{
  "index": "1957-B-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "3. For \\( f(x) \\) a positive, monotone decreasing function defined in \\( 0 \\leq x \\leq 1 \\) prove that\n\\[\n\\frac{\\int_{0}^{1} x f^{2}(x) d x}{\\int_{0}^{1} x f(x) d x} \\leq \\frac{\\int_{0}^{1} f^{2}(x) d x}{\\int_{0}^{1} f(x) d x}\n\\]",
  "solution": "Solution. The desired inequality is equivalent to\n\\[\n\\int_{0}^{1} f^{2}(x) d x \\int_{0}^{1} y f(y) d y-\\int_{0}^{1} y f^{2}(y) d y \\int_{0}^{1} f(x) d x \\geq 0\n\\]\nwhich can be rewritten\n\\[\n\\int_{0}^{1} \\int_{0}^{1} f(x) f(y) y[f(x)-f(y)] d x d y \\geq 0\n\\]\n\nDenote the left member of (1) by \\( I \\). Then\n\\[\nI=\\int_{0}^{1} \\int_{0}^{1} f(x) f(y) x[f(y)-f(x)] d x d y .\n\\]\n(We have interchanged the variables of integration and then the order of integration.) Hence\n\\[\n2 I=\\int_{0}^{1} \\int_{0}^{1} f(x) f(y)(y-x)[f(x)-f(y)] d x d y .\n\\]\n\nBecause \\( f \\) is decreasing, \\( (y-x)[f(x)-f(y)] \\geq 0 \\) for all \\( x \\) and \\( y \\). Then since \\( f \\) is everywhere positive, it is clear that \\( 2 I \\geq 0 \\). This proves (1).\n\nRemark. The argument given generalizes to prove\n\\[\n\\int_{0}^{1} f(x) g(x) w(x) d x \\int_{0}^{1} w(x) d x \\leq \\int_{0}^{1} f(x) w(x) d x \\int_{0}^{1} g(x) w(x) d x\n\\]\nwhenever \\( f \\) is decreasing, \\( g \\) is increasing, and \\( w \\) is a non-negative weight function (assuming, of course, that the integrals exist). Moreover, the inequality is strict unless either \\( f \\) or \\( g \\) is constant over the support of \\( w \\) (i.e., the closure of the set \\( \\{x: w(x)>0\\} \\) ). The present problem is the special case \\( g(x)=x, w(x)=f(x) \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "f",
    "g",
    "w",
    "I"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "f": "functionf",
        "g": "functiong",
        "w": "weightw",
        "I": "integralv"
      },
      "question": "3. For \\( functionf(variablex) \\) a positive, monotone decreasing function defined in \\( 0 \\leq variablex \\leq 1 \\) prove that\n\\[\n\\frac{\\int_{0}^{1} variablex functionf^{2}(variablex) d variablex}{\\int_{0}^{1} variablex functionf(variablex) d variablex} \\leq \\frac{\\int_{0}^{1} functionf^{2}(variablex) d variablex}{\\int_{0}^{1} functionf(variablex) d variablex}\n\\]",
      "solution": "Solution. The desired inequality is equivalent to\n\\[\n\\int_{0}^{1} functionf^{2}(variablex) d variablex \\int_{0}^{1} variabley functionf(variabley) d variabley-\\int_{0}^{1} variabley functionf^{2}(variabley) d variabley \\int_{0}^{1} functionf(variablex) d variablex \\geq 0\n\\]\nwhich can be rewritten\n\\[\n\\int_{0}^{1} \\int_{0}^{1} functionf(variablex) functionf(variabley) variabley[functionf(variablex)-functionf(variabley)] d variablex d variabley \\geq 0\n\\]\n\nDenote the left member of (1) by \\( integralv \\). Then\n\\[\nintegralv=\\int_{0}^{1} \\int_{0}^{1} functionf(variablex) functionf(variabley) variablex[functionf(variabley)-functionf(variablex)] d variablex d variabley .\n\\]\n(We have interchanged the variables of integration and then the order of integration.) Hence\n\\[\n2 integralv=\\int_{0}^{1} \\int_{0}^{1} functionf(variablex) functionf(variabley)(variabley-variablex)[functionf(variablex)-functionf(variabley)] d variablex d variabley .\n\\]\n\nBecause \\( functionf \\) is decreasing, \\( (variabley-variablex)[functionf(variablex)-functionf(variabley)] \\geq 0 \\) for all \\( variablex \\) and \\( variabley \\). Then since \\( functionf \\) is everywhere positive, it is clear that \\( 2 integralv \\geq 0 \\). This proves (1).\n\nRemark. The argument given generalizes to prove\n\\[\n\\int_{0}^{1} functionf(variablex) functiong(variablex) weightw(variablex) d variablex \\int_{0}^{1} weightw(variablex) d variablex \\leq \\int_{0}^{1} functionf(variablex) weightw(variablex) d variablex \\int_{0}^{1} functiong(variablex) weightw(variablex) d variablex\n\\]\nwhenever \\( functionf \\) is decreasing, \\( functiong \\) is increasing, and \\( weightw \\) is a non-negative weight function (assuming, of course, that the integrals exist). Moreover, the inequality is strict unless either \\( functionf \\) or \\( functiong \\) is constant over the support of \\( weightw \\) (i.e., the closure of the set \\( \\{variablex: weightw(variablex)>0\\} \\) ). The present problem is the special case \\( functiong(variablex)=variablex, weightw(variablex)=functionf(variablex) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandwich",
        "y": "mountain",
        "f": "landscape",
        "g": "crossroad",
        "w": "framework",
        "I": "indicator"
      },
      "question": "3. For \\( landscape(sandwich) \\) a positive, monotone decreasing function defined in \\( 0 \\leq sandwich \\leq 1 \\) prove that\n\\[\n\\frac{\\int_{0}^{1} sandwich\\, landscape^{2}(sandwich) d sandwich}{\\int_{0}^{1} sandwich\\, landscape(sandwich) d sandwich} \\leq \\frac{\\int_{0}^{1} landscape^{2}(sandwich) d sandwich}{\\int_{0}^{1} landscape(sandwich) d sandwich}\n\\]",
      "solution": "Solution. The desired inequality is equivalent to\n\\[\n\\int_{0}^{1} landscape^{2}(sandwich) d sandwich \\int_{0}^{1} mountain\\, landscape(mountain) d mountain-\\int_{0}^{1} mountain\\, landscape^{2}(mountain) d mountain \\int_{0}^{1} landscape(sandwich) d sandwich \\geq 0\n\\]\nwhich can be rewritten\n\\[\n\\int_{0}^{1} \\int_{0}^{1} landscape(sandwich) landscape(mountain) mountain[landscape(sandwich)-landscape(mountain)] d sandwich d mountain \\geq 0\n\\]\n\nDenote the left member of (1) by \\( indicator \\). Then\n\\[\nindicator=\\int_{0}^{1} \\int_{0}^{1} landscape(sandwich) landscape(mountain) sandwich[landscape(mountain)-landscape(sandwich)] d sandwich d mountain .\n\\]\n(We have interchanged the variables of integration and then the order of integration.) Hence\n\\[\n2 indicator=\\int_{0}^{1} \\int_{0}^{1} landscape(sandwich) landscape(mountain) (mountain-sandwich)[landscape(sandwich)-landscape(mountain)] d sandwich d mountain .\n\\]\n\nBecause \\( landscape \\) is decreasing, \\( (mountain-sandwich)[landscape(sandwich)-landscape(mountain)] \\geq 0 \\) for all \\( sandwich \\) and \\( mountain \\). Then since \\( landscape \\) is everywhere positive, it is clear that \\( 2 indicator \\geq 0 \\). This proves (1).\n\nRemark. The argument given generalizes to prove\n\\[\n\\int_{0}^{1} landscape(sandwich) crossroad(sandwich) framework(sandwich) d sandwich \\int_{0}^{1} framework(sandwich) d sandwich \\leq \\int_{0}^{1} landscape(sandwich) framework(sandwich) d sandwich \\int_{0}^{1} crossroad(sandwich) framework(sandwich) d sandwich\n\\]\nwhenever \\( landscape \\) is decreasing, \\( crossroad \\) is increasing, and \\( framework \\) is a non-negative weight function (assuming, of course, that the integrals exist). Moreover, the inequality is strict unless either \\( landscape \\) or \\( crossroad \\) is constant over the support of \\( framework \\) (i.e., the closure of the set \\( \\{sandwich: framework(sandwich)>0\\} \\) ). The present problem is the special case \\( crossroad(sandwich)=sandwich, framework(sandwich)=landscape(sandwich) \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "f": "ascendingvoid",
        "g": "descendingmap",
        "w": "weightless",
        "I": "emptiness"
      },
      "question": "3. For \\( ascendingvoid(verticalaxis) \\) a positive, monotone decreasing function defined in \\( 0 \\leq verticalaxis \\leq 1 \\) prove that\n\\[\n\\frac{\\int_{0}^{1} verticalaxis\\, ascendingvoid^{2}(verticalaxis) d verticalaxis}{\\int_{0}^{1} verticalaxis\\, ascendingvoid(verticalaxis) d verticalaxis} \\leq \\frac{\\int_{0}^{1} ascendingvoid^{2}(verticalaxis) d verticalaxis}{\\int_{0}^{1} ascendingvoid(verticalaxis) d verticalaxis}\n\\]\n",
      "solution": "Solution. The desired inequality is equivalent to\n\\[\n\\int_{0}^{1} ascendingvoid^{2}(verticalaxis) d verticalaxis \\int_{0}^{1} horizontalaxis\\, ascendingvoid(horizontalaxis) d horizontalaxis-\\int_{0}^{1} horizontalaxis\\, ascendingvoid^{2}(horizontalaxis) d horizontalaxis \\int_{0}^{1} ascendingvoid(verticalaxis) d verticalaxis \\geq 0\n\\]\nwhich can be rewritten\n\\[\n\\int_{0}^{1} \\int_{0}^{1} ascendingvoid(verticalaxis) ascendingvoid(horizontalaxis) horizontalaxis[ascendingvoid(verticalaxis)-ascendingvoid(horizontalaxis)] d verticalaxis d horizontalaxis \\geq 0\n\\]\n\nDenote the left member of (1) by \\( emptiness \\). Then\n\\[\nemptiness=\\int_{0}^{1} \\int_{0}^{1} ascendingvoid(verticalaxis) ascendingvoid(horizontalaxis) verticalaxis[ascendingvoid(horizontalaxis)-ascendingvoid(verticalaxis)] d verticalaxis d horizontalaxis .\n\\]\n(We have interchanged the variables of integration and then the order of integration.) Hence\n\\[\n2\\,emptiness=\\int_{0}^{1} \\int_{0}^{1} ascendingvoid(verticalaxis) ascendingvoid(horizontalaxis)(horizontalaxis-verticalaxis)[ascendingvoid(verticalaxis)-ascendingvoid(horizontalaxis)] d verticalaxis d horizontalaxis .\n\\]\n\nBecause \\( ascendingvoid \\) is decreasing, \\( (horizontalaxis-verticalaxis)[ascendingvoid(verticalaxis)-ascendingvoid(horizontalaxis)] \\geq 0 \\) for all \\( verticalaxis \\) and \\( horizontalaxis \\). Then since \\( ascendingvoid \\) is everywhere positive, it is clear that \\( 2\\,emptiness \\geq 0 \\). This proves (1).\n\nRemark. The argument given generalizes to prove\n\\[\n\\int_{0}^{1} ascendingvoid(verticalaxis) descendingmap(verticalaxis) weightless(verticalaxis) d verticalaxis \\int_{0}^{1} weightless(verticalaxis) d verticalaxis \\leq \\int_{0}^{1} ascendingvoid(verticalaxis) weightless(verticalaxis) d verticalaxis \\int_{0}^{1} descendingmap(verticalaxis) weightless(verticalaxis) d verticalaxis\n\\]\nwhenever \\( ascendingvoid \\) is decreasing, \\( descendingmap \\) is increasing, and \\( weightless \\) is a non-negative weight function (assuming, of course, that the integrals exist). Moreover, the inequality is strict unless either \\( ascendingvoid \\) or \\( descendingmap \\) is constant over the support of \\( weightless \\) (i.e., the closure of the set \\( \\{verticalaxis: weightless(verticalaxis)>0\\} \\) ). The present problem is the special case \\( descendingmap(verticalaxis)=verticalaxis, \\; weightless(verticalaxis)=ascendingvoid(verticalaxis) \\).\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "f": "mndkelsu",
        "g": "rjopavzi",
        "w": "cthymlqe",
        "I": "uvbzswko"
      },
      "question": "3. For \\( mndkelsu(qzxwvtnp) \\) a positive, monotone decreasing function defined in \\( 0 \\leq qzxwvtnp \\leq 1 \\) prove that\n\\[\n\\frac{\\int_{0}^{1} qzxwvtnp mndkelsu^{2}(qzxwvtnp) d qzxwvtnp}{\\int_{0}^{1} qzxwvtnp mndkelsu(qzxwvtnp) d qzxwvtnp} \\leq \\frac{\\int_{0}^{1} mndkelsu^{2}(qzxwvtnp) d qzxwvtnp}{\\int_{0}^{1} mndkelsu(qzxwvtnp) d qzxwvtnp}\n\\]\n",
      "solution": "Solution. The desired inequality is equivalent to\n\\[\n\\int_{0}^{1} mndkelsu^{2}(qzxwvtnp) d qzxwvtnp \\int_{0}^{1} hjgrksla\\, mndkelsu(hjgrksla) d hjgrksla-\\int_{0}^{1} hjgrksla\\, mndkelsu^{2}(hjgrksla) d hjgrksla \\int_{0}^{1} mndkelsu(qzxwvtnp) d qzxwvtnp \\geq 0\n\\]\nwhich can be rewritten\n\\[\n\\int_{0}^{1} \\int_{0}^{1} mndkelsu(qzxwvtnp) mndkelsu(hjgrksla) hjgrksla[mndkelsu(qzxwvtnp)-mndkelsu(hjgrksla)] d qzxwvtnp d hjgrksla \\geq 0\n\\]\n\nDenote the left member of (1) by \\( uvbzswko \\). Then\n\\[\nuvbzswko=\\int_{0}^{1} \\int_{0}^{1} mndkelsu(qzxwvtnp) mndkelsu(hjgrksla) qzxwvtnp[mndkelsu(hjgrksla)-mndkelsu(qzxwvtnp)] d qzxwvtnp d hjgrksla .\n\\]\n(We have interchanged the variables of integration and then the order of integration.) Hence\n\\[\n2\\,uvbzswko=\\int_{0}^{1} \\int_{0}^{1} mndkelsu(qzxwvtnp) mndkelsu(hjgrksla)(hjgrksla-qzxwvtnp)[mndkelsu(qzxwvtnp)-mndkelsu(hjgrksla)] d qzxwvtnp d hjgrksla .\n\\]\n\nBecause \\( mndkelsu \\) is decreasing, \\( (hjgrksla-qzxwvtnp)[mndkelsu(qzxwvtnp)-mndkelsu(hjgrksla)] \\geq 0 \\) for all \\( qzxwvtnp \\) and \\( hjgrksla \\). Then since \\( mndkelsu \\) is everywhere positive, it is clear that \\( 2\\,uvbzswko \\geq 0 \\). This proves (1).\n\nRemark. The argument given generalizes to prove\n\\[\n\\int_{0}^{1} mndkelsu(qzxwvtnp) rjopavzi(qzxwvtnp) cthymlqe(qzxwvtnp) d qzxwvtnp \\int_{0}^{1} cthymlqe(qzxwvtnp) d qzxwvtnp \\leq \\int_{0}^{1} mndkelsu(qzxwvtnp) cthymlqe(qzxwvtnp) d qzxwvtnp \\int_{0}^{1} rjopavzi(qzxwvtnp) cthymlqe(qzxwvtnp) d qzxwvtnp\n\\]\nwhenever \\( mndkelsu \\) is decreasing, \\( rjopavzi \\) is increasing, and \\( cthymlqe \\) is a non-negative weight function (assuming, of course, that the integrals exist). Moreover, the inequality is strict unless either \\( mndkelsu \\) or \\( rjopavzi \\) is constant over the support of \\( cthymlqe \\) (i.e., the closure of the set \\( \\{qzxwvtnp: cthymlqe(qzxwvtnp)>0\\} \\) ). The present problem is the special case \\( rjopavzi(qzxwvtnp)=qzxwvtnp,\\; cthymlqe(qzxwvtnp)=mndkelsu(qzxwvtnp) \\)."
    },
    "kernel_variant": {
      "question": "Let an integer $n\\ge 2$ be fixed and consider the $n$–dimensional cube  \n\\[\n\\mathcal I_n=[0,\\pi]^n ,\\qquad \nx=(x_1,\\dots ,x_n), \\qquad \nS(x):=x_1+\\dots +x_n .\n\\]\n\nData  \n\n$\\bullet$ coordinate–wise factorising weight  \n\\[\nw(x)=\\prod_{k=1}^{n}\\bigl(1+x_k^{2}\\bigr)>0 ,\\qquad x\\in\\mathcal I_n;\n\\]\n\n$\\bullet$ one–parameter family of strictly increasing weights  \n\\[\ng_\\alpha(x):=e^{\\alpha S(x)},\\qquad \\alpha\\ge 0;\n\\]\n\n$\\bullet$ a measurable map $f:\\mathcal I_n\\to(0,\\infty)$ that  \n\n (i) belongs to $L^{1}\\!\\bigl(\\mathcal I_n\\bigr)$;  \n\n (ii) is monotone \\emph{decreasing} in each coordinate;  \n\n (iii) is right–continuous at the boundary point $(\\pi,\\dots ,\\pi)$.\n\nFor $\\alpha\\ge 0$ introduce the normalised averages  \n\\[\nF(\\alpha):=\\frac{\\displaystyle\\int_{\\mathcal I_n} g_\\alpha(x)\\,w(x)\\,f(x)\\,dx}\n                 {\\displaystyle\\int_{\\mathcal I_n} g_\\alpha(x)\\,w(x)\\,dx}.\n\\]\n\nProve\n\n1. (Strict monotonicity)  The map $F:[0,\\infty)\\to\\mathbb R$ is strictly decreasing, i.e. for every $0\\le\\alpha_1<\\alpha_2$\n\\[\n\\frac{\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_2S(x)}\\,w(x)\\,f(x)\\,dx}\n     {\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_2S(x)}\\,w(x)\\,dx}\n<\n\\frac{\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_1S(x)}\\,w(x)\\,f(x)\\,dx}\n     {\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_1S(x)}\\,w(x)\\,dx},\n\\tag{$\\star$}\n\\]\nunless $f$ is (Lebesgue-)a.e. constant on $\\mathcal I_n$, in which case equality holds for all $\\alpha_1<\\alpha_2$.\n\n2. (Large–parameter limit)  $\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)=f(\\pi,\\dots ,\\pi).$\n\n3. (Rigidity)  If equality occurs in $(\\star)$ for one pair $0\\le\\alpha_1<\\alpha_2$, then $f$ must be a.e. constant on $\\mathcal I_n$; conversely, every a.e.–constant $f$ turns $(\\star)$ into an equality for \\emph{all} $\\alpha_1<\\alpha_2$.\n\n\n--------------------------------------------------------------------",
      "solution": "\\textbf{Step 0.  Probability representation.}  \nPut\n\\[\nZ(\\alpha):=\\int_{\\mathcal I_n} e^{\\alpha S(x)}\\,w(x)\\,dx ,\n\\qquad \n\\mu_\\alpha(dx):=\\frac{e^{\\alpha S(x)}\\,w(x)}{Z(\\alpha)}\\,dx ,\n\\]\nso that $\\mu_\\alpha$ is a probability measure on $\\mathcal I_n$ and  \n\\[\nF(\\alpha)=\\mathbb E_\\alpha[f(X)],\\qquad \nS:=S(X)=\\sum_{k=1}^{n}X_k ,\\qquad \nm_\\alpha:=\\mathbb E_\\alpha[S].\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 1.  Product structure of $\\mu_\\alpha$.}\n\nSince\n\\[\ne^{\\alpha S(x)}w(x)\n=\\prod_{k=1}^{n}\\Bigl(e^{\\alpha x_k}\\bigl(1+x_k^{2}\\bigr)\\Bigr),\n\\]\nwe have $\\mu_\\alpha=\\bigotimes_{k=1}^{n}\\mu_\\alpha^{(1)}$, where  \n\\[\n\\mu_\\alpha^{(1)}(du):=\\frac{e^{\\alpha u}(1+u^{2})\\,du}\n                           {\\displaystyle\\int_{0}^{\\pi}e^{\\alpha v}(1+v^{2})\\,dv},\n\\qquad 0\\le u\\le\\pi .\n\\]\nHence $X_1,\\dots ,X_n$ are independent under $\\mu_\\alpha$.\n\n--------------------------------------------------------------------\n\\textbf{Step 2.  Differentiation with respect to $\\alpha$.}\n\nFor every integrable $h\\colon\\mathcal I_n\\to\\mathbb R$\n\\[\n\\frac{d}{d\\alpha}\\mathbb E_\\alpha[h(X)]\n   =\\operatorname{Cov}_\\alpha\\!\\bigl(h(X),S\\bigr),\n\\tag{2.1}\n\\]\nobtained by differentiating under the integral sign and using the quotient rule.  \nApplying (2.1) with $h=f$ gives\n\\[\nF'(\\alpha)=\\operatorname{Cov}_\\alpha\\!\\bigl(f(X),S\\bigr).\n\\tag{2.2}\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 3.  Sign of the derivative and rigidity.}\n\n\\emph{3(a) Sign.}  \n$f$ is decreasing in every coordinate, while each $X_k$ and $S$ are increasing.  \nThe Harris–FKG inequality for product measures therefore yields\n\\[\n\\operatorname{Cov}_\\alpha\\!\\bigl(f(X),S\\bigr)\\le 0\n\\quad(\\alpha\\ge 0),\n\\tag{3.1}\n\\]\nso $F'(\\alpha)\\le 0$; thus $F$ is non–increasing.\n\n\\emph{3(b) Vanishing covariance forces constancy.}  \nSuppose $\\operatorname{Cov}_\\alpha(f,S)=0$ for some $\\alpha$.  \nBecause $S=\\sum_{k=1}^{n}X_k$ and the covariance is linear,\n\\[\n0=\\operatorname{Cov}_\\alpha(f,S)=\\sum_{k=1}^{n}\\operatorname{Cov}_\\alpha(f,X_k)\n\\]\nwith each summand $\\le 0$ by (3.1); hence\n\\[\n\\operatorname{Cov}_\\alpha(f,X_k)=0\\qquad(k=1,\\dots ,n).\n\\tag{3.2}\n\\]\n\nFix $k$.  Independence of the coordinates gives  \n$\\mu_\\alpha=\\mu_\\alpha^{(-k)}\\otimes\\mu_\\alpha^{(1)}$, where $\\mu_\\alpha^{(-k)}$ is the product of the $n-1$ other marginals.  \nSet\n\\[\nh_k(u):=\\mathbb E_\\alpha\\!\\bigl[f(X)\\mid X_k=u\\bigr]\n       =\\int f(y_1,\\dots ,y_{k-1},u,y_{k+1},\\dots ,y_n)\\,\n         \\mu_\\alpha^{(-k)}(dy),\\qquad 0\\le u\\le\\pi.\n\\]\nBecause $f$ is decreasing in $x_k$, $h_k$ is decreasing in $u$.  \nFurthermore, by conditioning,\n\\[\n\\operatorname{Cov}_\\alpha(f,X_k)=\\operatorname{Cov}_{\\mu_\\alpha^{(1)}}\\bigl(h_k,U\\bigr)=0,\n\\]\nwhere $U\\sim\\mu_\\alpha^{(1)}$.  \n\n\\medskip\n\\textbf{Lemma A (one–dimensional monotone rigidity).}  \nLet $\\nu$ be a probability measure on $[0,\\pi]$ with a strictly positive density, $U\\sim\\nu$, and $\\psi:[0,\\pi]\\to\\mathbb R$ be integrable and decreasing.  \nIf $\\operatorname{Cov}_\\nu(\\psi,U)=0$, then $\\psi$ is $\\nu$–a.e. constant.\n\n\\emph{Proof of Lemma A.}  Introduce two i.i.d. copies $U_1,U_2\\sim\\nu$ and write\n\\[\n\\operatorname{Cov}_\\nu(\\psi,U)\n=\\frac12\\mathbb E\\bigl[(\\psi(U_1)-\\psi(U_2))(U_1-U_2)\\bigr].\n\\]\nFor decreasing $\\psi$ the integrand is non–positive and strictly negative on a set of positive measure unless $\\psi$ is a.e. constant.  \nHence the covariance can vanish only in the constant case. \\qed\n\nApplying Lemma A with $\\psi=h_k$ and $\\nu=\\mu_\\alpha^{(1)}$ gives  \n\\[\nh_k(u)\\equiv c_k\\quad\\mu_\\alpha^{(1)}\\text{-a.e. on }[0,\\pi].\n\\tag{3.3}\n\\]\n\n\\emph{From constant $h_k$ to constant $f$.}  \nFix $u_1<u_2$ in $[0,\\pi]$.  \nBecause $h_k(u)$ is constant a.e.,\n\\[\n0=h_k(u_1)-h_k(u_2)\n  =\\mathbb E_{Y}\\!\\bigl[f(Y,u_1)-f(Y,u_2)\\bigr],\n\\tag{3.4}\n\\]\nwhere $Y:=(X_1,\\dots ,X_{k-1},X_{k+1},\\dots ,X_n)\\sim\\mu_\\alpha^{(-k)}$.  \nThe integrand in (3.4) is \\emph{non–negative} (monotonicity in the $k$–th coordinate) and its expectation equals $0$; hence\n\\[\nf(y_1,\\dots ,y_{k-1},u_1,y_{k+1},\\dots ,y_n)\n=f(y_1,\\dots ,y_{k-1},u_2,y_{k+1},\\dots ,y_n)\n\\]\nfor $\\mu_\\alpha^{(-k)}$–a.e. $y$.  \nBecause $u_1,u_2$ were arbitrary and the density of $\\mu_\\alpha^{(-k)}$ is strictly positive on the open cube,  \n\\[\nf \\text{ is independent of } x_k \\text{ on a set of full Lebesgue measure.}\n\\]\nRepeating the argument for every $k=1,\\dots ,n$ shows that $f$ is constant Lebesgue–a.e. on $\\mathcal I_n$.  \n\nConsequently\n\\[\n\\operatorname{Cov}_\\alpha(f,S)<0\\quad\\text{unless $f$ is a.e. constant.}\n\\]\nBy (2.2) we obtain  \n\n$\\bullet$ $F'(\\alpha)<0$ for every $\\alpha\\ge 0$ whenever $f$ is not a.e. constant;  \n\n$\\bullet$ $F'(\\alpha)=0$ for all $\\alpha$ if $f$ is a.e. constant.\n\nThis proves Statement $1$ (strict monotonicity and the equality case).\n\n--------------------------------------------------------------------\n\\textbf{Step 4.  Concentration as $\\alpha\\to\\infty$.}\n\nLet $M:=n\\pi$ and fix $\\delta\\in(0,\\pi/2)$.  \nDefine\n\\[\nA_\\delta:=\\{x\\in\\mathcal I_n:S(x)\\ge M-\\delta\\},\n\\qquad \nB_\\varepsilon:=\\{x\\in\\mathcal I_n:S(x)\\le M-\\varepsilon\\},\n\\]\nwhere $\\varepsilon\\in(\\delta,\\pi)$ is arbitrary.  \nBecause $S(x)\\le M-\\varepsilon$ on $B_\\varepsilon$ and $S(x)\\ge M-\\delta$ on $A_\\delta$,\n\\[\n\\mu_\\alpha(B_\\varepsilon)\n  \\le\n  \\frac{e^{\\alpha(M-\\varepsilon)}\\int_{\\mathcal I_n}w(x)\\,dx}\n       {e^{\\alpha(M-\\delta)}\\int_{A_\\delta}w(x)\\,dx}\n  =K_1 e^{-\\alpha(\\varepsilon-\\delta)},\n\\tag{4.1}\n\\]\nand, writing $C:=\\int_{\\mathcal I_n}w(x)f(x)\\,dx$,\n\\[\n\\mathbb E_\\alpha\\!\\bigl[f(X)\\mathbf 1_{B_\\varepsilon}\\bigr]\n  \\le K_2 e^{-\\alpha(\\varepsilon-\\delta)},\n\\quad\nK_2:=C/(\\int_{A_\\delta}w).\n\\tag{4.2}\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 5.  Limit of $F(\\alpha)$.}\n\nSplit\n\\[\n|F(\\alpha)-f(\\pi,\\dots ,\\pi)|\n\\le\nI_1(\\varepsilon)+I_2(\\varepsilon,\\alpha),\n\\]\nwhere\n\\[\nI_1(\\varepsilon):=\\bigl|\\mathbb E_\\alpha[(f(X)-f(\\pi,\\dots ,\\pi))\\mathbf 1_{A_\\varepsilon}]\\bigr|,\n\\quad\nI_2(\\varepsilon,\\alpha):=\\bigl|\\mathbb E_\\alpha[(f(X)-f(\\pi,\\dots ,\\pi))\\mathbf 1_{B_\\varepsilon}]\\bigr|.\n\\]\n\n\\emph{Continuity term.}  \nOn $A_\\varepsilon$ each coordinate lies in $[\\pi-\\varepsilon,\\pi]$; monotonicity gives\n\\[\n0\\le f(X)-f(\\pi,\\dots ,\\pi)\\le f(\\pi-\\varepsilon,\\dots ,\\pi-\\varepsilon)-f(\\pi,\\dots ,\\pi).\n\\]\nThe right–hand side tends to $0$ as $\\varepsilon\\downarrow 0$ by right–continuity, so $I_1(\\varepsilon)<\\eta/2$ for sufficiently small $\\varepsilon$.\n\n\\emph{Tail term.}  \nUsing (4.1)–(4.2),\n\\[\nI_2(\\varepsilon,\\alpha)\\le\nK_2 e^{-\\alpha(\\varepsilon-\\delta)}\n+f(\\pi,\\dots ,\\pi)\\,K_1 e^{-\\alpha(\\varepsilon-\\delta)}\n\\le\nK e^{-\\alpha(\\varepsilon-\\delta)}.\n\\]\nFor the fixed $\\varepsilon$ choose $\\alpha$ large enough that $I_2(\\varepsilon,\\alpha)<\\eta/2$.\n\nSince $\\eta>0$ was arbitrary,\n\\[\n\\boxed{\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)=f(\\pi,\\dots ,\\pi).}\n\\]\nThis proves Statement $2$.\n\n--------------------------------------------------------------------\n\\textbf{Step 6.  Equality discussion.}\n\nIf equality holds in $(\\star)$ for some $\\alpha_1<\\alpha_2$, then $F(\\alpha_1)=F(\\alpha_2)$.  \nBecause $F$ is non–increasing, it must be constant on $[\\alpha_1,\\alpha_2]$, so $F'(\\alpha)=0$ for all $\\alpha\\in(\\alpha_1,\\alpha_2)$.  \nBy Step $3(b)$ this forces $f$ to be a.e. constant on $\\mathcal I_n$.  \nConversely, if $f\\equiv c$ a.e., both sides of $(\\star)$ equal $c$ for every $\\alpha$, so equality holds for all pairs $\\alpha_1<\\alpha_2$.  \nStatement $3$ is proved.\n\nAll three assertions are now completely established. \\qed\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.489435",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional setting: the problem moves from one variable on \\([0,\\pi]\\) to an \\(n\\)-variable function on the cube \\([0,\\pi]^n\\) with \\(n\\ge 2\\).  \n\n• Parameter family: instead of comparing two fixed weights, the problem studies an *entire one-parameter* family \\(g_\\alpha=e^{\\alpha S}\\) and the *monotonicity* of the resulting averages, requiring analysis of derivatives with respect to \\(\\alpha\\).  \n\n• Measure–theoretic tools: the proof employs probability-measure normalisation, covariance calculus, differentiation under the integral sign, and an \\(n\\)-dimensional version of the Chebyshev/FGK inequality.  \n\n• Asymptotic analysis: determining \\(\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)\\) needs a Laplace-type localisation argument on a high-dimensional domain.  \n\n• Equality characterisation: identifying *all* equality cases demands a careful interplay between the sign of the covariance and the behaviour of \\(F'(\\alpha)\\).\n\nAll these features—higher dimension, a variable parameter, deeper probabilistic machinery, and asymptotic as well as equality analyses—make the enhanced variant substantially more technical and conceptually harder than both the original problem and the existing kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let an integer $n\\ge 2$ be fixed and consider the $n$–dimensional cube  \n\\[\n\\mathcal I_n=[0,\\pi]^n ,\\qquad \nx=(x_1,\\dots ,x_n), \\qquad \nS(x):=x_1+\\dots +x_n .\n\\]\n\nData  \n\n$\\bullet$ coordinate–wise factorising weight  \n\\[\nw(x)=\\prod_{k=1}^{n}\\bigl(1+x_k^{2}\\bigr)>0 ,\\qquad x\\in\\mathcal I_n;\n\\]\n\n$\\bullet$ one–parameter family of strictly increasing weights  \n\\[\ng_\\alpha(x):=e^{\\alpha S(x)},\\qquad \\alpha\\ge 0;\n\\]\n\n$\\bullet$ a measurable map $f:\\mathcal I_n\\to(0,\\infty)$ that  \n\n (i) belongs to $L^{1}\\!\\bigl(\\mathcal I_n\\bigr)$;  \n\n (ii) is monotone \\emph{decreasing} in each coordinate;  \n\n (iii) is right–continuous at the boundary point $(\\pi,\\dots ,\\pi)$.\n\nFor $\\alpha\\ge 0$ introduce the normalised averages  \n\\[\nF(\\alpha):=\\frac{\\displaystyle\\int_{\\mathcal I_n} g_\\alpha(x)\\,w(x)\\,f(x)\\,dx}\n                 {\\displaystyle\\int_{\\mathcal I_n} g_\\alpha(x)\\,w(x)\\,dx}.\n\\]\n\nProve\n\n1. (Strict monotonicity)  The map $F:[0,\\infty)\\to\\mathbb R$ is strictly decreasing, i.e. for every $0\\le\\alpha_1<\\alpha_2$\n\\[\n\\frac{\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_2S(x)}\\,w(x)\\,f(x)\\,dx}\n     {\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_2S(x)}\\,w(x)\\,dx}\n<\n\\frac{\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_1S(x)}\\,w(x)\\,f(x)\\,dx}\n     {\\displaystyle\\int_{\\mathcal I_n} e^{\\alpha_1S(x)}\\,w(x)\\,dx},\n\\tag{$\\star$}\n\\]\nunless $f$ is (Lebesgue-)a.e. constant on $\\mathcal I_n$, in which case equality holds for all $\\alpha_1<\\alpha_2$.\n\n2. (Large–parameter limit)  $\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)=f(\\pi,\\dots ,\\pi).$\n\n3. (Rigidity)  If equality occurs in $(\\star)$ for one pair $0\\le\\alpha_1<\\alpha_2$, then $f$ must be a.e. constant on $\\mathcal I_n$; conversely, every a.e.–constant $f$ turns $(\\star)$ into an equality for \\emph{all} $\\alpha_1<\\alpha_2$.\n\n\n--------------------------------------------------------------------",
      "solution": "\\textbf{Step 0.  Probability representation.}  \nPut\n\\[\nZ(\\alpha):=\\int_{\\mathcal I_n} e^{\\alpha S(x)}\\,w(x)\\,dx ,\n\\qquad \n\\mu_\\alpha(dx):=\\frac{e^{\\alpha S(x)}\\,w(x)}{Z(\\alpha)}\\,dx ,\n\\]\nso that $\\mu_\\alpha$ is a probability measure on $\\mathcal I_n$ and  \n\\[\nF(\\alpha)=\\mathbb E_\\alpha[f(X)],\\qquad \nS:=S(X)=\\sum_{k=1}^{n}X_k ,\\qquad \nm_\\alpha:=\\mathbb E_\\alpha[S].\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 1.  Product structure of $\\mu_\\alpha$.}\n\nSince\n\\[\ne^{\\alpha S(x)}w(x)\n=\\prod_{k=1}^{n}\\Bigl(e^{\\alpha x_k}\\bigl(1+x_k^{2}\\bigr)\\Bigr),\n\\]\nwe have $\\mu_\\alpha=\\bigotimes_{k=1}^{n}\\mu_\\alpha^{(1)}$, where  \n\\[\n\\mu_\\alpha^{(1)}(du):=\\frac{e^{\\alpha u}(1+u^{2})\\,du}\n                           {\\displaystyle\\int_{0}^{\\pi}e^{\\alpha v}(1+v^{2})\\,dv},\n\\qquad 0\\le u\\le\\pi .\n\\]\nHence $X_1,\\dots ,X_n$ are independent under $\\mu_\\alpha$.\n\n--------------------------------------------------------------------\n\\textbf{Step 2.  Differentiation with respect to $\\alpha$.}\n\nFor every integrable $h\\colon\\mathcal I_n\\to\\mathbb R$\n\\[\n\\frac{d}{d\\alpha}\\mathbb E_\\alpha[h(X)]\n   =\\operatorname{Cov}_\\alpha\\!\\bigl(h(X),S\\bigr),\n\\tag{2.1}\n\\]\nobtained by differentiating under the integral sign and using the quotient rule.  \nApplying (2.1) with $h=f$ gives\n\\[\nF'(\\alpha)=\\operatorname{Cov}_\\alpha\\!\\bigl(f(X),S\\bigr).\n\\tag{2.2}\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 3.  Sign of the derivative and rigidity.}\n\n\\emph{3(a) Sign.}  \n$f$ is decreasing in every coordinate, while each $X_k$ and $S$ are increasing.  \nThe Harris–FKG inequality for product measures therefore yields\n\\[\n\\operatorname{Cov}_\\alpha\\!\\bigl(f(X),S\\bigr)\\le 0\n\\quad(\\alpha\\ge 0),\n\\tag{3.1}\n\\]\nso $F'(\\alpha)\\le 0$; thus $F$ is non–increasing.\n\n\\emph{3(b) Vanishing covariance forces constancy.}  \nSuppose $\\operatorname{Cov}_\\alpha(f,S)=0$ for some $\\alpha$.  \nBecause $S=\\sum_{k=1}^{n}X_k$ and the covariance is linear,\n\\[\n0=\\operatorname{Cov}_\\alpha(f,S)=\\sum_{k=1}^{n}\\operatorname{Cov}_\\alpha(f,X_k)\n\\]\nwith each summand $\\le 0$ by (3.1); hence\n\\[\n\\operatorname{Cov}_\\alpha(f,X_k)=0\\qquad(k=1,\\dots ,n).\n\\tag{3.2}\n\\]\n\nFix $k$.  Independence of the coordinates gives  \n$\\mu_\\alpha=\\mu_\\alpha^{(-k)}\\otimes\\mu_\\alpha^{(1)}$, where $\\mu_\\alpha^{(-k)}$ is the product of the $n-1$ other marginals.  \nSet\n\\[\nh_k(u):=\\mathbb E_\\alpha\\!\\bigl[f(X)\\mid X_k=u\\bigr]\n       =\\int f(y_1,\\dots ,y_{k-1},u,y_{k+1},\\dots ,y_n)\\,\n         \\mu_\\alpha^{(-k)}(dy),\\qquad 0\\le u\\le\\pi.\n\\]\nBecause $f$ is decreasing in $x_k$, $h_k$ is decreasing in $u$.  \nFurthermore, by conditioning,\n\\[\n\\operatorname{Cov}_\\alpha(f,X_k)=\\operatorname{Cov}_{\\mu_\\alpha^{(1)}}\\bigl(h_k,U\\bigr)=0,\n\\]\nwhere $U\\sim\\mu_\\alpha^{(1)}$.  \n\n\\medskip\n\\textbf{Lemma A (one–dimensional monotone rigidity).}  \nLet $\\nu$ be a probability measure on $[0,\\pi]$ with a strictly positive density, $U\\sim\\nu$, and $\\psi:[0,\\pi]\\to\\mathbb R$ be integrable and decreasing.  \nIf $\\operatorname{Cov}_\\nu(\\psi,U)=0$, then $\\psi$ is $\\nu$–a.e. constant.\n\n\\emph{Proof of Lemma A.}  Introduce two i.i.d. copies $U_1,U_2\\sim\\nu$ and write\n\\[\n\\operatorname{Cov}_\\nu(\\psi,U)\n=\\frac12\\mathbb E\\bigl[(\\psi(U_1)-\\psi(U_2))(U_1-U_2)\\bigr].\n\\]\nFor decreasing $\\psi$ the integrand is non–positive and strictly negative on a set of positive measure unless $\\psi$ is a.e. constant.  \nHence the covariance can vanish only in the constant case. \\qed\n\nApplying Lemma A with $\\psi=h_k$ and $\\nu=\\mu_\\alpha^{(1)}$ gives  \n\\[\nh_k(u)\\equiv c_k\\quad\\mu_\\alpha^{(1)}\\text{-a.e. on }[0,\\pi].\n\\tag{3.3}\n\\]\n\n\\emph{From constant $h_k$ to constant $f$.}  \nFix $u_1<u_2$ in $[0,\\pi]$.  \nBecause $h_k(u)$ is constant a.e.,\n\\[\n0=h_k(u_1)-h_k(u_2)\n  =\\mathbb E_{Y}\\!\\bigl[f(Y,u_1)-f(Y,u_2)\\bigr],\n\\tag{3.4}\n\\]\nwhere $Y:=(X_1,\\dots ,X_{k-1},X_{k+1},\\dots ,X_n)\\sim\\mu_\\alpha^{(-k)}$.  \nThe integrand in (3.4) is \\emph{non–negative} (monotonicity in the $k$–th coordinate) and its expectation equals $0$; hence\n\\[\nf(y_1,\\dots ,y_{k-1},u_1,y_{k+1},\\dots ,y_n)\n=f(y_1,\\dots ,y_{k-1},u_2,y_{k+1},\\dots ,y_n)\n\\]\nfor $\\mu_\\alpha^{(-k)}$–a.e. $y$.  \nBecause $u_1,u_2$ were arbitrary and the density of $\\mu_\\alpha^{(-k)}$ is strictly positive on the open cube,  \n\\[\nf \\text{ is independent of } x_k \\text{ on a set of full Lebesgue measure.}\n\\]\nRepeating the argument for every $k=1,\\dots ,n$ shows that $f$ is constant Lebesgue–a.e. on $\\mathcal I_n$.  \n\nConsequently\n\\[\n\\operatorname{Cov}_\\alpha(f,S)<0\\quad\\text{unless $f$ is a.e. constant.}\n\\]\nBy (2.2) we obtain  \n\n$\\bullet$ $F'(\\alpha)<0$ for every $\\alpha\\ge 0$ whenever $f$ is not a.e. constant;  \n\n$\\bullet$ $F'(\\alpha)=0$ for all $\\alpha$ if $f$ is a.e. constant.\n\nThis proves Statement $1$ (strict monotonicity and the equality case).\n\n--------------------------------------------------------------------\n\\textbf{Step 4.  Concentration as $\\alpha\\to\\infty$.}\n\nLet $M:=n\\pi$ and fix $\\delta\\in(0,\\pi/2)$.  \nDefine\n\\[\nA_\\delta:=\\{x\\in\\mathcal I_n:S(x)\\ge M-\\delta\\},\n\\qquad \nB_\\varepsilon:=\\{x\\in\\mathcal I_n:S(x)\\le M-\\varepsilon\\},\n\\]\nwhere $\\varepsilon\\in(\\delta,\\pi)$ is arbitrary.  \nBecause $S(x)\\le M-\\varepsilon$ on $B_\\varepsilon$ and $S(x)\\ge M-\\delta$ on $A_\\delta$,\n\\[\n\\mu_\\alpha(B_\\varepsilon)\n  \\le\n  \\frac{e^{\\alpha(M-\\varepsilon)}\\int_{\\mathcal I_n}w(x)\\,dx}\n       {e^{\\alpha(M-\\delta)}\\int_{A_\\delta}w(x)\\,dx}\n  =K_1 e^{-\\alpha(\\varepsilon-\\delta)},\n\\tag{4.1}\n\\]\nand, writing $C:=\\int_{\\mathcal I_n}w(x)f(x)\\,dx$,\n\\[\n\\mathbb E_\\alpha\\!\\bigl[f(X)\\mathbf 1_{B_\\varepsilon}\\bigr]\n  \\le K_2 e^{-\\alpha(\\varepsilon-\\delta)},\n\\quad\nK_2:=C/(\\int_{A_\\delta}w).\n\\tag{4.2}\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Step 5.  Limit of $F(\\alpha)$.}\n\nSplit\n\\[\n|F(\\alpha)-f(\\pi,\\dots ,\\pi)|\n\\le\nI_1(\\varepsilon)+I_2(\\varepsilon,\\alpha),\n\\]\nwhere\n\\[\nI_1(\\varepsilon):=\\bigl|\\mathbb E_\\alpha[(f(X)-f(\\pi,\\dots ,\\pi))\\mathbf 1_{A_\\varepsilon}]\\bigr|,\n\\quad\nI_2(\\varepsilon,\\alpha):=\\bigl|\\mathbb E_\\alpha[(f(X)-f(\\pi,\\dots ,\\pi))\\mathbf 1_{B_\\varepsilon}]\\bigr|.\n\\]\n\n\\emph{Continuity term.}  \nOn $A_\\varepsilon$ each coordinate lies in $[\\pi-\\varepsilon,\\pi]$; monotonicity gives\n\\[\n0\\le f(X)-f(\\pi,\\dots ,\\pi)\\le f(\\pi-\\varepsilon,\\dots ,\\pi-\\varepsilon)-f(\\pi,\\dots ,\\pi).\n\\]\nThe right–hand side tends to $0$ as $\\varepsilon\\downarrow 0$ by right–continuity, so $I_1(\\varepsilon)<\\eta/2$ for sufficiently small $\\varepsilon$.\n\n\\emph{Tail term.}  \nUsing (4.1)–(4.2),\n\\[\nI_2(\\varepsilon,\\alpha)\\le\nK_2 e^{-\\alpha(\\varepsilon-\\delta)}\n+f(\\pi,\\dots ,\\pi)\\,K_1 e^{-\\alpha(\\varepsilon-\\delta)}\n\\le\nK e^{-\\alpha(\\varepsilon-\\delta)}.\n\\]\nFor the fixed $\\varepsilon$ choose $\\alpha$ large enough that $I_2(\\varepsilon,\\alpha)<\\eta/2$.\n\nSince $\\eta>0$ was arbitrary,\n\\[\n\\boxed{\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)=f(\\pi,\\dots ,\\pi).}\n\\]\nThis proves Statement $2$.\n\n--------------------------------------------------------------------\n\\textbf{Step 6.  Equality discussion.}\n\nIf equality holds in $(\\star)$ for some $\\alpha_1<\\alpha_2$, then $F(\\alpha_1)=F(\\alpha_2)$.  \nBecause $F$ is non–increasing, it must be constant on $[\\alpha_1,\\alpha_2]$, so $F'(\\alpha)=0$ for all $\\alpha\\in(\\alpha_1,\\alpha_2)$.  \nBy Step $3(b)$ this forces $f$ to be a.e. constant on $\\mathcal I_n$.  \nConversely, if $f\\equiv c$ a.e., both sides of $(\\star)$ equal $c$ for every $\\alpha$, so equality holds for all pairs $\\alpha_1<\\alpha_2$.  \nStatement $3$ is proved.\n\nAll three assertions are now completely established. \\qed\n\n\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.409495",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimensional setting: the problem moves from one variable on \\([0,\\pi]\\) to an \\(n\\)-variable function on the cube \\([0,\\pi]^n\\) with \\(n\\ge 2\\).  \n\n• Parameter family: instead of comparing two fixed weights, the problem studies an *entire one-parameter* family \\(g_\\alpha=e^{\\alpha S}\\) and the *monotonicity* of the resulting averages, requiring analysis of derivatives with respect to \\(\\alpha\\).  \n\n• Measure–theoretic tools: the proof employs probability-measure normalisation, covariance calculus, differentiation under the integral sign, and an \\(n\\)-dimensional version of the Chebyshev/FGK inequality.  \n\n• Asymptotic analysis: determining \\(\\displaystyle\\lim_{\\alpha\\to\\infty}F(\\alpha)\\) needs a Laplace-type localisation argument on a high-dimensional domain.  \n\n• Equality characterisation: identifying *all* equality cases demands a careful interplay between the sign of the covariance and the behaviour of \\(F'(\\alpha)\\).\n\nAll these features—higher dimension, a variable parameter, deeper probabilistic machinery, and asymptotic as well as equality analyses—make the enhanced variant substantially more technical and conceptually harder than both the original problem and the existing kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}