{
  "index": "2013-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "For any continuous real-valued function $f$ defined on the interval $[0,1]$, let\n\\begin{gather*}\n\\mu(f) = \\int_0^1 f(x)\\,dx, \\,\n\\mathrm{Var}(f) = \\int_0^1 (f(x) - \\mu(f))^2\\,dx, \\\\\nM(f) = \\max_{0 \\leq x \\leq 1} \\left| f(x) \\right|.\n\\end{gather*}\nShow that if $f$ and $g$ are continuous real-valued functions\ndefined on the interval $[0,1]$,\nthen\n\\[\n\\mathrm{Var}(fg) \\leq 2 \\mathrm{Var}(f) M(g)^2 + 2 \\mathrm{Var}(g) M(f)^2.\n\\]",
  "solution": "\\newcommand{\\Var}{\\mathrm{Var}}\n\nWrite $f_0(x) = f(x)-\\mu(f)$ and $g_0(x) = g(x)-\\mu(g)$, so that $\\int_0^1 f_0(x)^2\\,dx = \\Var(f)$, $\\int_0^1 g_0(x)^2\\,dx = \\Var(g)$, and $\\int_0^1 f_0(x)\\,dx = \\int_0^1 g_0(x)\\,dx = 0$. Now since $|g(x)| \\leq M(g)$ for all $x$, $0\\leq \\int_0^1 f_0(x)^2(M(g)^2-g(x)^2)\\,dx = \\Var(f) M(g)^2-\\int_0^1 f_0(x)^2g(x)^2\\,dx$, and similarly $0 \\leq \\Var(g)M(f)^2-\\int_0^1 f(x)^2g_0(x)^2\\,dx$. Summing gives\n\\begin{equation}\n\\Var(f)M(g)^2+\\Var(g)M(f)^2\n\\label{eq:1}\n\\geq \\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2)\\,dx.\n\\end{equation}\nNow\n\\begin{align*}\n&\\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2)\\,dx-\\Var(fg) \\\\&= \\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2-(f(x)g(x)-\\int_0^1 f(y)g(y)\\,dy)^2)\\,dx;\n\\end{align*}\nsubstituting $f_0(x)+\\mu(f)$ for $f(x)$ everywhere and $g_0(x)+\\mu(g)$ for $g(x)$ everywhere, and using the fact that $\\int_0^1 f_0(x)\\,dx = \\int_0^1 g_0(x)\\,dx = 0$, we can expand and simplify the right hand side of this equation to obtain\n\\begin{align*}\n&\\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2)\\,dx-\\Var(fg) \\\\\n&= \\int_0^1 f_0(x)^2g_0(x)^2\\,dx \\\\\n&-2\\mu(f)\\mu(g)\\int_0^1 f_0(x)g_0(x)\\,dx +(\\int_0^1 f_0(x)g_0(x)\\,dx)^2 \\\\\n&\\geq -2\\mu(f)\\mu(g)\\int_0^1 f_0(x)g_0(x)\\,dx.\n\\end{align*}\nBecause of \\eqref{eq:1}, it thus suffices to show that\n\\begin{equation}\n2\\mu(f)\\mu(g)\\int_0^1 f_0(x)g_0(x)\\,dx\n\\label{eq:3} \\leq \\Var(f)M(g)^2+\\Var(g)M(f)^2.\n\\end{equation}\nNow since $(\\mu(g) f_0(x)-\\mu(f) g_0(x))^2 \\geq 0$ for all $x$, we have\n\\begin{align*}\n2\\mu(f)\\mu(g) \\int_0^1 f_0(x)g_0(x)\\,dx\n& \\leq \\int_0^1 (\\mu(g)^2 f_0(x)^2 + \\mu(f)^2 g_0(x)^2) dx \\\\\n& = \\Var(f) \\mu(g)^2 + \\Var(g) \\mu(f)^2 \\\\\n& \\leq \\Var(f) M(g)^2 + \\Var(g) M(f)^2,\n\\end{align*}\nestablishing \\eqref{eq:3} and completing the proof.",
  "vars": [
    "f",
    "g",
    "f_0",
    "g_0",
    "x",
    "y"
  ],
  "params": [
    "M",
    "\\\\mu"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "functionf",
        "g": "functiong",
        "f_0": "zeromeanf",
        "g_0": "zeromeang",
        "x": "variablex",
        "y": "variabley",
        "M": "maxbound",
        "\\mu": "meanval"
      },
      "question": "For any continuous real-valued function $functionf$ defined on the interval $[0,1]$, let\n\\begin{gather*}\nmeanval(functionf) = \\int_0^1 functionf(variablex)\\,dvariablex, \\,\n\\mathrm{Var}(functionf) = \\int_0^1 (functionf(variablex) - meanval(functionf))^2\\,dvariablex, \\\\\nmaxbound(functionf) = \\max_{0 \\leq variablex \\leq 1} \\left| functionf(variablex) \\right|.\n\\end{gather*}\nShow that if $functionf$ and $functiong$ are continuous real-valued functions\ndefined on the interval $[0,1]$,\nthen\n\\[\n\\mathrm{Var}(functionf functiong) \\leq 2 \\, \\mathrm{Var}(functionf) \\, maxbound(functiong)^2 + 2 \\, \\mathrm{Var}(functiong) \\, maxbound(functionf)^2.\n\\]",
      "solution": "\\newcommand{\\Var}{\\mathrm{Var}}\n\nWrite $zeromeanf(variablex) = functionf(variablex)-meanval(functionf)$ and $zeromeang(variablex) = functiong(variablex)-meanval(functiong)$, so that $\\int_0^1 zeromeanf(variablex)^2\\,dvariablex = \\Var(functionf)$, $\\int_0^1 zeromeang(variablex)^2\\,dvariablex = \\Var(functiong)$, and $\\int_0^1 zeromeanf(variablex)\\,dvariablex = \\int_0^1 zeromeang(variablex)\\,dvariablex = 0$. Now since $|functiong(variablex)| \\leq maxbound(functiong)$ for all $variablex$, $0\\leq \\int_0^1 zeromeanf(variablex)^2(maxbound(functiong)^2-functiong(variablex)^2)\\,dvariablex = \\Var(functionf) \\, maxbound(functiong)^2-\\int_0^1 zeromeanf(variablex)^2functiong(variablex)^2\\,dvariablex$, and similarly $0 \\leq \\Var(functiong)\\,maxbound(functionf)^2-\\int_0^1 functionf(variablex)^2zeromeang(variablex)^2\\,dvariablex$. Summing gives\n\\begin{equation}\n\\Var(functionf)\\,maxbound(functiong)^2+\\Var(functiong)\\,maxbound(functionf)^2\n\\label{eq:1}\n\\geq \\int_0^1 (zeromeanf(variablex)^2functiong(variablex)^2+functionf(variablex)^2zeromeang(variablex)^2)\\,dvariablex.\n\\end{equation}\nNow\n\\begin{align*}\n&\\int_0^1 (zeromeanf(variablex)^2functiong(variablex)^2+functionf(variablex)^2zeromeang(variablex)^2)\\,dvariablex-\\Var(functionf functiong) \\\\&= \\int_0^1 (zeromeanf(variablex)^2functiong(variablex)^2+functionf(variablex)^2zeromeang(variablex)^2-(functionf(variablex)functiong(variablex)-\\int_0^1 functionf(variabley)functiong(variabley)\\,dvariabley)^2)\\,dvariablex;\n\\end{align*}\nsubstituting $zeromeanf(variablex)+meanval(functionf)$ for $functionf(variablex)$ everywhere and $zeromeang(variablex)+meanval(functiong)$ for $functiong(variablex)$ everywhere, and using the fact that $\\int_0^1 zeromeanf(variablex)\\,dvariablex = \\int_0^1 zeromeang(variablex)\\,dvariablex = 0$, we can expand and simplify the right hand side of this equation to obtain\n\\begin{align*}\n&\\int_0^1 (zeromeanf(variablex)^2functiong(variablex)^2+functionf(variablex)^2zeromeang(variablex)^2)\\,dvariablex-\\Var(functionf functiong) \\\\\n&= \\int_0^1 zeromeanf(variablex)^2zeromeang(variablex)^2\\,dvariablex \\\\\n&-2meanval(functionf)meanval(functiong)\\int_0^1 zeromeanf(variablex)zeromeang(variablex)\\,dvariablex +(\\int_0^1 zeromeanf(variablex)zeromeang(variablex)\\,dvariablex)^2 \\\\\n&\\geq -2meanval(functionf)meanval(functiong)\\int_0^1 zeromeanf(variablex)zeromeang(variablex)\\,dvariablex.\n\\end{align*}\nBecause of \\eqref{eq:1}, it thus suffices to show that\n\\begin{equation}\n2meanval(functionf)meanval(functiong)\\int_0^1 zeromeanf(variablex)zeromeang(variablex)\\,dvariablex\n\\label{eq:3} \\leq \\Var(functionf)\\,maxbound(functiong)^2+\\Var(functiong)\\,maxbound(functionf)^2.\n\\end{equation}\nNow since $(meanval(functiong) \\; zeromeanf(variablex)-meanval(functionf) \\; zeromeang(variablex))^2 \\geq 0$ for all $variablex$, we have\n\\begin{align*}\n2meanval(functionf)meanval(functiong) \\int_0^1 zeromeanf(variablex)zeromeang(variablex)\\,dvariablex\n& \\leq \\int_0^1 (meanval(functiong)^2 zeromeanf(variablex)^2 + meanval(functionf)^2 zeromeang(variablex)^2) \\, dvariablex \\\\\n& = \\Var(functionf) \\, meanval(functiong)^2 + \\Var(functiong) \\, meanval(functionf)^2 \\\\\n& \\leq \\Var(functionf) \\, maxbound(functiong)^2 + \\Var(functiong) \\, maxbound(functionf)^2,\n\\end{align*}\nestablishing \\eqref{eq:3} and completing the proof."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "hummingbird",
        "g": "raincloud",
        "f_0": "sheepdog",
        "g_0": "paintbrush",
        "x": "lanternfish",
        "y": "driftwood",
        "M": "waterfall",
        "\\mu": "sunflower"
      },
      "question": "For any continuous real-valued function $hummingbird$ defined on the interval $[0,1]$, let\n\\begin{gather*}\nsunflower(hummingbird) = \\int_0^1 hummingbird(lanternfish)\\,d lanternfish, \\,\n\\mathrm{Var}(hummingbird) = \\int_0^1 (hummingbird(lanternfish) - sunflower(hummingbird))^2\\,d lanternfish, \\\\\nwaterfall(hummingbird) = \\max_{0 \\leq lanternfish \\leq 1} \\left| hummingbird(lanternfish) \\right|.\n\\end{gather*}\nShow that if $hummingbird$ and $raincloud$ are continuous real-valued functions\ndefined on the interval $[0,1]$, then\n\\[\n\\mathrm{Var}(hummingbird\\,raincloud) \\leq 2\\,\\mathrm{Var}(hummingbird)\\,waterfall(raincloud)^2 + 2\\,\\mathrm{Var}(raincloud)\\,waterfall(hummingbird)^2.\n\\]",
      "solution": "\\newcommand{\\Var}{\\mathrm{Var}}\n\nWrite $sheepdog(lanternfish)=hummingbird(lanternfish)-sunflower(hummingbird)$ and $paintbrush(lanternfish)=raincloud(lanternfish)-sunflower(raincloud)$, so that $\\int_0^1 sheepdog(lanternfish)^2\\,d lanternfish = \\Var(hummingbird)$, $\\int_0^1 paintbrush(lanternfish)^2\\,d lanternfish = \\Var(raincloud)$, and $\\int_0^1 sheepdog(lanternfish)\\,d lanternfish = \\int_0^1 paintbrush(lanternfish)\\,d lanternfish = 0$.  Now since $|raincloud(lanternfish)| \\leq waterfall(raincloud)$ for all $lanternfish$, we have\n$0 \\leq \\int_0^1 sheepdog(lanternfish)^2(\\,waterfall(raincloud)^2-raincloud(lanternfish)^2)\\,d lanternfish = \\Var(hummingbird)\\,waterfall(raincloud)^2-\\int_0^1 sheepdog(lanternfish)^2 raincloud(lanternfish)^2\\,d lanternfish$, and similarly $0 \\leq \\Var(raincloud)\\,waterfall(hummingbird)^2-\\int_0^1 hummingbird(lanternfish)^2 paintbrush(lanternfish)^2\\,d lanternfish$.  Summing gives\n\\begin{equation}\n\\Var(hummingbird)\\,waterfall(raincloud)^2+\\Var(raincloud)\\,waterfall(hummingbird)^2\\label{eq:1}\n\\geq \\int_0^1 \\bigl(sheepdog(lanternfish)^2 raincloud(lanternfish)^2+hummingbird(lanternfish)^2 paintbrush(lanternfish)^2\\bigr)\\,d lanternfish.\n\\end{equation}\n\nNow\n\\begin{align*}\n&\\int_0^1 \\bigl(sheepdog(lanternfish)^2 raincloud(lanternfish)^2+hummingbird(lanternfish)^2 paintbrush(lanternfish)^2\\bigr)\\,d lanternfish-\\Var(hummingbird\\,raincloud)\\\\\n&=\\int_0^1 \\Bigl(sheepdog(lanternfish)^2 raincloud(lanternfish)^2+hummingbird(lanternfish)^2 paintbrush(lanternfish)^2-\\bigl(hummingbird(lanternfish) raincloud(lanternfish)-\\int_0^1 hummingbird(driftwood) raincloud(driftwood)\\,d driftwood\\bigr)^2\\Bigr)\\,d lanternfish;\n\\end{align*}\nsubstituting $sheepdog(lanternfish)+sunflower(hummingbird)$ for $hummingbird(lanternfish)$ and $paintbrush(lanternfish)+sunflower(raincloud)$ for $raincloud(lanternfish)$, and using $\\int_0^1 sheepdog(lanternfish)\\,d lanternfish = \\int_0^1 paintbrush(lanternfish)\\,d lanternfish = 0$, we obtain\n\\begin{align*}\n&\\int_0^1 \\bigl(sheepdog(lanternfish)^2 raincloud(lanternfish)^2+hummingbird(lanternfish)^2 paintbrush(lanternfish)^2\\bigr)\\,d lanternfish-\\Var(hummingbird\\,raincloud)\\\\\n&= \\int_0^1 sheepdog(lanternfish)^2 paintbrush(lanternfish)^2\\,d lanternfish \\\\\n&\\quad -2\\,sunflower(hummingbird)\\,sunflower(raincloud)\\int_0^1 sheepdog(lanternfish) paintbrush(lanternfish)\\,d lanternfish +\\Bigl(\\int_0^1 sheepdog(lanternfish) paintbrush(lanternfish)\\,d lanternfish\\Bigr)^2 \\\\\n&\\ge -2\\,sunflower(hummingbird)\\,sunflower(raincloud)\\int_0^1 sheepdog(lanternfish) paintbrush(lanternfish)\\,d lanternfish.\n\\end{align*}\nBecause of \\eqref{eq:1}, it suffices to show that\n\\begin{equation}\n2\\,sunflower(hummingbird)\\,sunflower(raincloud)\\int_0^1 sheepdog(lanternfish) paintbrush(lanternfish)\\,d lanternfish\\label{eq:3}\n\\le \\Var(hummingbird)\\,waterfall(raincloud)^2+\\Var(raincloud)\\,waterfall(hummingbird)^2.\n\\end{equation}\nNow, since $(sunflower(raincloud)\\,sheepdog(lanternfish)-sunflower(hummingbird)\\,paintbrush(lanternfish))^2 \\ge 0$ for all $lanternfish$, we have\n\\begin{align*}\n2\\,sunflower(hummingbird)\\,sunflower(raincloud) \\int_0^1 sheepdog(lanternfish) paintbrush(lanternfish)\\,d lanternfish\n&\\le \\int_0^1 \\bigl(sunflower(raincloud)^2 sheepdog(lanternfish)^2 + sunflower(hummingbird)^2 paintbrush(lanternfish)^2\\bigr)\\,d lanternfish\\\\\n&= \\Var(hummingbird)\\,sunflower(raincloud)^2 + \\Var(raincloud)\\,sunflower(hummingbird)^2\\\\\n&\\le \\Var(hummingbird)\\,waterfall(raincloud)^2 + \\Var(raincloud)\\,waterfall(hummingbird)^2,\n\\end{align*}\nwhich establishes \\eqref{eq:3} and completes the proof."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "discontinuous",
        "g": "discrete",
        "f_0": "unbalanced",
        "g_0": "lopsided",
        "x": "constant",
        "y": "staticval",
        "M": "minvalue",
        "\\mu": "extremum"
      },
      "question": "For any continuous real-valued function $discontinuous$ defined on the interval $[0,1]$, let\n\\begin{gather*}\nextremum(discontinuous) = \\int_0^1 discontinuous(constant)\\,dconstant, \\,\n\\mathrm{Var}(discontinuous) = \\int_0^1 (discontinuous(constant) - extremum(discontinuous))^2\\,dconstant, \\\\\nminvalue(discontinuous) = \\max_{0 \\leq constant \\leq 1} \\left| discontinuous(constant) \\right|.\n\\end{gather*}\nShow that if $discontinuous$ and $discrete$ are continuous real-valued functions\ndefined on the interval $[0,1]$, then\n\\[\n\\mathrm{Var}(discontinuousdiscrete) \\leq 2 \\mathrm{Var}(discontinuous) minvalue(discrete)^2 + 2 \\mathrm{Var}(discrete) minvalue(discontinuous)^2.\n\\]\n",
      "solution": "\\newcommand{\\Var}{\\mathrm{Var}}\n\nWrite $unbalanced(constant) = discontinuous(constant)-extremum(discontinuous)$ and $lopsided(constant) = discrete(constant)-extremum(discrete)$, so that $\\int_0^1 unbalanced(constant)^2\\,dconstant = \\Var(discontinuous)$, $\\int_0^1 lopsided(constant)^2\\,dconstant = \\Var(discrete)$, and $\\int_0^1 unbalanced(constant)\\,dconstant = \\int_0^1 lopsided(constant)\\,dconstant = 0$. Now since $|discrete(constant)| \\leq minvalue(discrete)$ for all $constant$, $0\\leq \\int_0^1 unbalanced(constant)^2(minvalue(discrete)^2-discrete(constant)^2)\\,dconstant = \\Var(discontinuous) minvalue(discrete)^2-\\int_0^1 unbalanced(constant)^2discrete(constant)^2\\,dconstant$, and similarly $0 \\leq \\Var(discrete)minvalue(discontinuous)^2-\\int_0^1 discontinuous(constant)^2lopsided(constant)^2\\,dconstant$. Summing gives\n\\begin{equation}\n\\Var(discontinuous)minvalue(discrete)^2+\\Var(discrete)minvalue(discontinuous)^2\n\\label{eq:1}\n\\geq \\int_0^1 (unbalanced(constant)^2discrete(constant)^2+discontinuous(constant)^2lopsided(constant)^2)\\,dconstant.\n\\end{equation}\nNow\n\\begin{align*}\n&\\int_0^1 (unbalanced(constant)^2discrete(constant)^2+discontinuous(constant)^2lopsided(constant)^2)\\,dconstant-\\Var(discontinuousdiscrete) \\\\&= \\int_0^1 (unbalanced(constant)^2discrete(constant)^2+discontinuous(constant)^2lopsided(constant)^2-(discontinuous(constant)discrete(constant)-\\int_0^1 discontinuous(staticval)discrete(staticval)\\,dstaticval)^2)\\,dconstant;\n\\end{align*}\nsubstituting $unbalanced(constant)+extremum(discontinuous)$ for $discontinuous(constant)$ everywhere and $lopsided(constant)+extremum(discrete)$ for $discrete(constant)$ everywhere, and using the fact that $\\int_0^1 unbalanced(constant)\\,dconstant = \\int_0^1 lopsided(constant)\\,dconstant = 0$, we can expand and simplify the right hand side of this equation to obtain\n\\begin{align*}\n&\\int_0^1 (unbalanced(constant)^2discrete(constant)^2+discontinuous(constant)^2lopsided(constant)^2)\\,dconstant-\\Var(discontinuousdiscrete) \\\\\n&= \\int_0^1 unbalanced(constant)^2lopsided(constant)^2\\,dconstant \\\\\n&-2extremum(discontinuous)extremum(discrete)\\int_0^1 unbalanced(constant)lopsided(constant)\\,dconstant +(\\int_0^1 unbalanced(constant)lopsided(constant)\\,dconstant)^2 \\\\\n&\\geq -2extremum(discontinuous)extremum(discrete)\\int_0^1 unbalanced(constant)lopsided(constant)\\,dconstant.\n\\end{align*}\nBecause of \\eqref{eq:1}, it thus suffices to show that\n\\begin{equation}\n2extremum(discontinuous)extremum(discrete)\\int_0^1 unbalanced(constant)lopsided(constant)\\,dconstant\n\\label{eq:3} \\leq \\Var(discontinuous)minvalue(discrete)^2+\\Var(discrete)minvalue(discontinuous)^2.\n\\end{equation}\nNow since $(extremum(discrete) unbalanced(constant)-extremum(discontinuous) lopsided(constant))^2 \\geq 0$ for all $constant$, we have\n\\begin{align*}\n2extremum(discontinuous)extremum(discrete) \\int_0^1 unbalanced(constant)lopsided(constant)\\,dconstant\n& \\leq \\int_0^1 (extremum(discrete)^2 unbalanced(constant)^2 + extremum(discontinuous)^2 lopsided(constant)^2) dconstant \\\\\n& = \\Var(discontinuous) extremum(discrete)^2 + \\Var(discrete) extremum(discontinuous)^2 \\\\\n& \\leq \\Var(discontinuous) minvalue(discrete)^2 + \\Var(discrete) minvalue(discontinuous)^2,\n\\end{align*}\nestablishing \\eqref{eq:3} and completing the proof."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "g": "nvbshqel",
        "f_0": "kplmztwo",
        "g_0": "rdbfycuv",
        "x": "tslqjmrx",
        "y": "wzndecot",
        "M": "pbkrawmz",
        "\\mu": "hjgrksla"
      },
      "question": "For any continuous real-valued function $qzxwvtnp$ defined on the interval $[0,1]$, let\n\\begin{gather*}\nhjgrksla(qzxwvtnp) = \\int_0^1 qzxwvtnp(tslqjmrx)\\,d tslqjmrx, \\,\n\\mathrm{Var}(qzxwvtnp) = \\int_0^1 (qzxwvtnp(tslqjmrx) - hjgrksla(qzxwvtnp))^2\\,d tslqjmrx, \\\\\npbkrawmz(qzxwvtnp) = \\max_{0 \\leq tslqjmrx \\leq 1} \\left| qzxwvtnp(tslqjmrx) \\right|.\n\\end{gather*}\nShow that if $qzxwvtnp$ and $nvbshqel$ are continuous real-valued functions\ndefined on the interval $[0,1]$,\nthen\n\\[\n\\mathrm{Var}(qzxwvtnp nvbshqel) \\leq 2 \\mathrm{Var}(qzxwvtnp) pbkrawmz(nvbshqel)^2 + 2 \\mathrm{Var}(nvbshqel) pbkrawmz(qzxwvtnp)^2.\n\\]",
      "solution": "\\newcommand{\\Var}{\\mathrm{Var}}\n\nWrite $kplmztwo(tslqjmrx) = qzxwvtnp(tslqjmrx)-hjgrksla(qzxwvtnp)$ and $rdbfycuv(tslqjmrx) = nvbshqel(tslqjmrx)-hjgrksla(nvbshqel)$, so that $\\int_0^1 kplmztwo(tslqjmrx)^2\\,d tslqjmrx = \\Var(qzxwvtnp)$, $\\int_0^1 rdbfycuv(tslqjmrx)^2\\,d tslqjmrx = \\Var(nvbshqel)$, and $\\int_0^1 kplmztwo(tslqjmrx)\\,d tslqjmrx = \\int_0^1 rdbfycuv(tslqjmrx)\\,d tslqjmrx = 0$. Now since $|nvbshqel(tslqjmrx)| \\leq pbkrawmz(nvbshqel)$ for all $tslqjmrx$, $0\\leq \\int_0^1 kplmztwo(tslqjmrx)^2(pbkrawmz(nvbshqel)^2-nvbshqel(tslqjmrx)^2)\\,d tslqjmrx = \\Var(qzxwvtnp) pbkrawmz(nvbshqel)^2-\\int_0^1 kplmztwo(tslqjmrx)^2 nvbshqel(tslqjmrx)^2\\,d tslqjmrx$, and similarly $0 \\leq \\Var(nvbshqel) pbkrawmz(qzxwvtnp)^2-\\int_0^1 qzxwvtnp(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2\\,d tslqjmrx$. Summing gives\n\\begin{equation}\n\\Var(qzxwvtnp) pbkrawmz(nvbshqel)^2+\\Var(nvbshqel) pbkrawmz(qzxwvtnp)^2\n\\label{eq:1}\n\\geq \\int_0^1 (kplmztwo(tslqjmrx)^2 nvbshqel(tslqjmrx)^2+qzxwvtnp(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2)\\,d tslqjmrx.\n\\end{equation}\nNow\n\\begin{align*}\n&\\int_0^1 (kplmztwo(tslqjmrx)^2 nvbshqel(tslqjmrx)^2+qzxwvtnp(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2)\\,d tslqjmrx-\\Var(qzxwvtnp nvbshqel) \\\\&= \\int_0^1 (kplmztwo(tslqjmrx)^2 nvbshqel(tslqjmrx)^2+qzxwvtnp(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2-(qzxwvtnp(tslqjmrx) nvbshqel(tslqjmrx)-\\int_0^1 qzxwvtnp(wzndecot) nvbshqel(wzndecot)\\,d wzndecot)^2)\\,d tslqjmrx;\n\\end{align*}\nsubstituting $kplmztwo(tslqjmrx)+hjgrksla(qzxwvtnp)$ for $qzxwvtnp(tslqjmrx)$ everywhere and $rdbfycuv(tslqjmrx)+hjgrksla(nvbshqel)$ for $nvbshqel(tslqjmrx)$ everywhere, and using the fact that $\\int_0^1 kplmztwo(tslqjmrx)\\,d tslqjmrx = \\int_0^1 rdbfycuv(tslqjmrx)\\,d tslqjmrx = 0$, we can expand and simplify the right hand side of this equation to obtain\n\\begin{align*}\n&\\int_0^1 (kplmztwo(tslqjmrx)^2 nvbshqel(tslqjmrx)^2+qzxwvtnp(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2)\\,d tslqjmrx-\\Var(qzxwvtnp nvbshqel) \\\\\n&= \\int_0^1 kplmztwo(tslqjmrx)^2 rdbfycuv(tslqjmrx)^2\\,d tslqjmrx \\\\\n&-2 hjgrksla(qzxwvtnp) hjgrksla(nvbshqel)\\int_0^1 kplmztwo(tslqjmrx) rdbfycuv(tslqjmrx)\\,d tslqjmrx +\\left(\\int_0^1 kplmztwo(tslqjmrx) rdbfycuv(tslqjmrx)\\,d tslqjmrx\\right)^2 \\\\\n&\\geq -2 hjgrksla(qzxwvtnp) hjgrksla(nvbshqel)\\int_0^1 kplmztwo(tslqjmrx) rdbfycuv(tslqjmrx)\\,d tslqjmrx.\n\\end{align*}\nBecause of \\eqref{eq:1}, it thus suffices to show that\n\\begin{equation}\n2 hjgrksla(qzxwvtnp) hjgrksla(nvbshqel)\\int_0^1 kplmztwo(tslqjmrx) rdbfycuv(tslqjmrx)\\,d tslqjmrx\n\\label{eq:3} \\leq \\Var(qzxwvtnp) pbkrawmz(nvbshqel)^2+\\Var(nvbshqel) pbkrawmz(qzxwvtnp)^2.\n\\end{equation}\nNow since $(hjgrksla(nvbshqel) kplmztwo(tslqjmrx)-hjgrksla(qzxwvtnp) rdbfycuv(tslqjmrx))^2 \\geq 0$ for all $tslqjmrx$, we have\n\\begin{align*}\n2 hjgrksla(qzxwvtnp) hjgrksla(nvbshqel) \\int_0^1 kplmztwo(tslqjmrx) rdbfycuv(tslqjmrx)\\,d tslqjmrx\n& \\leq \\int_0^1 (hjgrksla(nvbshqel)^2 kplmztwo(tslqjmrx)^2 + hjgrksla(qzxwvtnp)^2 rdbfycuv(tslqjmrx)^2) d tslqjmrx \\\\\n& = \\Var(qzxwvtnp) hjgrksla(nvbshqel)^2 + \\Var(nvbshqel) hjgrksla(qzxwvtnp)^2 \\\\\n& \\leq \\Var(qzxwvtnp) pbkrawmz(nvbshqel)^2 + \\Var(nvbshqel) pbkrawmz(qzxwvtnp)^2,\n\\end{align*}\nestablishing \\eqref{eq:3} and completing the proof."
    },
    "kernel_variant": {
      "question": "Let $(M,g)$ be a connected, compact, $n$–dimensional Riemannian manifold without boundary and normalise the Riemannian volume so that  \n\\[\n\\int_{M}\\mathrm dV = 1 .\n\\]\n\nFor every $u\\in W^{1,2}(M)\\cap L^{\\infty}(M)$ define  \n\\[\n\\mu(u)=\\int_{M}u\\,\\mathrm dV ,\\qquad\n\\operatorname{Var}(u)=\\int_{M}\\bigl(u-\\mu(u)\\bigr)^{2}\\,\\mathrm dV ,\\qquad\n\\operatorname{Dir}(u)=\\int_{M}\\lvert\\nabla u\\rvert^{2}\\,\\mathrm dV ,\n\\]\nand put $M(u)=\\operatorname*{ess\\,sup}_{x\\in M}\\lvert u(x)\\rvert$.\n\nDenote by $\\lambda_{1}>0$ the first non–zero eigenvalue of the\n(positive) Laplace–Beltrami operator $-\\Delta$; equivalently,\n$\\lambda_{1}$ is the best constant in the Poincar\\'e inequality  \n\\[\n\\operatorname{Var}(u)\\le \\lambda_{1}^{-1}\\operatorname{Dir}(u)\n\\qquad(u\\in W^{1,2}(M)).\\tag{$\\star$}\n\\]\n\n1.\\;(\\emph{Product estimates})  \nShow that for all $f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)$\n\\begin{align*}\n\\text{\\rm(a)}\\quad &\\operatorname{Var}(fg)\\le 2M(g)^{2}\\operatorname{Var}(f)\n                                   +2M(f)^{2}\\operatorname{Var}(g),\\\\[2mm]\n\\text{\\rm(b)}\\quad &\\operatorname{Dir}(fg)\\le 2M(g)^{2}\\operatorname{Dir}(f)\n                                   +2M(f)^{2}\\operatorname{Dir}(g).\n\\end{align*}\n\n2.\\;Deduce the coupled variance–energy estimate\n\\[\n\\operatorname{Var}(fg)+\\lambda_{1}^{-1}\\operatorname{Dir}(fg)\n     \\le 2M(g)^{2}\\Bigl[\\operatorname{Var}(f)+\\lambda_{1}^{-1}\\operatorname{Dir}(f)\\Bigr]\n         +2M(f)^{2}\\Bigl[\\operatorname{Var}(g)+\\lambda_{1}^{-1}\\operatorname{Dir}(g)\\Bigr].\n\\tag{1}\n\\]\n\n3.\\;Introduce the functional\n\\[\n\\mathcal Q(u)=\\operatorname{Var}(u)+\\lambda_{1}^{-1}\\operatorname{Dir}(u)\n\\qquad(u\\in W^{1,2}(M)\\cap L^{\\infty}(M))\n\\]\nand show\n\\[\n\\mathcal Q(fg)\\le 2\\bigl[M(g)^{2}\\mathcal Q(f)+M(f)^{2}\\mathcal Q(g)\\bigr],\\tag{2a}\n\\]\n\\[\n\\mathcal Q(fg)\\le 2\\max\\{M(f)^{2},M(g)^{2}\\}\\,\\bigl[\\mathcal Q(f)+\\mathcal Q(g)\\bigr].\n\\tag{2b}\n\\]\n\n4.\\;(\\emph{Equality analysis under additional regularity})  \nAssume in addition that $f,g\\in W^{1,2}(M)\\cap C(M)$.\nProve that equality holds in $(1)$ (equivalently, in $(2a)$)\nif and only if one of the following \\emph{disjoint} situations occurs:\n\\begin{itemize}\n\\item[(i)] Exactly one of the two functions is identically zero almost everywhere;\n\\item[(ii)] Neither function is identically zero almost everywhere and both are (essentially) constant on $M$  \n            (hence $\\operatorname{Var}(f)=\\operatorname{Var}(g)=\n            \\operatorname{Dir}(f)=\\operatorname{Dir}(g)=0$).\n\\end{itemize}\n\n5.\\;(\\emph{Sharpness of the constant})  \nShow that the coefficient “$2$” in $(1)$–$(2)$ cannot be replaced by any number\nstrictly smaller than $1$.  (Whether the optimal simultaneous\ncoefficient equals $1$ or $2$ remains open.)\n\n\\bigskip",
      "solution": "Throughout let $f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)$ and set  \n\\[\nf_{0}=f-\\mu(f), \\qquad g_{0}=g-\\mu(g).\n\\]\n\n\\paragraph{Step 1.  The variance bound (1a).}\nThe identity\n\\[\n\\operatorname{Var}(Z)=\\inf_{c\\in\\mathbb R}\\int_{M}(Z-c)^{2}\\,\\mathrm dV\n\\tag{3}\n\\]\nimplies, upon choosing $c=\\mu(f)\\mu(g)$ for $Z=fg$,\n\\[\n\\operatorname{Var}(fg)\n      \\le \\int_{M}\\bigl(fg-\\mu(f)\\mu(g)\\bigr)^{2}\\,\\mathrm dV .\n\\tag{4}\n\\]\nBecause $fg-\\mu(f)\\mu(g)=f_{0}g+\\mu(f)g_{0}$, expanding the square gives the integral of\n\\[\nf_{0}^{2}g^{2}+\\mu(f)^{2}g_{0}^{2}+2\\mu(f)f_{0}gg_{0}.\n\\]\nEstimating the three terms separately,\n\n\\[\n\\int_{M}f_{0}^{2}g^{2}\\,\\mathrm dV\\le M(g)^{2}\\operatorname{Var}(f),\\tag{5}\n\\]\n\\[\n\\mu(f)^{2}\\int_{M}g_{0}^{2}\\,\\mathrm dV\\le M(f)^{2}\\operatorname{Var}(g),\\tag{6}\n\\]\n\\[\n\\bigl|2\\mu(f)\\!\\int_{M}f_{0}gg_{0}\\,\\mathrm dV\\bigr|\n  \\le 2\\lvert\\mu(f)\\rvert M(g)\\sqrt{\\operatorname{Var}(f)\\operatorname{Var}(g)}\n  \\le M(g)^{2}\\operatorname{Var}(f)+M(f)^{2}\\operatorname{Var}(g).\\tag{7}\n\\]\nAdding (5)–(7) and inserting in (4) yields (1a).\n\n\\paragraph{Step 2.  The Dirichlet bound (1b).}\nThe product rule and the elementary inequality $\\lvert A+B\\rvert^{2}\\le 2(\\lvert A\\rvert^{2}+\\lvert B\\rvert^{2})$ give\n\\[\n\\lvert\\nabla(fg)\\rvert^{2}\n\\le 2\\lvert g\\rvert^{2}\\lvert\\nabla f\\rvert^{2}\n  +2\\lvert f\\rvert^{2}\\lvert\\nabla g\\rvert^{2}\n\\le 2M(g)^{2}\\lvert\\nabla f\\rvert^{2}\n  +2M(f)^{2}\\lvert\\nabla g\\rvert^{2}.\n\\]\nIntegrating proves (1b).\n\n\\paragraph{Step 3.  The coupled inequality (1) and the $\\mathcal Q$–bounds (2a), (2b).}\nAdd (1a) to $\\lambda_{1}^{-1}$ times (1b).  Writing the result with $\\mathcal Q(u)=\\operatorname{Var}(u)+\\lambda_{1}^{-1}\\operatorname{Dir}(u)$ gives (2a); the bound (2b) is immediate from  \n$M(f)^{2},M(g)^{2}\\le\\max\\{M(f)^{2},M(g)^{2}\\}$.\n\n\\paragraph{Step 4.  Equality analysis under $f,g\\in W^{1,2}(M)\\cap C(M)$.}\nAssume equality holds in (1).\\\\[-2mm]\n\n\\emph{4.1 Propagation of equalities.}\nAll inequalities used so far are non–negative; hence each of them must in fact be an equality.  \nBesides (5)–(7) and the Cauchy–Schwarz estimate, two further\nequalities are needed:\n\n\\begin{itemize}\n\\item[(E1)] equality in (3) with $Z=fg$ and $c=\\mu(f)\\mu(g)$, i.e.\n\\[\n\\mu(fg)=\\mu(f)\\mu(g); \\tag{8}\n\\]\n\\item[(E2)] equality in the pointwise estimate\n$\\lvert g\\nabla f+f\\nabla g\\rvert^{2}\\le 2\\lvert g\\rvert^{2}\\lvert\\nabla f\\rvert^{2}+2\\lvert f\\rvert^{2}\\lvert\\nabla g\\rvert^{2}$, namely\n\\[\ng\\nabla f = f\\nabla g\\qquad\\text{a.e.\\ on }M.\\tag{9}\n\\]\n\\end{itemize}\n\n\\emph{4.2 Consequences of (9): proportionality.}\nIf $g\\equiv 0$ almost everywhere we are in case (i); similarly if $f\\equiv 0$.  \nAssume henceforth $f,g$ are both non–zero.  \nOn $\\{g\\neq 0\\}$ equation (9) implies $\\nabla (f/g)=0$ a.e.; continuity and connectedness yield  \n\\[\nf=c\\,g\\qquad\\text{for some constant }c\\in\\mathbb R. \\tag{10}\n\\]\n\n\\emph{4.3 Elimination of non–constant proportional pairs.}\nWith $f=cg$ we have $\\mu(f)=c\\mu(g)$ and $fg=cg^{2}$.  \nCondition (8) therefore becomes\n\\[\nc\\mu(g^{2}) = c\\mu(g)^{2}. \\tag{11}\n\\]\nIf $c\\neq 0$ we conclude $\\mu(g^{2})=\\mu(g)^{2}$; hence $\\operatorname{Var}(g)=0$.  \nSince $g$ is continuous on $M$ and has zero variance, $g$ is constant, and so is $f=cg$.  \nIf $c=0$, then $f\\equiv 0$ and we are back in case (i).\n\n\\emph{4.4 Conclusion.}\nEquality in (1) implies either alternative (i) or (ii).  \nConversely, if (i) or (ii) holds, then either $fg\\equiv 0$ or both\n$\\operatorname{Var}(fg)$ and $\\operatorname{Dir}(fg)$ vanish, making every step above an equality; thus (1) holds with equality.\n\n\\paragraph{Step 5.  Sharpness of the coefficient “$2$”.}\nAssume that for some $0<\\gamma<1$\n\\[\n\\mathcal Q(fg)\\le 2\\gamma\n      \\bigl[M(g)^{2}\\mathcal Q(f)+M(f)^{2}\\mathcal Q(g)\\bigr]\n      \\qquad(\\forall f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)).\n\\tag{12}\n\\]\nChoose $g\\equiv 1$ (so $M(g)=1$ and $\\mathcal Q(g)=0$) and let $f=\\varphi$\nbe a non–constant first eigenfunction of $-\\Delta$, normalised by $M(\\varphi)=1$.\nThen $fg=\\varphi$ and $\\mathcal Q(\\varphi)>0$, so (12) gives\n\\[\n\\mathcal Q(\\varphi)\\le 2\\gamma\\,\\mathcal Q(\\varphi),\n\\]\nhence $2\\gamma\\ge 1$.  Therefore the universal constant cannot be\nsmaller than $1$.\n\n\\bigskip\nAll items of the corrected problem are rigorously proved.\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.830342",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher‐dimensional geometric setting: the domain is now an arbitrary compact Riemannian manifold, demanding familiarity with differential geometry and the Laplace–Beltrami operator.  \n2. Additional functional analytic structure: the inequality involves both the variance (an L² quantity) and the Dirichlet energy (an H¹–seminorm), linked by the Poincaré inequality and by the first eigenvalue λ₁.  \n3. Interaction of several analytical tools: the proof requires the product rule on manifolds, the sharp Poincaré inequality, L^{∞}–control, and careful bookkeeping of constants.  \n4. Sharpened constants and equality discussion: obtaining the factor 4 (instead of 2) obliges a second use of Poincaré and a subtle argument; characterising equality needs spectral theory of −Δ.  \n5. Non-Euclidean phenomena: nothing can be reduced to simple one-dimensional calculus by a change of variables—curvature plays no role explicitly but the manifold framework prevents easy “pattern-matching” with standard Euclidean proofs.\n\nAll these features make the enhanced variant significantly harder than both the original problem and the current kernel variant, requiring deeper insight into Sobolev spaces on manifolds, spectral theory, and careful handling of geometric constants."
      }
    },
    "original_kernel_variant": {
      "question": "Let \\((M,g)\\) be a connected, compact, \\(n\\)-dimensional Riemannian manifold without boundary and normalise the Riemannian volume so that  \n\\[\n\\int_{M}\\mathrm dV=1 .\n\\]\nFor every \\(u\\in W^{1,2}(M)\\cap L^{\\infty}(M)\\) define  \n\\[\n\\mu(u)=\\int_{M}u\\,\\mathrm dV ,\\qquad\n\\operatorname{Var}(u)=\\int_{M}(u-\\mu(u))^{2}\\,\\mathrm dV ,\\qquad\n\\operatorname{Dir}(u)=\\int_{M}|\\nabla u|^{2}\\,\\mathrm dV ,\n\\]\nand put \\(M(u)=\\operatorname*{ess\\,sup}_{x\\in M}|u(x)|\\).\nDenote by \\(\\lambda_{1}>0\\) the first non–zero eigenvalue of the\n(positive) Laplace–Beltrami operator \\(-\\Delta\\); equivalently,\n\\(\\lambda_{1}\\) is the best constant in the Poincaré inequality  \n\\[\n\\operatorname{Var}(u)\\le\\lambda_{1}^{-1}\\operatorname{Dir}(u)\n\\qquad(u\\in W^{1,2}(M)).\\tag{$\\star$}\n\\]\n\n1.\\;(\\emph{Product estimates})  \nShow that for all \\(f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)\\)\n\\begin{align*}\n\\text{\\rm(a)}\\quad &\\operatorname{Var}(fg)\\le 2M(g)^{2}\\operatorname{Var}(f)\n                                   +2M(f)^{2}\\operatorname{Var}(g),\\\\[2mm]\n\\text{\\rm(b)}\\quad &\\operatorname{Dir}(fg)\\le 2M(g)^{2}\\operatorname{Dir}(f)\n                                   +2M(f)^{2}\\operatorname{Dir}(g).\n\\end{align*}\n\n2.\\;Deduce the coupled variance–energy estimate\n\\[\n\\operatorname{Var}(fg)+\\lambda_{1}^{-1}\\operatorname{Dir}(fg)\n     \\le 2M(g)^{2}\\Bigl[\\operatorname{Var}(f)+\\lambda_{1}^{-1}\\operatorname{Dir}(f)\\Bigr]\n         +2M(f)^{2}\\Bigl[\\operatorname{Var}(g)+\\lambda_{1}^{-1}\\operatorname{Dir}(g)\\Bigr].\n\\tag{1}\n\\]\n\n3.\\;Introduce the functional\n\\[\n\\mathcal Q(u)=\\operatorname{Var}(u)+\\lambda_{1}^{-1}\\operatorname{Dir}(u)\n\\qquad(u\\in W^{1,2}(M)\\cap L^{\\infty}(M))\n\\]\nand show\n\\[\n\\mathcal Q(fg)\\le 2\\bigl[M(g)^{2}\\mathcal Q(f)+M(f)^{2}\\mathcal Q(g)\\bigr],\\tag{2a}\n\\]\n\\[\n\\mathcal Q(fg)\\le 2\\max\\{M(f)^{2},M(g)^{2}\\}\\,\\bigl[\\mathcal Q(f)+\\mathcal Q(g)\\bigr].\n\\tag{2b}\n\\]\n\n4.\\;(\\emph{Equality analysis under additional regularity})  \nAssume in addition that \\(f,g\\in W^{1,2}(M)\\cap C(M)\\).\nProve that equality holds in \\((1)\\) (equivalently, in \\((2a)\\))\nif and only if one of the following mutually exclusive situations occurs:\n\\begin{itemize}\n\\item[(i)] \\(f\\equiv 0\\) almost everywhere or \\(g\\equiv 0\\) almost everywhere;\n\\item[(ii)] both \\(f\\) and \\(g\\) are (essentially) constant on \\(M\\)  \n            (hence \\(\\operatorname{Var}(f)=\\operatorname{Var}(g)=\n            \\operatorname{Dir}(f)=\\operatorname{Dir}(g)=0\\)).\n\\end{itemize}\n\n5.\\;(\\emph{Sharpness of the constant})  \nShow that the coefficient “\\(2\\)” in \\((1)\\)–\\((2)\\) cannot be replaced by any number\nstrictly smaller than \\(1\\).  (Whether the optimal simultaneous\ncoefficient equals \\(1\\) or \\(2\\) remains open.)\n\n\\bigskip",
      "solution": "Throughout let \\(f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)\\) be fixed and denote  \n\\(f_{0}=f-\\mu(f)\\), \\(g_{0}=g-\\mu(g)\\).\n\n\\paragraph{Step 1.  The variance bound (1a).}\nRecall the elementary identity\n\\[\n\\operatorname{Var}(Z)=\\inf_{c\\in\\mathbb R}\\int_{M}(Z-c)^{2}\\,\\mathrm dV .\n\\tag{3}\n\\]\nChoosing \\(c=\\mu(f)\\mu(g)\\) for \\(Z=fg\\) gives\n\\[\n\\operatorname{Var}(fg)\\le\\int_{M}(fg-\\mu(f)\\mu(g))^{2}\\,\\mathrm dV .\n\\tag{4}\n\\]\nBecause \\(fg-\\mu(f)\\mu(g)=f_{0}g+\\mu(f)g_{0}\\), expanding the square yields\n\\[\n\\int_{M}f_{0}^{2}g^{2}\\,\\mathrm dV\n+\\mu(f)^{2}\\int_{M}g_{0}^{2}\\,\\mathrm dV\n+2\\mu(f)\\int_{M}f_{0}gg_{0}\\,\\mathrm dV .\n\\tag{5}\n\\]\nFirst term: \\(\\displaystyle \n\\int_{M}f_{0}^{2}g^{2}\\,\\mathrm dV\\le M(g)^{2}\\operatorname{Var}(f)\\).\n\\tag{6}\n\nSecond term: since \\(|\\mu(f)|\\le M(f)\\),\n\\(\\displaystyle\n\\mu(f)^{2}\\int_{M}g_{0}^{2}\\,\\mathrm dV\\le M(f)^{2}\\operatorname{Var}(g).\\)\n\\tag{7}\n\nCross term: by Cauchy–Schwarz,\n\\[\n\\Bigl|\\int_{M}f_{0}gg_{0}\\,\\mathrm dV\\Bigr|\n      \\le M(g)\\sqrt{\\operatorname{Var}(f)\\operatorname{Var}(g)} .\n\\]\nWith\n\\(a=M(g)\\sqrt{\\operatorname{Var}(f)}\\) and \n\\(b=|\\mu(f)|\\sqrt{\\operatorname{Var}(g)}\\),\n\\[\n2|ab|\\le a^{2}+b^{2}\\le M(g)^{2}\\operatorname{Var}(f)+M(f)^{2}\\operatorname{Var}(g).\n\\tag{8}\n\\]\n\nAdding \\((6)\\)–\\((8)\\) and inserting in \\((4)\\) gives the claimed\nvariance inequality.\n\n\\paragraph{Step 2.  The Dirichlet bound (1b).}\nUsing the product rule and the elementary bound\n\\(|A+B|^{2}\\le 2(|A|^{2}+|B|^{2})\\),\n\\[\n|\\nabla(fg)|^{2}=|g\\nabla f+f\\nabla g|^{2}\n\\le 2|g|^{2}|\\nabla f|^{2}+2|f|^{2}|\\nabla g|^{2}\n\\le 2M(g)^{2}|\\nabla f|^{2}+2M(f)^{2}|\\nabla g|^{2}.\n\\]\nIntegrating completes the proof of (1b).\n\n\\paragraph{Step 3.  The coupled estimate (1).}\nAdd (1a) to \\(\\lambda_{1}^{-1}\\) times (1b).  The right–hand side\nfactorises exactly as in the statement, proving (1).\n\n\\paragraph{Step 4.  Inequalities (2a)–(2b).}\nWith \\(\\mathcal Q(u)=\\operatorname{Var}(u)+\\lambda_{1}^{-1}\\operatorname{Dir}(u)\\),\nrelation (1) reads \\(\\mathcal Q(fg)\\le 2\\bigl[M(g)^{2}\\mathcal Q(f)+M(f)^{2}\\mathcal Q(g)\\bigr]\\),\ni.e.\\ (2a).  Clearly\n\\(M(g)^{2}\\le\\max\\{M(f)^{2},M(g)^{2}\\}\\) and similarly for \\(M(f)\\),\ngiving (2b).\n\n\\paragraph{Step 5.  Equality analysis (additional continuity).}\nAssume now \\(f,g\\in W^{1,2}(M)\\cap C(M)\\) and equality occurs in (1).\nThe argument is organised in three sub–steps.\n\n\\medskip\\noindent\n\\emph{5.1.  Equalities propagate.}\nAll differences introduced in Steps 1–3 are \\emph{non–negative}. \nHence equality in the \\emph{sum} can occur only if \\emph{each}\ninequality used along the way is an equality.  In particular  \n\\begin{itemize}\n\\item[(a)] equalities must hold in \\((6)\\) and \\((7)\\);  \n\\item[(b)] the Cauchy–Schwarz inequality and the elementary estimate\n\\(2ab\\le a^{2}+b^{2}\\) must be sharp;  \n\\item[(c)] equality must hold in the pointwise estimate\n\\(|g\\nabla f+f\\nabla g|^{2}\\le 2|g|^{2}|\\nabla f|^{2}+2|f|^{2}|\\nabla g|^{2}\\).\n\\end{itemize}\n\n\\medskip\\noindent\n\\emph{5.2.  Consequences of (a) and (b).}\nEquality in \\((6)\\) implies\n\\[\n|g(x)|=M(g)\\quad\\text{for a.e.\\ }x\\text{ with }f_{0}(x)\\neq 0,\n\\tag{9}\n\\]\nwhile equality in \\((7)\\) yields\n\\[\n|f(x)|=M(f)\\quad\\text{for a.e.\\ }x\\text{ with }g_{0}(x)\\neq 0.\\tag{10}\n\\]\nSharpness of Cauchy–Schwarz forces\n\\[\nf_{0}g\\quad\\text{and}\\quad g_{0}\\ \\text{to be proportional in }L^{2},\\tag{11}\n\\]\nand \\(2ab=a^{2}+b^{2}\\) gives the numerical condition\n\\[\nM(g)^{2}\\operatorname{Var}(f)=\\mu(f)^{2}\\operatorname{Var}(g).\\tag{12}\n\\]\n\n\\medskip\\noindent\n\\emph{5.3.  Consequences of (c).}\nEquality in the vector inequality\n\\(|A+B|^{2}\\le 2(|A|^{2}+|B|^{2})\\) holds iff \\(A=B\\).  Thus\n\\[\ng\\nabla f=f\\nabla g\\qquad\\text{a.e.\\ on }M.\\tag{13}\n\\]\nOn the set where \\(g\\neq 0\\) we may rewrite \\((13)\\) as\n\\[\n\\nabla\\!\\Bigl(\\frac{f}{g}\\Bigr)=0\\qquad\\text{a.e.}\\tag{14}\n\\]\nBecause \\(M\\) is connected and \\(f/g\\) is continuous,\n\\((14)\\) implies\n\\[\n\\frac{f}{g}\\equiv c\\quad\\text{on }M\\setminus\\{g=0\\}\\quad\\text{for some constant }c\\in\\mathbb R.\\tag{15}\n\\]\nIf \\(g\\equiv 0\\) a.e.\\ we are in case (i), and there is nothing\nto prove.  Hence assume that\n\\(g\\not\\equiv 0\\).\nEquation \\((15)\\) shows \\(f=cg\\) everywhere; substituting in \\((13)\\)\nwe obtain \\(g\\nabla(cg)=cg\\nabla g\\), whence\n\\[\n(1-c^{2})\\,g\\nabla g=0\\qquad\\text{a.e.}\\tag{16}\n\\]\nIf \\(c^{2}\\neq 1\\) then \\((16)\\) gives \\(\\nabla g=0\\) a.e., hence\n\\(\\operatorname{Dir}(g)=0\\), which together with \\((\\star)\\) implies\n\\(\\operatorname{Var}(g)=0\\); thus \\(g\\) is constant and \\(f=cg\\) is\nconstant as well.  \nIf \\(c^{2}=1\\) then \\(f=\\pm g\\).  Equality in \\((9)\\)–\\((10)\\) and\ncontinuity of \\(g\\) force \\(|g|\\equiv M(g)\\), i.e.\\ \\(g\\) itself is\nconstant (sign may be \\(\\pm\\)).  Again \\(f\\) is constant.\n\nConsequently case (ii) is obtained.  The converse (both constant or one\nfunction identically zero implies equality) is immediate because then\n\\(\\operatorname{Var}(fg)=\\operatorname{Dir}(fg)=0\\).\n\n\\paragraph{Step 6.  Sharpness of the factor “\\(2\\)”.}\nAssume that for some \\(0<\\gamma<1\\)\n\\[\n\\mathcal Q(fg)\\le 2\\gamma\\,\n      \\bigl[M(g)^{2}\\mathcal Q(f)+M(f)^{2}\\mathcal Q(g)\\bigr]\n      \\qquad(\\forall f,g\\in W^{1,2}(M)\\cap L^{\\infty}(M)).\n\\tag{17}\n\\]\nChoose \\(g\\equiv 1\\) (hence \\(M(g)=1\\) and \\(\\mathcal Q(g)=0\\))\nand let \\(f=\\varphi\\) be any (non–constant) first eigenfunction of\n\\(-\\Delta\\) normalised so that \\(M(\\varphi)=1\\).\nThen \\(fg=\\varphi\\) and \\(\\mathcal Q(\\varphi)>0\\), so \\((17)\\) yields\n\\[\n\\mathcal Q(\\varphi)\\le 2\\gamma\\,\\mathcal Q(\\varphi),\n\\]\nhence \\(2\\gamma\\ge 1\\).  Therefore the universal coefficient cannot be\npushed below \\(1\\).\n\n\\bigskip\nAll items of the corrected problem are now rigorously proved.\n\n\\bigskip",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.634165",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher‐dimensional geometric setting: the domain is now an arbitrary compact Riemannian manifold, demanding familiarity with differential geometry and the Laplace–Beltrami operator.  \n2. Additional functional analytic structure: the inequality involves both the variance (an L² quantity) and the Dirichlet energy (an H¹–seminorm), linked by the Poincaré inequality and by the first eigenvalue λ₁.  \n3. Interaction of several analytical tools: the proof requires the product rule on manifolds, the sharp Poincaré inequality, L^{∞}–control, and careful bookkeeping of constants.  \n4. Sharpened constants and equality discussion: obtaining the factor 4 (instead of 2) obliges a second use of Poincaré and a subtle argument; characterising equality needs spectral theory of −Δ.  \n5. Non-Euclidean phenomena: nothing can be reduced to simple one-dimensional calculus by a change of variables—curvature plays no role explicitly but the manifold framework prevents easy “pattern-matching” with standard Euclidean proofs.\n\nAll these features make the enhanced variant significantly harder than both the original problem and the current kernel variant, requiring deeper insight into Sobolev spaces on manifolds, spectral theory, and careful handling of geometric constants."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}