{
  "index": "1999-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Prove that there is a constant $C$ such that, if $p(x)$ is a polynomial\nof degree 1999, then\n\\[|p(0)|\\leq C \\int_{-1}^1 |p(x)|\\,dx.\\]",
  "solution": "First solution: (by Reid Barton)\nLet $r_1, \\dots, r_{1999}$ be the roots of $P$. Draw a disc of radius\n$\\epsilon$ around each $r_i$, where $\\epsilon < 1/3998$; this\ndisc covers a subinterval of $[-1/2,1/2]$\nof length at most $2\\epsilon$, and so of the 2000 (or fewer) uncovered\nintervals in $[-1/2,1/2]$,\none, which we call $I$, has length at least $\\delta = (1-3998\\epsilon)/2000\n> 0$.\nWe will exhibit an explicit lower bound for the integral of $|P(x)|/P(0)$ over this\ninterval, which will yield such a bound for the entire integral.\n\nNote that\n\\[\n\\frac{|P(x)|}{|P(0)|} = \\prod_{i=1}^{1999} \\frac{|x-r_i|}{|r_i|}.\n\\]\nAlso note that by construction, $|x - r_i| \\geq \\epsilon$ for each $x \\in I$.\nIf $|r_i| \\leq 1$, then we have $\\frac{|x-r_i|}{|r_i|} \\geq \\epsilon$. If $|r_i| > 1$, then\n\\[\n\\frac{|x-r_i|}{|r_i|} = |1 - x/r_i| \\geq 1 - |x/r_i| \\geq = 1/2\n> \\epsilon.\n\\]\nWe conclude that $\\int_I |P(x)/P(0)|\\,dx \\geq \\delta \\epsilon$,\nindependent of $P$.\n\nSecond solution:\nIt will be a bit more convenient to assume $P(0) = 1$ (which we\nmay achieve by rescaling unless $P(0)=0$, in which case\nthere is nothing to prove) and to prove that there exists $D>0$\nsuch that $\\int_{-1}^1 |P(x)|\\,dx \\geq D$, or even\nsuch that $\\int_0^1 |P(x)|\\,dx \\geq D$.\n\nWe first reduce to the case where $P$ has all of its roots in\n$[0,1]$. If this is not the case, we can\nfactor $P(x)$ as $Q(x) R(x)$, where $Q$ has all roots in\nthe interval and $R$ has none. Then $R$ is either always\npositive or always negative on $[0,1]$; assume the former.\nLet $k$ be the largest positive real number such that\n$R(x) - kx \\geq 0$ on $[0,1]$; then\n\\begin{align*}\n\\int_{-1}^1 |P(x)|\\,dx &= \\int_{-1}^1 |Q(x)R(x)|\\,dx \\\\\n&> \\int_{-1}^1 |Q(x)(R(x)-kx)|\\,dx,\n\\end{align*}\nand $Q(x)(R(x)-kx)$ has more roots in $[0,1]$ than does $P$\n(and has the same value at 0).\nRepeating this argument shows that $\\int_{0}^1 |P(x)|\\,dx$\nis greater than the corresponding integral for some polynomial\nwith all of its roots in $[0,1]$.\n\nUnder this assumption, we have\n\\[\nP(x) = c \\prod_{i=1}^{1999} (x-r_i)\n\\]\nfor some $r_i \\in (0,1]$. Since\n\\[\nP(0) = -c \\prod r_i = 1,\n\\]\nwe have\n\\[\n|c| \\geq \\prod |r_i^{-1}| \\geq 1.\n\\]\n\nThus it suffices to prove that if $Q(x)$ is a \\emph{monic} polynomial\nof degree 1999 with all of its roots in $[0,1]$,\nthen $\\int_0^1 |Q(x)|\\,dx \\geq D$ for some constant $D>0$. But the\nintegral of $\\int_0^1 \\prod_{i=1}^{1999} |x-r_i|\\,dx$ is a continuous\nfunction for $r_i \\in [0,1]$. The product of all of these intervals is\ncompact, so the integral achieves a minimum value for some $r_i$. This\nminimum is the desired $D$.\n\nThird solution (by Abe Kunin):\nIt suffices to prove the stronger inequality\n\\[\n\\sup_{x \\in [-1,1]} |P(x)| \\leq C \\int_{-1}^1 |P(x)|\\,dx\n\\]\nholds for some $C$.\nBut this follows immediately from the following standard fact: any two\nnorms on a finite-dimensional vector space (here the polynomials of\ndegree at most 1999) are equivalent. (The proof of this statement is also\na compactness argument: $C$ can be taken to be the maximum of the\nL1-norm divided by the sup norm over the set of polynomials with\nL1-norm 1.)\n\nNote: combining the first two approaches gives a constructive solution with\na constant that is better than that given by the first solution,\nbut is still far from optimal. I don't know\noffhand whether it is even known what the optimal constant and/or the\npolynomials achieving that constant are.",
  "vars": [
    "x",
    "p",
    "P",
    "Q",
    "R",
    "k",
    "c",
    "r_i"
  ],
  "params": [
    "C",
    "D",
    "\\\\epsilon",
    "\\\\delta"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "axisvar",
        "p": "polyfunc",
        "P": "polymain",
        "Q": "factorone",
        "R": "factortwo",
        "k": "scalefac",
        "c": "leadcoef",
        "r_i": "rootitem",
        "C": "constcap",
        "D": "constdee",
        "\\epsilon": "varepsil",
        "\\delta": "vardelta"
      },
      "question": "Prove that there is a constant $constcap$ such that, if $polyfunc(axisvar)$ is a polynomial\nof degree 1999, then\n\\[|polyfunc(0)|\\leq constcap \\int_{-1}^1 |polyfunc(axisvar)|\\,daxisvar.\\]",
      "solution": "First solution: (by Reid Barton)\nLet $rootitem_1, \\dots, rootitem_{1999}$ be the roots of $polymain$. Draw a disc of radius\n$varepsil$ around each $rootitem_i$, where $varepsil < 1/3998$; this\ndisc covers a subinterval of $[-1/2,1/2]$\nof length at most $2varepsil$, and so of the 2000 (or fewer) uncovered\nintervals in $[-1/2,1/2]$,\none, which we call $I$, has length at least $vardelta = (1-3998varepsil)/2000\n> 0$.\nWe will exhibit an explicit lower bound for the integral of $|polymain(axisvar)|/polymain(0)$ over this\ninterval, which will yield such a bound for the entire integral.\n\nNote that\n\\[\n\\frac{|polymain(axisvar)|}{|polymain(0)|} = \\prod_{i=1}^{1999} \\frac{|axisvar-rootitem|}{|rootitem|}.\n\\]\nAlso note that by construction, $|axisvar - rootitem| \\geq varepsil$ for each $axisvar \\in I$.\nIf $|rootitem| \\leq 1$, then we have $\\frac{|axisvar-rootitem|}{|rootitem|} \\geq varepsil$. If $|rootitem| > 1$, then\n\\[\n\\frac{|axisvar-rootitem|}{|rootitem|} = |1 - axisvar/rootitem| \\geq 1 - |axisvar/rootitem| \\geq = 1/2\n> varepsil.\n\\]\nWe conclude that $\\int_I |polymain(axisvar)/polymain(0)|\\,daxisvar \\geq vardelta varepsil$,\nindependent of $polymain$.\n\nSecond solution:\nIt will be a bit more convenient to assume $polymain(0) = 1$ (which we\nmay achieve by rescaling unless $polymain(0)=0$, in which case\nthere is nothing to prove) and to prove that there exists $constdee>0$\nsuch that $\\int_{-1}^1 |polymain(axisvar)|\\,daxisvar \\geq constdee$, or even\nsuch that $\\int_0^1 |polymain(axisvar)|\\,daxisvar \\geq constdee$.\n\nWe first reduce to the case where $polymain$ has all of its roots in\n$[0,1]$. If this is not the case, we can\nfactor $polymain(axisvar)$ as $factorone(axisvar) factortwo(axisvar)$, where $factorone$ has all roots in\nthe interval and $factortwo$ has none. Then $factortwo$ is either always\npositive or always negative on $[0,1]$; assume the former.\nLet $scalefac$ be the largest positive real number such that\n$factortwo(axisvar) - scalefac axisvar \\geq 0$ on $[0,1]$; then\n\\begin{align*}\n\\int_{-1}^1 |polymain(axisvar)|\\,daxisvar &= \\int_{-1}^1 |factorone(axisvar)factortwo(axisvar)|\\,daxisvar \\\\\n&> \\int_{-1}^1 |factorone(axisvar)(factortwo(axisvar)-scalefac axisvar)|\\,daxisvar,\n\\end{align*}\nand $factorone(axisvar)(factortwo(axisvar)-scalefac axisvar)$ has more roots in $[0,1]$ than does $polymain$\n(and has the same value at 0).\nRepeating this argument shows that $\\int_{0}^1 |polymain(axisvar)|\\,daxisvar$\nis greater than the corresponding integral for some polynomial\nwith all of its roots in $[0,1]$.\n\nUnder this assumption, we have\n\\[\npolymain(axisvar) = leadcoef \\prod_{i=1}^{1999} (axisvar-rootitem)\n\\]\nfor some $rootitem \\in (0,1]$. Since\n\\[\npolymain(0) = -leadcoef \\prod rootitem = 1,\n\\]\nwe have\n\\[\n|leadcoef| \\geq \\prod |rootitem^{-1}| \\geq 1.\n\\]\n\nThus it suffices to prove that if $factorone(axisvar)$ is a \\emph{monic} polynomial\nof degree 1999 with all of its roots in $[0,1]$,\nthen $\\int_0^1 |factorone(axisvar)|\\,daxisvar \\geq constdee$ for some constant $constdee>0$. But the\nintegral of $\\int_0^1 \\prod_{i=1}^{1999} |axisvar-rootitem|\\,daxisvar$ is a continuous\nfunction for $rootitem \\in [0,1]$. The product of all of these intervals is\ncompact, so the integral achieves a minimum value for some $rootitem$. This\nminimum is the desired $constdee$.\n\nThird solution (by Abe Kunin):\nIt suffices to prove the stronger inequality\n\\[\n\\sup_{axisvar \\in [-1,1]} |polymain(axisvar)| \\leq constcap \\int_{-1}^1 |polymain(axisvar)|\\,daxisvar\n\\]\nholds for some $constcap$.\nBut this follows immediately from the following standard fact: any two\nnorms on a finite-dimensional vector space (here the polynomials of\ndegree at most 1999) are equivalent. (The proof of this statement is also\na compactness argument: $constcap$ can be taken to be the maximum of the\nL1-norm divided by the sup norm over the set of polynomials with\nL1-norm 1.)\n\nNote: combining the first two approaches gives a constructive solution with\na constant that is better than that given by the first solution,\nbut is still far from optimal. I don't know\noffhand whether it is even known what the optimal constant and/or the\npolynomials achieving that constant are."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orchardview",
        "p": "lanternfish",
        "P": "moonlitsea",
        "Q": "cobblestone",
        "R": "glimmerpath",
        "k": "tangerine",
        "c": "honeybadger",
        "r_i": "buttercup",
        "C": "whirlwind",
        "D": "plaintiff",
        "\\epsilon": "dragonfly",
        "\\delta": "silhouette"
      },
      "question": "Prove that there is a constant $whirlwind$ such that, if $lanternfish(orchardview)$ is a polynomial\nof degree 1999, then\n\\[|lanternfish(0)|\\leq whirlwind \\int_{-1}^1 |lanternfish(orchardview)|\\,dorchardview.\\]",
      "solution": "First solution: (by Reid Barton)\nLet $r_1, \\dots, r_{1999}$ be the roots of $moonlitsea$. Draw a disc of radius\n$\\dragonfly$ around each $buttercup$, where $\\dragonfly < 1/3998$; this\ndisc covers a subinterval of $[-1/2,1/2]$\nof length at most $2\\dragonfly$, and so of the 2000 (or fewer) uncovered\nintervals in $[-1/2,1/2]$,\none, which we call $I$, has length at least $\\silhouette = (1-3998\\dragonfly)/2000\n> 0$.\nWe will exhibit an explicit lower bound for the integral of $|moonlitsea(orchardview)|/moonlitsea(0)$ over this\ninterval, which will yield such a bound for the entire integral.\n\nNote that\n\\[\n\\frac{|moonlitsea(orchardview)|}{|moonlitsea(0)|} = \\prod_{i=1}^{1999} \\frac{|orchardview-buttercup|}{|buttercup|}.\n\\]\nAlso note that by construction, $|orchardview - buttercup| \\geq \\dragonfly$ for each $orchardview \\in I$.\nIf $|buttercup| \\leq 1$, then we have $\\frac{|orchardview-buttercup|}{|buttercup|} \\geq \\dragonfly$. If $|buttercup| > 1$, then\n\\[\n\\frac{|orchardview-buttercup|}{|buttercup|} = |1 - orchardview/buttercup| \\geq 1 - |orchardview/buttercup| \\geq = 1/2\n> \\dragonfly.\n\\]\nWe conclude that $\\int_I |moonlitsea(orchardview)/moonlitsea(0)|\\,dorchardview \\geq \\silhouette \\dragonfly$,\nindependent of $moonlitsea$.\n\nSecond solution:\nIt will be a bit more convenient to assume $moonlitsea(0) = 1$ (which we\nmay achieve by rescaling unless $moonlitsea(0)=0$, in which case\nthere is nothing to prove) and to prove that there exists $plaintiff>0$\nsuch that $\\int_{-1}^1 |moonlitsea(orchardview)|\\,dorchardview \\geq plaintiff$, or even\nsuch that $\\int_0^1 |moonlitsea(orchardview)|\\,dorchardview \\geq plaintiff$.\n\nWe first reduce to the case where $moonlitsea$ has all of its roots in\n$[0,1]$. If this is not the case, we can\nfactor $moonlitsea(orchardview)$ as $cobblestone(orchardview) glimmerpath(orchardview)$, where $cobblestone$ has all roots in\nthe interval and $glimmerpath$ has none. Then $glimmerpath$ is either always\npositive or always negative on $[0,1]$; assume the former.\nLet $tangerine$ be the largest positive real number such that\n$glimmerpath(orchardview) - tangerine orchardview \\geq 0$ on $[0,1]$; then\n\\begin{align*}\n\\int_{-1}^1 |moonlitsea(orchardview)|\\,dorchardview &= \\int_{-1}^1 |cobblestone(orchardview)glimmerpath(orchardview)|\\,dorchardview \\\\\n&> \\int_{-1}^1 |cobblestone(orchardview)(glimmerpath(orchardview)-tangerine orchardview)|\\,dorchardview,\n\\end{align*}\nand $cobblestone(orchardview)(glimmerpath(orchardview)-tangerine orchardview)$ has more roots in $[0,1]$ than does $moonlitsea$\n(and has the same value at 0).\nRepeating this argument shows that $\\int_{0}^1 |moonlitsea(orchardview)|\\,dorchardview$\nis greater than the corresponding integral for some polynomial\nwith all of its roots in $[0,1]$.\n\nUnder this assumption, we have\n\\[\nmoonlitsea(orchardview) = honeybadger \\prod_{i=1}^{1999} (orchardview-buttercup)\n\\]\nfor some $buttercup \\in (0,1]$. Since\n\\[\nmoonlitsea(0) = -honeybadger \\prod buttercup = 1,\n\\]\nwe have\n\\[\n|honeybadger| \\geq \\prod |buttercup^{-1}| \\geq 1.\n\\]\n\nThus it suffices to prove that if $cobblestone(orchardview)$ is a \\emph{monic} polynomial\nof degree 1999 with all of its roots in $[0,1]$,\nthen $\\int_0^1 |cobblestone(orchardview)|\\,dorchardview \\geq plaintiff$ for some constant $plaintiff>0$. But the\nintegral of $\\int_0^1 \\prod_{i=1}^{1999} |orchardview-buttercup|\\,dorchardview$ is a continuous\nfunction for $buttercup \\in [0,1]$. The product of all of these intervals is\ncompact, so the integral achieves a minimum value for some $buttercup$. This\nminimum is the desired $plaintiff$.\n\nThird solution (by Abe Kunin):\nIt suffices to prove the stronger inequality\n\\[\n\\sup_{orchardview \\in [-1,1]} |moonlitsea(orchardview)| \\leq whirlwind \\int_{-1}^1 |moonlitsea(orchardview)|\\,dorchardview\n\\]\nholds for some $whirlwind$.\nBut this follows immediately from the following standard fact: any two\nnorms on a finite-dimensional vector space (here the polynomials of\ndegree at most 1999) are equivalent. (The proof of this statement is also\na compactness argument: $whirlwind$ can be taken to be the maximum of the\nL1-norm divided by the sup norm over the set of polynomials with\nL1-norm 1.)\n\nNote: combining the first two approaches gives a constructive solution with\na constant that is better than that given by the first solution,\nbut is still far from optimal. I don't know\noffhand whether it is even known what the optimal constant and/or the\npolynomials achieving that constant are."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "p": "nonpolynom",
        "P": "nonfunction",
        "Q": "irrelevant",
        "R": "inclusive",
        "k": "minimalvalue",
        "c": "variablecoef",
        "r_i": "peakvalue",
        "C": "unbounded",
        "D": "smallvalue",
        "\\epsilon": "\\largeness",
        "\\delta": "\\vastness"
      },
      "question": "Prove that there is a constant $unbounded$ such that, if $nonpolynom(constantvalue)$ is a polynomial\nof degree 1999, then\n\\[|nonpolynom(0)|\\leq unbounded \\int_{-1}^1 |nonpolynom(constantvalue)|\\,dconstantvalue.\\]",
      "solution": "First solution: (by Reid Barton)\nLet $r_1, \\dots, r_{1999}$ be the roots of $nonfunction$. Draw a disc of radius\n$\\largeness$ around each $peakvalue$, where $\\largeness < 1/3998$; this\ndisc covers a subinterval of $[-1/2,1/2]$\nof length at most $2\\largeness$, and so of the 2000 (or fewer) uncovered\nintervals in $[-1/2,1/2]$,\none, which we call $I$, has length at least $\\vastness = (1-3998\\largeness)/2000\n> 0$.\nWe will exhibit an explicit lower bound for the integral of $|nonfunction(constantvalue)|/nonfunction(0)$ over this\ninterval, which will yield such a bound for the entire integral.\n\nNote that\n\\[\n\\frac{|nonfunction(constantvalue)|}{|nonfunction(0)|} = \\prod_{i=1}^{1999} \\frac{|constantvalue-peakvalue|}{|peakvalue|}.\n\\]\nAlso note that by construction, $|constantvalue - peakvalue| \\geq \\largeness$ for each $constantvalue \\in I$.\nIf $|peakvalue| \\leq 1$, then we have $\\frac{|constantvalue-peakvalue|}{|peakvalue|} \\geq \\largeness$. If $|peakvalue| > 1$, then\n\\[\n\\frac{|constantvalue-peakvalue|}{|peakvalue|} = |1 - constantvalue/peakvalue| \\geq 1 - |constantvalue/peakvalue| \\geq = 1/2\n> \\largeness.\n\\]\nWe conclude that $\\int_I |nonfunction(constantvalue)/nonfunction(0)|\\,dconstantvalue \\geq \\vastness \\largeness$,\nindependent of $nonfunction$.\n\nSecond solution:\nIt will be a bit more convenient to assume $nonfunction(0) = 1$ (which we\nmay achieve by rescaling unless $nonfunction(0)=0$, in which case\nthere is nothing to prove) and to prove that there exists $smallvalue>0$\nsuch that $\\int_{-1}^1 |nonfunction(constantvalue)|\\,dconstantvalue \\geq smallvalue$, or even\nsuch that $\\int_0^1 |nonfunction(constantvalue)|\\,dconstantvalue \\geq smallvalue$.\n\nWe first reduce to the case where $nonfunction$ has all of its roots in\n$[0,1]$. If this is not the case, we can\nfactor $nonfunction(constantvalue)$ as $irrelevant(constantvalue) inclusive(constantvalue)$, where $irrelevant$ has all roots in\nthe interval and $inclusive$ has none. Then $inclusive(constantvalue)$ is either always\npositive or always negative on $[0,1]$; assume the former.\nLet $minimalvalue$ be the largest positive real number such that\n$inclusive(constantvalue) - minimalvalue constantvalue \\geq 0$ on $[0,1]$; then\n\\begin{align*}\n\\int_{-1}^1 |nonfunction(constantvalue)|\\,dconstantvalue &= \\int_{-1}^1 |irrelevant(constantvalue)inclusive(constantvalue)|\\,dconstantvalue \\\\\n&> \\int_{-1}^1 |irrelevant(constantvalue)(inclusive(constantvalue)-minimalvalue constantvalue)|\\,dconstantvalue,\n\\end{align*}\nand $irrelevant(constantvalue)(inclusive(constantvalue)-minimalvalue constantvalue)$ has more roots in $[0,1]$ than does $nonfunction$\n(and has the same value at 0).\nRepeating this argument shows that $\\int_{0}^1 |nonfunction(constantvalue)|\\,dconstantvalue$\nis greater than the corresponding integral for some polynomial\nwith all of its roots in $[0,1]$.\n\nUnder this assumption, we have\n\\[\nnonfunction(constantvalue) = variablecoef \\prod_{i=1}^{1999} (constantvalue-peakvalue)\n\\]\nfor some $peakvalue \\in (0,1]$. Since\n\\[\nnonfunction(0) = -variablecoef \\prod peakvalue = 1,\n\\]\nwe have\n\\[\n|variablecoef| \\geq \\prod |peakvalue^{-1}| \\geq 1.\n\\]\n\nThus it suffices to prove that if $irrelevant(constantvalue)$ is a \\emph{monic} polynomial\nof degree 1999 with all of its roots in $[0,1]$,\nthen $\\int_0^1 |irrelevant(constantvalue)|\\,dconstantvalue \\geq smallvalue$ for some constant $smallvalue>0$. But the\nintegral of $\\int_0^1 \\prod_{i=1}^{1999} |constantvalue-peakvalue|\\,dconstantvalue$ is a continuous\nfunction for $peakvalue \\in [0,1]$. The product of all of these intervals is\ncompact, so the integral achieves a minimum value for some $peakvalue$. This\nminimum is the desired $smallvalue$.\n\nThird solution (by Abe Kunin):\nIt suffices to prove the stronger inequality\n\\[\n\\sup_{constantvalue \\in [-1,1]} |nonfunction(constantvalue)| \\leq unbounded \\int_{-1}^1 |nonfunction(constantvalue)|\\,dconstantvalue\n\\]\nholds for some $unbounded$.\nBut this follows immediately from the following standard fact: any two\nnorms on a finite-dimensional vector space (here the polynomials of\ndegree at most 1999) are equivalent. (The proof of this statement is also\na compactness argument: $unbounded$ can be taken to be the maximum of the\nL1-norm divided by the sup norm over the set of polynomials with\nL1-norm 1.)\n\nNote: combining the first two approaches gives a constructive solution with\na constant that is better than that given by the first solution,\nbut is still far from optimal. I don't know\noffhand whether it is even known what the optimal constant and/or the\npolynomials achieving that constant are."
    },
    "garbled_string": {
      "map": {
        "x": "mcltzrga",
        "p": "zgqxvndh",
        "P": "rqwxlypdf",
        "Q": "kjsnmbtq",
        "R": "flhdpzsv",
        "k": "dpqvrmxe",
        "c": "hqtsrlyn",
        "r_i": "ypmvkeja",
        "C": "vbnkqzfs",
        "D": "qlwmsteh",
        "\\epsilon": "\\mnlwczva",
        "\\delta": "\\qvzxpokr"
      },
      "question": "Prove that there is a constant $vbnkqzfs$ such that, if $zgqxvndh(mcltzrga)$ is a polynomial\nof degree 1999, then\n\\[|zgqxvndh(0)|\\leq vbnkqzfs \\int_{-1}^1 |zgqxvndh(mcltzrga)|\\,d mcltzrga.\\]",
      "solution": "First solution: (by Reid Barton)\nLet $r_1, \\dots, r_{1999}$ be the roots of $rqwxlypdf$. Draw a disc of radius\n$\\mnlwczva$ around each $ypmvkeja$, where $\\mnlwczva < 1/3998$; this\ndisc covers a subinterval of $[-1/2,1/2]$\nof length at most $2\\mnlwczva$, and so of the 2000 (or fewer) uncovered\nintervals in $[-1/2,1/2]$,\none, which we call $I$, has length at least $\\qvzxpokr = (1-3998\\mnlwczva)/2000\n> 0$.\nWe will exhibit an explicit lower bound for the integral of $|rqwxlypdf(mcltzrga)|/rqwxlypdf(0)$ over this\ninterval, which will yield such a bound for the entire integral.\n\nNote that\n\\[\n\\frac{|rqwxlypdf(mcltzrga)|}{|rqwxlypdf(0)|} = \\prod_{i=1}^{1999} \\frac{|mcltzrga-ypmvkeja|}{|ypmvkeja|}.\n\\]\nAlso note that by construction, $|mcltzrga - ypmvkeja| \\geq \\mnlwczva$ for each $mcltzrga \\in I$.\nIf $|ypmvkeja| \\leq 1$, then we have $\\frac{|mcltzrga-ypmvkeja|}{|ypmvkeja|} \\geq \\mnlwczva$. If $|ypmvkeja| > 1$, then\n\\[\n\\frac{|mcltzrga-ypmvkeja|}{|ypmvkeja|} = |1 - mcltzrga/ypmvkeja| \\geq 1 - |mcltzrga/ypmvkeja| \\geq 1/2\n> \\mnlwczva.\n\\]\nWe conclude that $\\int_I |rqwxlypdf(mcltzrga)/rqwxlypdf(0)|\\,d mcltzrga \\geq \\qvzxpokr \\mnlwczva$,\nindependent of $rqwxlypdf$.\n\nSecond solution:\nIt will be a bit more convenient to assume $rqwxlypdf(0) = 1$ (which we\nmay achieve by rescaling unless $rqwxlypdf(0)=0$, in which case\nthere is nothing to prove) and to prove that there exists $qlwmsteh>0$\nsuch that $\\int_{-1}^1 |rqwxlypdf(mcltzrga)|\\,d mcltzrga \\geq qlwmsteh$, or even\nsuch that $\\int_0^1 |rqwxlypdf(mcltzrga)|\\,d mcltzrga \\geq qlwmsteh$.\n\nWe first reduce to the case where $rqwxlypdf$ has all of its roots in\n$[0,1]$. If this is not the case, we can\nfactor $rqwxlypdf(mcltzrga)$ as $kjsnmbtq(mcltzrga) flhdpzsv(mcltzrga)$, where $kjsnmbtq$ has all roots in\nthe interval and $flhdpzsv$ has none. Then $flhdpzsv$ is either always\npositive or always negative on $[0,1]$; assume the former.\nLet $dpqvrmxe$ be the largest positive real number such that\n$flhdpzsv(mcltzrga) - dpqvrmxe mcltzrga \\geq 0$ on $[0,1]$; then\n\\begin{align*}\n\\int_{-1}^1 |rqwxlypdf(mcltzrga)|\\,d mcltzrga &= \\int_{-1}^1 |kjsnmbtq(mcltzrga)flhdpzsv(mcltzrga)|\\,d mcltzrga \\\\\n&> \\int_{-1}^1 |kjsnmbtq(mcltzrga)(flhdpzsv(mcltzrga)-dpqvrmxe mcltzrga)|\\,d mcltzrga,\n\\end{align*}\nand $kjsnmbtq(mcltzrga)(flhdpzsv(mcltzrga)-dpqvrmxe mcltzrga)$ has more roots in $[0,1]$ than does $rqwxlypdf$\n(and has the same value at 0).\nRepeating this argument shows that $\\int_{0}^1 |rqwxlypdf(mcltzrga)|\\,d mcltzrga$\nis greater than the corresponding integral for some polynomial\nwith all of its roots in $[0,1]$.\n\nUnder this assumption, we have\n\\[\nrqwxlypdf(mcltzrga) = hqtsrlyn \\prod_{i=1}^{1999} (mcltzrga-ypmvkeja)\n\\]\nfor some $ypmvkeja \\in (0,1]$. Since\n\\[\nrqwxlypdf(0) = -hqtsrlyn \\prod ypmvkeja = 1,\n\\]\nwe have\n\\[\n|hqtsrlyn| \\geq \\prod |ypmvkeja^{-1}| \\geq 1.\n\\]\n\nThus it suffices to prove that if $kjsnmbtq(mcltzrga)$ is a \\emph{monic} polynomial\nof degree 1999 with all of its roots in $[0,1]$,\nthen $\\int_0^1 |kjsnmbtq(mcltzrga)|\\,d mcltzrga \\geq qlwmsteh$ for some constant $qlwmsteh>0$. But the\nintegral $\\int_0^1 \\prod_{i=1}^{1999} |mcltzrga-ypmvkeja|\\,d mcltzrga$ is a continuous\nfunction for $ypmvkeja \\in [0,1]$. The product of all of these intervals is\ncompact, so the integral achieves a minimum value for some $ypmvkeja$. This\nminimum is the desired $qlwmsteh$.\n\nThird solution (by Abe Kunin):\nIt suffices to prove the stronger inequality\n\\[\n\\sup_{mcltzrga \\in [-1,1]} |rqwxlypdf(mcltzrga)| \\leq vbnkqzfs \\int_{-1}^1 |rqwxlypdf(mcltzrga)|\\,d mcltzrga\n\\]\nholds for some $vbnkqzfs$.\nBut this follows immediately from the following standard fact: any two\nnorms on a finite-dimensional vector space (here the polynomials of\ndegree at most 1999) are equivalent. (The proof of this statement is also\na compactness argument: $vbnkqzfs$ can be taken to be the maximum of the\nL1-norm divided by the sup norm over the set of polynomials with\nL1-norm 1.)\n\nNote: combining the first two approaches gives a constructive solution with\na constant that is better than that given by the first solution,\nbut is still far from optimal. I don't know\noffhand whether it is even known what the optimal constant and/or the\npolynomials achieving that constant are."
    },
    "kernel_variant": {
      "question": "Let n = 2024 and fix a non–negative integer k = 50.  Show that there exists an absolute constant C (depending only on n and k, but independent of the particular polynomial) with the following property:\n\nFor every real polynomial P of degree at most n one has  \n  max_{0 ≤ j ≤ k} |P^{(j)}(2)|  ≤  C · ∫₁³ |P(x)| dx.                                        (★)\n\nIn words: not only the value of P at the interior point 2 but the first fifty derivatives at that point are all controlled by the single L¹–norm of P on the interval [1, 3].\n\n——————————————————————————————————",
      "solution": "(≈ 560 words)\n\nWe break the argument into four transparent steps, each mirroring the logical layering of the reference solution while adding a derivative component.\n\nStep 1.  Two auxiliary norms on a finite–dimensional space.  \nLet V := {real polynomials of degree ≤ n}.  This is a real vector space of dimension N = n+1.  On V introduce\n\n ‖P‖₁ := ∫₁³ |P(x)| dx,                (1)\n\n ‖P‖∞ := sup_{x∈[1,3]} |P(x)|,            (2)\n\n ‖P‖₍k₎ := max_{0≤j≤k} |P^{(j)}(2)|.          (3)\n\nThe quantity (3) is the norm that appears on the left–hand side of (★).  Observe that (1) and (2) were already present in the easier kernel, whereas (3) is new.\n\nStep 2.  Equivalence of norms implies a crude bound.  \nBecause V is finite–dimensional, any two norms on V are equivalent.  In particular, there is a constant A=A(n,k)>0 such that  \n ‖P‖₍k₎ ≤ A ‖P‖∞ for all P∈V.                                           (4)\n\n(Indeed, take the unit sphere S = {P : ‖P‖∞ = 1}; the mapping P ↦ ‖P‖₍k₎ is continuous on S and S is compact, so the maximum value of that map is finite; call it A.)\n\nStep 3.  A Markov-type inequality converting derivatives into sup-norm.  \nClassical Markov inequalities on a closed interval J of length 2 give, for every 0≤j≤k and every polynomial of degree ≤ n,\n\n  |P^{(j)}(x)| ≤ (n)^{j} · 2^{−j} · (n+1 choose j) ‖P‖∞  for all x∈J.    (5)\n\nSpecialising (5) to x=2 and taking maxima over j≤k we get\n\n ‖P‖₍k₎ ≤ M(n,k) ‖P‖∞,                                                 (6)\n\nwhere an explicit choice is M(n,k) = max_{0≤j≤k} (n)^j 2^{−j}(n+1 choose j).  Compare (4): (6) is still of the form “derivatives ≤ const·sup”, but with a concrete constant whose growth we completely control; this will matter later when estimating how the overall C depends on n and k.\n\nStep 4.  Transfer from sup-norm to L¹-norm.  \nReturn to the L¹–norm introduced in (1).  As in the reference solution, equivalence of norms on V yields a constant B=B(n)>0 satisfying\n\n ‖P‖∞ ≤ B ‖P‖₁  for all P∈V.                                           (7)\n\nCombine (6) and (7):\n\n ‖P‖₍k₎ ≤ M(n,k) ‖P‖∞ ≤ M(n,k) B ‖P‖₁.                         (8)\n\nDefine  \n\n C := M(n,k) B,                                                         (9)\n\na constant depending solely on the fixed integers n and k.  Inequality (8) is exactly statement (★), completing the proof.\n\nRemark 1.  Compactness was employed twice: first to bound evaluation of derivatives by the sup-norm (Step 2), second to relate the sup- and L¹-norms (Step 4).  The new ingredient vis-à-vis the original kernel is Step 3, which shows how classical approximation theory blends with linear-algebraic norm equivalence to control an entire jet (value and fifty derivatives) at a single point.\n\nRemark 2.  Nothing essential changes if we replace the point x=2 by any interior point of [1,3] or adjust the interval endpoints: only the constants in (5) and hence in (9) would alter slightly.\n\n——————————————————————————————————",
      "_replacement_note": {
        "replaced_at": "2025-07-05T22:17:12.043384",
        "reason": "Original kernel variant was too easy compared to the original problem"
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}