_n(x)$ is smaller than $(2/\pi)\log n$ along all large $n$.

WORK

Lemma 1132.1 (universal pointwise lower bound). For any $n\ge 1$, any distinct nodes $x_1,\dots,x_n$, and any $x\in[-1,1]$, \[ L_n(x)=\sum_{k=1}^n |l_k^{(n)}(x)|\ge 1. \]

Proof. For fixed $n$, the Lagrange polynomials satisfy the partition of unity \[ \sum_{k=1}^n l_k^{(n)}(x)=1 \] for all $x$, by the same argument as Lemma 1129.1. Applying the triangle inequality yields \[ 1=\left|\sum_{k=1}^n l_k^{(n)}(x)\right|\le \sum_{k=1}^n |l_k^{(n)}(x)|=L_n(x). \] $\square$

Lemma 1132.2 (values at nodes). For each $n\ge 1$ and each $1\le j\le n$, one has \[ L_n(x_j)=1. \]

Proof. By definition of the Lagrange basis, $l_k^{(n)}(x_j)=\delta_{kj}$. Therefore \[ L_n(x_j)=\sum_{k=1}^n |\delta_{kj}|=1. \square \]

Lemma 1132.3 (continuity and attainment of maxima). For fixed $n$ and fixed distinct nodes $x_1,\dots,x_n$, the function $x\mapsto L_n(x)$ is continuous on $[-1,1]$ and hence attains a maximum on $[-1,1]$.

Proof. Each $l_k^{(n)}(x)$ is a polynomial, hence continuous. The absolute value of a continuous function is continuous, and finite sums of continuous functions are continuous. Thus $L_n$ is continuous on the compact set $[-1,1]$, so it attains its maximum by the extreme value theorem. $\square$

FAST REALITY CHECK (a concrete infinite sequence). Take the explicit (clustered) sequence $x_k=1-2^{-k}$. For this sequence I computed the exact values of $L_n(0)$ for $2\le n\le 10$.

Sequence x_k = 1-2^{-k}
  n=2:  L_n(0) = 5
  n=3:  L_n(0) = 29
  n=4:  L_n(0) = 269
  n=5:  L_n(0) = 4589
  n=6:  L_n(0) = 151469
  n=7:  L_n(0) = 9845549
  n=8:  L_n(0) = 1270075949
  n=9:  L_n(0) = 326409519149
  n=10: L_n(0) = 167448083323949

This shows that for some sequences, $L_n(x)$ can grow extremely quickly at a fixed interior point $x=0$.

VERIFICATION

• Lemmas 1132.1--1132.2 rely only on the basic Lagrange interpolation identities and are correct whenever the nodes are distinct.
• The exact $L_n(0)$ values for $x_k=1-2^{-k}$ were computed using rational arithmetic (Fractions), so the integers listed are exact.
• The example does not address the problem's lower-bound order $(2/\pi)\log n$ in a meaningful way (growth here is much faster), but it sanity-checks that no pointwise upper bound of order $\log n$ can hold without additional assumptions on $(x_k)$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. For each $n$ and node set, $L_n(x)\ge 1$ for all $x$ and $L_n(x_j)=1$ at each node (Lemmas 1132.1--1132.2). For a concrete clustered sequence $x_k=1-2^{-k}$, one can have $L_n(0)$ extremely large even for moderate $n$ (exact values listed above).

(ii) First gap (crisp). Prove or disprove that for every infinite sequence $(x_k)$ there exists a single point $x\in(-1,1)$ such that $L_n(x)\ge \tfrac{2}{\pi}\log n - O(1)$ for infinitely many $n$.

(iii) Top 3 next moves.

• Attempt a compactness/diagonal lemma: if $\max_{x}L_n(x)$ is large for each $n$, prove existence of a point $x$ where $L_n(x)$ is frequently large by controlling the modulus of continuity of $L_n$.
• Look for an "energy" inequality relating averages of $L_n(x)$ over $x$ (or $\int L_n(x)\,dx$) to $\max_x L_n(x)$, to transfer large maxima into a large limsup set.
• Explore candidate counterexamples: sequences $(x_k)$ designed so that the locations of maxima of $L_n$ move around rapidly with $n$, potentially preventing a fixed $x$ from seeing large values infinitely often.

(iv) Minimal counterexample structure. A counterexample would be a sequence $(x_k)$ such that for every fixed $x\in(-1,1)$, $L_n(x)$ stays below $(2/\pi-\delta)\log n$ (up to $O(1)$) for all sufficiently large $n$, while still allowing $\max_x L_n(x)$ to be large for each $n$.