{
  "index": "2018-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $f = (f_1, f_2)$ be a function from $\\mathbb{R}^2$ to $\\mathbb{R}^2$ with continuous partial derivatives\n$\\frac{\\partial f_i}{\\partial x_j}$ that are positive everywhere. Suppose that\n\\[\n\\frac{\\partial f_1}{\\partial x_1} \\frac{\\partial f_2}{\\partial x_2} - \\frac{1}{4} \\left( \\frac{\\partial f_1}{\\partial x_2}  + \\frac{\\partial f_2}{\\partial x_1} \\right)^2 > 0\n\\]\neverywhere. Prove that $f$ is one-to-one.",
  "solution": "Let $(a_1,a_2)$ and $(a_1',a_2')$ be distinct points in $\\mathbb{R}^2$; we want to show that $f(a_1,a_2) \\neq f(a_1',a_2')$. Write $(v_1,v_2) = (a_1',a_2')-(a_1,a_2)$, and let $\\gamma(t) = (a_1,a_2)+t(v_1,v_2)$, $t \\in [0,1]$, be the path between $(a_1,a_2)$ and $(a_1',a_2')$. Define a real-valued function $g$ by $g(t) = (v_1,v_2) \\cdot f(\\gamma(t))$.\nBy the Chain Rule, \n\\[\nf'(\\gamma(t)) = \\left( \\begin{matrix} \\partial f_1/\\partial x_1 & \\partial f_1/\\partial x_2 \\\\ \\partial f_2/\\partial x_1 & \\partial f_2/\\partial x_2 \\end{matrix} \\right) \\left(\n\\begin{matrix} v_1 \\\\ v_2 \\end{matrix} \\right). \n\\]\nAbbreviate $\\partial f_i/\\partial x_j$ by $f_{ij}$; then\n\\begin{align*}\ng'(t) &= \\left( \\begin{matrix} v_1 & v_2 \\end{matrix} \\right) \\left( \\begin{matrix} f_{11} & f_{12} \\\\ f_{21} & f_{22} \\end{matrix} \\right) \\left( \\begin{matrix} v_1 \\\\ v_2 \\end{matrix} \\right) \\\\\n&= f_{11} v_1^2 + (f_{12}+f_{21})v_1v_2+f_{22} v_2^2 \\\\\n&= f_{11} \\left(v_1+\\frac{f_{12}+f_{21}}{2f_{11}} v_2 \\right)^2 + \\frac{4f_{11}f_{22}-(f_{12}+f_{21})^2}{4f_{11}} v_2^2 \\\\\n& \\geq 0\n\\end{align*}\nsince $f_{11}$ and $f_{11}f_{22}-(f_{12}+f_{21})^2/4$ are positive by assumption. Since the only way that equality could hold is if $v_1$ and $v_2$ are both $0$, we in fact have $g'(t)>0$ for all $t$. But if $f(a_1,a_2) = f(a_1',a_2')$, then $g(0) = g(1)$, a contradiction.\n\n\\noindent\n\\textbf{Remark.}\nA similar argument shows more generally that $f:\\thinspace \\mathbb{R}^n \\to \\mathbb{R}^n$ is injective if at all points in $\\mathbb{R}^n$, the Jacobian matrix $Df$ satisfies the following property: the quadratic form associated to the bilinear form with matrix $Df$ (or the symmetrized bilinear form with matrix $(Df+(Df)^T)/2$) is positive definite. In the setting of the problem, the symmetrized matrix is\n\\[\n\\left( \\begin{matrix} f_{11} & (f_{12}+f_{21})/2 \\\\ (f_{12}+f_{21})/2 & f_{22} \\end{matrix} \\right),\n\\]\nand this is positive definite if and only if $f_{11}$ and the determinant of the matrix are both positive\n(Sylvester's criterion). Note that the assumptions that $f_{12},f_{21}>0$ are unnecessary for the argument;\nit is also easy to see that the hypotheses $f_{11}, f_{12} > 0$ are also superfluous.\n(The assumption $f_{11}f_{22}-(f_{12}+f_{21})^2 > 0$ implies $f_{11} f_{22} > 0$, so both are nonzero and of the same sign; by continuity, this common sign must be constant over all of $\\mathbb{R}^2$. If it is negative, then\napply the same logic to $(-f_1, -f_2)$.)",
  "vars": [
    "f",
    "f_1",
    "f_2",
    "x_j",
    "f_i",
    "f_ij",
    "f_11",
    "f_12",
    "f_21",
    "f_22",
    "a_1",
    "a_2",
    "v_1",
    "v_2",
    "t",
    "g",
    "\\\\gamma",
    "n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "vectorfunc",
        "f_1": "imageone",
        "f_2": "imagetwo",
        "x_j": "coordjvar",
        "f_i": "imageindex",
        "f_ij": "jacijpair",
        "f_11": "jaconeone",
        "f_12": "jaconetwo",
        "f_21": "jactwoone",
        "f_22": "jactwotwo",
        "a_1": "pointone",
        "a_2": "pointtwo",
        "v_1": "diffone",
        "v_2": "difftwo",
        "t": "paramtime",
        "g": "pathscalar",
        "\\\\gamma": "segmentpath",
        "n": "dimension"
      },
      "question": "Let $vectorfunc = (imageone, imagetwo)$ be a function from \\mathbb{R}^2 to \\mathbb{R}^2 with continuous partial derivatives \\frac{\\partial imageindex}{\\partial coordjvar} that are positive everywhere. Suppose that\n\\[\n\\frac{\\partial imageone}{\\partial x_1} \\frac{\\partial imagetwo}{\\partial x_2} - \\frac{1}{4} \\left( \\frac{\\partial imageone}{\\partial x_2}  + \\frac{\\partial imagetwo}{\\partial x_1} \\right)^2 > 0\n\\]\neverywhere. Prove that $vectorfunc$ is one-to-one.",
      "solution": "Let $(pointone,pointtwo)$ and $(a_1',a_2')$ be distinct points in \\mathbb{R}^2; we want to show that $vectorfunc(pointone,pointtwo) \\neq vectorfunc(a_1',a_2')$. Write $(diffone,difftwo) = (a_1',a_2')-(pointone,pointtwo)$, and let $segmentpath(paramtime) = (pointone,pointtwo)+paramtime(diffone,difftwo)$, $paramtime \\in [0,1]$, be the path between $(pointone,pointtwo)$ and $(a_1',a_2')$. Define a real-valued function $pathscalar$ by $pathscalar(paramtime) = (diffone,difftwo) \\cdot vectorfunc(segmentpath(paramtime))$.\nBy the Chain Rule,\n\\[\nvectorfunc'(segmentpath(paramtime)) = \\begin{pmatrix} \\partial imageone/\\partial x_1 & \\partial imageone/\\partial x_2 \\\\ \\partial imagetwo/\\partial x_1 & \\partial imagetwo/\\partial x_2 \\end{pmatrix} \\begin{pmatrix} diffone \\\\ difftwo \\end{pmatrix}.\n\\]\nAbbreviate $\\partial imageindex/\\partial coordjvar$ by jacijpair; then\n\\begin{align*}\npathscalar'(paramtime) &= \\begin{pmatrix} diffone & difftwo \\end{pmatrix} \\begin{pmatrix} jaconeone & jaconetwo \\\\ jactwoone & jactwotwo \\end{pmatrix} \\begin{pmatrix} diffone \\\\ difftwo \\end{pmatrix} \\\\\n&= jaconeone\\, diffone^2 + (jaconetwo+jactwoone)diffone\\,difftwo + jactwotwo\\, difftwo^2 \\\\\n&= jaconeone \\left(diffone+\\frac{jaconetwo+jactwoone}{2jaconeone} difftwo \\right)^2 + \\frac{4jaconeone jactwotwo-(jaconetwo+jactwoone)^2}{4jaconeone} \\, difftwo^2 \\\\\n& \\ge 0\n\\end{align*}\nsince jaconeone and $jaconeone jactwotwo-(jaconetwo+jactwoone)^2/4$ are positive by assumption. Since the only way that equality could hold is if diffone and difftwo are both $0$, we in fact have $pathscalar'(paramtime)>0$ for all $paramtime$. But if $vectorfunc(pointone,pointtwo) = vectorfunc(a_1',a_2')$, then $pathscalar(0) = pathscalar(1)$, a contradiction.\n\n\\noindent\\textbf{Remark.}\nA similar argument shows more generally that $vectorfunc:\\thinspace \\mathbb{R}^{dimension} \\to \\mathbb{R}^{dimension}$ is injective if at all points in \\mathbb{R}^{dimension}, the Jacobian matrix $Dvectorfunc$ satisfies the following property: the quadratic form associated to the bilinear form with matrix $Dvectorfunc$ (or the symmetrized bilinear form with matrix $(Dvectorfunc+(Dvectorfunc)^T)/2$) is positive definite. In the setting of the problem, the symmetrized matrix is\n\\[\n\\begin{pmatrix} jaconeone & (jaconetwo+jactwoone)/2 \\\\ (jaconetwo+jactwoone)/2 & jactwotwo \\end{pmatrix},\n\\]\nand this is positive definite if and only if jaconeone and the determinant of the matrix are both positive (Sylvester's criterion). Note that the assumptions that jaconetwo,jactwoone>0 are unnecessary for the argument; it is also easy to see that the hypotheses jaconeone, jaconetwo > 0 are also superfluous. (The assumption jaconeone jactwotwo-(jaconetwo+jactwoone)^2 > 0 implies jaconeone jactwotwo > 0, so both are nonzero and of the same sign; by continuity, this common sign must be constant over all of \\mathbb{R}^2. If it is negative, then apply the same logic to $(-imageone, -imagetwo)$.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "waterfall",
        "f_1": "teaspoon",
        "f_2": "drumstick",
        "x_j": "sculpture",
        "f_i": "rainstorm",
        "f_ij": "blackbird",
        "f_11": "arrowhead",
        "f_12": "moonlight",
        "f_21": "snowflake",
        "f_22": "pineapple",
        "a_1": "lakeshore",
        "a_2": "firmament",
        "v_1": "sandstorm",
        "v_2": "motorway",
        "t": "hinterland",
        "g": "sunflower",
        "\\\\gamma": "riverbed",
        "n": "cinnamon"
      },
      "question": "Let $waterfall = (teaspoon, drumstick)$ be a function from $\\mathbb{R}^2$ to $\\mathbb{R}^2$ with continuous partial derivatives\n$\\frac{\\partial rainstorm}{\\partial sculpture}$ that are positive everywhere. Suppose that\n\\[\n\\frac{\\partial teaspoon}{\\partial x_1} \\frac{\\partial drumstick}{\\partial x_2} - \\frac{1}{4} \\left( \\frac{\\partial teaspoon}{\\partial x_2}  + \\frac{\\partial drumstick}{\\partial x_1} \\right)^2 > 0\n\\]\neverywhere. Prove that $waterfall$ is one-to-one.",
      "solution": "Let $(lakeshore,firmament)$ and $(lakeshore',firmament')$ be distinct points in $\\mathbb{R}^2$; we want to show that $waterfall(lakeshore,firmament) \\neq waterfall(lakeshore',firmament')$. Write $(sandstorm,motorway) = (lakeshore',firmament')-(lakeshore,firmament)$, and let $riverbed(hinterland) = (lakeshore,firmament)+hinterland(sandstorm,motorway)$, $hinterland \\in [0,1]$, be the path between $(lakeshore,firmament)$ and $(lakeshore',firmament')$. Define a real-valued function $sunflower$ by $sunflower(hinterland) = (sandstorm,motorway) \\cdot waterfall(riverbed(hinterland))$.\nBy the Chain Rule,\n\\[\nwaterfall'(riverbed(hinterland)) = \\left( \\begin{matrix} \\partial teaspoon/\\partial x_1 & \\partial teaspoon/\\partial x_2 \\\\ \\partial drumstick/\\partial x_1 & \\partial drumstick/\\partial x_2 \\end{matrix} \\right) \\left(\\begin{matrix} sandstorm \\\\ motorway \\end{matrix}\\right).\n\\]\nAbbreviate $\\partial rainstorm/\\partial sculpture$ by $blackbird$; then\n\\begin{align*}\nsunflower'(hinterland) &= \\left( \\begin{matrix} sandstorm & motorway \\end{matrix} \\right) \\left( \\begin{matrix} arrowhead & moonlight \\\\ snowflake & pineapple \\end{matrix} \\right) \\left( \\begin{matrix} sandstorm \\\\ motorway \\end{matrix} \\right) \\\\\n&= arrowhead\\, sandstorm^2 + (moonlight+snowflake)\\, sandstorm\\, motorway + pineapple\\, motorway^2 \\\\\n&= arrowhead \\left(sandstorm+\\frac{moonlight+snowflake}{2\\,arrowhead}\\, motorway \\right)^2 + \\frac{4\\,arrowhead\\, pineapple-(moonlight+snowflake)^2}{4\\,arrowhead} \\, motorway^2 \\\\\n& \\ge 0\n\\end{align*}\nsince arrowhead and $arrowhead\\, pineapple-(moonlight+snowflake)^2/4$ are positive by assumption. Since the only way that equality could hold is if sandstorm and motorway are both $0$, we in fact have $sunflower'(hinterland)>0$ for all $hinterland$. But if $waterfall(lakeshore,firmament) = waterfall(lakeshore',firmament')$, then $sunflower(0) = sunflower(1)$, a contradiction.\n\n\\noindent\\textbf{Remark.}\nA similar argument shows more generally that $waterfall:\\thinspace \\mathbb{R}^{cinnamon} \\to \\mathbb{R}^{cinnamon}$ is injective if at all points in $\\mathbb{R}^{cinnamon}$, the Jacobian matrix $Dwaterfall$ satisfies the following property: the quadratic form associated to the bilinear form with matrix $Dwaterfall$ (or the symmetrized bilinear form with matrix $(Dwaterfall+(Dwaterfall)^T)/2$) is positive definite. In the setting of the problem, the symmetrized matrix is\n\\[\n\\left( \\begin{matrix} arrowhead & (moonlight+snowflake)/2 \\\\ (moonlight+snowflake)/2 & pineapple \\end{matrix} \\right),\n\\]\nand this is positive definite if and only if arrowhead and the determinant of the matrix are both positive (Sylvester's criterion). Note that the assumptions that moonlight,snowflake>0 are unnecessary for the argument; it is also easy to see that the hypotheses arrowhead, moonlight > 0 are also superfluous. (The assumption $arrowhead\\, pineapple-(moonlight+snowflake)^2 > 0$ implies $arrowhead\\, pineapple > 0$, so both are nonzero and of the same sign; by continuity, this common sign must be constant over all of $\\mathbb{R}^2$. If it is negative, then apply the same logic to $(-teaspoon, -drumstick)$.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "malfunction",
        "f_1": "secondpart",
        "f_2": "firstpart",
        "x_j": "outputspot",
        "f_i": "staticvalue",
        "f_ij": "emptyfill",
        "f_11": "hollowcell",
        "f_12": "solidcell",
        "f_21": "liquidcell",
        "f_22": "gascell",
        "a_1": "endpointtwo",
        "a_2": "endpointone",
        "v_1": "stillone",
        "v_2": "stilltwo",
        "t": "spaceparam",
        "g": "antifunc",
        "\\\\gamma": "blockade",
        "n": "zeroindex"
      },
      "question": "Let $malfunction = (secondpart, firstpart)$ be a function from $\\mathbb{R}^2$ to $\\mathbb{R}^2$ with continuous partial derivatives\n$\\frac{\\partial staticvalue}{\\partial outputspot}$ that are positive everywhere. Suppose that\n\\[\n\\frac{\\partial secondpart}{\\partial x_1} \\frac{\\partial firstpart}{\\partial x_2} - \\frac{1}{4} \\left( \\frac{\\partial secondpart}{\\partial x_2}  + \\frac{\\partial firstpart}{\\partial x_1} \\right)^2 > 0\n\\]\neverywhere. Prove that $malfunction$ is one-to-one.",
      "solution": "Let $(endpointtwo,endpointone)$ and $(endpointtwo',endpointone')$ be distinct points in $\\mathbb{R}^2$; we want to show that $malfunction(endpointtwo,endpointone) \\neq malfunction(endpointtwo',endpointone')$. Write $(stillone,stilltwo) = (endpointtwo',endpointone')-(endpointtwo,endpointone)$, and let $blockade(spaceparam) = (endpointtwo,endpointone)+spaceparam(stillone,stilltwo)$, $spaceparam \\in [0,1]$, be the path between $(endpointtwo,endpointone)$ and $(endpointtwo',endpointone')$. Define a real-valued function $antifunc$ by $antifunc(spaceparam) = (stillone,stilltwo) \\cdot malfunction(blockade(spaceparam))$.\nBy the Chain Rule, \n\\[\nmalfunction'(blockade(spaceparam)) = \\left( \\begin{matrix} \\partial secondpart/\\partial x_1 & \\partial secondpart/\\partial x_2 \\\\ \\partial firstpart/\\partial x_1 & \\partial firstpart/\\partial x_2 \\end{matrix} \\right) \\left(\n\\begin{matrix} stillone \\\\ stilltwo \\end{matrix} \\right). \n\\]\nAbbreviate $\\partial staticvalue/\\partial outputspot$ by emptyfill; then\n\\begin{align*}\nantifunc'(spaceparam) &= \\left( \\begin{matrix} stillone & stilltwo \\end{matrix} \\right) \\left( \\begin{matrix} hollowcell & solidcell \\\\ liquidcell & gascell \\end{matrix} \\right) \\left( \\begin{matrix} stillone \\\\ stilltwo \\end{matrix} \\right) \\\\\n&= hollowcell \\, stillone^2 + (solidcell+liquidcell)\\,stillone\\,stilltwo+gascell \\, stilltwo^2 \\\\\n&= hollowcell \\left(stillone+\\frac{solidcell+liquidcell}{2\\,hollowcell} \\, stilltwo \\right)^2 + \\frac{4\\,hollowcell\\,gascell-(solidcell+liquidcell)^2}{4\\,hollowcell} \\, stilltwo^2 \\\\\n& \\geq 0\n\\end{align*}\nsince $hollowcell$ and $hollowcell\\,gascell-(solidcell+liquidcell)^2/4$ are positive by assumption. Since the only way that equality could hold is if $stillone$ and $stilltwo$ are both $0$, we in fact have $antifunc'(spaceparam)>0$ for all $spaceparam$. But if $malfunction(endpointtwo,endpointone) = malfunction(endpointtwo',endpointone')$, then $antifunc(0) = antifunc(1)$, a contradiction.\n\n\\noindent\n\\textbf{Remark.}\nA similar argument shows more generally that $malfunction:\\thinspace \\mathbb{R}^{zeroindex} \\to \\mathbb{R}^{zeroindex}$ is injective if at all points in $\\mathbb{R}^{zeroindex}$, the Jacobian matrix $Dmalfunction$ satisfies the following property: the quadratic form associated to the bilinear form with matrix $Dmalfunction$ (or the symmetrized bilinear form with matrix $(Dmalfunction+(Dmalfunction)^T)/2$) is positive definite. In the setting of the problem, the symmetrized matrix is\n\\[\n\\left( \\begin{matrix} hollowcell & (solidcell+liquidcell)/2 \\\\ (solidcell+liquidcell)/2 & gascell \\end{matrix} \\right),\n\\]\nand this is positive definite if and only if $hollowcell$ and the determinant of the matrix are both positive\n(Sylvester's criterion). Note that the assumptions that solidcell,liquidcell>0 are unnecessary for the argument;\nit is also easy to see that the hypotheses hollowcell, solidcell > 0 are also superfluous.\n(The assumption $hollowcell\\,gascell-(solidcell+liquidcell)^2 > 0$ implies $hollowcell \\, gascell > 0$, so both are nonzero and of the same sign; by continuity, this common sign must be constant over all of $\\mathbb{R}^2$. If it is negative, then\napply the same logic to $(-secondpart, -firstpart)$.)"
    },
    "garbled_string": {
      "map": {
        "f": "ujhnrpex",
        "f_1": "kqzmbgsu",
        "f_2": "vxotdlpw",
        "x_j": "qsdnmvza",
        "f_i": "pbxrafyc",
        "f_ij": "nrgzpskt",
        "f_11": "zasvrdlu",
        "f_12": "tqmwyhvk",
        "f_21": "gbfrplhe",
        "f_22": "hijsknvd",
        "a_1": "fcdlqzme",
        "a_2": "wkgrsnpo",
        "v_1": "uzxhajqm",
        "v_2": "rybsvnpk",
        "t": "lpqweihz",
        "g": "ljmtrvka",
        "\\\\gamma": "ztqvkcni",
        "n": "ksyhocfv"
      },
      "question": "Let $ujhnrpex = (kqzmbgsu, vxotdlpw)$ be a function from $\\mathbb{R}^2$ to $\\mathbb{R}^2$ with continuous partial derivatives\n$\\frac{\\partial pbxrafyc}{\\partial qsdnmvza}$ that are positive everywhere. Suppose that\n\\[\n\\frac{\\partial kqzmbgsu}{\\partial x_1} \\frac{\\partial vxotdlpw}{\\partial x_2} - \\frac{1}{4} \\left( \\frac{\\partial kqzmbgsu}{\\partial x_2}  + \\frac{\\partial vxotdlpw}{\\partial x_1} \\right)^2 > 0\n\\]\neverywhere. Prove that $ujhnrpex$ is one-to-one.",
      "solution": "Let $(fcdlqzme,wkgrsnpo)$ and $(fcdlqzme',wkgrsnpo')$ be distinct points in $\\mathbb{R}^2$; we want to show that $ujhnrpex(fcdlqzme,wkgrsnpo) \\neq ujhnrpex(fcdlqzme',wkgrsnpo')$. Write $(uzxhajqm,rybsvnpk) = (fcdlqzme',wkgrsnpo')-(fcdlqzme,wkgrsnpo)$, and let $ztqvkcni(lpqweihz) = (fcdlqzme,wkgrsnpo)+lpqweihz(uzxhajqm,rybsvnpk)$, $lpqweihz \\in [0,1]$, be the path between $(fcdlqzme,wkgrsnpo)$ and $(fcdlqzme',wkgrsnpo')$. Define a real-valued function $ljmtrvka$ by $ljmtrvka(lpqweihz) = (uzxhajqm,rybsvnpk) \\cdot ujhnrpex(ztqvkcni(lpqweihz))$.\nBy the Chain Rule, \n\\[\nujhnrpex'(ztqvkcni(lpqweihz)) = \\left( \\begin{matrix} \\partial kqzmbgsu/\\partial x_1 & \\partial kqzmbgsu/\\partial x_2 \\\\ \\partial vxotdlpw/\\partial x_1 & \\partial vxotdlpw/\\partial x_2 \\end{matrix} \\right) \\left(\n\\begin{matrix} uzxhajqm \\\\ rybsvnpk \\end{matrix} \\right). \n\\]\nAbbreviate $\\partial pbxrafyc/\\partial qsdnmvza$ by $nrgzpskt$; then\n\\begin{align*}\nljmtrvka'(lpqweihz) &= \\left( \\begin{matrix} uzxhajqm & rybsvnpk \\end{matrix} \\right) \\left( \\begin{matrix} zasvrdlu & tqmwyhvk \\\\ gbfrplhe & hijsknvd \\end{matrix} \\right) \\left( \\begin{matrix} uzxhajqm \\\\ rybsvnpk \\end{matrix} \\right) \\\\\n&= zasvrdlu\\, uzxhajqm^2 + (tqmwyhvk+gbfrplhe) uzxhajqm\\, rybsvnpk + hijsknvd\\, rybsvnpk^2 \\\\\n&= zasvrdlu \\left( uzxhajqm + \\frac{tqmwyhvk+gbfrplhe}{2zasvrdlu} rybsvnpk \\right)^2 + \\frac{4zasvrdlu\\, hijsknvd-(tqmwyhvk+gbfrplhe)^2}{4zasvrdlu} rybsvnpk^2 \\\\\n& \\geq 0\n\\end{align*}\nsince zasvrdlu and $zasvrdlu\\, hijsknvd-(tqmwyhvk+gbfrplhe)^2/4$ are positive by assumption. Since the only way that equality could hold is if uzxhajqm and rybsvnpk are both $0$, we in fact have $ljmtrvka'(lpqweihz)>0$ for all lpqweihz. But if $ujhnrpex(fcdlqzme,wkgrsnpo) = ujhnrpex(fcdlqzme',wkgrsnpo')$, then $ljmtrvka(0) = ljmtrvka(1)$, a contradiction.\n\n\\noindent\n\\textbf{Remark.}\nA similar argument shows more generally that $ujhnrpex:\\thinspace \\mathbb{R}^{ksyhocfv} \\to \\mathbb{R}^{ksyhocfv}$ is injective if at all points in $\\mathbb{R}^{ksyhocfv}$, the Jacobian matrix $Dujhnrpex$ satisfies the following property: the quadratic form associated to the bilinear form with matrix $Dujhnrpex$ (or the symmetrized bilinear form with matrix $(Dujhnrpex+(Dujhnrpex)^T)/2$) is positive definite. In the setting of the problem, the symmetrized matrix is\n\\[\n\\left( \\begin{matrix} zasvrdlu & (tqmwyhvk+gbfrplhe)/2 \\\\ (tqmwyhvk+gbfrplhe)/2 & hijsknvd \\end{matrix} \\right),\n\\]\nand this is positive definite if and only if zasvrdlu and the determinant of the matrix are both positive\n(Sylvester's criterion). Note that the assumptions that tqmwyhvk,gbfrplhe>0 are unnecessary for the argument;\nit is also easy to see that the hypotheses zasvrdlu, tqmwyhvk > 0 are also superfluous.\n(The assumption $zasvrdlu\\, hijsknvd-(tqmwyhvk+gbfrplhe)^2 > 0$ implies $zasvrdlu\\, hijsknvd > 0$, so both are nonzero and of the same sign; by continuity, this common sign must be constant over all of $\\mathbb{R}^2$. If it is negative, then\napply the same logic to $(-kqzmbgsu, -vxotdlpw)$.)"
    },
    "kernel_variant": {
      "question": "Let H be a (possibly infinite–dimensional) real separable Hilbert space with inner product ⟨·,·⟩ and norm ‖·‖.  \nLet F : H→H be a C¹ map (Fréchet differentiable with continuous derivative).  \nFor every x∈H set  \n\n  S(x):=½(DF(x)+DF(x)*) and A(x):=½(DF(x)−DF(x)*) so that DF(x)=S(x)+A(x).\n\nAssume  \n\n(1)  (Uniform strong monotonicity) There is a constant λ>0 with  \n   ⟨S(x)v,v⟩ ≥ λ‖v‖² for all x,v∈H.  \n\n(2)  (Uniform Lipschitz bound on the derivative) There exists L with 0≤L<2λ such that  \n   ‖DF(x)−DF(y)‖ ≤ L‖x−y‖ for all x,y∈H.  \n\n(3)  (Global boundedness of the derivative) M:=sup_{x∈H}‖DF(x)‖<∞.\n\nProve that\n\n(a) F is injective.\n\n(b) F is surjective, hence a bijection H→H.\n\n(c) F is a C¹–diffeomorphism whose inverse is globally Lipschitz.  In particular  \n\n  λ‖x−y‖ ≤ ‖F(x)−F(y)‖ ≤ M‖x−y‖,     (1)  \n  ‖F⁻¹(u)−F⁻¹(v)‖ ≤ 1/(λ−L/2)·‖u−v‖.  (2)\n\n(d)  (Topological degree) For every bounded open set Ω⊂H with 0∈Ω the Browder–Minty degree deg(F,Ω,0) is defined and equals +1.  In particular F is of (positive) degree +1 on H.\n\n\n────────────────────────────────────────────────────",
      "solution": "Throughout write h:=x−y for x,y∈H.\n\nStep 0.  Integral representation.  \nSince DF is continuous,\n\n  F(x)−F(y)=∫₀¹DF(y+th)h dt.  (3)\n\n\n\n────────────────────────────────────────  \n1.  Uniform invertibility of DF(x)  \n────────────────────────────────────────  \n\nLemma 1.  For every x∈H the operator DF(x) is a bounded linear isomorphism and  \n\n  ‖DF(x)⁻¹‖ ≤ 1/λ.  (4)\n\nProof.  For v∈H,\n\n ‖DF(x)v‖‖v‖ ≥ ⟨DF(x)v,v⟩=⟨S(x)v,v⟩ ≥ λ‖v‖²,\n\nso ‖DF(x)v‖ ≥ λ‖v‖.  Hence DF(x) is injective, has closed range and satisfies (4).  \nIf w⊥Ran DF(x) then 0=⟨DF(x)ᵗw,v⟩=⟨(S(x)−A(x))w,v⟩ for every v, whence (taking v=w)\n\n 0=⟨S(x)w,w⟩ ≥ λ‖w‖²⇒w=0.\n\nThus Ran DF(x)=H, so DF(x) is surjective and therefore invertible. ∎\n\n\n\n────────────────────────────────────────  \n2.  Injectivity of F and the left inequality in (1)  \n────────────────────────────────────────  \n\nTaking inner product of (3) with h and using (1),\n\n ⟨F(x)−F(y),h⟩\n  =∫₀¹⟨S(y+th)h,h⟩dt ≥ λ‖h‖². (5)\n\nConsequently h≠0⇒⟨F(x)−F(y),h⟩>0, so F is strictly (λ-)monotone and injective.  \nBy Cauchy–Schwarz, (5) gives the first inequality in (1):\n\n  ‖F(x)−F(y)‖ ≥ λ‖x−y‖.\n\n\n\n────────────────────────────────────────  \n3.  Surjectivity of F  \n────────────────────────────────────────  \n\nPut R:=F(H).  \n\n(i) Openness.  Lemma 1 implies every DF(x) is a Banach–space isomorphism; the inverse–function theorem yields local C¹-diffeomorphisms, hence R is open.\n\n(ii) Closedness.  Let (yₙ)⊂R with yₙ→y.  Choose xₙ with F(xₙ)=yₙ.  Using (1),\n\n ‖xₙ−x_m‖ ≤ (1/λ)‖yₙ−y_m‖,\n\nso (xₙ) is Cauchy and converges to some x.  Continuity of F gives F(x)=y, hence y∈R.  Thus R is closed.\n\nBecause H is connected and R is non-empty, open, and closed, R=H; surjectivity follows.\n\n\n\n────────────────────────────────────────  \n4.  Global C¹–diffeomorphism and quantitative bounds  \n────────────────────────────────────────  \n\nWe know F is bijective and C¹ with everywhere invertible derivative.  \nTo see that the locally defined inverses patch to a global C¹ inverse, note that any two local inverses coincide on the (non-empty) connected set where they are both defined—their images there are obtained by composing both maps with F—so they glue to a single C¹ map F⁻¹:H→H.\n\nRight inequality in (1).  From (3) and (3),\n\n ‖F(x)−F(y)‖ ≤ ∫₀¹‖DF(y+th)‖‖h‖dt ≤ M‖x−y‖.\n\nInverse Lipschitz bound (2).  \nGiven u,v set x:=F⁻¹(u), y:=F⁻¹(v), h:=x−y.  Rewrite (3) at y:\n\n u−v=DF(y)h+∫₀¹(DF(y+th)−DF(y))h dt. (6)\n\nApply DF(y)⁻¹, use (4) and assumption (2):\n\n ‖h‖ ≤ (1/λ)‖u−v‖+(1/λ)∫₀¹‖DF(y+th)−DF(y)‖‖h‖dt  \n   ≤ (1/λ)‖u−v‖+(L‖h‖)/(2λ).\n\nSince L<2λ, transferring the second term to the left gives (2).\n\nHence F is a global C¹-diffeomorphism satisfying the announced two-sided Lipschitz estimates.\n\n\n\n────────────────────────────────────────  \n5.  Degree of F equals +1  \n────────────────────────────────────────  \n\nFix R>‖F(0)‖/λ and set B_R:={x:‖x‖<R}.  Using (5) with y=0,\n\n ⟨F(x),x⟩ = ⟨F(x)−F(0),x⟩+⟨F(0),x⟩ ≥ λ‖x‖²−‖F(0)‖‖x‖  \n  > λR²−‖F(0)‖R > 0 for ‖x‖=R. (7)\n\nDefine the homotopy\n\n H_t(x):=(1−t)λx+tF(x), t∈[0,1].\n\nFor ‖x‖=R,\n\n ⟨H_t(x),x⟩=(1−t)λR²+t⟨F(x),x⟩≥(1−t)λR²+t(λR²−‖F(0)‖R)>0.\n\nThus 0∉H_t(∂B_R) for every t, so the Browder–Minty degree deg(H_t,B_R,0) is defined.  \nBy homotopy invariance of the degree (see, e.g., Zeidler, Nonlinear Functional Analysis, Vol. I, Thm. 26.B),  \n\n deg(F,B_R,0)=deg(H_1,B_R,0)=deg(H_0,B_R,0).\n\nBut H_0(x)=λx is the radial expansion by the positive factor λ, hence deg(H_0,B_R,0)=+1.  \nBecause any other bounded open Ω containing 0 can be embedded in some ball B_R, naturality of the degree gives deg(F,Ω,0)=+1 for every such Ω. ∎\n\n\n\n────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.857063",
        "was_fixed": false,
        "difficulty_analysis": "1. Infinite-dimensional setting.  The original problems live in ℝ² or ℝ³; passing to an arbitrary separable Hilbert space removes compactness, requires Fréchet derivatives, and forbids reliance on elementary matrix algebra.\n\n2. Interaction of several advanced concepts.  One must combine  \n   – strong monotonicity,  \n   – Lipschitz continuity of the derivative,  \n   – the Browder–Minty theorem for monotone operators,  \n   – the Banach-space inverse–function theorem, and  \n   – topological (degree) theory.  \n   None of these appear in the original statement.\n\n3. Additional quantitative demands.  Besides injectivity the solver must prove surjectivity, C¹ regularity of the inverse, bi-Lipschitz estimates (with explicit constants), and orientation preservation.\n\n4. New technical obstacles.  \n   • Mean-value formulas and eigenvalue arguments must be recast with path integrals of bounded linear operators.  \n   • Coercivity and properness are subtle in infinite dimensions; compactness arguments no longer work.  \n   • Controlling the inverse requires non-trivial estimates on (DF)⁻¹ that involve both λ and the global Lipschitz constant L.\n\n5. More steps and deeper insights.  The solution forces the competitor to navigate functional analysis, operator theory, nonlinear PDE tools (monotone operators), and global nonlinear analysis, far exceeding the differential-calculus reasoning that sufficed for the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let H be a (possibly infinite–dimensional) real separable Hilbert space with inner product ⟨·,·⟩ and norm ‖·‖.  \nLet F : H→H be a C¹ map (Fréchet differentiable with continuous derivative).  \nFor every x∈H set  \n\n  S(x):=½(DF(x)+DF(x)*) and A(x):=½(DF(x)−DF(x)*) so that DF(x)=S(x)+A(x).\n\nAssume  \n\n(1)  (Uniform strong monotonicity) There is a constant λ>0 with  \n   ⟨S(x)v,v⟩ ≥ λ‖v‖² for all x,v∈H.  \n\n(2)  (Uniform Lipschitz bound on the derivative) There exists L with 0≤L<2λ such that  \n   ‖DF(x)−DF(y)‖ ≤ L‖x−y‖ for all x,y∈H.  \n\n(3)  (Global boundedness of the derivative) M:=sup_{x∈H}‖DF(x)‖<∞.\n\nProve that\n\n(a) F is injective.\n\n(b) F is surjective, hence a bijection H→H.\n\n(c) F is a C¹–diffeomorphism whose inverse is globally Lipschitz.  In particular  \n\n  λ‖x−y‖ ≤ ‖F(x)−F(y)‖ ≤ M‖x−y‖,     (1)  \n  ‖F⁻¹(u)−F⁻¹(v)‖ ≤ 1/(λ−L/2)·‖u−v‖.  (2)\n\n(d)  (Topological degree) For every bounded open set Ω⊂H with 0∈Ω the Browder–Minty degree deg(F,Ω,0) is defined and equals +1.  In particular F is of (positive) degree +1 on H.\n\n\n────────────────────────────────────────────────────",
      "solution": "Throughout write h:=x−y for x,y∈H.\n\nStep 0.  Integral representation.  \nSince DF is continuous,\n\n  F(x)−F(y)=∫₀¹DF(y+th)h dt.  (3)\n\n\n\n────────────────────────────────────────  \n1.  Uniform invertibility of DF(x)  \n────────────────────────────────────────  \n\nLemma 1.  For every x∈H the operator DF(x) is a bounded linear isomorphism and  \n\n  ‖DF(x)⁻¹‖ ≤ 1/λ.  (4)\n\nProof.  For v∈H,\n\n ‖DF(x)v‖‖v‖ ≥ ⟨DF(x)v,v⟩=⟨S(x)v,v⟩ ≥ λ‖v‖²,\n\nso ‖DF(x)v‖ ≥ λ‖v‖.  Hence DF(x) is injective, has closed range and satisfies (4).  \nIf w⊥Ran DF(x) then 0=⟨DF(x)ᵗw,v⟩=⟨(S(x)−A(x))w,v⟩ for every v, whence (taking v=w)\n\n 0=⟨S(x)w,w⟩ ≥ λ‖w‖²⇒w=0.\n\nThus Ran DF(x)=H, so DF(x) is surjective and therefore invertible. ∎\n\n\n\n────────────────────────────────────────  \n2.  Injectivity of F and the left inequality in (1)  \n────────────────────────────────────────  \n\nTaking inner product of (3) with h and using (1),\n\n ⟨F(x)−F(y),h⟩\n  =∫₀¹⟨S(y+th)h,h⟩dt ≥ λ‖h‖². (5)\n\nConsequently h≠0⇒⟨F(x)−F(y),h⟩>0, so F is strictly (λ-)monotone and injective.  \nBy Cauchy–Schwarz, (5) gives the first inequality in (1):\n\n  ‖F(x)−F(y)‖ ≥ λ‖x−y‖.\n\n\n\n────────────────────────────────────────  \n3.  Surjectivity of F  \n────────────────────────────────────────  \n\nPut R:=F(H).  \n\n(i) Openness.  Lemma 1 implies every DF(x) is a Banach–space isomorphism; the inverse–function theorem yields local C¹-diffeomorphisms, hence R is open.\n\n(ii) Closedness.  Let (yₙ)⊂R with yₙ→y.  Choose xₙ with F(xₙ)=yₙ.  Using (1),\n\n ‖xₙ−x_m‖ ≤ (1/λ)‖yₙ−y_m‖,\n\nso (xₙ) is Cauchy and converges to some x.  Continuity of F gives F(x)=y, hence y∈R.  Thus R is closed.\n\nBecause H is connected and R is non-empty, open, and closed, R=H; surjectivity follows.\n\n\n\n────────────────────────────────────────  \n4.  Global C¹–diffeomorphism and quantitative bounds  \n────────────────────────────────────────  \n\nWe know F is bijective and C¹ with everywhere invertible derivative.  \nTo see that the locally defined inverses patch to a global C¹ inverse, note that any two local inverses coincide on the (non-empty) connected set where they are both defined—their images there are obtained by composing both maps with F—so they glue to a single C¹ map F⁻¹:H→H.\n\nRight inequality in (1).  From (3) and (3),\n\n ‖F(x)−F(y)‖ ≤ ∫₀¹‖DF(y+th)‖‖h‖dt ≤ M‖x−y‖.\n\nInverse Lipschitz bound (2).  \nGiven u,v set x:=F⁻¹(u), y:=F⁻¹(v), h:=x−y.  Rewrite (3) at y:\n\n u−v=DF(y)h+∫₀¹(DF(y+th)−DF(y))h dt. (6)\n\nApply DF(y)⁻¹, use (4) and assumption (2):\n\n ‖h‖ ≤ (1/λ)‖u−v‖+(1/λ)∫₀¹‖DF(y+th)−DF(y)‖‖h‖dt  \n   ≤ (1/λ)‖u−v‖+(L‖h‖)/(2λ).\n\nSince L<2λ, transferring the second term to the left gives (2).\n\nHence F is a global C¹-diffeomorphism satisfying the announced two-sided Lipschitz estimates.\n\n\n\n────────────────────────────────────────  \n5.  Degree of F equals +1  \n────────────────────────────────────────  \n\nFix R>‖F(0)‖/λ and set B_R:={x:‖x‖<R}.  Using (5) with y=0,\n\n ⟨F(x),x⟩ = ⟨F(x)−F(0),x⟩+⟨F(0),x⟩ ≥ λ‖x‖²−‖F(0)‖‖x‖  \n  > λR²−‖F(0)‖R > 0 for ‖x‖=R. (7)\n\nDefine the homotopy\n\n H_t(x):=(1−t)λx+tF(x), t∈[0,1].\n\nFor ‖x‖=R,\n\n ⟨H_t(x),x⟩=(1−t)λR²+t⟨F(x),x⟩≥(1−t)λR²+t(λR²−‖F(0)‖R)>0.\n\nThus 0∉H_t(∂B_R) for every t, so the Browder–Minty degree deg(H_t,B_R,0) is defined.  \nBy homotopy invariance of the degree (see, e.g., Zeidler, Nonlinear Functional Analysis, Vol. I, Thm. 26.B),  \n\n deg(F,B_R,0)=deg(H_1,B_R,0)=deg(H_0,B_R,0).\n\nBut H_0(x)=λx is the radial expansion by the positive factor λ, hence deg(H_0,B_R,0)=+1.  \nBecause any other bounded open Ω containing 0 can be embedded in some ball B_R, naturality of the degree gives deg(F,Ω,0)=+1 for every such Ω. ∎\n\n\n\n────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.652933",
        "was_fixed": false,
        "difficulty_analysis": "1. Infinite-dimensional setting.  The original problems live in ℝ² or ℝ³; passing to an arbitrary separable Hilbert space removes compactness, requires Fréchet derivatives, and forbids reliance on elementary matrix algebra.\n\n2. Interaction of several advanced concepts.  One must combine  \n   – strong monotonicity,  \n   – Lipschitz continuity of the derivative,  \n   – the Browder–Minty theorem for monotone operators,  \n   – the Banach-space inverse–function theorem, and  \n   – topological (degree) theory.  \n   None of these appear in the original statement.\n\n3. Additional quantitative demands.  Besides injectivity the solver must prove surjectivity, C¹ regularity of the inverse, bi-Lipschitz estimates (with explicit constants), and orientation preservation.\n\n4. New technical obstacles.  \n   • Mean-value formulas and eigenvalue arguments must be recast with path integrals of bounded linear operators.  \n   • Coercivity and properness are subtle in infinite dimensions; compactness arguments no longer work.  \n   • Controlling the inverse requires non-trivial estimates on (DF)⁻¹ that involve both λ and the global Lipschitz constant L.\n\n5. More steps and deeper insights.  The solution forces the competitor to navigate functional analysis, operator theory, nonlinear PDE tools (monotone operators), and global nonlinear analysis, far exceeding the differential-calculus reasoning that sufficed for the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}