{
  "index": "1955-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "2. Suppose that \\( f \\) is a function with two continuous derivatives and \\( f(0)= \\) 0 . Prove that the function \\( g \\), defined by \\( g(0)=f^{\\prime}(0), g(x)=f(x)^{\\prime} x \\) for \\( x \\neq \\) 0 , has a continuous derivative.",
  "solution": "First Solution. It is clear that \\( g \\) has a continuous derivative (even two continuous derivatives) on \\( \\mathbf{R}-\\{0\\} \\) with\n\\[\ng^{\\prime}(x)=\\frac{x f^{\\prime}(x)-f(x)}{x^{2}}\n\\]\n\nNow \\( f \\) is differentiable at 0 , so\n\\[\n\\lim _{x \\rightarrow 0} \\frac{f(x)-f(0)}{x}=\\lim _{x \\rightarrow 0} g(x)=f^{\\prime}(0)\n\\]\n\nThus \\( g \\) is continuous at 0 .\nTo find \\( g^{\\prime}(0) \\), we must consider\n\\[\n\\lim _{x \\rightarrow 0} \\frac{g(x)-g(0)}{x}=\\lim _{x \\rightarrow 0} \\frac{f(x)-x f^{\\prime}(0)}{x^{2}}\n\\]\n\nBy the extended mean value theorem, for each \\( x \\) there is a \\( \\theta \\in(0,1) \\) such that\n\\[\nf(x)=f(0)+x f^{\\prime}(0)+\\frac{1}{2} x^{2} f^{\\prime \\prime}(\\theta x)\n\\]\nso\n\\[\n\\lim _{x \\rightarrow 0} \\frac{f(x)-x f^{\\prime}(0)}{x^{2}}=\\lim _{x \\rightarrow 0} \\frac{1}{2} f^{\\prime \\prime}(\\theta x)=\\frac{1}{2} f^{\\prime \\prime}(0)\n\\]\nsince \\( f^{\\prime \\prime} \\) is continuous. Thus \\( g \\) is differentiable at 0 and \\( g^{\\prime}(0)=f^{\\prime \\prime}(0) / 2 \\).\nBy the ordinary mean value theorem, for each \\( x \\) there is an \\( \\eta \\in(0,1) \\) such that\n\\[\nf^{\\prime}(x)-f^{\\prime}(0)=x f^{\\prime \\prime}(\\eta x)\n\\]\n\nHence we have\n\\[\n\\begin{aligned}\n\\lim _{x \\rightarrow 0} g^{\\prime}(x) & =\\lim _{x \\rightarrow 0} \\frac{x f^{\\prime}(x)-f(x)}{x^{2}} \\\\\n& =\\lim _{x \\rightarrow 0}\\left(f^{\\prime \\prime}(\\eta x)-\\frac{1}{2} f^{\\prime \\prime}(\\theta x)\\right)=\\frac{1}{2} f^{\\prime \\prime}(0)\n\\end{aligned}\n\\]\nand so \\( g^{\\prime} \\) is continuous at 0 .\nThe explicit references to the mean value theorem can be avoided by using L'Hospital's rule.\n\nSecond Solution. Note that\n\\[\ng(x)=\\int_{0}^{1} f^{\\prime}(x t) d t\n\\]\n\nSince \\( f^{\\prime} \\) has a continuous derivative, we can differentiate under the integral sign and conclude that \\( g^{\\prime} \\) exists and is given by\n\\[\ng^{\\prime}(x)=\\int_{0}^{1} t f^{\\prime \\prime}(x t) d t\n\\]\n\nThen, since \\( f^{\\prime \\prime} \\) is continuous, \\( g^{\\prime} \\) is continuous.",
  "vars": [
    "x",
    "t"
  ],
  "params": [
    "f",
    "g",
    "\\\\theta",
    "\\\\eta"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "t": "paramvar",
        "f": "basefun",
        "g": "derivedfun",
        "\\theta": "thetavalue",
        "\\eta": "etavalue"
      },
      "question": "2. Suppose that \\( basefun \\) is a function with two continuous derivatives and \\( basefun(0)=0 \\). Prove that the function \\( derivedfun \\), defined by \\( derivedfun(0)=basefun^{\\prime}(0),\\; derivedfun(inputvar)=basefun(inputvar)^{\\prime}\\, inputvar \\) for \\( inputvar\\neq 0 \\), has a continuous derivative.",
      "solution": "First Solution. It is clear that \\( derivedfun \\) has a continuous derivative (even two continuous derivatives) on \\( \\mathbf{R}-\\{0\\} \\) with\n\\[\nderivedfun^{\\prime}(inputvar)=\\frac{inputvar\\, basefun^{\\prime}(inputvar)-basefun(inputvar)}{inputvar^{2}}\n\\]\nNow \\( basefun \\) is differentiable at 0 , so\n\\[\n\\lim _{inputvar \\rightarrow 0} \\frac{basefun(inputvar)-basefun(0)}{inputvar}=\\lim _{inputvar \\rightarrow 0} derivedfun(inputvar)=basefun^{\\prime}(0)\n\\]\nThus \\( derivedfun \\) is continuous at 0 .\nTo find \\( derivedfun^{\\prime}(0) \\), we must consider\n\\[\n\\lim _{inputvar \\rightarrow 0} \\frac{derivedfun(inputvar)-derivedfun(0)}{inputvar}=\\lim _{inputvar \\rightarrow 0} \\frac{basefun(inputvar)-inputvar\\, basefun^{\\prime}(0)}{inputvar^{2}}\n\\]\nBy the extended mean value theorem, for each \\( inputvar \\) there is a \\( thetavalue \\in(0,1) \\) such that\n\\[\nbasefun(inputvar)=basefun(0)+inputvar\\, basefun^{\\prime}(0)+\\frac{1}{2} inputvar^{2} basefun^{\\prime \\prime}(thetavalue\\, inputvar)\n\\]\nso\n\\[\n\\lim _{inputvar \\rightarrow 0} \\frac{basefun(inputvar)-inputvar\\, basefun^{\\prime}(0)}{inputvar^{2}}=\\lim _{inputvar \\rightarrow 0} \\frac{1}{2} basefun^{\\prime \\prime}(thetavalue\\, inputvar)=\\frac{1}{2} basefun^{\\prime \\prime}(0)\n\\]\nsince \\( basefun^{\\prime \\prime} \\) is continuous. Thus \\( derivedfun \\) is differentiable at 0 and \\( derivedfun^{\\prime}(0)=basefun^{\\prime \\prime}(0) / 2 \\).\nBy the ordinary mean value theorem, for each \\( inputvar \\) there is an \\( etavalue \\in(0,1) \\) such that\n\\[\nbasefun^{\\prime}(inputvar)-basefun^{\\prime}(0)=inputvar\\, basefun^{\\prime \\prime}(etavalue\\, inputvar)\n\\]\nHence we have\n\\[\n\\begin{aligned}\n\\lim _{inputvar \\rightarrow 0} derivedfun^{\\prime}(inputvar) & =\\lim _{inputvar \\rightarrow 0} \\frac{inputvar\\, basefun^{\\prime}(inputvar)-basefun(inputvar)}{inputvar^{2}} \\\\\n& =\\lim _{inputvar \\rightarrow 0}\\left(basefun^{\\prime \\prime}(etavalue\\, inputvar)-\\frac{1}{2} basefun^{\\prime \\prime}(thetavalue\\, inputvar)\\right)=\\frac{1}{2} basefun^{\\prime \\prime}(0)\n\\end{aligned}\n\\]\nand so \\( derivedfun^{\\prime} \\) is continuous at 0 .\nThe explicit references to the mean value theorem can be avoided by using L'Hospital's rule.\n\nSecond Solution. Note that\n\\[\nderivedfun(inputvar)=\\int_{0}^{1} basefun^{\\prime}(inputvar\\, paramvar) d paramvar\n\\]\nSince \\( basefun^{\\prime} \\) has a continuous derivative, we can differentiate under the integral sign and conclude that \\( derivedfun^{\\prime} \\) exists and is given by\n\\[\nderivedfun^{\\prime}(inputvar)=\\int_{0}^{1} paramvar\\, basefun^{\\prime \\prime}(inputvar\\, paramvar) d paramvar\n\\]\nThen, since \\( basefun^{\\prime \\prime} \\) is continuous, \\( derivedfun^{\\prime} \\) is continuous."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "magnitude",
        "t": "velocity",
        "f": "diameter",
        "g": "radiance",
        "\\theta": "longitude",
        "\\eta": "gradient"
      },
      "question": "Suppose that \\( diameter \\) is a function with two continuous derivatives and \\( diameter(0)= \\) 0 . Prove that the function \\( radiance \\), defined by \\( radiance(0)=diameter^{\\prime}(0), radiance(magnitude)=diameter(magnitude)^{\\prime} magnitude \\) for \\( magnitude \\neq \\) 0 , has a continuous derivative.",
      "solution": "First Solution. It is clear that \\( radiance \\) has a continuous derivative (even two continuous derivatives) on \\( \\mathbf{R}-\\{0\\} \\) with\n\\[\nradiance^{\\prime}(magnitude)=\\frac{magnitude \\, diameter^{\\prime}(magnitude)-diameter(magnitude)}{magnitude^{2}}\n\\]\n\nNow \\( diameter \\) is differentiable at 0 , so\n\\[\n\\lim _{magnitude \\rightarrow 0} \\frac{diameter(magnitude)-diameter(0)}{magnitude}=\\lim _{magnitude \\rightarrow 0} radiance(magnitude)=diameter^{\\prime}(0)\n\\]\n\nThus \\( radiance \\) is continuous at 0 .\nTo find \\( radiance^{\\prime}(0) \\), we must consider\n\\[\n\\lim _{magnitude \\rightarrow 0} \\frac{radiance(magnitude)-radiance(0)}{magnitude}=\\lim _{magnitude \\rightarrow 0} \\frac{diameter(magnitude)-magnitude \\, diameter^{\\prime}(0)}{magnitude^{2}}\n\\]\n\nBy the extended mean value theorem, for each \\( magnitude \\) there is a \\( longitude \\in(0,1) \\) such that\n\\[\ndiameter(magnitude)=diameter(0)+magnitude \\, diameter^{\\prime}(0)+\\frac{1}{2} magnitude^{2} \\, diameter^{\\prime \\prime}(longitude \\, magnitude)\n\\]\nso\n\\[\n\\lim _{magnitude \\rightarrow 0} \\frac{diameter(magnitude)-magnitude \\, diameter^{\\prime}(0)}{magnitude^{2}}=\\lim _{magnitude \\rightarrow 0} \\frac{1}{2} diameter^{\\prime \\prime}(longitude \\, magnitude)=\\frac{1}{2} diameter^{\\prime \\prime}(0)\n\\]\nsince \\( diameter^{\\prime \\prime} \\) is continuous. Thus \\( radiance \\) is differentiable at 0 and \\( radiance^{\\prime}(0)=diameter^{\\prime \\prime}(0) / 2 \\).\nBy the ordinary mean value theorem, for each \\( magnitude \\) there is an \\( gradient \\in(0,1) \\) such that\n\\[\ndiameter^{\\prime}(magnitude)-diameter^{\\prime}(0)=magnitude \\, diameter^{\\prime \\prime}(gradient \\, magnitude)\n\\]\n\nHence we have\n\\[\n\\begin{aligned}\n\\lim _{magnitude \\rightarrow 0} radiance^{\\prime}(magnitude) & =\\lim _{magnitude \\rightarrow 0} \\frac{magnitude \\, diameter^{\\prime}(magnitude)-diameter(magnitude)}{magnitude^{2}} \\\\\n& =\\lim _{magnitude \\rightarrow 0}\\left(diameter^{\\prime \\prime}(gradient \\, magnitude)-\\frac{1}{2} diameter^{\\prime \\prime}(longitude \\, magnitude)\\right)=\\frac{1}{2} diameter^{\\prime \\prime}(0)\n\\end{aligned}\n\\]\nand so \\( radiance^{\\prime} \\) is continuous at 0 .\nThe explicit references to the mean value theorem can be avoided by using L'Hospital's rule.\n\nSecond Solution. Note that\n\\[\nradiance(magnitude)=\\int_{0}^{1} diameter^{\\prime}(magnitude \\, velocity) d \\, velocity\n\\]\n\nSince \\( diameter^{\\prime} \\) has a continuous derivative, we can differentiate under the integral sign and conclude that \\( radiance^{\\prime} \\) exists and is given by\n\\[\nradiance^{\\prime}(magnitude)=\\int_{0}^{1} velocity \\, diameter^{\\prime \\prime}(magnitude \\, velocity) d \\, velocity\n\\]\n\nThen, since \\( diameter^{\\prime \\prime} \\) is continuous, \\( radiance^{\\prime} \\) is continuous."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "antifunc",
        "g": "staticfunc",
        "x": "constantval",
        "t": "nontemporal",
        "\\theta": "nonanglevar",
        "\\eta": "straightcoef"
      },
      "question": "2. Suppose that \\( antifunc \\) is a function with two continuous derivatives and \\( antifunc(0)= \\) 0 . Prove that the function \\( staticfunc \\), defined by \\( staticfunc(0)=antifunc^{\\prime}(0), staticfunc(constantval)=antifunc(constantval)^{\\prime} constantval \\) for \\( constantval \\neq \\) 0 , has a continuous derivative.",
      "solution": "First Solution. It is clear that \\( staticfunc \\) has a continuous derivative (even two continuous derivatives) on \\( \\mathbf{R}-\\{0\\} \\) with\n\\[\nstaticfunc^{\\prime}(constantval)=\\frac{constantval antifunc^{\\prime}(constantval)-antifunc(constantval)}{constantval^{2}}\n\\]\n\nNow antifunc is differentiable at 0 , so\n\\[\n\\lim _{constantval \\rightarrow 0} \\frac{antifunc(constantval)-antifunc(0)}{constantval}=\\lim _{constantval \\rightarrow 0} staticfunc(constantval)=antifunc^{\\prime}(0)\n\\]\n\nThus staticfunc is continuous at 0.\nTo find staticfunc^{\\prime}(0), we must consider\n\\[\n\\lim _{constantval \\rightarrow 0} \\frac{staticfunc(constantval)-staticfunc(0)}{constantval}=\\lim _{constantval \\rightarrow 0} \\frac{antifunc(constantval)-constantval\\,antifunc^{\\prime}(0)}{constantval^{2}}\n\\]\n\nBy the extended mean value theorem, for each \\( constantval \\) there is a \\( nonanglevar \\in(0,1) \\) such that\n\\[\nantifunc(constantval)=antifunc(0)+constantval\\,antifunc^{\\prime}(0)+\\frac{1}{2} constantval^{2} antifunc^{\\prime\\prime}(nonanglevar constantval)\n\\]\nso\n\\[\n\\lim _{constantval \\rightarrow 0} \\frac{antifunc(constantval)-constantval\\,antifunc^{\\prime}(0)}{constantval^{2}}=\\lim _{constantval \\rightarrow 0} \\frac{1}{2} antifunc^{\\prime\\prime}(nonanglevar constantval)=\\frac{1}{2} antifunc^{\\prime\\prime}(0)\n\\]\nsince antifunc^{\\prime\\prime} is continuous. Thus staticfunc is differentiable at 0 and staticfunc^{\\prime}(0)=antifunc^{\\prime\\prime}(0)/2.\nBy the ordinary mean value theorem, for each \\( constantval \\) there is an \\( straightcoef \\in(0,1) \\) such that\n\\[\nantifunc^{\\prime}(constantval)-antifunc^{\\prime}(0)=constantval\\,antifunc^{\\prime\\prime}(straightcoef constantval)\n\\]\n\nHence we have\n\\[\n\\begin{aligned}\n\\lim _{constantval \\rightarrow 0} staticfunc^{\\prime}(constantval) & =\\lim _{constantval \\rightarrow 0} \\frac{constantval antifunc^{\\prime}(constantval)-antifunc(constantval)}{constantval^{2}} \\\\\n& =\\lim _{constantval \\rightarrow 0}\\left(antifunc^{\\prime\\prime}(straightcoef constantval)-\\frac{1}{2} antifunc^{\\prime\\prime}(nonanglevar constantval)\\right)=\\frac{1}{2} antifunc^{\\prime\\prime}(0)\n\\end{aligned}\n\\]\nand so staticfunc^{\\prime} is continuous at 0.\nThe explicit references to the mean value theorem can be avoided by using L'Hospital's rule.\n\nSecond Solution. Note that\n\\[\nstaticfunc(constantval)=\\int_{0}^{1} antifunc^{\\prime}(constantval nontemporal) d\\,nontemporal\n\\]\n\nSince antifunc^{\\prime} has a continuous derivative, we can differentiate under the integral sign and conclude that staticfunc^{\\prime} exists and is given by\n\\[\nstaticfunc^{\\prime}(constantval)=\\int_{0}^{1} nontemporal\\,antifunc^{\\prime\\prime}(constantval nontemporal) d\\,nontemporal\n\\]\n\nThen, since antifunc^{\\prime\\prime} is continuous, staticfunc^{\\prime} is continuous."
    },
    "garbled_string": {
      "map": {
        "x": "bqrjmkls",
        "t": "wzdnphqc",
        "f": "plwdkqms",
        "g": "nzcuxdle",
        "\\theta": "hdsqkbnm",
        "\\eta": "vczmpxla"
      },
      "question": "2. Suppose that \\( plwdkqms \\) is a function with two continuous derivatives and \\( plwdkqms(0)= \\) 0 . Prove that the function \\( nzcuxdle \\), defined by \\( nzcuxdle(0)=plwdkqms^{\\prime}(0), nzcuxdle(bqrjmkls)=plwdkqms(bqrjmkls)^{\\prime} bqrjmkls \\) for \\( bqrjmkls \\neq \\) 0 , has a continuous derivative.",
      "solution": "First Solution. It is clear that \\( nzcuxdle \\) has a continuous derivative (even two continuous derivatives) on \\( \\mathbf{R}-\\{0\\} \\) with\n\\[\nnzcuxdle^{\\prime}(bqrjmkls)=\\frac{bqrjmkls plwdkqms^{\\prime}(bqrjmkls)-plwdkqms(bqrjmkls)}{bqrjmkls^{2}}\n\\]\n\nNow \\( plwdkqms \\) is differentiable at 0 , so\n\\[\n\\lim _{bqrjmkls \\rightarrow 0} \\frac{plwdkqms(bqrjmkls)-plwdkqms(0)}{bqrjmkls}=\\lim _{bqrjmkls \\rightarrow 0} nzcuxdle(bqrjmkls)=plwdkqms^{\\prime}(0)\n\\]\n\nThus \\( nzcuxdle \\) is continuous at 0 .\nTo find \\( nzcuxdle^{\\prime}(0) \\), we must consider\n\\[\n\\lim _{bqrjmkls \\rightarrow 0} \\frac{nzcuxdle(bqrjmkls)-nzcuxdle(0)}{bqrjmkls}=\\lim _{bqrjmkls \\rightarrow 0} \\frac{plwdkqms(bqrjmkls)-bqrjmkls plwdkqms^{\\prime}(0)}{bqrjmkls^{2}}\n\\]\n\nBy the extended mean value theorem, for each \\( bqrjmkls \\) there is a \\( hdsqkbnm \\in(0,1) \\) such that\n\\[\nplwdkqms(bqrjmkls)=plwdkqms(0)+bqrjmkls plwdkqms^{\\prime}(0)+\\frac{1}{2} bqrjmkls^{2} plwdkqms^{\\prime \\prime}(hdsqkbnm bqrjmkls)\n\\]\nso\n\\[\n\\lim _{bqrjmkls \\rightarrow 0} \\frac{plwdkqms(bqrjmkls)-bqrjmkls plwdkqms^{\\prime}(0)}{bqrjmkls^{2}}=\\lim _{bqrjmkls \\rightarrow 0} \\frac{1}{2} plwdkqms^{\\prime \\prime}(hdsqkbnm bqrjmkls)=\\frac{1}{2} plwdkqms^{\\prime \\prime}(0)\n\\]\nsince \\( plwdkqms^{\\prime \\prime} \\) is continuous. Thus \\( nzcuxdle \\) is differentiable at 0 and \\( nzcuxdle^{\\prime}(0)=plwdkqms^{\\prime \\prime}(0) / 2 \\).\nBy the ordinary mean value theorem, for each \\( bqrjmkls \\) there is an \\( vczmpxla \\in(0,1) \\) such that\n\\[\nplwdkqms^{\\prime}(bqrjmkls)-plwdkqms^{\\prime}(0)=bqrjmkls plwdkqms^{\\prime \\prime}(vczmpxla bqrjmkls)\n\\]\n\nHence we have\n\\[\n\\begin{aligned}\n\\lim _{bqrjmkls \\rightarrow 0} nzcuxdle^{\\prime}(bqrjmkls) & =\\lim _{bqrjmkls \\rightarrow 0} \\frac{bqrjmkls plwdkqms^{\\prime}(bqrjmkls)-plwdkqms(bqrjmkls)}{bqrjmkls^{2}} \\\\\n& =\\lim _{bqrjmkls \\rightarrow 0}\\left(plwdkqms^{\\prime \\prime}(vczmpxla bqrjmkls)-\\frac{1}{2} plwdkqms^{\\prime \\prime}(hdsqkbnm bqrjmkls)\\right)=\\frac{1}{2} plwdkqms^{\\prime \\prime}(0)\n\\end{aligned}\n\\]\nand so \\( nzcuxdle^{\\prime} \\) is continuous at 0 .\nThe explicit references to the mean value theorem can be avoided by using L'Hospital's rule.\n\nSecond Solution. Note that\n\\[\nnzcuxdle(bqrjmkls)=\\int_{0}^{1} plwdkqms^{\\prime}(bqrjmkls wzdnphqc) d wzdnphqc\n\\]\n\nSince \\( plwdkqms^{\\prime} \\) has a continuous derivative, we can differentiate under the integral sign and conclude that \\( nzcuxdle^{\\prime} \\) exists and is given by\n\\[\nnzcuxdle^{\\prime}(bqrjmkls)=\\int_{0}^{1} wzdnphqc plwdkqms^{\\prime \\prime}(bqrjmkls wzdnphqc) d wzdnphqc\n\\]\n\nThen, since \\( plwdkqms^{\\prime \\prime} \\) is continuous, \\( nzcuxdle^{\\prime} \\) is continuous."
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 and let Ω ⊂ ℝⁿ be an open set that is star–shaped with respect to the origin, i.e.  \n  tx ∈ Ω for every x ∈ Ω and every t ∈ [0,1].\n\nLet  \n\n  f:Ω→ℝ  with f∈C³(Ω) and f(0)=0.\n\nFor x≠0 define the matrix–valued map  \n\n  G(x)=∫₀¹ (1−t) Hess f(tx) dt ∈ ℝ^{n×n},      (★)\n\nand set G(0)=½ Hess f(0).\n\n(a)  Show that G extends continuously to x=0 (hence G∈C⁰(Ω;ℝ^{n×n})).\n\n(b)  Prove that G is differentiable at x=0 and that for every v∈ℝⁿ  \n\n  (DG(0)) v = 1⁄6 D³f(0)[v, ·, ·],                     (†)\n\ni.e. the directional derivative of G at 0 is the bilinear form\n(y,z)↦⅙ D³f(0)[v,y,z].\n\n(c)  Show that DG is continuous at 0 and conclude that G∈C¹(Ω;ℝ^{n×n}).\n\n(d)  (Harder extension) Assume in addition that f∈C⁴(Ω).  \nProve that G is twice differentiable at 0 and that, for all v,w∈ℝⁿ,\n\n  (D²G(0))[v,w] = 1⁄12 D⁴f(0)[v,w, ·, ·].\n\nYou may NOT invoke Taylor’s theorem with explicit remainder or any version of l’Hospital’s rule.  \nOnly the mean–value theorem, the fundamental theorem of calculus and basic facts about continuously differentiable functions may be used.",
      "solution": "Step 0. Notation.  \nFor a 3–linear form T on ℝⁿ we write T[h] for the bilinear form (y,z)↦T(h,y,z).  \nAll norms are Euclidean operator norms.\n\n———————————————————————————\nStep 1. Continuity of G at 0 (part (a)).  \nBecause Hess f is continuous on Ω and Ω is star–shaped,  \n\n  lim_{(t,x)→(0,0)} Hess f(tx)=Hess f(0).\n\nGiven ε>0, choose δ>0 such that ‖Hess f(y)−Hess f(0)‖<ε whenever ‖y‖<δ.  \nIf ‖x‖<δ then ‖tx‖<δ for every t∈[0,1], whence  \n\n‖G(x)−G(0)‖ = ‖∫₀¹(1−t)(Hess f(tx)−Hess f(0))dt‖  \n      ≤ ∫₀¹(1−t)‖Hess f(tx)−Hess f(0)‖dt < ε∫₀¹(1−t)dt = ε/2.\n\nThus lim_{x→0}G(x)=G(0); G is continuous at 0 and therefore on Ω. ∎\n\n———————————————————————————\nStep 2. A first–order integral identity.  \nFix h∈ℝⁿ and set Φ(t)=Hess f(th).  Because f∈C³, Φ is C¹, and the 1–d fundamental theorem of calculus yields\n\n  Φ(t)−Φ(0)=∫₀ᵗΦ′(s)ds = ∫₀ᵗ D(Hess f)(sh)[h] ds  \n      = ∫₀ᵗ D³f(sh)[h,·,·] ds.                             (1)\n\n———————————————————————————\nStep 3. Existence and value of DG(0) (part (b)).  \nLet h be small.  From (★) and (1):\n\nG(h)−G(0)\n  =∫₀¹(1−t)(Hess f(th)−Hess f(0))dt\n  =∫₀¹(1−t)∫₀ᵗ D³f(sh)[h,·,·] ds dt.                        (2)\n\nExchange the order of integration (Fubini; integrand continuous):\n\n=∫₀¹(∫_{s}^{1}(1−t)dt) D³f(sh)[h,·,·] ds\n =∫₀¹(½−s+½s²) D³f(sh)[h,·,·] ds.                           (3)\n\nBecause D³f is continuous at 0, D³f(sh)=D³f(0)+o(1) uniformly in s∈[0,1].  Hence\n\nG(h)−G(0)=D³f(0)[h,·,·]∫₀¹(½−s+½s²)ds + o(‖h‖).\n\nA routine calculation gives ∫₀¹(½−s+½s²)ds=1/6, so\n\nG(h)−G(0) = (1/6) D³f(0)[h,·,·] + o(‖h‖).                  (4)\n\nTherefore G is Fréchet–differentiable at 0 and DG(0) is the linear map stated in (†). ∎\n\n———————————————————————————\nStep 4. Continuity of DG at 0 (part (c)).  \nFor x≠0 and v∈ℝⁿ we may differentiate under the integral sign (the map t↦Hess f(tx) is C¹) to obtain\n\n  DG(x)·v = ∫₀¹(1−t)t D³f(tx)[v,·,·] dt.                 (5)\n\nContinuity of D³f implies uniform convergence of the integrand in (5) to (1−t)t D³f(0)[v,·,·] as x→0.  Dominated convergence then gives\n\nlim_{x→0}DG(x)·v = ∫₀¹(1−t)t D³f(0)[v,·,·]dt = (1/6)D³f(0)[v,·,·] = DG(0)·v.\n\nThus DG is continuous at 0; continuity on Ω\\{0} is evident from (5).  Hence G∈C¹(Ω;ℝ^{n×n}). ∎\n\n———————————————————————————\nStep 5. Second derivative at 0 (part (d)).  \nAssume now f∈C⁴(Ω).  Fix v∈ℝⁿ and write, for small h,\n\nDG(h)·v = ∫₀¹(1−t)t D³f(th)[v,·,·] dt.                    (6)\n\nSince D³f is C¹, for each t\n\n  D³f(th) = D³f(0) + ∫₀ᵗ D⁴f(sh)[h,·,·,·] ds.             (7)\n\nInsert (7) into (6) and subtract DG(0)·v = (1/6)D³f(0)[v,·,·]:\n\nDG(h)·v − DG(0)·v\n  = ∫₀¹(1−t)t∫₀ᵗ D⁴f(sh)[h,v,·,·] ds dt.                  (8)\n\nInterchange the s– and t–integrals:\n\n= ∫₀¹(∫_{s}^{1}(1−t)t dt) D⁴f(sh)[h,v,·,·] ds.            (9)\n\nThe inner integral equals\n  ∫_{s}^{1}(t−t²)dt = [t²/2 − t³/3]_{s}^{1} = 1/6 − s²/2 + s³/3.\n\nHence (9) becomes\n\nDG(h)·v − DG(0)·v\n  = ∫₀¹(1/6 − s²/2 + s³/3) D⁴f(sh)[h,v,·,·] ds.           (10)\n\nBecause D⁴f is continuous at 0, D⁴f(sh)=D⁴f(0)+o(1) uniformly in s, and\n\n∫₀¹(1/6 − s²/2 + s³/3) ds = 1/12.\n\nTherefore\n\nDG(h)·v − DG(0)·v = (1/12) D⁴f(0)[h,v,·,·] + o(‖h‖).     (11)\n\nThe map (h,v)↦(1/12)D⁴f(0)[h,v,·,·] is bilinear, so (11) shows that DG is Fréchet–differentiable at 0 with\n\n(D²G(0))[h,v] = 1/12 D⁴f(0)[h,v,·,·].\n\nSince D⁴f is continuous, an argument identical to Step 4 yields continuity of D²G at 0. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.474648",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is set in ℝⁿ with n≥2 and the unknown is matrix-valued rather than scalar.  \n2. Higher-order derivatives: the solution requires handling third-order derivatives and bilinear/trilinear forms, far beyond the second-order work in the original variant.  \n3. Integral representation: proving differentiability relies on a double integral identity and Fubini’s theorem, demanding careful order-of-integration arguments.  \n4. Abstract linear-operator algebra: the derivative DG(0) is itself a linear map taking values in a space of matrices; interpreting and manipulating such objects is substantially more sophisticated than treating a single real derivative.  \n5. No Taylor remainder allowed: the proof must construct everything from the mean–value theorem and the fundamental theorem of calculus, forcing delicate estimates instead of a one-line Taylor expansion.  \n\nThese layers of added technicality—higher dimensions, matrix-valued functions, third-order tensors, and intricate integral manipulations—make the enhanced kernel variant markedly harder than both the original problem and its current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 and let Ω ⊂ ℝⁿ be an open set that is star–shaped with respect to the origin, i.e.  \n  tx ∈ Ω for every x ∈ Ω and every t ∈ [0,1].\n\nLet  \n\n  f:Ω→ℝ  with f∈C³(Ω) and f(0)=0.\n\nFor x≠0 define the matrix–valued map  \n\n  G(x)=∫₀¹ (1−t) Hess f(tx) dt ∈ ℝ^{n×n},      (★)\n\nand set G(0)=½ Hess f(0).\n\n(a)  Show that G extends continuously to x=0 (hence G∈C⁰(Ω;ℝ^{n×n})).\n\n(b)  Prove that G is differentiable at x=0 and that for every v∈ℝⁿ  \n\n  (DG(0)) v = 1⁄6 D³f(0)[v, ·, ·],                     (†)\n\ni.e. the directional derivative of G at 0 is the bilinear form\n(y,z)↦⅙ D³f(0)[v,y,z].\n\n(c)  Show that DG is continuous at 0 and conclude that G∈C¹(Ω;ℝ^{n×n}).\n\n(d)  (Harder extension) Assume in addition that f∈C⁴(Ω).  \nProve that G is twice differentiable at 0 and that, for all v,w∈ℝⁿ,\n\n  (D²G(0))[v,w] = 1⁄12 D⁴f(0)[v,w, ·, ·].\n\nYou may NOT invoke Taylor’s theorem with explicit remainder or any version of l’Hospital’s rule.  \nOnly the mean–value theorem, the fundamental theorem of calculus and basic facts about continuously differentiable functions may be used.",
      "solution": "Step 0. Notation.  \nFor a 3–linear form T on ℝⁿ we write T[h] for the bilinear form (y,z)↦T(h,y,z).  \nAll norms are Euclidean operator norms.\n\n———————————————————————————\nStep 1. Continuity of G at 0 (part (a)).  \nBecause Hess f is continuous on Ω and Ω is star–shaped,  \n\n  lim_{(t,x)→(0,0)} Hess f(tx)=Hess f(0).\n\nGiven ε>0, choose δ>0 such that ‖Hess f(y)−Hess f(0)‖<ε whenever ‖y‖<δ.  \nIf ‖x‖<δ then ‖tx‖<δ for every t∈[0,1], whence  \n\n‖G(x)−G(0)‖ = ‖∫₀¹(1−t)(Hess f(tx)−Hess f(0))dt‖  \n      ≤ ∫₀¹(1−t)‖Hess f(tx)−Hess f(0)‖dt < ε∫₀¹(1−t)dt = ε/2.\n\nThus lim_{x→0}G(x)=G(0); G is continuous at 0 and therefore on Ω. ∎\n\n———————————————————————————\nStep 2. A first–order integral identity.  \nFix h∈ℝⁿ and set Φ(t)=Hess f(th).  Because f∈C³, Φ is C¹, and the 1–d fundamental theorem of calculus yields\n\n  Φ(t)−Φ(0)=∫₀ᵗΦ′(s)ds = ∫₀ᵗ D(Hess f)(sh)[h] ds  \n      = ∫₀ᵗ D³f(sh)[h,·,·] ds.                             (1)\n\n———————————————————————————\nStep 3. Existence and value of DG(0) (part (b)).  \nLet h be small.  From (★) and (1):\n\nG(h)−G(0)\n  =∫₀¹(1−t)(Hess f(th)−Hess f(0))dt\n  =∫₀¹(1−t)∫₀ᵗ D³f(sh)[h,·,·] ds dt.                        (2)\n\nExchange the order of integration (Fubini; integrand continuous):\n\n=∫₀¹(∫_{s}^{1}(1−t)dt) D³f(sh)[h,·,·] ds\n =∫₀¹(½−s+½s²) D³f(sh)[h,·,·] ds.                           (3)\n\nBecause D³f is continuous at 0, D³f(sh)=D³f(0)+o(1) uniformly in s∈[0,1].  Hence\n\nG(h)−G(0)=D³f(0)[h,·,·]∫₀¹(½−s+½s²)ds + o(‖h‖).\n\nA routine calculation gives ∫₀¹(½−s+½s²)ds=1/6, so\n\nG(h)−G(0) = (1/6) D³f(0)[h,·,·] + o(‖h‖).                  (4)\n\nTherefore G is Fréchet–differentiable at 0 and DG(0) is the linear map stated in (†). ∎\n\n———————————————————————————\nStep 4. Continuity of DG at 0 (part (c)).  \nFor x≠0 and v∈ℝⁿ we may differentiate under the integral sign (the map t↦Hess f(tx) is C¹) to obtain\n\n  DG(x)·v = ∫₀¹(1−t)t D³f(tx)[v,·,·] dt.                 (5)\n\nContinuity of D³f implies uniform convergence of the integrand in (5) to (1−t)t D³f(0)[v,·,·] as x→0.  Dominated convergence then gives\n\nlim_{x→0}DG(x)·v = ∫₀¹(1−t)t D³f(0)[v,·,·]dt = (1/6)D³f(0)[v,·,·] = DG(0)·v.\n\nThus DG is continuous at 0; continuity on Ω\\{0} is evident from (5).  Hence G∈C¹(Ω;ℝ^{n×n}). ∎\n\n———————————————————————————\nStep 5. Second derivative at 0 (part (d)).  \nAssume now f∈C⁴(Ω).  Fix v∈ℝⁿ and write, for small h,\n\nDG(h)·v = ∫₀¹(1−t)t D³f(th)[v,·,·] dt.                    (6)\n\nSince D³f is C¹, for each t\n\n  D³f(th) = D³f(0) + ∫₀ᵗ D⁴f(sh)[h,·,·,·] ds.             (7)\n\nInsert (7) into (6) and subtract DG(0)·v = (1/6)D³f(0)[v,·,·]:\n\nDG(h)·v − DG(0)·v\n  = ∫₀¹(1−t)t∫₀ᵗ D⁴f(sh)[h,v,·,·] ds dt.                  (8)\n\nInterchange the s– and t–integrals:\n\n= ∫₀¹(∫_{s}^{1}(1−t)t dt) D⁴f(sh)[h,v,·,·] ds.            (9)\n\nThe inner integral equals\n  ∫_{s}^{1}(t−t²)dt = [t²/2 − t³/3]_{s}^{1} = 1/6 − s²/2 + s³/3.\n\nHence (9) becomes\n\nDG(h)·v − DG(0)·v\n  = ∫₀¹(1/6 − s²/2 + s³/3) D⁴f(sh)[h,v,·,·] ds.           (10)\n\nBecause D⁴f is continuous at 0, D⁴f(sh)=D⁴f(0)+o(1) uniformly in s, and\n\n∫₀¹(1/6 − s²/2 + s³/3) ds = 1/12.\n\nTherefore\n\nDG(h)·v − DG(0)·v = (1/12) D⁴f(0)[h,v,·,·] + o(‖h‖).     (11)\n\nThe map (h,v)↦(1/12)D⁴f(0)[h,v,·,·] is bilinear, so (11) shows that DG is Fréchet–differentiable at 0 with\n\n(D²G(0))[h,v] = 1/12 D⁴f(0)[h,v,·,·].\n\nSince D⁴f is continuous, an argument identical to Step 4 yields continuity of D²G at 0. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.398959",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the problem is set in ℝⁿ with n≥2 and the unknown is matrix-valued rather than scalar.  \n2. Higher-order derivatives: the solution requires handling third-order derivatives and bilinear/trilinear forms, far beyond the second-order work in the original variant.  \n3. Integral representation: proving differentiability relies on a double integral identity and Fubini’s theorem, demanding careful order-of-integration arguments.  \n4. Abstract linear-operator algebra: the derivative DG(0) is itself a linear map taking values in a space of matrices; interpreting and manipulating such objects is substantially more sophisticated than treating a single real derivative.  \n5. No Taylor remainder allowed: the proof must construct everything from the mean–value theorem and the fundamental theorem of calculus, forcing delicate estimates instead of a one-line Taylor expansion.  \n\nThese layers of added technicality—higher dimensions, matrix-valued functions, third-order tensors, and intricate integral manipulations—make the enhanced kernel variant markedly harder than both the original problem and its current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}