{
  "index": "1996-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "Let $(a_1, b_1), (a_2, b_2), \\ldots, (a_n, b_n)$ be the vertices of a\nconvex polygon which contains the origin in its interior. Prove that\nthere exist positive real numbers $x$ and $y$ such that\n\\begin{gather*}\n(a_1, b_1)x^{a_1} y^{b_1} + (a_2, b_2)x^{a_2}y^{b_2} + \\cdots \\\\\n+ (a_n, b_n)x^{a_n}y^{b_n} = (0,0).\n\\end{gather*}\n\\end{itemize}\n\n\\end{document}",
  "solution": "We will prove the claim assuming only that the convex hull of the\npoints $(a_{i}, b_{i})$ contains the origin in its interior. (Thanks\nto Marshall Buck for pointing out that the last three words are\nnecessary in the previous sentence!) Let $u = \\log x, v = \\log\ny$ so that the left-hand side of the given equation is\n\\begin{multline}\n(a_1, b_1) \\exp(a_1 u + b_1 v) + (a_2, b_2) \\exp(a_2 u + b_2 v) + \\\\\n\\cdots + (a_n, b_n) \\exp(a_n u + b_n v).\n\\end{multline}\nNow note that (1) is the gradient of the function\n\\begin{gather*}\nf(u,v) = exp(a_1 u + b_1 v) +\nexp(a_2 u + b_2 v) + \\\\\n\\cdots + exp(a_n u + b_n v),\n\\end{gather*}\nand so it suffices to show $f$ has a critical point. We will in fact\nshow $f$ has a global minimum.\n\nClearly we have\n\\[\nf(u,v) \\geq \\exp\\left( \\max_i (a_i u + b_i v) \\right).\n\\]\nNote that this maximum is positive for $(u,v) \\neq (0,0)$:  if we had\n$a_i u + b_i v < 0$ for all $i$, then the subset $ur + vs < 0$ of the\n$rs$-plane would be a half-plane containing all of the points $(a_i,\nb_i)$, whose convex hull would then not contain the origin, a\ncontradiction.\n\nThe function $\\max_{i} (a_{i}u + b_{i}v)$ is clearly\ncontinuous on the unit circle $u^{2} + v^{2} = 1$, which is compact.\nHence it has a global minimum $M > 0$, and so for all $u,v$,\n\\[\n\\max_{i} (a_{i} u + b_{i} v) \\geq M \\sqrt{u^{2} + v^{2}}.\n\\]\nIn particular, $f \\geq n+1$ on the disk of radius $\\sqrt{(n+1)/M}$.\nSince $f(0,0) = n$, the infimum of $f$ is the same over the entire\n$uv$-plane as over this disk, which again is compact.\nHence $f$ attains its infimal value at some point in the disk,\nwhich is the desired global minimum.\n\nNoam Elkies has suggested an alternate solution as follows: for $r >\n0$, draw the loop traced by (1) as $(u,v)$ travels\ncounterclockwise around the circle $u^2 + v^2 = r^2$. For $r=0$, this\nof course has winding number 0 about any point, but for $r$ large, one\ncan show this loop has winding number 1 about the origin, so somewhere in\nbetween the loop must pass through the origin. (Proving this latter\nfact is a little tricky.)\n\n\\end{itemize}\n\\end{document}",
  "vars": [
    "x",
    "y",
    "u",
    "v",
    "r",
    "s"
  ],
  "params": [
    "a_1",
    "a_2",
    "a_n",
    "a_i",
    "b_1",
    "b_2",
    "b_n",
    "b_i",
    "n",
    "M"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "varxpos",
        "y": "varypos",
        "u": "loguvar",
        "v": "logvvar",
        "r": "radialvr",
        "s": "slopevar",
        "a_1": "coeffaone",
        "a_2": "coeffatwo",
        "a_n": "coeffanmax",
        "a_i": "coeffaidx",
        "b_1": "coeffbone",
        "b_2": "coeffbtwo",
        "b_n": "coeffbnmax",
        "b_i": "coeffbidx",
        "n": "verticesn",
        "M": "minbound"
      },
      "question": "Let $(coeffaone, coeffbone), (coeffatwo, coeffbtwo), \\ldots, (coeffanmax, coeffbnmax)$ be the vertices of a\nconvex polygon which contains the origin in its interior. Prove that\nthere exist positive real numbers $varxpos$ and $varypos$ such that\n\\begin{gather*}\n(coeffaone, coeffbone)varxpos^{coeffaone} varypos^{coeffbone} + (coeffatwo, coeffbtwo)varxpos^{coeffatwo}varypos^{coeffbtwo} + \\cdots \\\\\n+ (coeffanmax, coeffbnmax)varxpos^{coeffanmax}varypos^{coeffbnmax} = (0,0).\n\\end{gather*}\n\\end{itemize}\n\n\\end{document}",
      "solution": "We will prove the claim assuming only that the convex hull of the\npoints $(coeffaidx, coeffbidx)$ contains the origin in its interior. (Thanks\nto Marshall Buck for pointing out that the last three words are\nnecessary in the previous sentence!) Let $loguvar = \\log varxpos, logvvar = \\log\nvarypos$ so that the left-hand side of the given equation is\n\\begin{multline}\n(coeffaone, coeffbone) \\exp(coeffaone loguvar + coeffbone logvvar) + (coeffatwo, coeffbtwo) \\exp(coeffatwo loguvar + coeffbtwo logvvar) + \\\\\n\\cdots + (coeffanmax, coeffbnmax) \\exp(coeffanmax loguvar + coeffbnmax logvvar).\n\\end{multline}\nNow note that (1) is the gradient of the function\n\\begin{gather*}\nf(loguvar,logvvar) = \\exp(coeffaone loguvar + coeffbone logvvar) +\n\\exp(coeffatwo loguvar + coeffbtwo logvvar) + \\\\\n\\cdots + \\exp(coeffanmax loguvar + coeffbnmax logvvar),\n\\end{gather*}\nand so it suffices to show $f$ has a critical point. We will in fact\nshow $f$ has a global minimum.\n\nClearly we have\n\\[\nf(loguvar,logvvar) \\geq \\exp\\left( \\max_i (coeffaidx loguvar + coeffbidx logvvar) \\right).\n\\]\nNote that this maximum is positive for $(loguvar,logvvar) \\neq (0,0)$:  if we had\n$coeffaidx loguvar + coeffbidx logvvar < 0$ for all $i$, then the subset $loguvar radialvr + logvvar slopevar < 0$ of the\n$radialvr slopevar$-plane would be a half-plane containing all of the points $(coeffaidx,\ncoeffbidx)$, whose convex hull would then not contain the origin, a\ncontradiction.\n\nThe function $\\max_{i} (coeffaidx loguvar + coeffbidx logvvar)$ is clearly\ncontinuous on the unit circle $loguvar^{2} + logvvar^{2} = 1$, which is compact.\nHence it has a global minimum $minbound > 0$, and so for all $loguvar,logvvar$,\n\\[\n\\max_{i} (coeffaidx loguvar + coeffbidx logvvar) \\geq minbound \\sqrt{loguvar^{2} + logvvar^{2}}.\n\\]\nIn particular, $f \\geq verticesn+1$ on the disk of radius $\\sqrt{(verticesn+1)/minbound}$.\nSince $f(0,0) = verticesn$, the infimum of $f$ is the same over the entire\nloguvar logvvar-plane as over this disk, which again is compact.\nHence $f$ attains its infimal value at some point in the disk,\nwhich is the desired global minimum.\n\nNoam Elkies has suggested an alternate solution as follows: for $radialvr >\n0$, draw the loop traced by (1) as $(loguvar,logvvar)$ travels\ncounterclockwise around the circle $loguvar^{2} + logvvar^{2} = radialvr^{2}$. For $radialvr=0$, this\nof course has winding number 0 about any point, but for $radialvr$ large, one\ncan show this loop has winding number 1 about the origin, so somewhere in\nbetween the loop must pass through the origin. (Proving this latter\nfact is a little tricky.)\n\n\\end{itemize}\n\\end{document}"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lighthouse",
        "y": "snowflake",
        "u": "breadcrumb",
        "v": "paintbrush",
        "r": "cloudscape",
        "s": "moonstone",
        "a_1": "eclipseone",
        "a_2": "eclipsetwo",
        "a_n": "eclipsezen",
        "a_i": "eclipseeach",
        "b_1": "nebulaone",
        "b_2": "nebulatwo",
        "b_n": "nebulazen",
        "b_i": "nebulaeach",
        "n": "driftwood",
        "M": "starflower"
      },
      "question": "Let $(eclipseone, nebulaone), (eclipsetwo, nebulatwo), \\ldots, (eclipsezen, nebulazen)$ be the vertices of a\nconvex polygon which contains the origin in its interior. Prove that\nthere exist positive real numbers $lighthouse$ and $snowflake$ such that\n\\begin{gather*}\n(eclipseone, nebulaone)lighthouse^{eclipseone} snowflake^{nebulaone} + (eclipsetwo, nebulatwo)lighthouse^{eclipsetwo}snowflake^{nebulatwo} + \\cdots \\\\\n+ (eclipsezen, nebulazen)lighthouse^{eclipsezen}snowflake^{nebulazen} = (0,0).\n\\end{gather*}\n\\end{itemize}\n\n\\end{document}",
      "solution": "We will prove the claim assuming only that the convex hull of the\npoints $(eclipseeach, nebulaeach)$ contains the origin in its interior. (Thanks\nto Marshall Buck for pointing out that the last three words are\nnecessary in the previous sentence!) Let $breadcrumb = \\log lighthouse, paintbrush = \\log\nsnowflake$ so that the left-hand side of the given equation is\n\\begin{multline}\n(eclipseone, nebulaone) \\exp(eclipseone \\, breadcrumb + nebulaone \\, paintbrush) + (eclipsetwo, nebulatwo) \\exp(eclipsetwo \\, breadcrumb + nebulatwo \\, paintbrush) + \\\\\n\\cdots + (eclipsezen, nebulazen) \\exp(eclipsezen \\, breadcrumb + nebulazen \\, paintbrush).\n\\end{multline}\nNow note that (1) is the gradient of the function\n\\begin{gather*}\nf(breadcrumb,paintbrush) = \\exp(eclipseone \\, breadcrumb + nebulaone \\, paintbrush) +\n\\exp(eclipsetwo \\, breadcrumb + nebulatwo \\, paintbrush) + \\\\\n\\cdots + \\exp(eclipsezen \\, breadcrumb + nebulazen \\, paintbrush),\n\\end{gather*}\nand so it suffices to show $f$ has a critical point. We will in fact\nshow $f$ has a global minimum.\n\nClearly we have\n\\[\nf(breadcrumb,paintbrush) \\geq \\exp\\left( \\max_i (eclipseeach \\, breadcrumb + nebulaeach \\, paintbrush) \\right).\n\\]\nNote that this maximum is positive for $(breadcrumb,paintbrush) \\neq (0,0)$:  if we had\n$eclipseeach \\, breadcrumb + nebulaeach \\, paintbrush < 0$ for all $i$, then the subset $breadcrumb \\, cloudscape + paintbrush \\, moonstone < 0$ of the\n$cloudscape moonstone$-plane would be a half-plane containing all of the points $(eclipseeach,\nnebulaeach)$, whose convex hull would then not contain the origin, a\ncontradiction.\n\nThe function $\\max_{i} (eclipseeach \\, breadcrumb + nebulaeach \\, paintbrush)$ is clearly\ncontinuous on the unit circle $breadcrumb^{2} + paintbrush^{2} = 1$, which is compact.\nHence it has a global minimum $starflower > 0$, and so for all $breadcrumb,paintbrush$,\n\\[\n\\max_{i} (eclipseeach \\, breadcrumb + nebulaeach \\, paintbrush) \\geq starflower \\sqrt{breadcrumb^{2} + paintbrush^{2}}.\n\\]\nIn particular, $f \\geq driftwood+1$ on the disk of radius $\\sqrt{(driftwood+1)/starflower}$.\nSince $f(0,0) = driftwood$, the infimum of $f$ is the same over the entire\n$breadcrumb paintbrush$-plane as over this disk, which again is compact.\nHence $f$ attains its infimal value at some point in the disk,\nwhich is the desired global minimum.\n\nNoam Elkies has suggested an alternate solution as follows: for $cloudscape >\n0$, draw the loop traced by (1) as $(breadcrumb,paintbrush)$ travels\ncounterclockwise around the circle $breadcrumb^{2} + paintbrush^{2} = cloudscape^{2}$. For $cloudscape=0$, this\nof course has winding number 0 about any point, but for $cloudscape$ large, one\ncan show this loop has winding number 1 about the origin, so somewhere in\nbetween the loop must pass through the origin. (Proving this latter\nfact is a little tricky.)\n\n\\end{itemize}\n\\end{document}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoor",
        "y": "horizontalcoor",
        "u": "compression",
        "v": "stretching",
        "r": "diameter",
        "s": "stillness",
        "a_1": "verticalone",
        "a_2": "verticaltwo",
        "a_n": "verticaln",
        "a_i": "verticali",
        "b_1": "horizontalone",
        "b_2": "horizontaltwo",
        "b_n": "horizontaln",
        "b_i": "horizontali",
        "n": "continuum",
        "M": "globalmax"
      },
      "question": "Let $(verticalone, horizontalone), (verticaltwo, horizontaltwo), \\ldots, (verticaln, horizontaln)$ be the vertices of a\nconvex polygon which contains the origin in its interior. Prove that\nthere exist positive real numbers $verticalcoor$ and $horizontalcoor$ such that\n\\begin{gather*}\n(verticalone, horizontalone)verticalcoor^{verticalone} horizontalcoor^{horizontalone} + (verticaltwo, horizontaltwo)verticalcoor^{verticaltwo}horizontalcoor^{horizontaltwo} + \\cdots \\\\\n+ (verticaln, horizontaln)verticalcoor^{verticaln}horizontalcoor^{horizontaln} = (0,0).\n\\end{gather*}\n",
      "solution": "We will prove the claim assuming only that the convex hull of the\npoints $(verticali, horizontali)$ contains the origin in its interior. (Thanks\nto Marshall Buck for pointing out that the last three words are\nnecessary in the previous sentence!) Let $compression = \\log verticalcoor,\\; stretching = \\log\nhorizontalcoor$ so that the left-hand side of the given equation is\n\\begin{multline}\n(verticalone, horizontalone) \\exp(verticalone compression + horizontalone stretching) + (verticaltwo, horizontaltwo) \\exp(verticaltwo compression + horizontaltwo stretching) + \\\\\n\\cdots + (verticaln, horizontaln) \\exp(verticaln compression + horizontaln stretching).\n\\end{multline}\nNow note that (1) is the gradient of the function\n\\begin{gather*}\nf(compression,stretching) = exp(verticalone compression + horizontalone stretching) +\nexp(verticaltwo compression + horizontaltwo stretching) + \\\\\n\\cdots + exp(verticaln compression + horizontaln stretching),\n\\end{gather*}\nand so it suffices to show $f$ has a critical point. We will in fact\nshow $f$ has a global minimum.\n\nClearly we have\n\\[\nf(compression,stretching) \\geq \\exp\\left( \\max_i (verticali compression + horizontali stretching) \\right).\n\\]\nNote that this maximum is positive for $(compression,stretching) \\neq (0,0)$:  if we had\n$verticali compression + horizontali stretching < 0$ for all $i$, then the subset $compression diameter + stretching stillness < 0$ of the\n$diameter stillness$-plane would be a half-plane containing all of the points $(verticali,\nhorizontali)$, whose convex hull would then not contain the origin, a\ncontradiction.\n\nThe function $\\max_{i} (verticali compression + horizontali stretching)$ is clearly\ncontinuous on the unit circle $compression^{2} + stretching^{2} = 1$, which is compact.\nHence it has a global minimum $globalmax > 0$, and so for all $compression,stretching$,\n\\[\n\\max_{i} (verticali compression + horizontali stretching) \\geq globalmax \\sqrt{compression^{2} + stretching^{2}}.\n\\]\nIn particular, $f \\geq continuum+1$ on the disk of radius $\\sqrt{(continuum+1)/globalmax}$.\nSince $f(0,0) = continuum$, the infimum of $f$ is the same over the entire\ncompressionstretching-plane as over this disk, which again is compact.\nHence $f$ attains its infimal value at some point in the disk,\nwhich is the desired global minimum.\n\nNoam Elkies has suggested an alternate solution as follows: for $diameter >\n0$, draw the loop traced by (1) as $(compression,stretching)$ travels\ncounterclockwise around the circle $compression^2 + stretching^2 = diameter^2$. For $diameter=0$, this\nof course has winding number 0 about any point, but for $diameter$ large, one\ncan show this loop has winding number 1 about the origin, so somewhere in\nbetween the loop must pass through the origin. (Proving this latter\nfact is a little tricky.)\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "u": "mclpzrqe",
        "v": "nskdwtuy",
        "r": "fjgkrnsa",
        "s": "plhmdzqo",
        "a_1": "zvxcrlpe",
        "a_2": "bjkstemq",
        "a_n": "lsdowprm",
        "a_i": "kqjtmzre",
        "b_1": "gfjqdmzo",
        "b_2": "wlyhkpra",
        "b_n": "crmsqvtx",
        "b_i": "tpvdzwla",
        "n": "xrcpsgqa",
        "M": "hrfbwzeq"
      },
      "question": "Let $(zvxcrlpe, gfjqdmzo), (bjkstemq, wlyhkpra), \\ldots, (lsdowprm, crmsqvtx)$ be the vertices of a\nconvex polygon which contains the origin in its interior. Prove that\nthere exist positive real numbers $qzxwvtnp$ and $hjgrksla$ such that\n\\begin{gather*}\n(zvxcrlpe, gfjqdmzo)qzxwvtnp^{zvxcrlpe} hjgrksla^{gfjqdmzo} + (bjkstemq, wlyhkpra)qzxwvtnp^{bjkstemq}hjgrksla^{wlyhkpra} + \\cdots \\\\\n+ (lsdowprm, crmsqvtx)qzxwvtnp^{lsdowprm}hjgrksla^{crmsqvtx} = (0,0).\n\\end{gather*}\n\\end{itemize}\n\n\\end{document}",
      "solution": "We will prove the claim assuming only that the convex hull of the\npoints $(kqjtmzre, tpvdzwla)$ contains the origin in its interior. (Thanks\nto Marshall Buck for pointing out that the last three words are\nnecessary in the previous sentence!) Let $mclpzrqe = \\log qzxwvtnp, nskdwtuy = \\log\nhjgrksla$ so that the left-hand side of the given equation is\n\\begin{multline}\n(zvxcrlpe, gfjqdmzo) \\exp(zvxcrlpe mclpzrqe + gfjqdmzo nskdwtuy) + (bjkstemq, wlyhkpra) \\exp(bjkstemq mclpzrqe + wlyhkpra nskdwtuy) + \\\\\n\\cdots + (lsdowprm, crmsqvtx) \\exp(lsdowprm mclpzrqe + crmsqvtx nskdwtuy).\n\\end{multline}\nNow note that (1) is the gradient of the function\n\\begin{gather*}\nf(mclpzrqe,nskdwtuy) = \\exp(zvxcrlpe mclpzrqe + gfjqdmzo nskdwtuy) +\n\\exp(bjkstemq mclpzrqe + wlyhkpra nskdwtuy) + \\\\\n\\cdots + \\exp(lsdowprm mclpzrqe + crmsqvtx nskdwtuy),\n\\end{gather*}\nand so it suffices to show $f$ has a critical point. We will in fact\nshow $f$ has a global minimum.\n\nClearly we have\n\\[\nf(mclpzrqe,nskdwtuy) \\geq \\exp\\left( \\max_i (kqjtmzre mclpzrqe + tpvdzwla nskdwtuy) \\right).\n\\]\nNote that this maximum is positive for $(mclpzrqe,nskdwtuy) \\neq (0,0)$:  if we had\n$kqjtmzre mclpzrqe + tpvdzwla nskdwtuy < 0$ for all $i$, then the subset $mclpzrqe fjgkrnsa + nskdwtuy plhmdzqo < 0$ of the\n$fjgkrnsa plhmdzqo$-plane would be a half-plane containing all of the points $(kqjtmzre,\n tpvdzwla)$, whose convex hull would then not contain the origin, a\ncontradiction.\n\nThe function $\\max_{i} (kqjtmzre mclpzrqe + tpvdzwla nskdwtuy)$ is clearly\ncontinuous on the unit circle $mclpzrqe^{2} + nskdwtuy^{2} = 1$, which is compact.\nHence it has a global minimum $hrfbwzeq > 0$, and so for all $mclpzrqe,nskdwtuy$,\n\\[\n\\max_{i} (kqjtmzre mclpzrqe + tpvdzwla nskdwtuy) \\geq hrfbwzeq \\sqrt{mclpzrqe^{2} + nskdwtuy^{2}}.\n\\]\nIn particular, $f \\geq xrcpsgqa+1$ on the disk of radius $\\sqrt{(xrcpsgqa+1)/hrfbwzeq}$.\nSince $f(0,0) = xrcpsgqa$, the infimum of $f$ is the same over the entire\n$mclpzrqe nskdwtuy$-plane as over this disk, which again is compact.\nHence $f$ attains its infimal value at some point in the disk,\nwhich is the desired global minimum.\n\nNoam Elkies has suggested an alternate solution as follows: for $fjgkrnsa >\n0$, draw the loop traced by (1) as $(mclpzrqe,nskdwtuy)$ travels\ncounterclockwise around the circle $mclpzrqe^2 + nskdwtuy^2 = fjgkrnsa^2$. For $fjgkrnsa=0$, this\nof course has winding number 0 about any point, but for $fjgkrnsa$ large, one\ncan show this loop has winding number 1 about the origin, so somewhere in\nbetween the loop must pass through the origin. (Proving this latter\nfact is a little tricky.)\n\n\\end{itemize}\n\\end{document}"
    },
    "kernel_variant": {
      "question": "Let $d\\ge 3$ and let $K\\subset\\mathbb R^{d}$ be a compact set whose convex hull ${\\operatorname{conv}}\\,K$ contains the origin in its interior.  \n\nLet $\\mu$ be a finite positive Borel measure on $K$ whose support is all of $K$ (equivalently, $\\mu$ is not supported by any proper affine subspace).\n\nFor $u\\in\\mathbb R^{d}$ define  \n\\[\nZ(u)=\\int_{K}e^{\\langle v,u\\rangle}\\,d\\mu(v),\\qquad\nG(u)=\\int_{K}v\\,e^{\\langle v,u\\rangle}\\,d\\mu(v),\\qquad\n\\Sigma(u)=\\int_{K}vv^{\\top}e^{\\langle v,u\\rangle}\\,d\\mu(v).\n\\]\n(Here $A^{\\top}$ denotes the transpose of a matrix $A$ and $\\langle\\cdot,\\cdot\\rangle$ is the Euclidean inner product.)\n\nProve the following assertions.\n\n1. (Central parameter)  \n   There exists a unique $u^{\\ast}\\in\\mathbb R^{d}$ such that $G(u^{\\ast})=0$.\n\n2. (Strict positivity of the covariance at $u^{\\ast}$)  \n   $\\Sigma(u^{\\ast})$ is positive–definite.\n\n3. (Global parametrisation of the interior of the polytope)  \n   The map  \n   \\[\n   \\Phi:=\\nabla\\log Z:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{conv}}\\,K\\bigr),\n   \\qquad \n   u\\longmapsto\\frac{G(u)}{Z(u)},\n   \\]\n   is a $\\mathcal C^{\\infty}$–diffeomorphism of $\\mathbb R^{d}$ onto $\\operatorname{int}(\\operatorname{conv}K)$.  \n   Equivalently, for every $p\\in\\operatorname{int}(\\operatorname{conv}K)$ there exists a unique $u(p)\\in\\mathbb R^{d}$ such that  \n   \\[\n   \\int_{K}v\\,e^{\\langle v,u(p)\\rangle}\\,d\\mu(v)=p\\,Z\\bigl(u(p)\\bigr).\n   \\]\n\n4. (The covariance map need not be injective)  \n   Give an explicit example showing that  \n   \\[\n   C:\\mathbb R^{d}\\longrightarrow\\operatorname{Sym}^{+}_{d},\\qquad \n   u\\longmapsto\\Sigma(u),\n   \\]\n   is not injective in general.  (Hint: take $d=3$, $K$ the vertices of the cube $\\{\\pm1\\}^{3}$, and $\\mu$ the uniform measure.)\n\n───────────────────────────────────────────────",
      "solution": "Throughout we use the notation introduced in the statement, write $f:=\\log Z$ and $\\|\\cdot\\|$ for the Euclidean norm.\n\nStep 0.  Elementary properties.  \nBecause $K$ is compact and $\\mu$ is finite, $Z,G,\\Sigma$ are finite–valued and $\\mathcal C^{\\infty}$ on $\\mathbb R^{d}$, and $Z(u)>0$ for every $u$.\n\n───────────────────────────────────────────────\nStep 1.  Coercivity of $f$.  \nSince $0\\in{\\operatorname{int}}({\\operatorname{conv}}\\,K)$, there exists $\\varepsilon>0$ such that for every unit vector $\\xi$ one can find $v_{\\xi}\\in K$ with $\\langle v_{\\xi},\\xi\\rangle\\ge\\varepsilon$.  Hence, for $t\\ge0$,\n\\[\nZ(t\\xi)\\;\\ge\\;e^{t\\varepsilon}\\,\\mu\\!\\bigl(\\{v\\in K:\\langle v,\\xi\\rangle\\ge\\varepsilon\\}\\bigr)\\;\\xrightarrow[t\\to\\infty]{}\\;\\infty,\n\\]\nand therefore $f(u)=\\log Z(u)\\xrightarrow[\\|u\\|\\to\\infty]{}\\infty$.  Thus $f$ is coercive and attains at least one global minimum.\n\n───────────────────────────────────────────────\nStep 2.  Strict convexity of $f$.  \nDifferentiating,\n\\[\n\\nabla f(u)=\\frac{G(u)}{Z(u)},\\qquad\n\\nabla^{2}f(u)=\\frac{\\Sigma(u)}{Z(u)}-\\frac{G(u)G(u)^{\\top}}{Z(u)^{2}}\n          =\\operatorname{Cov}_{\\mu_{u}}(v),\n\\]\nwhere $\\mu_{u}$ is the probability measure with density\n$d\\mu_{u}(v)=e^{\\langle v,u\\rangle}d\\mu(v)/Z(u)$.  \nBecause $\\operatorname{supp}\\mu_{u}=K$ is full–dimensional and $\\mu_{u}$ is not supported by any proper affine subspace, the variance of $\\langle w,v\\rangle$ under $\\mu_{u}$ is positive for every $0\\ne w$, whence $\\nabla^{2}f(u)$ is positive–definite.  Thus $f$ is strictly convex and its global minimiser is unique.\n\n───────────────────────────────────────────────\nStep 3.  The central parameter $u^{\\ast}$.  \nLet $u^{\\ast}$ be the unique minimiser of $f$.  The first–order condition $\\nabla f(u^{\\ast})=0$ is exactly $G(u^{\\ast})=0$, proving Item 1.\n\n───────────────────────────────────────────────\nStep 4.  Positive–definiteness of $\\Sigma(u^{\\ast})$.  \nSince $G(u^{\\ast})=0$,\n\\[\n\\nabla^{2}f(u^{\\ast})=\\frac{\\Sigma(u^{\\ast})}{Z(u^{\\ast})}.\n\\]\nThe left–hand side is positive–definite (Step 2) and $Z(u^{\\ast})>0$, so $\\Sigma(u^{\\ast})$ is positive–definite.  Item 2 is established.\n\n───────────────────────────────────────────────\nStep 5.  $\\Phi$ is a global $\\mathcal C^{\\infty}$–diffeomorphism.\n\n5a.  $\\Phi$ is a local diffeomorphism.  \nBecause $\\nabla^{2}f(u)$ is positive–definite everywhere, the Jacobian $D\\Phi(u)=\\nabla^{2}f(u)$ is invertible for every $u$.  The inverse–function theorem yields that $\\Phi$ is a local $\\mathcal C^{\\infty}$–diffeomorphism.\n\n5b.  Injectivity of $\\Phi$.  \nFor $u_{1}\\ne u_{2}$ strict convexity of $f$ gives\n\\[\n\\langle\\Phi(u_{1})-\\Phi(u_{2}),\\,u_{1}-u_{2}\\rangle\n=\\langle\\nabla f(u_{1})-\\nabla f(u_{2}),\\,u_{1}-u_{2}\\rangle>0,\n\\]\nso $\\Phi(u_{1})=\\Phi(u_{2})$ would imply $u_{1}=u_{2}$.  Thus $\\Phi$ is injective.\n\n5c.  Surjectivity onto ${\\operatorname{int}}({\\operatorname{conv}}\\,K)$ via Legendre–Fenchel duality.  \n\n(i)  The Legendre conjugate.  \nDefine the convex conjugate of $f$,\n\\[\nf^{\\ast}(p):=\\sup_{u\\in\\mathbb R^{d}}\\bigl(\\langle p,u\\rangle-f(u)\\bigr),\n\\qquad p\\in\\mathbb R^{d}.\n\\]\nBecause $f$ is finite everywhere and strictly convex, $f^{\\ast}$ is a proper, lower semicontinuous, strictly convex function.\n\n(ii)  Identification of $\\operatorname{dom}f^{\\ast}$.  \nLet $h_{K}$ be the support function of $K$,\n\\[\nh_{K}(\\xi):=\\max_{v\\in K}\\langle v,\\xi\\rangle,\\qquad\\xi\\in\\mathbb R^{d}.\n\\]\nCompactness of $K$ yields $|h_{K}(\\xi)|\\le\\|K\\|_{\\infty}\\|\\xi\\|$ and\n\\[\nf(u)=\\log\\!\\int_{K}e^{\\langle v,u\\rangle}\\,d\\mu(v)\n        \\le h_{K}(u)+\\log\\mu(K)\\qquad(u\\in\\mathbb R^{d}).\n\\]\nIf $p\\notin{\\operatorname{conv}}\\,K$, Hahn–Banach separation provides $\\xi$ with $\\langle p,\\xi\\rangle>h_{K}(\\xi)$.  Taking $u=t\\xi$ and $t\\to\\infty$ gives\n\\[\nf^{\\ast}(p)\\ge\\langle p,t\\xi\\rangle-f(t\\xi)\n            \\ge t\\bigl(\\langle p,\\xi\\rangle-h_{K}(\\xi)\\bigr)-\\log\\mu(K)\n            \\xrightarrow[t\\to\\infty]{}\\infty,\n\\]\nso $f^{\\ast}(p)=\\infty$.  Conversely, if $p\\in{\\operatorname{conv}}\\,K$, then $\\langle p,u\\rangle\\le h_{K}(u)$ for all $u$, whence $f^{\\ast}(p)\\le\\log\\mu(K)<\\infty$.  Thus\n\\[\n{\\operatorname{dom}}f^{\\ast}={\\operatorname{conv}}\\,K,\\qquad \n{\\operatorname{int}}({\\operatorname{dom}}f^{\\ast})={\\operatorname{int}}({\\operatorname{conv}}\\,K).\n\\]\n\n(iii)  Legendre type of $f$.  \nThe function $f$ is $\\mathcal C^{1}$ on all of $\\mathbb R^{d}$ and strictly convex, and its gradient blows up only at infinity (there is no boundary to its effective domain).  Hence $f$ is a \\emph{Legendre} (sometimes called \\emph{essentially smooth and essentially strictly convex}) function in the sense of Rockafellar, Theorem 26A.\n\n(iv)  Global bijectivity of the gradient.  \nFor Legendre functions one has (Rockafellar, Theorem 26.5):\n\\[\n\\nabla f:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{dom}}f^{\\ast}\\bigr)\n\\quad\\text{is a}\\;\\mathcal C^{\\infty}\\text{–diffeomorphism}.\n\\]\nBecause ${\\operatorname{dom}}f^{\\ast}={\\operatorname{conv}}\\,K$, this gives\n\\[\n\\Phi=\\nabla f:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{conv}}\\,K\\bigr)\n\\quad\\text{is onto}.\n\\]\nCombining surjectivity with 5a and 5b we obtain that $\\Phi$ is a global $\\mathcal C^{\\infty}$–diffeomorphism, completing Item 3.\n\n───────────────────────────────────────────────\nStep 6.  A non-injective covariance map (Item 4).  \nTake $d=3$, $K=\\{\\pm1\\}^{3}$ (the $8$ vertices of the unit cube) and let $\\mu$ be the uniform measure (mass $1/8$ at each vertex).  Writing $u=(u_{1},u_{2},u_{3})$,\n\\[\nZ(u)=8\\prod_{j=1}^{3}\\cosh u_{j}.\n\\]\nBecause $\\cosh$ is an even function, $\\Sigma(-u)=\\Sigma(u)$ for every $u$.  For $u\\ne0$ we have $u\\ne -u$, yet $C(u)=C(-u)$.  Hence the covariance map $C:u\\mapsto\\Sigma(u)$ is not injective.\n\n───────────────────────────────────────────────\nAll four assertions are therefore proved.\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.749356",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension and continuum:  The original statement dealt with finitely many points in two or three dimensions; the enhanced problem handles {\\em arbitrary} dimension $d\\ge3$ and a {\\em continuum} of points via an integral over a Borel measure.  \n• Multiple objectives:  Not only must one find $u$ with vanishing first moment, one must also prove uniqueness, global diffeomorphism properties, and the realisation of {\\em every} admissible covariance matrix.  \n• Advanced tools:  The solution invokes strict convexity of log-Laplace transforms, coercivity arguments, the inverse-function theorem, proper maps, covering-space theory, and a touch of degree theory—well beyond the elementary gradient-vanishing argument of the original.  \n• Interacting concepts:  Convex geometry, differential topology, and probability (moments and covariance) interact intricately.  \nThese additions raise the problem far above the complexity of both the original and the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $d\\ge 3$ and let $K\\subset\\mathbb R^{d}$ be a compact set whose convex hull ${\\operatorname{conv}}\\,K$ contains the origin in its interior.  \n\nLet $\\mu$ be a finite positive Borel measure on $K$ whose support is all of $K$ (equivalently, $\\mu$ is not supported by any proper affine subspace).\n\nFor $u\\in\\mathbb R^{d}$ define  \n\\[\nZ(u)=\\int_{K}e^{\\langle v,u\\rangle}\\,d\\mu(v),\\qquad\nG(u)=\\int_{K}v\\,e^{\\langle v,u\\rangle}\\,d\\mu(v),\\qquad\n\\Sigma(u)=\\int_{K}vv^{\\top}e^{\\langle v,u\\rangle}\\,d\\mu(v).\n\\]\n(Here $A^{\\top}$ denotes the transpose of a matrix $A$ and $\\langle\\cdot,\\cdot\\rangle$ is the Euclidean inner product.)\n\nProve the following assertions.\n\n1. (Central parameter)  \n   There exists a unique $u^{\\ast}\\in\\mathbb R^{d}$ such that $G(u^{\\ast})=0$.\n\n2. (Strict positivity of the covariance at $u^{\\ast}$)  \n   $\\Sigma(u^{\\ast})$ is positive–definite.\n\n3. (Global parametrisation of the interior of the polytope)  \n   The map  \n   \\[\n   \\Phi:=\\nabla\\log Z:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{conv}}\\,K\\bigr),\n   \\qquad \n   u\\longmapsto\\frac{G(u)}{Z(u)},\n   \\]\n   is a $\\mathcal C^{\\infty}$–diffeomorphism of $\\mathbb R^{d}$ onto $\\operatorname{int}(\\operatorname{conv}K)$.  \n   Equivalently, for every $p\\in\\operatorname{int}(\\operatorname{conv}K)$ there exists a unique $u(p)\\in\\mathbb R^{d}$ such that  \n   \\[\n   \\int_{K}v\\,e^{\\langle v,u(p)\\rangle}\\,d\\mu(v)=p\\,Z\\bigl(u(p)\\bigr).\n   \\]\n\n4. (The covariance map need not be injective)  \n   Give an explicit example showing that  \n   \\[\n   C:\\mathbb R^{d}\\longrightarrow\\operatorname{Sym}^{+}_{d},\\qquad \n   u\\longmapsto\\Sigma(u),\n   \\]\n   is not injective in general.  (Hint: take $d=3$, $K$ the vertices of the cube $\\{\\pm1\\}^{3}$, and $\\mu$ the uniform measure.)\n\n───────────────────────────────────────────────",
      "solution": "Throughout we use the notation introduced in the statement, write $f:=\\log Z$ and $\\|\\cdot\\|$ for the Euclidean norm.\n\nStep 0.  Elementary properties.  \nBecause $K$ is compact and $\\mu$ is finite, $Z,G,\\Sigma$ are finite–valued and $\\mathcal C^{\\infty}$ on $\\mathbb R^{d}$, and $Z(u)>0$ for every $u$.\n\n───────────────────────────────────────────────\nStep 1.  Coercivity of $f$.  \nSince $0\\in{\\operatorname{int}}({\\operatorname{conv}}\\,K)$, there exists $\\varepsilon>0$ such that for every unit vector $\\xi$ one can find $v_{\\xi}\\in K$ with $\\langle v_{\\xi},\\xi\\rangle\\ge\\varepsilon$.  Hence, for $t\\ge0$,\n\\[\nZ(t\\xi)\\;\\ge\\;e^{t\\varepsilon}\\,\\mu\\!\\bigl(\\{v\\in K:\\langle v,\\xi\\rangle\\ge\\varepsilon\\}\\bigr)\\;\\xrightarrow[t\\to\\infty]{}\\;\\infty,\n\\]\nand therefore $f(u)=\\log Z(u)\\xrightarrow[\\|u\\|\\to\\infty]{}\\infty$.  Thus $f$ is coercive and attains at least one global minimum.\n\n───────────────────────────────────────────────\nStep 2.  Strict convexity of $f$.  \nDifferentiating,\n\\[\n\\nabla f(u)=\\frac{G(u)}{Z(u)},\\qquad\n\\nabla^{2}f(u)=\\frac{\\Sigma(u)}{Z(u)}-\\frac{G(u)G(u)^{\\top}}{Z(u)^{2}}\n          =\\operatorname{Cov}_{\\mu_{u}}(v),\n\\]\nwhere $\\mu_{u}$ is the probability measure with density\n$d\\mu_{u}(v)=e^{\\langle v,u\\rangle}d\\mu(v)/Z(u)$.  \nBecause $\\operatorname{supp}\\mu_{u}=K$ is full–dimensional and $\\mu_{u}$ is not supported by any proper affine subspace, the variance of $\\langle w,v\\rangle$ under $\\mu_{u}$ is positive for every $0\\ne w$, whence $\\nabla^{2}f(u)$ is positive–definite.  Thus $f$ is strictly convex and its global minimiser is unique.\n\n───────────────────────────────────────────────\nStep 3.  The central parameter $u^{\\ast}$.  \nLet $u^{\\ast}$ be the unique minimiser of $f$.  The first–order condition $\\nabla f(u^{\\ast})=0$ is exactly $G(u^{\\ast})=0$, proving Item 1.\n\n───────────────────────────────────────────────\nStep 4.  Positive–definiteness of $\\Sigma(u^{\\ast})$.  \nSince $G(u^{\\ast})=0$,\n\\[\n\\nabla^{2}f(u^{\\ast})=\\frac{\\Sigma(u^{\\ast})}{Z(u^{\\ast})}.\n\\]\nThe left–hand side is positive–definite (Step 2) and $Z(u^{\\ast})>0$, so $\\Sigma(u^{\\ast})$ is positive–definite.  Item 2 is established.\n\n───────────────────────────────────────────────\nStep 5.  $\\Phi$ is a global $\\mathcal C^{\\infty}$–diffeomorphism.\n\n5a.  $\\Phi$ is a local diffeomorphism.  \nBecause $\\nabla^{2}f(u)$ is positive–definite everywhere, the Jacobian $D\\Phi(u)=\\nabla^{2}f(u)$ is invertible for every $u$.  The inverse–function theorem yields that $\\Phi$ is a local $\\mathcal C^{\\infty}$–diffeomorphism.\n\n5b.  Injectivity of $\\Phi$.  \nFor $u_{1}\\ne u_{2}$ strict convexity of $f$ gives\n\\[\n\\langle\\Phi(u_{1})-\\Phi(u_{2}),\\,u_{1}-u_{2}\\rangle\n=\\langle\\nabla f(u_{1})-\\nabla f(u_{2}),\\,u_{1}-u_{2}\\rangle>0,\n\\]\nso $\\Phi(u_{1})=\\Phi(u_{2})$ would imply $u_{1}=u_{2}$.  Thus $\\Phi$ is injective.\n\n5c.  Surjectivity onto ${\\operatorname{int}}({\\operatorname{conv}}\\,K)$ via Legendre–Fenchel duality.  \n\n(i)  The Legendre conjugate.  \nDefine the convex conjugate of $f$,\n\\[\nf^{\\ast}(p):=\\sup_{u\\in\\mathbb R^{d}}\\bigl(\\langle p,u\\rangle-f(u)\\bigr),\n\\qquad p\\in\\mathbb R^{d}.\n\\]\nBecause $f$ is finite everywhere and strictly convex, $f^{\\ast}$ is a proper, lower semicontinuous, strictly convex function.\n\n(ii)  Identification of $\\operatorname{dom}f^{\\ast}$.  \nLet $h_{K}$ be the support function of $K$,\n\\[\nh_{K}(\\xi):=\\max_{v\\in K}\\langle v,\\xi\\rangle,\\qquad\\xi\\in\\mathbb R^{d}.\n\\]\nCompactness of $K$ yields $|h_{K}(\\xi)|\\le\\|K\\|_{\\infty}\\|\\xi\\|$ and\n\\[\nf(u)=\\log\\!\\int_{K}e^{\\langle v,u\\rangle}\\,d\\mu(v)\n        \\le h_{K}(u)+\\log\\mu(K)\\qquad(u\\in\\mathbb R^{d}).\n\\]\nIf $p\\notin{\\operatorname{conv}}\\,K$, Hahn–Banach separation provides $\\xi$ with $\\langle p,\\xi\\rangle>h_{K}(\\xi)$.  Taking $u=t\\xi$ and $t\\to\\infty$ gives\n\\[\nf^{\\ast}(p)\\ge\\langle p,t\\xi\\rangle-f(t\\xi)\n            \\ge t\\bigl(\\langle p,\\xi\\rangle-h_{K}(\\xi)\\bigr)-\\log\\mu(K)\n            \\xrightarrow[t\\to\\infty]{}\\infty,\n\\]\nso $f^{\\ast}(p)=\\infty$.  Conversely, if $p\\in{\\operatorname{conv}}\\,K$, then $\\langle p,u\\rangle\\le h_{K}(u)$ for all $u$, whence $f^{\\ast}(p)\\le\\log\\mu(K)<\\infty$.  Thus\n\\[\n{\\operatorname{dom}}f^{\\ast}={\\operatorname{conv}}\\,K,\\qquad \n{\\operatorname{int}}({\\operatorname{dom}}f^{\\ast})={\\operatorname{int}}({\\operatorname{conv}}\\,K).\n\\]\n\n(iii)  Legendre type of $f$.  \nThe function $f$ is $\\mathcal C^{1}$ on all of $\\mathbb R^{d}$ and strictly convex, and its gradient blows up only at infinity (there is no boundary to its effective domain).  Hence $f$ is a \\emph{Legendre} (sometimes called \\emph{essentially smooth and essentially strictly convex}) function in the sense of Rockafellar, Theorem 26A.\n\n(iv)  Global bijectivity of the gradient.  \nFor Legendre functions one has (Rockafellar, Theorem 26.5):\n\\[\n\\nabla f:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{dom}}f^{\\ast}\\bigr)\n\\quad\\text{is a}\\;\\mathcal C^{\\infty}\\text{–diffeomorphism}.\n\\]\nBecause ${\\operatorname{dom}}f^{\\ast}={\\operatorname{conv}}\\,K$, this gives\n\\[\n\\Phi=\\nabla f:\\mathbb R^{d}\\longrightarrow{\\operatorname{int}}\\bigl({\\operatorname{conv}}\\,K\\bigr)\n\\quad\\text{is onto}.\n\\]\nCombining surjectivity with 5a and 5b we obtain that $\\Phi$ is a global $\\mathcal C^{\\infty}$–diffeomorphism, completing Item 3.\n\n───────────────────────────────────────────────\nStep 6.  A non-injective covariance map (Item 4).  \nTake $d=3$, $K=\\{\\pm1\\}^{3}$ (the $8$ vertices of the unit cube) and let $\\mu$ be the uniform measure (mass $1/8$ at each vertex).  Writing $u=(u_{1},u_{2},u_{3})$,\n\\[\nZ(u)=8\\prod_{j=1}^{3}\\cosh u_{j}.\n\\]\nBecause $\\cosh$ is an even function, $\\Sigma(-u)=\\Sigma(u)$ for every $u$.  For $u\\ne0$ we have $u\\ne -u$, yet $C(u)=C(-u)$.  Hence the covariance map $C:u\\mapsto\\Sigma(u)$ is not injective.\n\n───────────────────────────────────────────────\nAll four assertions are therefore proved.\n\n───────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.578378",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension and continuum:  The original statement dealt with finitely many points in two or three dimensions; the enhanced problem handles {\\em arbitrary} dimension $d\\ge3$ and a {\\em continuum} of points via an integral over a Borel measure.  \n• Multiple objectives:  Not only must one find $u$ with vanishing first moment, one must also prove uniqueness, global diffeomorphism properties, and the realisation of {\\em every} admissible covariance matrix.  \n• Advanced tools:  The solution invokes strict convexity of log-Laplace transforms, coercivity arguments, the inverse-function theorem, proper maps, covering-space theory, and a touch of degree theory—well beyond the elementary gradient-vanishing argument of the original.  \n• Interacting concepts:  Convex geometry, differential topology, and probability (moments and covariance) interact intricately.  \nThese additions raise the problem far above the complexity of both the original and the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}