Reused writeup on Math-6092 (gpt pro 5.2)

Erdős Problem 1038 --- collection record

PROVENANCE AND STATUS

TeX syntax corrected for display; mathematical content unchanged. Official problem: https://www.erdosproblems.com/1038

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Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1038.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1038.tex.

Determine the infimum and supremum of\[\lvert \{ x\in \mathbb{R} : \lvert f(x)\rvert < 1\}\rvert\]as $f\in \mathbb{R}[x]$ ranges over all non-constant monic polynomials, all of whose roots are real and in the interval $[-1,1]$. A problem of Erdős, Herzog, and Piranian [EHP58], who proved that the measure of the set in question is always at most $2\sqrt{2}$ under the assumption that all the roots are in $\{-1,1\}$, and conjecture this is the best possible upper bound. They also note that the infimum of the set in question is less than $2$, as witnessed by $f(x)=(x+1)(x-1)^m$ for $m\geq 3$. They further note that if the roots are restricted to $[-2,2]$ then the infimum is zero, as witnessed by a small perturbation of the Chebyshev polynomials. They further conjectured that, if the roots are restricted to $[-2,2]$, then\[\lvert \{ x\in \mathbb{R} : \lvert f(x)\rvert < 1\}\rvert\geq n^{-c}\]for an absolute constant $c>0$. This was proved by Pommerenke [Po61], who in fact showed that this set must contain an interval of width $\gg n^{-4}$. The current best known bounds (see the discussion in the comments) are\[1.519\approx 2^{4/3}-1\leq \inf \leq 1.835\cdots\]and\[\sup = 2\sqrt{2}\approx 2.828.\] References [EHP58] Erdős, P. and Herzog, F. and Piranian, G., Metric properties of polynomials . J. Analyse Math. $1958$, 125-148. [Po61] Pommerenke, Ch., On metric properties of complex polynomials . Michigan Math. J. $1961$, 97-115.

FORMAL RESTATEMENT Let f be a non-constant monic real polynomial whose roots are all real and lie in [-1,1]. Define S(f) := { x in R : |f(x)| < 1 } and let m(f) be the Lebesgue measure (total length) of S(f). Determine inf_f m(f) and sup_f m(f) over all such f.

QUICK LITERATURE/CONTEXT CHECK Only facts explicitly present in the problem statement are treated as established here. In particular: * Under the extra restriction that all roots lie in {-1,1}, it is stated that m(f) <= 2*sqrt$2$ and this upper bound is conjectured sharp. * For roots restricted to [-2,2], an n^{-c} lower bound (and even an interval of width >> n^{-4}) is stated to be known. * The problem statement records best-known numeric bounds for the infimum and that sup = 2*sqrt$2$.

ATTACK PLAN $1$ Test explicit families f to produce lower bounds for the supremum and upper bounds for the infimum. $2$ For a given explicit f, reduce m(f) to locating the real roots of f(x)^2-1 and summing the lengths of the intervals on which the sign is negative. $3$ Use small-degree numerical checks as sanity.

WORK FAST REALITY CHECK (small degrees; computed by a minimal script that samples roots on grids / random and computes m(f) via the real roots of f^2-1). The script outputs: * degree 1: always m(f)=2. * degree 2: sampled min approx 2.000000, sampled max approx 2.828427 (achieved at roots -1,1, i.e. f=x^2-1). * degree 3: sampled min approx 2.000000, sampled max approx 2.735895 (random search; not exhaustive). Also for the family f(x)=$x+1$$x-1$^m: * m=3: m(f) approx 1.984305 * m=4: m(f) approx 1.903210 * m=5: m(f) approx 1.876260 (these are numerical; the lemma below gives a rigorous strict inequality m(f)<2 for m=3).

Lemma 1038.1 (Erdos-lemniscate example reaches length 2*sqrt$2$). For any integer m>=1, let f_m(x) = $x^2-1$^m. Then m$f_m$ = 2*sqrt$2$.

Proof. We have |f_m(x)|<1 iff |x^2-1|^m <1 iff |x^2-1|<1. The inequality |x^2-1|<1 is equivalent to -1 < x^2-1 < 1, which is equivalent to 0 < x^2 < 2. Thus S$f_m$ = (-sqrt$2$, sqrt$2$) with the two points x=0 and x=$\pm$ sqrt$2$ removed/added only at the boundary where |f_m|=1. These boundary points do not affect Lebesgue measure. Hence m$f_m$=2*sqrt$2$.

Lemma 1038.2 (a concrete f with m(f)<2). Let f(x) = $x+1$$x-1$^3. Then m(f) < 2.

Proof. Write f(x) = $x+1$$x-1$^3 and define the set S = {x: |f(x)|<1}. We analyze S on the three regions x>1, -1<=x<=1, and x<-1.

$1$ Region x>1. Put y=x-1>0. Then |f(x)| = $x+1$$x-1$^3 = $y+2$ y^3. Let y_+ be the unique positive solution to $y+2$y^3=1 (monotone increasing for y>0). Then $S\cap(1,\infty) = (1, 1+y_+)$, and this contributes length y_+. The equation is y^4+2y^3-1=0. Evaluate at the rationals 71/100 and 718/1000: q(y) := y^4+2y^3-1. Direct computation gives q$71/100$ < 0 and q$718/1000$ > 0, so 71/100 < y_+ < 718/1000. In particular, y_+ < 0.718.

$2$ Region -1<=x<=1. Put t=1-x in [0,2], so x=1-t and |x+1|=2-t. Then |f(x)| = $2-t$ t^3. Let t_* be the unique solution in $1,2$ of $2-t$t^3=1; this is unique because h(t)=$2-t$t^3 satisfies h$1$=1, increases for t in $1,3/2$, then decreases to 0 at t=2. Then h(t)<1 holds for t in $[0,1) \cup (t_*,2]$, i.e. x in $(0,1] \cup [-1, 1-t_*)$. So the contribution of $S\cap[-1,1]$ has length length$(0,1)$ + length$(-1, 1-t_*)$ = 1 + $2-t_*$. The equation $2-t$t^3=1 rearranges to p(t)=t^4-2t^3+1=0. Compute p$1839/1000$ < 0 and p$1840/1000$ > 0; hence 1839/1000 < t_* < 1840/1000. Therefore 2-t_* < 2-1839/1000 = 161/1000 = 0.161. So $S\cap[-1,1]$ has length < 1 + 0.161 = 1.161.

$3$ Region x<-1. Put t=1-x >2, so |x+1|=t-2 and |x-1|=t. Then |f(x)| = $t-2$t^3. Let t_+>2 be the unique solution of $t-2$t^3=1. Then $S\cap(-\infty,-1) = (1-t_+, -1)$, and this contributes length t_+-2. The equation is r(t)=t^4-2t^3-1=0. Compute r$2106/1000$ < 0 and r$2107/1000$ > 0; hence 2.106 < t_+ < 2.107 and thus t_+-2 < 0.107.

Putting $1$$2$$3$ together, we obtain m(f) = y_+ + $1 + (2-t_*)$ + $t_+-2$ < 0.718 + 1 + 0.161 + 0.107 = 1.986. In particular m(f) < 2.

VERIFICATION * Lemma 1038.1 is an exact algebraic reduction. * Lemma 1038.2 reduces the measure computation to three monotone boundary equations and brackets each boundary root by explicit sign evaluations at rationals.

FINAL UNRESOLVED (i) Strongest proved partial result here: explicit extremal-type example f_m=$x^2-1$^m gives m$f_m$=2*sqrt$2$ (Lemma 1038.1), and an explicit example f(x)=$x+1$$x-1$^3 has m(f)<2 (Lemma 1038.2), showing the global infimum is strictly below 2. (ii) First gap: determine the exact values of inf_f m(f) and sup_f m(f) under the full root constraint (all real roots in [-1,1]). (iii) Top 3 next moves: 1. Attempt to prove a universal upper bound m(f) <= 2*sqrt$2$ without the restriction roots in {-1,1}, perhaps by comparison with a symmetrized polynomial or an extremal inequality on level sets. 2. For the infimum, study candidate extremizers with many roots near 1 and a few near -1 (or vice versa), and attempt to turn the 1.5--1.8 numeric window into rigorous bounds by analytic inequalities. 3. Computational exploration for fixed degree n: optimize m(f) over root locations (e.g. gradient methods on the roots) to guess extremizers and then attempt to certify via inequalities. (iv) Minimal counterexample structure (if sup>2*sqrt$2$ or inf smaller than expected): a sequence of monic polynomials with roots in [-1,1] whose |f|<1 set either stretches far outside [-sqrt$2$,sqrt$2$] (for sup) or concentrates into very short neighbourhoods of roots with large gaps (for inf).

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.