gpt pro 5.2 attempt on Math-6092

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).