{
  "index": "1980-A-6",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Problem A-6\nLet \\( C \\) be the class of all real valued continuously differentiable functions \\( f \\) on the interval \\( 0<x<1 \\) with \\( f(0)=0 \\) and \\( f(1)=1 \\). Determine the largest real number \\( u \\) such that\n\\[\nu \\leqslant \\int_{0}^{1}\\left|f^{\\prime}(x)-f(x)\\right| d x\n\\]\nfor all \\( f \\) in \\( C \\).",
  "solution": "A-6.\nWe show that \\( u=1 / e \\). Since \\( f^{\\prime}-f=\\left(f e^{-x}\\right)^{\\prime} e^{x} \\) and \\( e^{x} \\geqslant 1 \\) for \\( x \\geqslant 0 \\),\n\\[\n\\int_{0}^{1}\\left|f^{\\prime}-f\\right| d x=\\int_{0}^{1}\\left|\\left(f e^{-x}\\right)^{\\prime} e^{x}\\right| d x \\geqslant \\int_{0}^{1}\\left(f e^{-x}\\right)^{\\prime} d x=\\left[f e^{-x}\\right]_{0}^{1}=1 / e\n\\]\n\nTo see that \\( 1 / e \\) is the largest lower bound, we use functions \\( f_{a}(x) \\) defined by\n\\[\nf_{a}(x)=\\left(e^{a-1} / a\\right) x \\text { for } 0 \\leqslant x \\leqslant a, \\quad f_{a}(x)=e^{x-1} \\quad \\text { for } a \\leqslant x \\leqslant 1\n\\]\n\nLet \\( m=e^{a-1} / a \\). Then\n\\[\n\\int_{0}^{1}\\left|f_{a}^{\\prime}(x)-f_{a}(x)\\right| d x=\\int_{0}^{a}|m-m x| d x=m\\left(a-\\frac{a^{2}}{2}\\right)=e^{a-1}\\left(1-\\frac{a}{2}\\right) .\n\\]\n\nAs \\( a \\rightarrow 0 \\), this expression approaches \\( 1 / e \\). The function \\( f_{a}(x) \\) does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than \\( 1 / e \\) can be an upper bound.",
  "vars": [
    "x",
    "f"
  ],
  "params": [
    "u",
    "a",
    "m",
    "C",
    "f_a"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "f": "function",
        "u": "lowerbound",
        "a": "cutoff",
        "m": "gradient",
        "C": "functionclass",
        "f_a": "piecewise"
      },
      "question": "Problem A-6\nLet \\( functionclass \\) be the class of all real valued continuously differentiable functions \\( function \\) on the interval \\( 0<variable<1 \\) with \\( function(0)=0 \\) and \\( function(1)=1 \\). Determine the largest real number \\( lowerbound \\) such that\n\\[\nlowerbound \\leqslant \\int_{0}^{1}\\left|function^{\\prime}(variable)-function(variable)\\right| d variable\n\\]\nfor all \\( function \\) in \\( functionclass \\).",
      "solution": "A-6.\nWe show that \\( lowerbound=1 / e \\). Since \\( function^{\\prime}-function=\\left(function e^{-variable}\\right)^{\\prime} e^{variable} \\) and \\( e^{variable} \\geqslant 1 \\) for \\( variable \\geqslant 0 \\),\n\\[\n\\int_{0}^{1}\\left|function^{\\prime}-function\\right| d variable=\\int_{0}^{1}\\left|\\left(function e^{-variable}\\right)^{\\prime} e^{variable}\\right| d variable \\geqslant \\int_{0}^{1}\\left(function e^{-variable}\\right)^{\\prime} d variable=\\left[function e^{-variable}\\right]_{0}^{1}=1 / e\n\\]\n\nTo see that \\( 1 / e \\) is the largest lower bound, we use functions \\( piecewise(variable) \\) defined by\n\\[\npiecewise(variable)=\\left(e^{cutoff-1} / cutoff\\right) variable \\text { for } 0 \\leqslant variable \\leqslant cutoff, \\quad piecewise(variable)=e^{variable-1} \\quad \\text { for } cutoff \\leqslant variable \\leqslant 1\n\\]\n\nLet \\( gradient=e^{cutoff-1} / cutoff \\). Then\n\\[\n\\int_{0}^{1}\\left|piecewise^{\\prime}(variable)-piecewise(variable)\\right| d variable=\\int_{0}^{cutoff}|gradient-gradient variable| d variable=gradient\\left(cutoff-\\frac{cutoff^{2}}{2}\\right)=e^{cutoff-1}\\left(1-\\frac{cutoff}{2}\\right) .\n\\]\n\nAs \\( cutoff \\rightarrow 0 \\), this expression approaches \\( 1 / e \\). The function \\( piecewise(variable) \\) does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than \\( 1 / e \\) can be an upper bound."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pineapple",
        "f": "suitcase",
        "u": "lighthouse",
        "a": "porcupine",
        "m": "binoculars",
        "C": "harmonica",
        "f_a": "suitcaseporc"
      },
      "question": "Problem A-6\nLet \\( harmonica \\) be the class of all real valued continuously differentiable functions \\( suitcase \\) on the interval \\( 0<pineapple<1 \\) with \\( suitcase(0)=0 \\) and \\( suitcase(1)=1 \\). Determine the largest real number \\( lighthouse \\) such that\n\\[\nlighthouse \\leqslant \\int_{0}^{1}\\left|suitcase^{\\prime}(pineapple)-suitcase(pineapple)\\right| d pineapple\n\\]\nfor all \\( suitcase \\) in \\( harmonica \\).",
      "solution": "A-6.\nWe show that \\( lighthouse=1 / e \\). Since \\( suitcase^{\\prime}-suitcase=\\left(suitcase e^{-pineapple}\\right)^{\\prime} e^{pineapple} \\) and \\( e^{pineapple} \\geqslant 1 \\) for \\( pineapple \\geqslant 0 \\),\n\\[\n\\int_{0}^{1}\\left|suitcase^{\\prime}-suitcase\\right| d pineapple=\\int_{0}^{1}\\left|\\left(suitcase e^{-pineapple}\\right)^{\\prime} e^{pineapple}\\right| d pineapple \\geqslant \\int_{0}^{1}\\left(suitcase e^{-pineapple}\\right)^{\\prime} d pineapple=\\left[suitcase e^{-pineapple}\\right]_{0}^{1}=1 / e\n\\]\n\nTo see that \\( 1 / e \\) is the largest lower bound, we use functions \\( suitcaseporc(pineapple) \\) defined by\n\\[\nsuitcaseporc(pineapple)=\\left(e^{porcupine-1} / porcupine\\right) pineapple \\text { for } 0 \\leqslant pineapple \\leqslant porcupine, \\quad suitcaseporc(pineapple)=e^{pineapple-1} \\quad \\text { for } porcupine \\leqslant pineapple \\leqslant 1\n\\]\n\nLet \\( binoculars=e^{porcupine-1} / porcupine \\). Then\n\\[\n\\int_{0}^{1}\\left|suitcaseporc^{\\prime}(pineapple)-suitcaseporc(pineapple)\\right| d pineapple=\\int_{0}^{porcupine}|binoculars-binoculars pineapple| d pineapple=binoculars\\left(porcupine-\\frac{porcupine^{2}}{2}\\right)=e^{porcupine-1}\\left(1-\\frac{porcupine}{2}\\right) .\n\\]\n\nAs \\( porcupine \\rightarrow 0 \\), this expression approaches \\( 1 / e \\). The function \\( suitcaseporc(pineapple) \\) does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than \\( 1 / e \\) can be an upper bound."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immobile",
        "f": "nonfunction",
        "u": "ceilingvalue",
        "a": "vastness",
        "m": "flatness",
        "C": "singleton",
        "f_a": "staticvalue"
      },
      "question": "Problem A-6\nLet \\( singleton \\) be the class of all real valued continuously differentiable functions \\( nonfunction \\) on the interval \\( 0<immobile<1 \\) with \\( nonfunction(0)=0 \\) and \\( nonfunction(1)=1 \\). Determine the largest real number \\( ceilingvalue \\) such that\n\\[\nceilingvalue \\leqslant \\int_{0}^{1}\\left|nonfunction^{\\prime}(immobile)-nonfunction(immobile)\\right| d immobile\n\\]\nfor all \\( nonfunction \\) in \\( singleton \\).",
      "solution": "A-6.\nWe show that \\( ceilingvalue=1 / e \\). Since \\( nonfunction^{\\prime}-nonfunction=\\left(nonfunction e^{-immobile}\\right)^{\\prime} e^{immobile} \\) and \\( e^{immobile} \\geqslant 1 \\) for \\( immobile \\geqslant 0 \\),\n\\[\n\\int_{0}^{1}\\left|nonfunction^{\\prime}-nonfunction\\right| d immobile=\\int_{0}^{1}\\left|\\left(nonfunction e^{-immobile}\\right)^{\\prime} e^{immobile}\\right| d immobile \\geqslant \\int_{0}^{1}\\left(nonfunction e^{-immobile}\\right)^{\\prime} d immobile=\\left[nonfunction e^{-immobile}\\right]_{0}^{1}=1 / e\n\\]\n\nTo see that \\( 1 / e \\) is the largest lower bound, we use functions \\( staticvalue(immobile) \\) defined by\n\\[\nstaticvalue(immobile)=\\left(e^{vastness-1} / vastness\\right) immobile \\text { for } 0 \\leqslant immobile \\leqslant vastness, \\quad staticvalue(immobile)=e^{immobile-1} \\quad \\text { for } vastness \\leqslant immobile \\leqslant 1\n\\]\n\nLet \\( flatness=e^{vastness-1} / vastness \\). Then\n\\[\n\\int_{0}^{1}\\left|staticvalue^{\\prime}(immobile)-staticvalue(immobile)\\right| d immobile=\\int_{0}^{vastness}|flatness-flatness immobile| d immobile=flatness\\left(vastness-\\frac{vastness^{2}}{2}\\right)=e^{vastness-1}\\left(1-\\frac{vastness}{2}\\right) .\n\\]\n\nAs \\( vastness \\rightarrow 0 \\), this expression approaches \\( 1 / e \\). The function \\( staticvalue(immobile) \\) does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than \\( 1 / e \\) can be an upper bound."
    },
    "garbled_string": {
      "map": {
        "x": "wdjqkefn",
        "f": "hslpgtam",
        "u": "qzxrvnap",
        "a": "mjqwerty",
        "m": "pxonlksa",
        "C": "ghetubyr",
        "f_a": "bnvmcsle"
      },
      "question": "Problem A-6\nLet \\( ghetubyr \\) be the class of all real valued continuously differentiable functions \\( hslpgtam \\) on the interval \\( 0<wdjqkefn<1 \\) with \\( hslpgtam(0)=0 \\) and \\( hslpgtam(1)=1 \\). Determine the largest real number \\( qzxrvnap \\) such that\n\\[\nqzxrvnap \\leqslant \\int_{0}^{1}\\left|hslpgtam^{\\prime}(wdjqkefn)-hslpgtam(wdjqkefn)\\right| d wdjqkefn\n\\]\nfor all \\( hslpgtam \\) in \\( ghetubyr \\).",
      "solution": "A-6.\nWe show that \\( qzxrvnap=1 / e \\). Since \\( hslpgtam^{\\prime}-hslpgtam=\\left(hslpgtam e^{-wdjqkefn}\\right)^{\\prime} e^{wdjqkefn} \\) and \\( e^{wdjqkefn} \\geqslant 1 \\) for \\( wdjqkefn \\geqslant 0 \\),\n\\[\n\\int_{0}^{1}\\left|hslpgtam^{\\prime}-hslpgtam\\right| d wdjqkefn=\\int_{0}^{1}\\left|\\left(hslpgtam e^{-wdjqkefn}\\right)^{\\prime} e^{wdjqkefn}\\right| d wdjqkefn \\geqslant \\int_{0}^{1}\\left(hslpgtam e^{-wdjqkefn}\\right)^{\\prime} d wdjqkefn=\\left[hslpgtam e^{-wdjqkefn}\\right]_{0}^{1}=1 / e\n\\]\n\nTo see that \\( 1 / e \\) is the largest lower bound, we use functions \\( bnvmcsle(wdjqkefn) \\) defined by\n\\[\nbnvmcsle(wdjqkefn)=\\left(e^{mjqwerty-1} / mjqwerty\\right) wdjqkefn \\text { for } 0 \\leqslant wdjqkefn \\leqslant mjqwerty, \\quad bnvmcsle(wdjqkefn)=e^{wdjqkefn-1} \\quad \\text { for } mjqwerty \\leqslant wdjqkefn \\leqslant 1\n\\]\n\nLet \\( pxonlksa=e^{mjqwerty-1} / mjqwerty \\). Then\n\\[\n\\int_{0}^{1}\\left|bnvmcsle^{\\prime}(wdjqkefn)-bnvmcsle(wdjqkefn)\\right| d wdjqkefn=\\int_{0}^{mjqwerty}|pxonlksa-pxonlksa wdjqkefn| d wdjqkefn=pxonlksa\\left(mjqwerty-\\frac{mjqwerty^{2}}{2}\\right)=e^{mjqwerty-1}\\left(1-\\frac{mjqwerty}{2}\\right) .\n\\]\n\nAs \\( mjqwerty \\rightarrow 0 \\), this expression approaches \\( 1 / e \\). The function \\( bnvmcsle(wdjqkefn) \\) does not have a continuous derivative, but one can smooth out the corner, keeping the change in the integral as small as one wishes, and thus show that no number greater than \\( 1 / e \\) can be an upper bound."
    },
    "kernel_variant": {
      "question": "Let C be the set of all real–valued functions f that are continuously differentiable on the closed interval [0,1] and satisfy the four‐point boundary conditions   \n f(0)=0, f(⅓)=3, f(⅔)=2, f(1)=1.  \nDetermine the largest real number u such that the inequality  \n u ≤ ∫₀¹ |f ′(x) − f(x)| dx                                       (★)  \nholds for every f in C.",
      "solution": "Step 1.  A convenient substitution.  \nSet  \n g(x)=f(x)e^{−x}.   \nBecause f is C¹, so is g.  Moreover  \n g ′(x)=e^{−x}\\bigl(f ′(x)−f(x)\\bigr).  \n\nHence  \n |f ′(x)−f(x)|=e^{x}|g ′(x)|,             (1)  \nand (★) becomes  \n u ≤ ∫₀¹ e^{x}|g ′(x)| dx.               (2)\n\nStep 2.  Fixing the values of g at the prescribed points.  \nFrom the boundary data for f we get\n\n g(0)=f(0)=0,  \n g(⅓)=3e^{−1/3},  \n g(⅔)=2e^{−2/3},  \n g(1)=e^{−1}.                              (3)\n\nDenote these four numbers by  \n G₀:=0,     G₁:=3e^{−1/3},     G₂:=2e^{−2/3},     G₃:=e^{−1}.\n\nStep 3.  Lower–bounding the integral on each subinterval.  \nSplit [0,1] into the three subintervals  \nI₁=[0,⅓], I₂=[⅓,⅔], I₃=[⅔,1].\n\nBecause the weight e^{x} is monotone increasing,\n\n min_{x∈I₁} e^{x}=e^{0}=1,  \n min_{x∈I₂} e^{x}=e^{1/3},  \n min_{x∈I₃} e^{x}=e^{2/3}.                (4)\n\nUsing (2), (4) and the triangle inequality,\n\n∫_{I₁}e^{x}|g′(x)| dx ≥ 1·|G₁−G₀|,  \n∫_{I₂}e^{x}|g′(x)| dx ≥ e^{1/3}|G₂−G₁|,  \n∫_{I₃}e^{x}|g′(x)| dx ≥ e^{2/3}|G₃−G₂|.  (5)\n\nAdding the three inequalities gives\n\n ∫₀¹e^{x}|g′(x)| dx ≥ |G₁−G₀| + e^{1/3}|G₂−G₁| + e^{2/3}|G₃−G₂|.  (6)\n\nStep 4.  Evaluating the right–hand side exactly.  \nDirect substitutions from (3) yield\n\n|G₁−G₀| = 3e^{−1/3},  \n\n|G₂−G₁| = |2e^{−2/3}−3e^{−1/3}|  \n        = (3e^{−1/3}−2e^{−2/3})  (because 3e^{−1/3}>2e^{−2/3}),  \n\n|G₃−G₂| = |e^{−1}−2e^{−2/3}|  \n        = (2e^{−2/3}−e^{−1}).                                     (7)\n\nMultiplying by the weights in (6),\n\ne^{1/3}|G₂−G₁| = (3e^{−1/3}−2e^{−2/3})·e^{1/3} = 3 − 2e^{−1/3},  \n\ne^{2/3}|G₃−G₂| = (2e^{−2/3}−e^{−1})·e^{2/3} = 2 − e^{−1/3}.      (8)\n\nNow sum the three contributions:\n\n|G₁−G₀| + e^{1/3}|G₂−G₁| + e^{2/3}|G₃−G₂|  \n = 3e^{−1/3} + (3 − 2e^{−1/3}) + (2 − e^{−1/3})  \n = 5.                                                              (9)\n\nCombining (2), (6) and (9) we have  \n u ≤ ∫₀¹e^{x}|g′(x)| dx ≥ 5,  \nso every admissible f satisfies  \n ∫₀¹|f′(x)−f(x)| dx ≥ 5.   Therefore u_max ≥ 5.                (10)\n\nStep 5.  Sharpness: constructing almost‐extremal functions.  \nFix a small ε>0.  Define a piecewise linear function g_ε on [0,1] as follows:\n\ng_ε(x)=⎧  \n 0         (0≤x≤ε),  \n ((G₁)/ε)(x−ε)  (ε≤x≤2ε),  \n G₁       (2ε≤x≤⅓),  \n G₁+((G₂−G₁)/ε)(x−⅓) (⅓≤x≤⅓+ε),  \n G₂       (⅓+ε≤x≤⅔),  \n G₂+((G₃−G₂)/ε)(x−⅔) (⅔≤x≤⅔+ε),  \n G₃       (⅔+ε≤x≤1).                                (11)\n\nDefine f_ε(x)=e^{x}g_ε(x).  \nBecause g_ε is continuous, has corners only at finitely many points, and g_ε(0),g_ε(⅓),g_ε(⅔),g_ε(1) equal G₀,G₁,G₂,G₃, the function f_ε lies in C after a standard smooth “rounding” of the corners inside intervals of length ε/2 (which changes the integral by o(1) as ε→0).\n\nOn each linear piece of g_ε the derivative g_ε′ is constant, so by (1)\n\n∫₀¹|f_ε′−f_ε| dx = ∑_{k=1}^{3} e^{x_k^*}|G_k−G_{k−1}| + O(ε),  (12)\n\nwhere each x_k^* is some point in the ε–wide transition layer lying in I_k, hence e^{x_k^*}→e^{\\min I_k} as ε→0.  Consequently\n\nlim_{ε→0}∫₀¹|f_ε′−f_ε| dx = |G₁−G₀|+e^{1/3}|G₂−G₁|+e^{2/3}|G₃−G₂|=5. (13)\n\nThus for every δ>0 there exists f∈C with  \n ∫₀¹|f′−f| dx < 5+δ,  \nso no number larger than 5 can serve as a universal lower bound.\n\nStep 6.  Conclusion.  \nFrom (10) and (13) we deduce that the largest real number u satisfying (★) for all f∈C is  \n\n u = 5.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.649900",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional boundary conditions: Unlike the original problem, which involved only two end–point constraints, this variant imposes four distinct point constraints, forcing the solver to track the function on three separate subintervals and to respect non-trivial interior data.\n\n2. Piecewise analysis with variable weights: The weight e^{x} in the transformed integral is no longer constant on the subintervals, so the solver must exploit minimum–value estimates on each piece and carefully compute the exact contribution of each jump in g.\n\n3. Exact algebraic cancellation: Arriving at the clean constant 5 requires non-obvious algebraic manipulations (see equations (7)–(9)) in which different exponential factors cancel to an integer.\n\n4. Construction of extremals: Producing almost-minimal functions now entails designing a three-stage piecewise construction and arguing that the inevitable corner smoothing can be done without enlarging the integral—considerably subtler than the single-corner adjustment in the original solution.\n\n5. Multiple interacting concepts:  Exponential change of variables, weighted triangle inequalities, sharp lower bounds, limiting constructions and delicate smoothing all interact, demanding a broader toolkit and a longer chain of reasoning than the original one-step estimate.\n\nFor these reasons the enhanced kernel variant is significantly harder and technically richer than both the original problem and the earlier kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let C be the set of all real–valued functions f that are continuously differentiable on the closed interval [0,1] and satisfy the four‐point boundary conditions   \n f(0)=0, f(⅓)=3, f(⅔)=2, f(1)=1.  \nDetermine the largest real number u such that the inequality  \n u ≤ ∫₀¹ |f ′(x) − f(x)| dx                                       (★)  \nholds for every f in C.",
      "solution": "Step 1.  A convenient substitution.  \nSet  \n g(x)=f(x)e^{−x}.   \nBecause f is C¹, so is g.  Moreover  \n g ′(x)=e^{−x}\\bigl(f ′(x)−f(x)\\bigr).  \n\nHence  \n |f ′(x)−f(x)|=e^{x}|g ′(x)|,             (1)  \nand (★) becomes  \n u ≤ ∫₀¹ e^{x}|g ′(x)| dx.               (2)\n\nStep 2.  Fixing the values of g at the prescribed points.  \nFrom the boundary data for f we get\n\n g(0)=f(0)=0,  \n g(⅓)=3e^{−1/3},  \n g(⅔)=2e^{−2/3},  \n g(1)=e^{−1}.                              (3)\n\nDenote these four numbers by  \n G₀:=0,     G₁:=3e^{−1/3},     G₂:=2e^{−2/3},     G₃:=e^{−1}.\n\nStep 3.  Lower–bounding the integral on each subinterval.  \nSplit [0,1] into the three subintervals  \nI₁=[0,⅓], I₂=[⅓,⅔], I₃=[⅔,1].\n\nBecause the weight e^{x} is monotone increasing,\n\n min_{x∈I₁} e^{x}=e^{0}=1,  \n min_{x∈I₂} e^{x}=e^{1/3},  \n min_{x∈I₃} e^{x}=e^{2/3}.                (4)\n\nUsing (2), (4) and the triangle inequality,\n\n∫_{I₁}e^{x}|g′(x)| dx ≥ 1·|G₁−G₀|,  \n∫_{I₂}e^{x}|g′(x)| dx ≥ e^{1/3}|G₂−G₁|,  \n∫_{I₃}e^{x}|g′(x)| dx ≥ e^{2/3}|G₃−G₂|.  (5)\n\nAdding the three inequalities gives\n\n ∫₀¹e^{x}|g′(x)| dx ≥ |G₁−G₀| + e^{1/3}|G₂−G₁| + e^{2/3}|G₃−G₂|.  (6)\n\nStep 4.  Evaluating the right–hand side exactly.  \nDirect substitutions from (3) yield\n\n|G₁−G₀| = 3e^{−1/3},  \n\n|G₂−G₁| = |2e^{−2/3}−3e^{−1/3}|  \n        = (3e^{−1/3}−2e^{−2/3})  (because 3e^{−1/3}>2e^{−2/3}),  \n\n|G₃−G₂| = |e^{−1}−2e^{−2/3}|  \n        = (2e^{−2/3}−e^{−1}).                                     (7)\n\nMultiplying by the weights in (6),\n\ne^{1/3}|G₂−G₁| = (3e^{−1/3}−2e^{−2/3})·e^{1/3} = 3 − 2e^{−1/3},  \n\ne^{2/3}|G₃−G₂| = (2e^{−2/3}−e^{−1})·e^{2/3} = 2 − e^{−1/3}.      (8)\n\nNow sum the three contributions:\n\n|G₁−G₀| + e^{1/3}|G₂−G₁| + e^{2/3}|G₃−G₂|  \n = 3e^{−1/3} + (3 − 2e^{−1/3}) + (2 − e^{−1/3})  \n = 5.                                                              (9)\n\nCombining (2), (6) and (9) we have  \n u ≤ ∫₀¹e^{x}|g′(x)| dx ≥ 5,  \nso every admissible f satisfies  \n ∫₀¹|f′(x)−f(x)| dx ≥ 5.   Therefore u_max ≥ 5.                (10)\n\nStep 5.  Sharpness: constructing almost‐extremal functions.  \nFix a small ε>0.  Define a piecewise linear function g_ε on [0,1] as follows:\n\ng_ε(x)=⎧  \n 0         (0≤x≤ε),  \n ((G₁)/ε)(x−ε)  (ε≤x≤2ε),  \n G₁       (2ε≤x≤⅓),  \n G₁+((G₂−G₁)/ε)(x−⅓) (⅓≤x≤⅓+ε),  \n G₂       (⅓+ε≤x≤⅔),  \n G₂+((G₃−G₂)/ε)(x−⅔) (⅔≤x≤⅔+ε),  \n G₃       (⅔+ε≤x≤1).                                (11)\n\nDefine f_ε(x)=e^{x}g_ε(x).  \nBecause g_ε is continuous, has corners only at finitely many points, and g_ε(0),g_ε(⅓),g_ε(⅔),g_ε(1) equal G₀,G₁,G₂,G₃, the function f_ε lies in C after a standard smooth “rounding” of the corners inside intervals of length ε/2 (which changes the integral by o(1) as ε→0).\n\nOn each linear piece of g_ε the derivative g_ε′ is constant, so by (1)\n\n∫₀¹|f_ε′−f_ε| dx = ∑_{k=1}^{3} e^{x_k^*}|G_k−G_{k−1}| + O(ε),  (12)\n\nwhere each x_k^* is some point in the ε–wide transition layer lying in I_k, hence e^{x_k^*}→e^{\\min I_k} as ε→0.  Consequently\n\nlim_{ε→0}∫₀¹|f_ε′−f_ε| dx = |G₁−G₀|+e^{1/3}|G₂−G₁|+e^{2/3}|G₃−G₂|=5. (13)\n\nThus for every δ>0 there exists f∈C with  \n ∫₀¹|f′−f| dx < 5+δ,  \nso no number larger than 5 can serve as a universal lower bound.\n\nStep 6.  Conclusion.  \nFrom (10) and (13) we deduce that the largest real number u satisfying (★) for all f∈C is  \n\n u = 5.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.514989",
        "was_fixed": false,
        "difficulty_analysis": "1. Additional boundary conditions: Unlike the original problem, which involved only two end–point constraints, this variant imposes four distinct point constraints, forcing the solver to track the function on three separate subintervals and to respect non-trivial interior data.\n\n2. Piecewise analysis with variable weights: The weight e^{x} in the transformed integral is no longer constant on the subintervals, so the solver must exploit minimum–value estimates on each piece and carefully compute the exact contribution of each jump in g.\n\n3. Exact algebraic cancellation: Arriving at the clean constant 5 requires non-obvious algebraic manipulations (see equations (7)–(9)) in which different exponential factors cancel to an integer.\n\n4. Construction of extremals: Producing almost-minimal functions now entails designing a three-stage piecewise construction and arguing that the inevitable corner smoothing can be done without enlarging the integral—considerably subtler than the single-corner adjustment in the original solution.\n\n5. Multiple interacting concepts:  Exponential change of variables, weighted triangle inequalities, sharp lower bounds, limiting constructions and delicate smoothing all interact, demanding a broader toolkit and a longer chain of reasoning than the original one-step estimate.\n\nFor these reasons the enhanced kernel variant is significantly harder and technically richer than both the original problem and the earlier kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}