{
  "index": "1967-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "A-4. Show that if \\( \\lambda>\\frac{1}{2} \\) there does not exist a real-valued function \\( u \\) such that for all \\( x \\) in the closed interval \\( 0 \\leqq x \\leqq 1, u(x)=1+\\lambda \\int_{x}^{1} u(y) u(y-x) d y \\).",
  "solution": "A-4 Assuming that there is a solution \\( u \\), then integrating with respect to \\( x \\) from 0 to 1 , one obtains \\( \\int_{0}^{1} u(x) d x=\\int_{0}^{1} 1 \\cdot d x+\\lambda \\int_{0}^{1}\\left\\{\\int_{x}^{1} u(y) u(y-x) d y\\right\\} d x \\). In the iterated integral, one can interchange the order of integration, and letting \\( \\int_{0}^{1} u(x) d x=\\alpha \\), one gets \\( \\alpha=1+\\lambda \\int_{0}^{1}\\left\\{u(y) \\int_{0}^{y} u(y-x) d x\\right\\} d y \\). Now, holding \\( y \\) fixed, let \\( z=y-x \\) to get \\( \\alpha=1+\\lambda \\int_{0}^{1}\\left\\{u(y) \\int_{0}^{y} u(z) d z\\right\\} d y \\). Set \\( f(y)=\\int_{0}^{y} u(z) d z \\). Then \\( \\alpha=1 \\) \\( +\\lambda \\int_{0}^{1} f^{\\prime}(y) f(y) d y=1+\\lambda\\left\\{\\frac{1}{2} f^{2}(1)-\\frac{1}{2} f^{2}(0)\\right\\}=1+\\lambda \\cdot \\frac{1}{2} \\alpha^{2} \\), or \\( \\lambda \\alpha^{2}-2 \\alpha+2=0 \\). The discriminant of this quadratic shows that if \\( \\lambda>\\frac{1}{2} \\) then the roots are imaginary.",
  "vars": [
    "x",
    "y",
    "u",
    "z",
    "f",
    "\\\\alpha"
  ],
  "params": [
    "\\\\lambda"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "u": "unknownu",
        "z": "variablez",
        "f": "integrand",
        "\\alpha": "meanval",
        "\\lambda": "lambdacoef"
      },
      "question": "A-4. Show that if \\( lambdacoef>\\frac{1}{2} \\) there does not exist a real-valued function \\( unknownu \\) such that for all \\( variablex \\) in the closed interval \\( 0 \\leqq variablex \\leqq 1, unknownu(variablex)=1+lambdacoef \\int_{variablex}^{1} unknownu(variabley) unknownu(variabley-variablex) d variabley \\).",
      "solution": "A-4 Assuming that there is a solution \\( unknownu \\), then integrating with respect to \\( variablex \\) from 0 to 1 , one obtains \\( \\int_{0}^{1} unknownu(variablex) d variablex=\\int_{0}^{1} 1 \\cdot d variablex+lambdacoef \\int_{0}^{1}\\left\\{\\int_{variablex}^{1} unknownu(variabley) unknownu(variabley-variablex) d variabley\\right\\} d variablex \\). In the iterated integral, one can interchange the order of integration, and letting \\( \\int_{0}^{1} unknownu(variablex) d variablex=meanval \\), one gets \\( meanval=1+lambdacoef \\int_{0}^{1}\\left\\{unknownu(variabley) \\int_{0}^{variabley} unknownu(variabley-variablex) d variablex\\right\\} d variabley \\). Now, holding \\( variabley \\) fixed, let \\( variablez=variabley-variablex \\) to get \\( meanval=1+lambdacoef \\int_{0}^{1}\\left\\{unknownu(variabley) \\int_{0}^{variabley} unknownu(variablez) d variablez\\right\\} d variabley \\). Set \\( integrand(variabley)=\\int_{0}^{variabley} unknownu(variablez) d variablez \\). Then \\( meanval=1 \\) \\( +lambdacoef \\int_{0}^{1} integrand^{\\prime}(variabley) integrand(variabley) d variabley=1+lambdacoef\\left\\{\\frac{1}{2} integrand^{2}(1)-\\frac{1}{2} integrand^{2}(0)\\right\\}=1+lambdacoef \\cdot \\frac{1}{2} meanval^{2} \\), or \\( lambdacoef meanval^{2}-2 meanval+2=0 \\). The discriminant of this quadratic shows that if \\( lambdacoef>\\frac{1}{2} \\) then the roots are imaginary."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "heronwing",
        "y": "falconcrest",
        "u": "lanternfish",
        "z": "riverstone",
        "f": "sugarplum",
        "\\\\alpha": "marigold",
        "\\\\lambda": "windchime"
      },
      "question": "A-4. Show that if \\( windchime>\\frac{1}{2} \\) there does not exist a real-valued function \\( lanternfish \\) such that for all \\( heronwing \\) in the closed interval \\( 0 \\leqq heronwing \\leqq 1, lanternfish(heronwing)=1+windchime \\int_{heronwing}^{1} lanternfish(falconcrest) lanternfish(falconcrest-heronwing) d falconcrest \\).",
      "solution": "A-4 Assuming that there is a solution \\( lanternfish \\), then integrating with respect to \\( heronwing \\) from 0 to 1 , one obtains \\( \\int_{0}^{1} lanternfish(heronwing) d heronwing=\\int_{0}^{1} 1 \\cdot d heronwing+windchime \\int_{0}^{1}\\left\\{\\int_{heronwing}^{1} lanternfish(falconcrest) lanternfish(falconcrest-heronwing) d falconcrest\\right\\} d heronwing \\). In the iterated integral, one can interchange the order of integration, and letting \\( \\int_{0}^{1} lanternfish(heronwing) d heronwing=marigold \\), one gets \\( marigold=1+windchime \\int_{0}^{1}\\left\\{lanternfish(falconcrest) \\int_{0}^{falconcrest} lanternfish(falconcrest-heronwing) d heronwing\\right\\} d falconcrest \\). Now, holding \\( falconcrest \\) fixed, let \\( riverstone=falconcrest-heronwing \\) to get \\( marigold=1+windchime \\int_{0}^{1}\\left\\{lanternfish(falconcrest) \\int_{0}^{falconcrest} lanternfish(riverstone) d riverstone\\right\\} d falconcrest \\). Set \\( sugarplum(falconcrest)=\\int_{0}^{falconcrest} lanternfish(riverstone) d riverstone \\). Then \\( marigold=1 \\) \\( +windchime \\int_{0}^{1} sugarplum^{\\prime}(falconcrest) sugarplum(falconcrest) d falconcrest=1+windchime\\left\\{\\frac{1}{2} sugarplum^{2}(1)-\\frac{1}{2} sugarplum^{2}(0)\\right\\}=1+windchime \\cdot \\frac{1}{2} marigold^{2} \\), or \\( windchime marigold^{2}-2 marigold+2=0 \\). The discriminant of this quadratic shows that if \\( windchime>\\frac{1}{2} \\) then the roots are imaginary."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "y": "staticvalue",
        "u": "knownconstant",
        "z": "stationary",
        "f": "nonfunctional",
        "\\alpha": "variablevalue",
        "\\lambda": "varyingcoeff"
      },
      "question": "A-4. Show that if \\( varyingcoeff>\\frac{1}{2} \\) there does not exist a real-valued function \\( knownconstant \\) such that for all \\( fixedvalue \\) in the closed interval \\( 0 \\leqq fixedvalue \\leqq 1, knownconstant(fixedvalue)=1+varyingcoeff \\int_{fixedvalue}^{1} knownconstant(staticvalue) knownconstant(staticvalue-fixedvalue) d staticvalue \\).",
      "solution": "A-4 Assuming that there is a solution \\( knownconstant \\), then integrating with respect to \\( fixedvalue \\) from 0 to 1 , one obtains \\( \\int_{0}^{1} knownconstant(fixedvalue) d fixedvalue=\\int_{0}^{1} 1 \\cdot d fixedvalue+varyingcoeff \\int_{0}^{1}\\left\\{\\int_{fixedvalue}^{1} knownconstant(staticvalue) knownconstant(staticvalue-fixedvalue) d staticvalue\\right\\} d fixedvalue \\). In the iterated integral, one can interchange the order of integration, and letting \\( \\int_{0}^{1} knownconstant(fixedvalue) d fixedvalue=variablevalue \\), one gets \\( variablevalue=1+varyingcoeff \\int_{0}^{1}\\left\\{knownconstant(staticvalue) \\int_{0}^{staticvalue} knownconstant(staticvalue-fixedvalue) d fixedvalue\\right\\} d staticvalue \\). Now, holding \\( staticvalue \\) fixed, let \\( stationary=staticvalue-fixedvalue \\) to get \\( variablevalue=1+varyingcoeff \\int_{0}^{1}\\left\\{knownconstant(staticvalue) \\int_{0}^{staticvalue} knownconstant(stationary) d stationary\\right\\} d staticvalue \\). Set \\( nonfunctional(staticvalue)=\\int_{0}^{staticvalue} knownconstant(stationary) d stationary \\). Then \\( variablevalue=1 \\) \\( +varyingcoeff \\int_{0}^{1} nonfunctional^{\\prime}(staticvalue) nonfunctional(staticvalue) d staticvalue=1+varyingcoeff\\left\\{\\frac{1}{2} nonfunctional^{2}(1)-\\frac{1}{2} nonfunctional^{2}(0)\\right\\}=1+varyingcoeff \\cdot \\frac{1}{2} variablevalue^{2} \\), or \\( varyingcoeff \\, variablevalue^{2}-2 \\, variablevalue+2=0 \\). The discriminant of this quadratic shows that if \\( varyingcoeff>\\frac{1}{2} \\) then the roots are imaginary."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "u": "rmdkpsov",
        "z": "flmqnrta",
        "f": "gzpdhkwe",
        "\\\\alpha": "xnbgvslq",
        "\\\\lambda": "lkjhdwqe"
      },
      "question": "A-4. Show that if \\( lkjhdwqe>\\frac{1}{2} \\) there does not exist a real-valued function \\( rmdkpsov \\) such that for all \\( qzxwvtnp \\) in the closed interval \\( 0 \\leqq qzxwvtnp \\leqq 1, rmdkpsov(qzxwvtnp)=1+lkjhdwqe \\int_{qzxwvtnp}^{1} rmdkpsov(hjgrksla) rmdkpsov(hjgrksla-qzxwvtnp) d hjgrksla \\).",
      "solution": "A-4 Assuming that there is a solution \\( rmdkpsov \\), then integrating with respect to \\( qzxwvtnp \\) from 0 to 1, one obtains \\( \\int_{0}^{1} rmdkpsov(qzxwvtnp) d qzxwvtnp=\\int_{0}^{1} 1 \\cdot d qzxwvtnp+lkjhdwqe \\int_{0}^{1}\\left\\{\\int_{qzxwvtnp}^{1} rmdkpsov(hjgrksla) rmdkpsov(hjgrksla-qzxwvtnp) d hjgrksla\\right\\} d qzxwvtnp \\). In the iterated integral, one can interchange the order of integration, and letting \\( \\int_{0}^{1} rmdkpsov(qzxwvtnp) d qzxwvtnp=xnbgvslq \\), one gets \\( xnbgvslq=1+lkjhdwqe \\int_{0}^{1}\\left\\{rmdkpsov(hjgrksla) \\int_{0}^{hjgrksla} rmdkpsov(hjgrksla-qzxwvtnp) d qzxwvtnp\\right\\} d hjgrksla \\). Now, holding \\( hjgrksla \\) fixed, let \\( flmqnrta=hjgrksla-qzxwvtnp \\) to get \\( xnbgvslq=1+lkjhdwqe \\int_{0}^{1}\\left\\{rmdkpsov(hjgrksla) \\int_{0}^{hjgrksla} rmdkpsov(flmqnrta) d flmqnrta\\right\\} d hjgrksla \\). Set \\( gzpdhkwe(hjgrksla)=\\int_{0}^{hjgrksla} rmdkpsov(flmqnrta) d flmqnrta \\). Then \\( xnbgvslq=1+lkjhdwqe \\int_{0}^{1} gzpdhkwe^{\\prime}(hjgrksla) gzpdhkwe(hjgrksla) d hjgrksla=1+lkjhdwqe\\left\\{\\frac{1}{2} gzpdhkwe^{2}(1)-\\frac{1}{2} gzpdhkwe^{2}(0)\\right\\}=1+lkjhdwqe \\cdot \\frac{1}{2} xnbgvslq^{2} \\), or \\( lkjhdwqe xnbgvslq^{2}-2 xnbgvslq+2=0 \\). The discriminant of this quadratic shows that if \\( lkjhdwqe>\\frac{1}{2} \\) then the roots are imaginary."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 1$ be an integer and put  \n\\[\nQ:=\\,[0,1]^n\\subset \\mathbb{R}^n .\n\\]\n\nFor vectors $\\mathbf{x}=(x_1,\\dots ,x_n)$ and $\\mathbf{y}=(y_1,\\dots ,y_n)$ write $\\mathbf{y}\\succeq \\mathbf{x}$ if $y_i\\ge x_i$ for every $i$, and set  \n\\[\n\\mathbf{y}-\\mathbf{x}:=(y_1-x_1,\\dots ,y_n-x_n).\n\\]\n\nLet $\\lambda\\in\\mathbb{R}$.  Suppose that there exists a (Lebesgue-)integrable one–variable function  \n\\[\nv:[0,1]\\longrightarrow \\mathbb{R},\\qquad v\\in L^{1}([0,1])\\cap L^{2}([0,1]),\n\\]\nand define, for every $\\mathbf{x}\\in Q$,  \n\\[\nu(\\mathbf{x}):=\\prod_{i=1}^{n} v(x_i).\n\\tag{$\\ast$}\n\\]\n\n(The map $u$ is multiplicatively separable and therefore symmetric in its coordinates.)\n\nAssume that $u$ satisfies the nonlinear multidimensional Volterra equation  \n\\[\nu(\\mathbf{x})\n \\;=\\;\n1+\\lambda\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n          u(\\mathbf{y})\\,u\\bigl(\\mathbf{y}-\\mathbf{x}\\bigr)\\,\n          d\\mathbf{y}\n\\qquad\\bigl(\\forall \\mathbf{x}\\in Q\\bigr),\n\\tag{1}\n\\]\nwhere, by definition,  \n\\[\n\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n:=\\int_{y_1=x_1}^{1}\\!\\!\\cdots\\!\\!\\int_{y_n=x_n}^{1}.\n\\]\n\nProve that if  \n\\[\n\\boxed{\\;\\lambda>2^{\\,n-2}\\;}\n\\tag{2}\n\\]\nthen no such function $v$—and hence no such function $u$ constructed by $(\\ast)$—can exist.",
      "solution": "Throughout the proof we repeatedly use Tonelli–Fubini; all applications are legitimate because $v\\in L^{1}\\cap L^{2}$ implies $u\\in L^{1}\\cap L^{2}(Q)$, whence $u(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\\in L^{1}(Q\\times Q)$.\n\n\\medskip\n\\textbf{Step 1.  Scalar quantities.}\nSet\n\\[\n\\alpha_1:=\\int_{0}^{1} v(t)\\,dt,\n\\qquad \n\\alpha:=\\int_{Q} u(\\mathbf{x})\\,d\\mathbf{x}\n       =\\bigl(\\alpha_1\\bigr)^{n}.\n\\tag{3}\n\\]\n\n\\medskip\n\\textbf{Step 2.  Reduction of equation (1).}\nIntegrate identity (1) over $Q$:\n\n\\[\n\\alpha\n=\\int_{Q}1\\,d\\mathbf{x}\n       +\\lambda\\int_{Q}\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n                      u(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\\,\n                      d\\mathbf{y}\\,d\\mathbf{x}\n=:1+\\lambda I .\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 3.  Factorisation of $I$.}\nBecause of $(\\ast)$,\n\\[\nu(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\n   =\\prod_{k=1}^{n} v(y_k)\\,v(y_k-x_k).\n\\]\nThe domain of the double integral in $I$ is the Cartesian product of $n$ copies of $0\\le x_k\\le y_k\\le 1$.  Fubini therefore factorises $I$ into a product of identical $2$-fold integrals:\n\n\\[\nI=\\prod_{k=1}^{n} J,\n\\qquad \nJ:=\\int_{0}^{1} v(y)\\left(\\int_{0}^{y} v(z)\\,dz\\right)dy .\n\\tag{5}\n\\]\n\n\\medskip\n\\textbf{Step 4.  Exact evaluation of $J$.}\nIntroduce $F(y):=\\int_{0}^{y} v(z)\\,dz$.  Then $F^{\\prime}=v$ a.e., so\n\n\\[\nJ=\\int_{0}^{1} F^{\\prime}(y)\\,F(y)\\,dy\n  =\\frac{1}{2}\\Bigl[F(y)^2\\Bigr]_{0}^{1}\n  =\\frac{1}{2}\\,\\alpha_1^{\\,2}.\n\\tag{6}\n\\]\n\n\\medskip\n\\textbf{Step 5.  Computation of $I$.}\nFrom (5)–(6) we get\n\\[\nI=\\left(\\frac{1}{2}\\,\\alpha_1^{\\,2}\\right)^{n}\n  =\\frac{\\alpha^{2}}{2^{\\,n}}.\n\\tag{7}\n\\]\n\n\\medskip\n\\textbf{Step 6.  A quadratic for $\\alpha$.}\nInsert (7) into (4):\n\n\\[\n\\alpha \\;=\\; 1+\\lambda\\frac{\\alpha^{2}}{2^{\\,n}}\n\\quad\\Longrightarrow\\quad\n\\lambda\\alpha^{2}-2^{\\,n}\\alpha+2^{\\,n}=0.\n\\tag{8}\n\\]\n\n\\medskip\n\\textbf{Step 7.  Discriminant analysis.}\nTreating (8) as a quadratic equation in the real variable $\\alpha$, its discriminant is\n\n\\[\n\\Delta\n=(2^{\\,n})^{2}-4\\lambda\\cdot 2^{\\,n}\n=2^{\\,n}\\bigl(2^{\\,n}-4\\lambda\\bigr).\n\\tag{9}\n\\]\n\nIf condition (2) holds, then $2^{\\,n}-4\\lambda<0$, hence $\\Delta<0$.  The quadratic (8) consequently has no real root, contradicting the fact that $\\alpha=\\int_{Q} u$ is real.  Therefore no real-valued function $v$ can satisfy the hypotheses when $\\lambda>2^{\\,n-2}$.\n\n\\[\n\\boxed{\\text{No such function }v\\text{ (resp.\\ }u) \\text{ exists under condition (2).}}\n\\]\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.570367",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is lifted from one variable to n ≥ 3 variables, turning a simple convolution‐type integral equation into a multi-dimensional Volterra equation.  \n2. Extra structure: coordinate-wise ordering in ℝⁿ and the indicator χ(x,y) demand careful measure-theoretic treatment and a combinatorial argument to compute probabilities in n dimensions.  \n3. Deeper theory: the proof uses Fubini’s theorem on Q×Q, symmetry considerations, and probability reasoning to evaluate a multi-dimensional nested integral, rather than a single elementary substitution.  \n4. Parameter dependence: the critical value 2^{\\,n-2} must be derived; it specializes to ½ when n=1 (the original problem) and to 1 when n=2 (a first kernel variant), but grows exponentially with n, showing how dimension alters the threshold.  \n5. More steps: the solution introduces several non-trivial transformations—rewriting the double integral, symmetrization, evaluation of a combinatorial probability, and discriminant analysis of the resulting quadratic—before the contradiction appears.\n\nConsequently, both technical complexity and conceptual depth are substantially greater than in the original and current kernel variants."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 1$ be an integer and put  \n\\[\nQ:=\\,[0,1]^n\\subset \\mathbb{R}^n .\n\\]\n\nFor vectors $\\mathbf{x}=(x_1,\\dots ,x_n)$ and $\\mathbf{y}=(y_1,\\dots ,y_n)$ write $\\mathbf{y}\\succeq \\mathbf{x}$ if $y_i\\ge x_i$ for every $i$, and set  \n\\[\n\\mathbf{y}-\\mathbf{x}:=(y_1-x_1,\\dots ,y_n-x_n).\n\\]\n\nLet $\\lambda\\in\\mathbb{R}$.  Suppose that there exists a (Lebesgue-)integrable one–variable function  \n\\[\nv:[0,1]\\longrightarrow \\mathbb{R},\\qquad v\\in L^{1}([0,1])\\cap L^{2}([0,1]),\n\\]\nand define, for every $\\mathbf{x}\\in Q$,  \n\\[\nu(\\mathbf{x}):=\\prod_{i=1}^{n} v(x_i).\n\\tag{$\\ast$}\n\\]\n\n(The map $u$ is multiplicatively separable and therefore symmetric in its coordinates.)\n\nAssume that $u$ satisfies the nonlinear multidimensional Volterra equation  \n\\[\nu(\\mathbf{x})\n \\;=\\;\n1+\\lambda\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n          u(\\mathbf{y})\\,u\\bigl(\\mathbf{y}-\\mathbf{x}\\bigr)\\,\n          d\\mathbf{y}\n\\qquad\\bigl(\\forall \\mathbf{x}\\in Q\\bigr),\n\\tag{1}\n\\]\nwhere, by definition,  \n\\[\n\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n:=\\int_{y_1=x_1}^{1}\\!\\!\\cdots\\!\\!\\int_{y_n=x_n}^{1}.\n\\]\n\nProve that if  \n\\[\n\\boxed{\\;\\lambda>2^{\\,n-2}\\;}\n\\tag{2}\n\\]\nthen no such function $v$—and hence no such function $u$ constructed by $(\\ast)$—can exist.",
      "solution": "Throughout the proof we repeatedly use Tonelli–Fubini; all applications are legitimate because $v\\in L^{1}\\cap L^{2}$ implies $u\\in L^{1}\\cap L^{2}(Q)$, whence $u(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\\in L^{1}(Q\\times Q)$.\n\n\\medskip\n\\textbf{Step 1.  Scalar quantities.}\nSet\n\\[\n\\alpha_1:=\\int_{0}^{1} v(t)\\,dt,\n\\qquad \n\\alpha:=\\int_{Q} u(\\mathbf{x})\\,d\\mathbf{x}\n       =\\bigl(\\alpha_1\\bigr)^{n}.\n\\tag{3}\n\\]\n\n\\medskip\n\\textbf{Step 2.  Reduction of equation (1).}\nIntegrate identity (1) over $Q$:\n\n\\[\n\\alpha\n=\\int_{Q}1\\,d\\mathbf{x}\n       +\\lambda\\int_{Q}\\int_{\\mathbf{y}\\succeq \\mathbf{x}}\n                      u(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\\,\n                      d\\mathbf{y}\\,d\\mathbf{x}\n=:1+\\lambda I .\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 3.  Factorisation of $I$.}\nBecause of $(\\ast)$,\n\\[\nu(\\mathbf{y})\\,u(\\mathbf{y}-\\mathbf{x})\n   =\\prod_{k=1}^{n} v(y_k)\\,v(y_k-x_k).\n\\]\nThe domain of the double integral in $I$ is the Cartesian product of $n$ copies of $0\\le x_k\\le y_k\\le 1$.  Fubini therefore factorises $I$ into a product of identical $2$-fold integrals:\n\n\\[\nI=\\prod_{k=1}^{n} J,\n\\qquad \nJ:=\\int_{0}^{1} v(y)\\left(\\int_{0}^{y} v(z)\\,dz\\right)dy .\n\\tag{5}\n\\]\n\n\\medskip\n\\textbf{Step 4.  Exact evaluation of $J$.}\nIntroduce $F(y):=\\int_{0}^{y} v(z)\\,dz$.  Then $F^{\\prime}=v$ a.e., so\n\n\\[\nJ=\\int_{0}^{1} F^{\\prime}(y)\\,F(y)\\,dy\n  =\\frac{1}{2}\\Bigl[F(y)^2\\Bigr]_{0}^{1}\n  =\\frac{1}{2}\\,\\alpha_1^{\\,2}.\n\\tag{6}\n\\]\n\n\\medskip\n\\textbf{Step 5.  Computation of $I$.}\nFrom (5)–(6) we get\n\\[\nI=\\left(\\frac{1}{2}\\,\\alpha_1^{\\,2}\\right)^{n}\n  =\\frac{\\alpha^{2}}{2^{\\,n}}.\n\\tag{7}\n\\]\n\n\\medskip\n\\textbf{Step 6.  A quadratic for $\\alpha$.}\nInsert (7) into (4):\n\n\\[\n\\alpha \\;=\\; 1+\\lambda\\frac{\\alpha^{2}}{2^{\\,n}}\n\\quad\\Longrightarrow\\quad\n\\lambda\\alpha^{2}-2^{\\,n}\\alpha+2^{\\,n}=0.\n\\tag{8}\n\\]\n\n\\medskip\n\\textbf{Step 7.  Discriminant analysis.}\nTreating (8) as a quadratic equation in the real variable $\\alpha$, its discriminant is\n\n\\[\n\\Delta\n=(2^{\\,n})^{2}-4\\lambda\\cdot 2^{\\,n}\n=2^{\\,n}\\bigl(2^{\\,n}-4\\lambda\\bigr).\n\\tag{9}\n\\]\n\nIf condition (2) holds, then $2^{\\,n}-4\\lambda<0$, hence $\\Delta<0$.  The quadratic (8) consequently has no real root, contradicting the fact that $\\alpha=\\int_{Q} u$ is real.  Therefore no real-valued function $v$ can satisfy the hypotheses when $\\lambda>2^{\\,n-2}$.\n\n\\[\n\\boxed{\\text{No such function }v\\text{ (resp.\\ }u) \\text{ exists under condition (2).}}\n\\]\n\n\\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.464255",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is lifted from one variable to n ≥ 3 variables, turning a simple convolution‐type integral equation into a multi-dimensional Volterra equation.  \n2. Extra structure: coordinate-wise ordering in ℝⁿ and the indicator χ(x,y) demand careful measure-theoretic treatment and a combinatorial argument to compute probabilities in n dimensions.  \n3. Deeper theory: the proof uses Fubini’s theorem on Q×Q, symmetry considerations, and probability reasoning to evaluate a multi-dimensional nested integral, rather than a single elementary substitution.  \n4. Parameter dependence: the critical value 2^{\\,n-2} must be derived; it specializes to ½ when n=1 (the original problem) and to 1 when n=2 (a first kernel variant), but grows exponentially with n, showing how dimension alters the threshold.  \n5. More steps: the solution introduces several non-trivial transformations—rewriting the double integral, symmetrization, evaluation of a combinatorial probability, and discriminant analysis of the resulting quadratic—before the contradiction appears.\n\nConsequently, both technical complexity and conceptual depth are substantially greater than in the original and current kernel variants."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}