{
  "index": "2009-A-6",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $f:[0,1]^2 \\to \\mathbb{R}$ be a continuous function on the closed unit\nsquare such that $\\frac{\\partial f}{\\partial x}$ and $\\frac{\\partial f}{\\partial y}$ exist\nand are continuous on the interior $(0,1)^2$. Let $a = \\int_0^1 f(0,y)\\,dy$,\n$b = \\int_0^1 f(1,y)\\,dy$, $c = \\int_0^1 f(x,0)\\,dx$, $d = \\int_0^1 f(x,1)\\,dx$.\nProve or disprove: There must be a point $(x_0,y_0)$ in $(0,1)^2$ such that\n\\[\n\\frac{\\partial f}{\\partial x} (x_0,y_0) = b - a\n\\quad \\mbox{and} \\quad\n\\frac{\\partial f}{\\partial y} (x_0,y_0) = d - c.\n\\]",
  "solution": "We disprove the assertion using the example\n\\[\nf(x,y) = 3(1 + y)(2x-1)^2 -y.\n\\]\nWe have $b-a = d-c = 0$ because the identity $f(x,y) = f(1-x,y)$ forces $a=b$,\nand because\n\\begin{align*}\nc &= \\int_0^1 3(2x-1)^2\\,dx = 1, \\\\\nd &= \\int_0^1 (6(2x-1)^2-1)\\,dx = 1.\n\\end{align*}\nMoreover, the partial derivatives\n\\begin{align*}\n \\frac{\\partial f}{\\partial x}(x_0,y_0) &= 3(1+y_0)(8x_0-4) \\\\\n\\frac{\\partial f}{\\partial y}(x_0,y_0) &= 3(2x_0-1)^2-1.\n\\end{align*}\nhave no common zero in $(0,1)^2$. Namely,\nfor the first partial to vanish, we must have $x_0 = 1/2$ since\n$1 + y_0$ is nowhere zero, but for $x_0 = 1/2$\nthe second partial cannot vanish.\n\n\\textbf{Remark.}\nThis problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate\nfunctions. Many counterexamples are possible. Kent Merryfield suggests\n$y \\sin(2 \\pi x)$, for which all four of the boundary integrals vanish;\nhere the partial derivatives are $2\\pi y \\cos (2 \\pi x)$ and $\\sin (2 \\pi x)$.\nCatalin Zara suggests $x^{1/3} y^{2/3}$. Qingchun Ren suggests $xy(1-y)$.",
  "vars": [
    "x",
    "y",
    "x_0",
    "y_0"
  ],
  "params": [
    "f",
    "a",
    "b",
    "c",
    "d"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "bifunct",
        "a": "leftsum",
        "b": "rightsum",
        "c": "bottomsum",
        "d": "topsum",
        "x": "xcoord",
        "y": "ycoord",
        "x_0": "xcenter",
        "y_0": "ycenter"
      },
      "question": "Let $bifunct:[0,1]^2 \\to \\mathbb{R}$ be a continuous function on the closed unit\nsquare such that $\\frac{\\partial bifunct}{\\partial xcoord}$ and $\\frac{\\partial bifunct}{\\partial ycoord}$ exist\nand are continuous on the interior $(0,1)^2$. Let $leftsum = \\int_0^1 bifunct(0,ycoord)\\,dycoord$,\n$rightsum = \\int_0^1 bifunct(1,ycoord)\\,dycoord$, $bottomsum = \\int_0^1 bifunct(xcoord,0)\\,dxcoord$, $topsum = \\int_0^1 bifunct(xcoord,1)\\,dxcoord$.\nProve or disprove: There must be a point $(xcenter,ycenter)$ in $(0,1)^2$ such that\n\\[\n\\frac{\\partial bifunct}{\\partial xcoord} (xcenter,ycenter) = rightsum - leftsum\n\\quad \\mbox{and} \\quad\n\\frac{\\partial bifunct}{\\partial ycoord} (xcenter,ycenter) = topsum - bottomsum.\n\\]",
      "solution": "We disprove the assertion using the example\n\\[\nbifunct(xcoord,ycoord) = 3(1 + ycoord)(2xcoord-1)^2 -ycoord.\n\\]\nWe have $rightsum-leftsum = topsum-bottomsum = 0$ because the identity $bifunct(xcoord,ycoord) = bifunct(1-xcoord,ycoord)$ forces $leftsum=rightsum$,\nand because\n\\begin{align*}\nbottomsum &= \\int_0^1 3(2xcoord-1)^2\\,dxcoord = 1, \\\\\ntopsum &= \\int_0^1 (6(2xcoord-1)^2-1)\\,dxcoord = 1.\n\\end{align*}\nMoreover, the partial derivatives\n\\begin{align*}\n \\frac{\\partial bifunct}{\\partial xcoord}(xcenter,ycenter) &= 3(1+ycenter)(8xcenter-4) \\\\\n\\frac{\\partial bifunct}{\\partial ycoord}(xcenter,ycenter) &= 3(2xcenter-1)^2-1.\n\\end{align*}\nhave no common zero in $(0,1)^2$. Namely,\nfor the first partial to vanish, we must have $xcenter = 1/2$ since\n$1 + ycenter$ is nowhere zero, but for $xcenter = 1/2$\nthe second partial cannot vanish.\n\n\\textbf{Remark.}\nThis problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate\nfunctions. Many counterexamples are possible. Kent Merryfield suggests\n$ycoord \\sin(2 \\pi xcoord)$, for which all four of the boundary integrals vanish;\nhere the partial derivatives are $2\\pi ycoord \\cos (2 \\pi xcoord)$ and $\\sin (2 \\pi xcoord)$.\nCatalin Zara suggests $xcoord^{1/3} ycoord^{2/3}$. Qingchun Ren suggests $xcoord ycoord(1-ycoord)$.",
      "errors": []
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "y": "tangerine",
        "x_0": "accordion",
        "y_0": "heliumgas",
        "f": "chandelier",
        "a": "watermelon",
        "b": "blacksmith",
        "c": "snowblower",
        "d": "lumberjack"
      },
      "question": "Let $chandelier:[0,1]^2 \\to \\mathbb{R}$ be a continuous function on the closed unit\nsquare such that $\\frac{\\partial chandelier}{\\partial pineapple}$ and $\\frac{\\partial chandelier}{\\partial tangerine}$ exist\nand are continuous on the interior $(0,1)^2$. Let $watermelon = \\int_0^1 chandelier(0,tangerine)\\,d tangerine$,\n$blacksmith = \\int_0^1 chandelier(1,tangerine)\\,d tangerine$, $snowblower = \\int_0^1 chandelier(pineapple,0)\\,d pineapple$, $lumberjack = \\int_0^1 chandelier(pineapple,1)\\,d pineapple$.\nProve or disprove: There must be a point $(accordion,heliumgas)$ in $(0,1)^2$ such that\n\\[\n\\frac{\\partial chandelier}{\\partial pineapple} (accordion,heliumgas) = blacksmith - watermelon\n\\quad \\mbox{and} \\quad\n\\frac{\\partial chandelier}{\\partial tangerine} (accordion,heliumgas) = lumberjack - snowblower.\n\\]",
      "solution": "We disprove the assertion using the example\n\\[\nchandelier(pineapple,tangerine) = 3(1 + tangerine)(2 pineapple-1)^2 -tangerine.\n\\]\nWe have $blacksmith-watermelon = lumberjack-snowblower = 0$ because the identity $chandelier(pineapple,tangerine) = chandelier(1-pineapple,tangerine)$ forces $watermelon=blacksmith$,\nand because\n\\begin{align*}\nsnowblower &= \\int_0^1 3(2 pineapple-1)^2\\,d pineapple = 1, \\\\\nlumberjack &= \\int_0^1 (6(2 pineapple-1)^2-1)\\,d pineapple = 1.\n\\end{align*}\nMoreover, the partial derivatives\n\\begin{align*}\n \\frac{\\partial chandelier}{\\partial pineapple}(accordion,heliumgas) &= 3(1+heliumgas)(8 accordion-4) \\\\\n\\frac{\\partial chandelier}{\\partial tangerine}(accordion,heliumgas) &= 3(2 accordion-1)^2-1.\n\\end{align*}\nhave no common zero in $(0,1)^2$. Namely,\nfor the first partial to vanish, we must have $accordion = 1/2$ since\n$1 + heliumgas$ is nowhere zero, but for $accordion = 1/2$\nthe second partial cannot vanish.\n\n\\textbf{Remark.}\nThis problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate\nfunctions. Many counterexamples are possible. Kent Merryfield suggests\n$tangerine \\sin(2 \\pi pineapple)$, for which all four of the boundary integrals vanish;\nhere the partial derivatives are $2\\pi tangerine \\cos (2 \\pi pineapple)$ and $\\sin (2 \\pi pineapple)$.\nCatalin Zara suggests $pineapple^{1/3} tangerine^{2/3}$. Qingchun Ren suggests $pineapple tangerine(1-tangerine)$.",
      "}": "}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoord",
        "y": "horizontalcoord",
        "x_0": "edgecoordinate",
        "y_0": "startcoordy",
        "f": "constantmap",
        "a": "centermean",
        "b": "interiorvalue",
        "c": "roofvalue",
        "d": "floorvalue"
      },
      "question": "Let $constantmap:[0,1]^2 \\to \\mathbb{R}$ be a continuous function on the closed unit\nsquare such that $\\frac{\\partial constantmap}{\\partial verticalcoord}$ and $\\frac{\\partial constantmap}{\\partial horizontalcoord}$ exist\nand are continuous on the interior $(0,1)^2$. Let $centermean = \\int_0^1 constantmap(0,horizontalcoord)\\,d horizontalcoord$,\n$interiorvalue = \\int_0^1 constantmap(1,horizontalcoord)\\,d horizontalcoord$, $roofvalue = \\int_0^1 constantmap(verticalcoord,0)\\,d verticalcoord$, $floorvalue = \\int_0^1 constantmap(verticalcoord,1)\\,d verticalcoord$.\nProve or disprove: There must be a point $(edgecoordinate,startcoordy)$ in $(0,1)^2$ such that\n\\[\n\\frac{\\partial constantmap}{\\partial verticalcoord} (edgecoordinate,startcoordy) = interiorvalue - centermean\n\\quad \\mbox{and} \\quad\n\\frac{\\partial constantmap}{\\partial horizontalcoord} (edgecoordinate,startcoordy) = floorvalue - roofvalue.\n\\]",
      "solution": "We disprove the assertion using the example\n\\[\nconstantmap(verticalcoord,horizontalcoord) = 3(1 + horizontalcoord)(2verticalcoord-1)^2 - horizontalcoord.\n\\]\nWe have $interiorvalue-centermean = floorvalue-roofvalue = 0$ because the identity $constantmap(verticalcoord,horizontalcoord) = constantmap(1-verticalcoord,horizontalcoord)$ forces $centermean=interiorvalue$,\nand because\n\\begin{align*}\nroofvalue &= \\int_0^1 3(2verticalcoord-1)^2\\,d verticalcoord = 1, \\\\\nfloorvalue &= \\int_0^1 (6(2verticalcoord-1)^2-1)\\,d verticalcoord = 1.\n\\end{align*}\nMoreover, the partial derivatives\n\\begin{align*}\n \\frac{\\partial constantmap}{\\partial verticalcoord}(edgecoordinate,startcoordy) &= 3(1+startcoordy)(8edgecoordinate-4) \\\\\n\\frac{\\partial constantmap}{\\partial horizontalcoord}(edgecoordinate,startcoordy) &= 3(2edgecoordinate-1)^2-1.\n\\end{align*}\nhave no common zero in $(0,1)^2$. Namely,\nfor the first partial to vanish, we must have $edgecoordinate = 1/2$ since\n$1 + startcoordy$ is nowhere zero, but for $edgecoordinate = 1/2$\nthe second partial cannot vanish.\n\n\\textbf{Remark.}\nThis problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate\nfunctions. Many counterexamples are possible. Kent Merryfield suggests\n$horizontalcoord \\sin(2 \\pi verticalcoord)$, for which all four of the boundary integrals vanish;\nhere the partial derivatives are $2\\pi horizontalcoord \\cos (2 \\pi verticalcoord)$ and $\\sin (2 \\pi verticalcoord)$.\nCatalin Zara suggests $verticalcoord^{1/3} horizontalcoord^{2/3}$. Qingchun Ren suggests $verticalcoord\\,horizontalcoord(1-horizontalcoord)$.",
      "confidence": 0.13
    },
    "garbled_string": {
      "map": {
        "x": "flarngto",
        "y": "wixobleu",
        "x_0": "smeftular",
        "y_0": "prandegmi",
        "f": "jubnexico",
        "a": "krunvipse",
        "b": "hladromex",
        "c": "tzarmuvel",
        "d": "vogliptre"
      },
      "question": "Let $jubnexico:[0,1]^2 \\to \\mathbb{R}$ be a continuous function on the closed unit square such that $\\frac{\\partial jubnexico}{\\partial flarngto}$ and $\\frac{\\partial jubnexico}{\\partial wixobleu}$ exist and are continuous on the interior $(0,1)^2$. Let $krunvipse = \\int_0^1 jubnexico(0,wixobleu)\\,d wixobleu$, $hladromex = \\int_0^1 jubnexico(1,wixobleu)\\,d wixobleu$, $tzarmuvel = \\int_0^1 jubnexico(flarngto,0)\\,d flarngto$, $vogliptre = \\int_0^1 jubnexico(flarngto,1)\\,d flarngto$. Prove or disprove: There must be a point $(smeftular,prandegmi)$ in $(0,1)^2$ such that\n\\[\n\\frac{\\partial jubnexico}{\\partial flarngto}(smeftular,prandegmi) = hladromex - krunvipse\n\\quad \\mbox{and} \\quad\n\\frac{\\partial jubnexico}{\\partial wixobleu}(smeftular,prandegmi) = vogliptre - tzarmuvel.\n\\]",
      "solution": "We disprove the assertion using the example\n\\[\njubnexico(flarngto,wixobleu) = 3(1 + wixobleu)(2flarngto-1)^2 - wixobleu.\n\\]\nWe have $hladromex-krunvipse = vogliptre-tzarmuvel = 0$ because the identity $jubnexico(flarngto,wixobleu) = jubnexico(1-flarngto,wixobleu)$ forces $krunvipse = hladromex$, and because\n\\begin{align*}\n tzarmuvel &= \\int_0^1 3(2flarngto-1)^2\\,d flarngto = 1, \\\\\n vogliptre &= \\int_0^1 (6(2flarngto-1)^2-1)\\,d flarngto = 1.\n\\end{align*}\nMoreover, the partial derivatives\n\\begin{align*}\n \\frac{\\partial jubnexico}{\\partial flarngto}(smeftular,prandegmi) &= 3(1+prandegmi)(8smeftular-4), \\\\\n \\frac{\\partial jubnexico}{\\partial wixobleu}(smeftular,prandegmi) &= 3(2smeftular-1)^2-1,\n\\end{align*}\nhave no common zero in $(0,1)^2$. Namely, for the first partial to vanish, we must have $smeftular = 1/2$ since $1 + prandegmi$ is nowhere zero, but for $smeftular = 1/2$ the second partial cannot vanish.\n\n\\textbf{Remark.} This problem amounts to refuting a potential generalization of the Mean Value Theorem to bivariate functions. Many counterexamples are possible. Kent Merryfield suggests $wixobleu \\sin(2 \\pi flarngto)$, for which all four of the boundary integrals vanish; here the partial derivatives are $2\\pi wixobleu \\cos (2 \\pi flarngto)$ and $\\sin (2 \\pi flarngto)$. Catalin Zara suggests $flarngto^{1/3} wixobleu^{2/3}$. Qingchun Ren suggests $flarngto wixobleu(1-wixobleu)$. "
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nf\\colon[0,1]^{3}\\longrightarrow\\mathbb{R}\n\\]\nbe continuous on the closed cube and of class $C^{1}$ on the open cube\n$(0,1)^{3}$.\n\n(E)  (Boundary–regularity hypothesis)  \nEach partial derivative admits a continuous extension to the whole cube:\n\\[\n\\widetilde{\\partial_{i}f}\\colon[0,1]^{3}\\longrightarrow\\mathbb{R},\n\\qquad\n\\widetilde{\\partial_{i}f}\\bigl|_{(0,1)^{3}}=\\partial_{i}f,\n\\qquad i=1,2,3.\n\\]\n\nFor every $i\\in\\{1,2,3\\}$ introduce the face–integrals  \n\\[\nA_{i}= \\iint_{[0,1]^{2}}\n      f\\!\\bigl(x_{1},\\dots ,x_{i-1},0,x_{i+1},x_{3}\\bigr)\\,\n      \\mathrm d\\sigma ,\\qquad  \nB_{i}= \\iint_{[0,1]^{2}}\n      f\\!\\bigl(x_{1},\\dots ,x_{i-1},1,x_{i+1},x_{3}\\bigr)\\,\n      \\mathrm d\\sigma ,\n\\]\nand set  \n\\[\nv:=(v_{1},v_{2},v_{3}),\\qquad v_{i}:=B_{i}-A_{i}\\quad(i=1,2,3).\n\\]\n\nFor later use define  \n\\[\nF(x):=\\nabla f(x)-v,\\qquad x=(x_{1},x_{2},x_{3}).\n\\]\n\nThe problem has two independent parts.\n\n(a)  (Existence under fibre–constant boundary gaps).  \nAssume that  \n\n(H)  For each $i\\in\\{1,2,3\\}$ there exists a constant\n$\\lambda_{i}\\in\\mathbb{R}$ such that  \n\\[\nf(1,y,z)-f(0,y,z)=\\lambda_{1}\\quad\\forall (y,z)\\in[0,1]^{2},\n\\]\n\\[\nf(x,1,z)-f(x,0,z)=\\lambda_{2}\\quad\\forall (x,z)\\in[0,1]^{2},\n\\]\n\\[\nf(x,y,1)-f(x,y,0)=\\lambda_{3}\\quad\\forall (x,y)\\in[0,1]^{2}.\n\\]\n\nIn addition $f$ is \\emph{separately convex}, i.e. for every $i$ and\nevery fixed pair of the remaining coordinates the map\n$t\\mapsto f(\\dots,t,\\dots)$ is convex on $[0,1]$.\n\nProve that under (E) and (H) there exists a point  \n\\[\n(x_{0},y_{0},z_{0})\\in(0,1)^{3}\\quad\\text{such that}\\quad\n\\nabla f(x_{0},y_{0},z_{0})=v.\n\\tag{$\\star$}\n\\]\n\n(b)  (Necessity of (H)).  \nConstruct a $C^{\\infty}$ separately convex function $f$ on $[0,1]^{3}$\nfor which $v=(0,0,0)$ but  \n\\[\n\\nabla f(x,y,z)\\neq(0,0,0)\\qquad\\forall(x,y,z)\\in(0,1)^{3},\n\\]\nthereby showing that (H) is indispensable.",
      "solution": "Throughout write $Q:=[0,1]^{3}$ and $\\operatorname{int}Q=(0,1)^{3}$.\n\n\\textbf{(a)  Proof of $(\\star)$}\n\n\\emph{Step 1:  Identification of the numbers $v_{i}$.}  \nFor $i=1$ and every $(y,z)\\in[0,1]^{2}$ the fundamental theorem of\ncalculus gives  \n\\[\n\\lambda_{1}=f(1,y,z)-f(0,y,z)=\\int_{0}^{1}\\partial_{x}f(t,y,z)\\,\n\\mathrm dt.\n\\]\nIntegrating in $(y,z)$ and using Fubini,\n\\[\nv_{1}=B_{1}-A_{1}\n     =\\int_{Q}\\partial_{x}f\n     =\\iint_{[0,1]^{2}}\\!\\bigl[f(1,y,z)-f(0,y,z)\\bigr]\\mathrm dy\\mathrm dz\n     =\\lambda_{1}.\n\\]\nExactly the same calculation yields  \n\\[\nv_{i}=\\lambda_{i}\\qquad(i=1,2,3).\n\\tag{1}\n\\]\n\n\\emph{Step 2:  One–sided limits of slice derivatives.}  \nFix $(y,z)\\in[0,1]^{2}$ and put\n$g_{y,z}(t):=\\partial_{x}f(t,y,z)$ for $t\\in(0,1)$.  \nSeparate convexity implies that $g_{y,z}$ is monotone non–decreasing.\nHence  \n\\[\ng_{y,z}(0^{+}):=\\lim_{t\\downarrow0}g_{y,z}(t),\\qquad\ng_{y,z}(1^{-}):=\\lim_{t\\uparrow1}g_{y,z}(t)\n\\]\nexist and, by (1),\n\\[\ng_{y,z}(0^{+})\\le v_{1}\\le g_{y,z}(1^{-}).\n\\tag{2}\n\\]\nBecause of (E) these limits coincide with\n$\\widetilde{\\partial_{x}f}(0,y,z)$ and $\\widetilde{\\partial_{x}f}(1,y,z)$,\nrespectively.  Analogous statements hold for $i=2,3$.\n\n\\emph{Step 3:  A vector field with signed faces.}  \nDefine  \n\\[\nG:Q\\longrightarrow\\mathbb{R}^{3},\\qquad\nG(x):=\\nabla f(x)-v.\n\\]\nUsing (2) and its analogues we have the face–sign pattern  \n\\[\n\\begin{cases}\nG_{1}(0,y,z)\\le0,\\quad &G_{1}(1,y,z)\\ge0,\\\\[4pt]\nG_{2}(x,0,z)\\le0,\\quad &G_{2}(x,1,z)\\ge0,\\\\[4pt]\nG_{3}(x,y,0)\\le0,\\quad &G_{3}(x,y,1)\\ge0,\n\\end{cases}\n\\qquad\\forall(x,y,z)\\in[0,1]^{3}.\n\\tag{3}\n\\]\n\n\\emph{Step 4:  Miranda gives a zero in the closed cube.}  \nThe $3$–dimensional Miranda theorem asserts that a continuous map\n$H:Q\\to\\mathbb{R}^{3}$ satisfying the sign conditions (3) possesses a\nzero.  Applied to $H:=G$ this yields  \n\\[\n\\xi:=(x_{*},y_{*},z_{*})\\in Q\\quad\\text{with}\\quad G(\\xi)=0.\n\\tag{4}\n\\]\n\n\\emph{Step 5:  The zero can be moved into the interior.}  \nWe prove that $G$ vanishes at \\emph{some} point of $\\operatorname{int}Q$.\n\nAssume first that $G(\\xi)=0$ with $x_{*}=0$ (the case\n$x_{*}=1$ is analogous).  \nFix $y_{*},z_{*}$ and set $g(t):=\\partial_{x}f(t,y_{*},z_{*})$\nfor $t\\in[0,1]$.  \nBy (E) and (4), $g$ is continuous on $[0,1]$,\n$g(0)=v_{1}$, and, by separate convexity, $g$ is monotone\nnon–decreasing.  Moreover  \n\\[\n\\int_{0}^{1}\\bigl[g(t)-v_{1}\\bigr]\\mathrm dt\n   =\\int_{0}^{1}g(t)\\,\\mathrm dt-\\lambda_{1}=v_{1}-v_{1}=0\n\\]\nusing (1).  Because a continuous non–decreasing function that\nnever drops below its initial value has non–negative integral excess,\nthe last equality forces  \n\\[\ng(t)\\equiv v_{1}\\qquad\\forall t\\in[0,1].\n\\tag{5}\n\\]\nConsequently $G_{1}(t,y_{*},z_{*})=0$ for \\emph{all} $t\\in[0,1]$.\nThus the whole line segment\n\\[\nL:=\\{(t,y_{*},z_{*})\\mid t\\in[0,1]\\}\n\\]\nlies in the zero–set of $G$.  Picking any $t_{0}\\in(0,1)$ we obtain the\ndesired interior zero $(t_{0},y_{*},z_{*})$.\n\nIf instead $\\xi$ happens to satisfy $y_{*}=0$ or $y_{*}=1$ (or\n$z_{*}=0,1$) the same argument, applied to the corresponding coordinate,\nproduces a segment in the zero–set parallel to the $y$– or\n$z$–axis, from which an interior point can be chosen.\n\nFinally, if $\\xi$ lies on an edge or a corner, then two or all three\ncoordinates of $G$ satisfy the constancy argument simultaneously; hence\nan entire rectangle (or the whole cube) is contained in the zero–set and\nagain an interior point is available.\n\nCombining all cases we conclude that  \n\\[\n\\exists(x_{0},y_{0},z_{0})\\in(0,1)^{3}\\quad\\text{with}\\quad G(x_{0},y_{0},z_{0})=0,\n\\]\ni.e. $\\nabla f(x_{0},y_{0},z_{0})=v$.  This proves $(\\star)$.\n\n\\bigskip\n\\textbf{(b)  A smooth separately convex counterexample when (H) fails.}\n\nDefine  \n\\[\n\\delta(y):=y-\\tfrac12,\\qquad\n\\lambda(y):=\\tfrac{17}{6}-10\\bigl(y-\\tfrac12\\bigr)^{2},\n\\]\nand set  \n\\[\n\\boxed{\nf(x,y,z):=x\\,\\delta(y)\n+\\delta(y)^{2}\n+\\bigl(z-3y\\bigr)^{2}\n-\\tfrac{13}{2}\\,y\n+\\lambda(y)\\,z\n}\\qquad(x,y,z)\\in[0,1]^{3}.\n\\tag{6}\n\\]\n\n1. \\emph{Regularity and separate convexity.}  \n$f$ is a polynomial, hence $C^{\\infty}$.  Slice second derivatives are  \n\\[\n\\partial_{xx}^{2}f\\equiv0,\\qquad\n\\partial_{zz}^{2}f\\equiv2,\\qquad\n\\partial_{yy}^{2}f=20(1-z)\\ge0\\quad(0\\le z\\le1),\n\\]\nso each one–variable slice is convex.\n\n2. \\emph{Evaluation of $v$.}\n\\[\n\\begin{aligned}\nf(1,y,z)-f(0,y,z)&=\\delta(y),\\\\\nf(x,1,z)-f(x,0,z)&=x+\\bigl(z-3\\bigr)^{2}-z^{2}-\\tfrac{13}{2},\\\\\nf(x,y,1)-f(x,y,0)&=1-6y+\\lambda(y).\n\\end{aligned}\n\\]\nEach expression has integral $0$ over its free variables, hence\n$v=(0,0,0)$.\n\n3. \\emph{The gradient never vanishes in the open cube.}\n\\[\n\\nabla f(x,y,z)=\n\\Bigl(\n\\delta(y),\\;\nx+2\\delta(y)-6\\bigl(z-3y\\bigr)-\\tfrac{13}{2}+\\lambda'(y)z,\\;\n2\\bigl(z-3y\\bigr)+\\lambda(y)\n\\Bigr),\n\\]\nwith $\\lambda'(y)=-20\\delta(y)$.  \nIf $\\nabla f(x,y,z)=\\mathbf 0$, the first component forces\n$y=\\tfrac12$; the third then yields $z=\\tfrac1{12}$, and the second\nbecomes $x+2\\neq0$ – a contradiction.  Thus $\\nabla f\\neq\\mathbf0$ on\n$(0,1)^{3}$.\n\n4. \\emph{Failure of (H).}  \nAlready\n\\[\nf(1,y,z)-f(0,y,z)=y-\\tfrac12\n\\]\ndepends on $y$, contradicting the first clause of (H).  \nTherefore (H) cannot be omitted.  \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.814685",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a 2-dimensional square to a 3-dimensional cube; the sought equality involves all three components of the gradient simultaneously.\n\n2. Additional structural hypothesis: the “separate convexity’’ condition in (a) forces sophisticated use of monotone operators inside an integral identity and necessitates an n-dimensional intermediate-value theorem (Poincaré–Miranda) rather than the one-dimensional argument used in the original kernel variant.\n\n3. Two-part nature: contestants must both (i) establish the positive result under a non-trivial analytic constraint and (ii) craft a completely explicit C^∞ counter-example once that constraint is removed.  The first part requires integrating monotone families of functions and handling boundary inequalities pointwise; the second part demands inventiveness to ensure all six face averages cancel while simultaneously blocking any interior solution.\n\n4. Interacting concepts: the solution mixes convex analysis, integral identities, monotonicity arguments, multivariate fixed-point theorems, and explicit algebraic construction, substantially deepening the theoretical toolkit compared with the original problem, where a single polynomial counter-example sufficed."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nf\\colon[0,1]^{3}\\longrightarrow\\mathbb R\n\\]\nbe continuous on the closed cube and of class $C^{1}$ on its interior.  \nFor $i\\in\\{1,2,3\\}$ put  \n\\[\nA_{i}= \\!\\!\\iint_{[0,1]^{2}}\n      f\\!\\bigl(x_{1},\\dots ,x_{i-1},0,x_{i+1},x_{3}\\bigr)\\,\n      \\mathrm d\\sigma ,\\qquad  \nB_{i}= \\!\\!\\iint_{[0,1]^{2}}\n      f\\!\\bigl(x_{1},\\dots ,x_{i-1},1,x_{i+1},x_{3}\\bigr)\\,\n      \\mathrm d\\sigma ,\n\\]\nand define the vector  \n\\[\nv=(v_{1},v_{2},v_{3}),\\qquad v_{i}:=B_{i}-A_{i}\\quad(i=1,2,3).\n\\]\n\nSet  \n\\[\nF(x):=\\nabla f(x)-v,\\qquad x=(x_{1},x_{2},x_{3}).\n\\]\n\nThe problem has two parts.\n\n(a) (Existence under fibre–constant boundary gaps).  \nAssume  \n\n(H)\\;For every $i\\in\\{1,2,3\\}$ there is a constant $\\lambda_{i}\\in\\mathbb R$ such that  \n\\[\nf(1,y,z)-f(0,y,z)=\\lambda_{1}\\quad\\forall (y,z)\\in[0,1]^{2},\n\\]\n\\[\nf(x,1,z)-f(x,0,z)=\\lambda_{2}\\quad\\forall (x,z)\\in[0,1]^{2},\n\\]\n\\[\nf(x,y,1)-f(x,y,0)=\\lambda_{3}\\quad\\forall (x,y)\\in[0,1]^{2}.\n\\]\n\nMoreover $f$ is {\\em separately convex}:\nfor each $i$ and every fixed pair of the remaining variables the map\n$t\\mapsto f(\\dots,t,\\dots)$ is convex on $[0,1]$.\n\nProve that there exists a point $(x_{0},y_{0},z_{0})\\in(0,1)^{3}$ such that  \n\\[\n\\nabla f(x_{0},y_{0},z_{0})=v.\n\\tag{$\\star$}\n\\]\n\n(b) (Necessity of (H)).  \nConstruct a $C^{\\infty}$ separately convex function on $[0,1]^{3}$\nfor which $v=(0,0,0)$ but  \n\\[\n\\nabla f(x,y,z)\\neq v\\qquad\\forall(x,y,z)\\in(0,1)^{3},\n\\]\nthereby showing that the additional assumption (H) cannot be omitted.",
      "solution": "(a) Proof of $(\\star)$.\n\n{\\bf Step 1: identification of the numbers $v_{i}$.}\n\nBecause of (H) we have for $i=1$\n\\[\nf(1,y,z)-f(0,y,z)=\\lambda_{1}\\quad\\forall(y,z)\\in[0,1]^{2}.\n\\]\nApplying the Fundamental Theorem of Calculus in the $x$–variable,\n\\[\n\\lambda_{1}\n      =\\int_{0}^{1}\\partial_{x}f(t,y,z)\\,\\mathrm dt,\\qquad\n      v_{1}=B_{1}-A_{1}\n      =\\int_{[0,1]^{3}}\\partial_{x}f\n      =\\int_{[0,1]^{2}}\\bigl[f(1,y,z)-f(0,y,z)\\bigr]\n      =\\lambda_{1}.\n\\]\nThe same computation for the other two coordinates yields  \n\\[\nv_{i}=\\lambda_{i}\\quad(i=1,2,3).\n\\tag{1}\n\\]\n\n{\\bf Step 2: sign information on the faces.}\n\nFor $(y,z)\\in[0,1]^{2}$ define  \n\\[\ng_{y,z}(t):=f(t,y,z).\n\\]\nSeparate convexity implies that $g_{y,z}$ is convex and hence\n$t\\mapsto\\partial_{t}g_{y,z}(t)=\\partial_{x}f(t,y,z)$ is\nnon–decreasing.  Using (1) we get\n\\[\n\\partial_{x}f(0,y,z)\\le v_{1}\\le\\partial_{x}f(1,y,z),\n\\]\nand therefore\n\\[\nF_{1}(0,y,z)\\le 0,\\qquad \nF_{1}(1,y,z)\\ge 0.\n\\]\nRepeating the argument for the other two variables gives\n\\[\n\\begin{aligned}\nF_{1}(0,y,z)&\\le 0,&\\quad F_{1}(1,y,z)&\\ge 0,\\\\\nF_{2}(x,0,z)&\\le 0,&\\quad F_{2}(x,1,z)&\\ge 0,\\\\\nF_{3}(x,y,0)&\\le 0,&\\quad F_{3}(x,y,1)&\\ge 0.\n\\end{aligned}\n\\tag{2}\n\\]\n\n{\\bf Step 3: a zero of $F$ in the closed cube.}\n\nConditions (2) are precisely those of the Poincaré–Miranda theorem,\nhence there is a point $p=(x_{*},y_{*},z_{*})\\in[0,1]^{3}$ with\n$F(p)=\\mathbf 0$.\n\n{\\bf Step 4: pushing the zero into the {\\em open} cube.}\n\nWe refine the construction of Step 4 in the draft by proving explicitly\nthat the required coordinate selectors are continuous.\n\n{\\em 4.1. The zero correspondence and its values lie in $(0,1)$.}\n\nFor $(y,z)\\in[0,1]^{2}$ consider\n\\[\nG_{1}(t;y,z):=F_{1}(t,y,z),\\qquad t\\in[0,1].\n\\]\nBy (2) the map $t\\mapsto G_{1}(t;y,z)$ is continuous,\nnon–decreasing, satisfies $G_{1}(0;y,z)\\le 0\\le G_{1}(1;y,z)$, and\n\\[\n\\int_{0}^{1}G_{1}(t;y,z)\\,\\mathrm dt\n      =\\int_{0}^{1}\\bigl(\\partial_{x}f(t,y,z)-v_{1}\\bigr)\\,\\mathrm dt\n      =f(1,y,z)-f(0,y,z)-\\lambda_{1}=0.\n\\]\nHence {\\em exactly} one of the following occurs:\n(a) $G_{1}\\equiv 0$,  \n(b) $G_{1}(0;y,z)<0<G_{1}(1;y,z)$.  \nIn case (b) the first zero\n\\[\n\\phi_{1}(y,z):=\\inf\\{\\,t\\in(0,1]\\mid G_{1}(t;y,z)=0\\,\\}\n\\]\nis {\\em strictly} positive; in case (a) we simply set\n$\\phi_{1}(y,z):=\\tfrac12$.  Thus\n\\[\n\\phi_{1}\\colon[0,1]^{2}\\longrightarrow(0,1).\n\\tag{3}\n\\]\n\n{\\em 4.2. Continuity of $\\phi_{1}$.}\n\nLet\n\\[\n\\Phi_{1}(y,z):=\\{\\,t\\in[0,1]\\mid G_{1}(t;y,z)=0\\,\\}.\n\\]\nBecause $G_{1}$ is continuous, $\\Phi_{1}(y,z)$ is always a closed\ninterval (possibly a singleton).  The set–valued map\n$(y,z)\\mapsto\\Phi_{1}(y,z)$ has a closed graph and compact values, so\nby Berge’s Maximum Theorem both the minimum and the maximum of\n$\\Phi_{1}(y,z)$ vary {\\em continuously} with $(y,z)$.  In particular\nthe choice\n\\[\n\\phi_{1}(y,z):=\\tfrac12\\bigl[\\min\\Phi_{1}(y,z)+\\max\\Phi_{1}(y,z)\\bigr]\n\\]\ndefines a continuous function that still satisfies (3) and\n$G_{1}(\\phi_{1}(y,z);y,z)=0$.  Repeating the argument for $F_{2}$ and\n$F_{3}$ we obtain continuous maps\n\\[\n\\phi_{2}\\colon[0,1]^{2}\\longrightarrow(0,1),\\qquad\n\\phi_{3}\\colon[0,1]^{2}\\longrightarrow(0,1),\n\\]\nwith\n\\[\nF_{2}(x,\\phi_{2}(x,z),z)=0,\\qquad\nF_{3}(x,y,\\phi_{3}(x,y))=0.\n\\tag{4}\n\\]\n\n{\\em 4.3. A continuous self–map of the cube and Brouwer’s fixed point.}\n\nDefine\n\\[\nT(x,y,z):=\\bigl(\\phi_{1}(y,z),\\ \\phi_{2}(x,z),\\ \\phi_{3}(x,y)\\bigr),\n         \\qquad (x,y,z)\\in[0,1]^{3}.\n\\]\nBy (3)–(4) the map $T$ is continuous and $T([0,1]^{3})\\subset(0,1)^{3}$.\nBrouwer’s fixed–point theorem yields\n\\[\n(x_{0},y_{0},z_{0})\\in[0,1]^{3}\\quad\\text{with}\\quad\nT(x_{0},y_{0},z_{0})=(x_{0},y_{0},z_{0}).\n\\]\nBecause $T$ takes its values in $(0,1)^{3}$, the fixed point lies in\nthe {\\em open} cube:\n\\[\n(x_{0},y_{0},z_{0})\\in(0,1)^{3}.\n\\]\n\n{\\em 4.4. Verification that $F(x_{0},y_{0},z_{0})=\\mathbf 0$.}\n\nThe first coordinate of the fixed–point identity gives\n$x_{0}=\\phi_{1}(y_{0},z_{0})$, whence\n$F_{1}(x_{0},y_{0},z_{0})=0$ by construction.  The other two\ncoordinates, together with (4), furnish\n$F_{2}(x_{0},y_{0},z_{0})=F_{3}(x_{0},y_{0},z_{0})=0$.\nConsequently $F(x_{0},y_{0},z_{0})=\\mathbf 0$, i.e.\\ $(\\star)$ holds.\n\n\\medskip\n(b) A smooth separately convex counter–example when (H) fails.\n\nIntroduce  \n\\[\n\\delta(y):=y-\\tfrac12,\\qquad \n\\lambda(y):=\\tfrac{17}{6}-10\\bigl(y-\\tfrac12\\bigr)^{2},\n\\]\nand set  \n\\[\n\\boxed{\nf(x,y,z)=x\\,\\delta(y)\n+\\delta(y)^{2}\n+\\bigl(z-3y\\bigr)^{2}\n-\\tfrac{13}{2}\\,y\n+\\lambda(y)\\,z\n}\\qquad(x,y,z\\in[0,1]).\n\\tag{5}\n\\]\n\n(1) {\\em Regularity and separate convexity.}  \nThe function $f$ is a polynomial, hence $C^{\\infty}$.  The slice second\nderivatives are  \n\\[\n\\partial_{xx}^{2}f\\equiv 0,\\qquad\n\\partial_{zz}^{2}f\\equiv 2,\\qquad\n\\partial_{yy}^{2}f = 20(1-z)\\ge 0\\quad(0\\le z\\le 1),\n\\]\nso each one–variable slice is convex; thus $f$ is separately convex.\n\n(2) {\\em Computation of $v$.}  \nThe boundary differences are\n\\[\n\\Delta_{x}f(y,z)=f(1,y,z)-f(0,y,z)=\\delta(y),\n\\]\n\\[\n\\Delta_{y}f(x,z)=f(x,1,z)-f(x,0,z)\n                =x+\\bigl(z-3\\bigr)^{2}-z^{2}-\\tfrac{13}{2},\n\\]\n\\[\n\\Delta_{z}f(x,y)=f(x,y,1)-f(x,y,0)=1-6y+\\lambda(y).\n\\]\nEach of these has zero average over its variables, hence\n$v_{1}=v_{2}=v_{3}=0$ and $v=\\mathbf 0$.\n\n(3) {\\em The gradient never equals $v$.}  \nA direct computation yields\n\\[\n\\nabla f(x,y,z)=\n\\Bigl(\n\\delta(y),\\;\nx+2\\delta(y)-6\\bigl(z-3y\\bigr)-\\tfrac{13}{2}+\\,\\lambda'(y)\\,z,\\;\n2\\bigl(z-3y\\bigr)+\\lambda(y)\n\\Bigr),\n\\]\nwith $\\lambda'(y)=-20\\bigl(y-\\tfrac12\\bigr)$.  \nIf the first component vanishes then $y=\\tfrac12$, which forces\n$z=\\tfrac1{12}$ from the third component; the second component then\nequals $x+2\\neq 0$ for $x\\in(0,1)$.  Hence\n\\[\n\\nabla f(x,y,z)\\neq\\mathbf 0=v\\qquad\\forall(x,y,z)\\in(0,1)^{3}.\n\\]\n\n(4) {\\em Failure of (H).}  \nAlready\n\\[\nf(1,y,z)-f(0,y,z)=y-\\tfrac12\n\\]\ndepends on $y$, so the first line in (H) is violated.\nTherefore $(\\star)$ can fail without the fibre–constant assumption,\nwhich proves its necessity. \\hfill$\\square$",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.623570",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem moves from a 2-dimensional square to a 3-dimensional cube; the sought equality involves all three components of the gradient simultaneously.\n\n2. Additional structural hypothesis: the “separate convexity’’ condition in (a) forces sophisticated use of monotone operators inside an integral identity and necessitates an n-dimensional intermediate-value theorem (Poincaré–Miranda) rather than the one-dimensional argument used in the original kernel variant.\n\n3. Two-part nature: contestants must both (i) establish the positive result under a non-trivial analytic constraint and (ii) craft a completely explicit C^∞ counter-example once that constraint is removed.  The first part requires integrating monotone families of functions and handling boundary inequalities pointwise; the second part demands inventiveness to ensure all six face averages cancel while simultaneously blocking any interior solution.\n\n4. Interacting concepts: the solution mixes convex analysis, integral identities, monotonicity arguments, multivariate fixed-point theorems, and explicit algebraic construction, substantially deepening the theoretical toolkit compared with the original problem, where a single polynomial counter-example sufficed."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}