gpt pro 5.2 attempt on Math-6187

FORMAL RESTATEMENT

Fix $C>0$. Question: does there exist $\epsilon>0$ such that for all sufficiently large integers $n$ the following holds? For every choice of nodes $x_1,\dots,x_n\in[-1,1]$, there exist target values $y_1,\dots,y_n\in[-1,1]$ such that: for every real polynomial $P$ of degree \[ \deg P = m < (1+\epsilon)n \] that satisfies \[ \lvert\{1\le i\le n: P(x_i)=y_i\}\rvert\ge (1-\epsilon)n, \] one has \[ \max_{x\in[-1,1]} |P(x)|>C. \]

QUICK LITERATURE/CONTEXT CHECK

I only record context explicitly stated in the problem file. The file states Erdős proved a related result: for any $C>0$ there exists $\epsilon>0$ such that for large $n$ and $m=\lfloor(1+\epsilon)n\rfloor$, for any $x_1,\dots,x_m\in[-1,1]$ there is a polynomial $P$ of degree $n$ with $|P(x_i)|\le 1$ at those $m$ points but $\max_{[-1,1]}|P|>C$. It also states Erdős could not prove the conjectured statement even for $m=n$.

ATTACK PLAN

Proof track ideas.

• Reduce to a statement about interpolation operators and Lebesgue constants: large Lebesgue constants allow choosing $y_i\in\{\pm 1\}$ forcing large sup norm for the interpolant.
• Try to "robustify" the exact interpolation forcing to handle matching only $(1-\epsilon)n$ points: show that any polynomial matching many prescribed signs must have large sup due to oscillation constraints.

Disproof track ideas.

• Attempt to show that for some configurations of points, any assignment $y_i\in[-1,1]$ can be matched on most points by a degree $<(1+\epsilon)n$ polynomial with controlled sup norm.

WORK

Lemma 1133.1 (explicit interpolation formula for degree $\le n-1$). Let $x_1,\dots,x_n$ be distinct points in $\mathbb{R}$ and let $y_1,\dots,y_n\in\mathbb{R}$. Define Lagrange polynomials $l_k$ from these nodes. Then the polynomial \[ P(x):=\sum_{k=1}^n y_k\,l_k(x) \] has degree at most $n-1$ and satisfies $P(x_i)=y_i$ for every $1\le i\le n$. Moreover, it is the unique polynomial of degree at most $n-1$ with these interpolation values.

Proof. Each $l_k$ has degree $n-1$, so $P$ has degree at most $n-1$. At a node $x_i$, we have $l_k(x_i)=\delta_{ki}$, hence \[ P(x_i)=\sum_{k=1}^n y_k\delta_{ki}=y_i. \] For uniqueness, suppose $Q$ is another polynomial of degree at most $n-1$ with $Q(x_i)=y_i$ for all $i$. Then $R:=P-Q$ has degree at most $n-1$ and has $n$ distinct roots $x_1,\dots,x_n$, so $R\equiv 0$. Therefore $P\equiv Q$. $\square$

Lemma 1133.2 (forcing large sup norm via the Lebesgue constant). Let $x_1,\dots,x_n\in[-1,1]$ be distinct nodes and let \[ \Lambda:=\max_{x\in[-1,1]}\sum_{k=1}^n |l_k(x)| \] be the Lebesgue constant associated to these nodes. Choose $x^*\in[-1,1]$ where the maximum is attained and define \[ y_k:=\begin{cases} \phantom{-}1,& l_k(x^*)\ge 0,\\ -1,& l_k(x^*)<0. \end{cases} \] Let $P$ be the degree $\le n-1$ interpolant through $(x_k,y_k)$ given by Lemma 1133.1. Then \[ P(x^*)=\sum_{k=1}^n |l_k(x^*)|=\Lambda, \] and hence $\max_{x\in[-1,1]}|P(x)|\ge \Lambda$.

Proof. By construction, $y_k=\operatorname{sign}(l_k(x^*))$ (with the convention $\operatorname{sign}(0)=1$). Therefore $y_k l_k(x^*)=|l_k(x^*)|$ for every $k$, and Lemma 1133.1 gives \[ P(x^*)=\sum_{k=1}^n y_k l_k(x^*)=\sum_{k=1}^n |l_k(x^*)|=\Lambda. \] Taking absolute values and the maximum over $x$ yields $\max_{[-1,1]}|P|\ge |P(x^*)|=\Lambda$. $\square$

Interpretation. Lemma 1133.2 shows a weaker (but relevant) forcing statement: for exact interpolation by degree $\le n-1$ polynomials (the case $\epsilon=0$ and $m=n-1$), one can always choose $y_k\in\{\pm 1\}$ so that any polynomial matching all $n$ constraints (necessarily the interpolant) must have sup norm at least $\Lambda$. Since the problem file states $\Lambda\gg\log n$ for every node set (Faber), this forcing exceeds any fixed $C$ for sufficiently large $n$. What remains open in the posed problem is robustness: allowing degree up to $(1+\epsilon)n$ and allowing the polynomial to ignore $\epsilon n$ constraints.

FAST REALITY CHECK (numerical illustration). For $n=5$ equally spaced nodes $[-1,-0.5,0,0.5,1]$, the above construction gives:

nodes [-1.  -0.5  0.   0.5  1. ]
Lambda approx 2.20782438320626 at x* -0.7919
y_k [ 1.  1. -1.  1. -1.]
P sup approx 2.20782438320626 attained near -0.7919
P(x*) 2.20782438320626

This empirically confirms Lemma 1133.2 in a concrete case.

VERIFICATION

• Lemma 1133.1 is standard and gap-free: interpolation property and uniqueness via $n$ roots.
• Lemma 1133.2 depends on existence of a maximizer $x^*$, which holds because $x\mapsto \sum_k|l_k(x)|$ is continuous on compact $[-1,1]$.
• The leap from Lemma 1133.2 to the conjectured statement is nontrivial because the conjecture permits dropping $\epsilon n$ constraints and increasing degree to $(1+\epsilon)n$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. For any node set $x_1,\dots,x_n$, one can choose $y_k\in\{\pm 1\}$ so that the unique interpolant of degree $\le n-1$ has $\max_{[-1,1]}|P|\ge \Lambda(x_1,\dots,x_n)$ (Lemma 1133.2).

(ii) First gap (crisp). Upgrade the exact-interpolation forcing (degree $n-1$ and matching all $n$ points) to the robust regime of the problem: degree $<(1+\epsilon)n$ and matching at least $(1-\epsilon)n$ points.

(iii) Top 3 next moves.

• Prove a robustness lemma: if a degree $m$ polynomial matches a sign pattern on many nodes, then its sup norm is bounded below in terms of a "restricted" Lebesgue constant for those nodes.
• Use quantitative bounds on how many alternations/oscillations a degree $m$ polynomial with bounded sup norm can have on $[-1,1]$, and choose $y_i$ to force more oscillation than allowed.
• Try probabilistic constructions for $y_i\in\{\pm 1\}$ and prove that with positive probability any degree-$m$ polynomial matching $>(1-\epsilon)n$ points must have large sup norm.

(iv) Minimal counterexample structure. A counterexample for a given $C$ would be a sequence of node sets $x^{(n)}_1,\dots,x^{(n)}_n$ such that for every assignment $y_i\in[-1,1]$ there exists a polynomial of degree $<(1+\epsilon)n$ matching $>(1-\epsilon)n$ constraints while keeping $\max_{[-1,1]}|P|\le C$.