{
  "index": "1946-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "4. Let \\( g(x) \\) be a function that has a continuous first derivative \\( g^{\\prime}(x) \\) for all values of \\( x \\). Suppose that the following conditions hold for every \\( x \\) : (i) \\( g(0)= \\) \\( 0 ; \\) (ii) \\( \\left|g^{\\prime}(x)\\right| \\leq|g(x)| \\). Prove that \\( g(x) \\) vanishes identically.",
  "solution": "First Solution. We first convert the differential inequality into an integral inequality. Suppose \\( x \\geq 0 \\). Using (i), (ii), and the continuity of \\( g^{\\prime} \\), we have\n\\[\n|g(x)|=\\left|\\int_{0}^{x} g^{\\prime}(t) d t\\right| \\leq \\int_{0}^{x}\\left|g^{\\prime}(t)\\right| d t \\leq \\int_{0}^{x}|g(t)| d t .\n\\]\n\nThus\n\\[\n|g(x)| \\leq \\int_{0}^{x}|g(t)| d t, \\quad \\text { for } x \\geq 0\n\\]\n\nLet \\( a \\geq 0 \\) be chosen arbitrarily. Since \\( g \\) is differentiable, it is continuous and therefore bounded on any finite interval. So there is a number \\( K \\) such that\n\\[\n|g(x)| \\leq K, \\text { for } 0 \\leq x \\leq a .\n\\]\n\nIf \\( 0 \\leq t \\leq x \\leq a \\), we have \\( |g(t)| \\leq K \\), so (1) gives\n\\[\n|g(x)| \\leq \\int_{0}^{x} K d t=K x, \\quad \\text { for } \\quad 0 \\leq x \\leq a .\n\\]\n\nNow if \\( 0 \\leq t \\leq x \\leq a \\), we have \\( |g(t)| \\leq K t \\), and (1) gives\n\\[\n|g(x)| \\leq \\int_{0}^{x} K t d t=\\frac{1}{2} K x^{2}, \\quad \\text { for } \\quad 0 \\leq x \\leq a .\n\\]\n\nContinuing in this way we find\n\\[\n|g(x)| \\leq \\frac{1}{n!} K x^{n}, \\text { for } 0 \\leq x \\leq a\n\\]\nand all positive integers \\( n \\).\nTo prove (2) formally, we use mathematical induction. We have shown that (2) is true for \\( n=1 \\). Suppose it is true for \\( n=p \\). Then for \\( 0 \\leq t \\leq \\) \\( x \\leq a \\), we have \\( |g(t)| \\leq K t^{p} / p! \\). Using this in (1), we find\n\\[\n|g(x)| \\leq \\int_{0}^{x} \\frac{1}{p!} K t^{p} d t=\\frac{1}{(p+1)!} K x^{p+1}, \\text { for } 0 \\leq x \\leq a .\n\\]\n\nThus (2) is true for \\( n=p+1 \\). We conclude it is true for all \\( n \\).\nSetting \\( x=a \\) in (2) and letting \\( n \\rightarrow \\infty \\), we have\n\\[\n|g(a)| \\leq \\lim _{n \\rightarrow \\infty} \\frac{1}{n!} K a^{n}=0 .\n\\]\n\nTherefore \\( g(a)=0 \\). But \\( a \\) was arbitrary, so \\( g \\) vanishes on all of \\( [0, \\infty) \\).\nConsider the function \\( h \\) defined by \\( h(x)=g(-x) \\). Since \\( h \\) has a continuous derivative and satisfies (i) and (ii), \\( h \\) vanishes on \\( [0, \\infty \\) ), by what we have already proved. Therefore \\( g \\) vanishes on \\( (-\\infty, 0 \\) ] as well.\n\nBy a slight variation of this reasoning, one can prove directly inequalities of the form\n\\[\n|g(x)| \\leq \\frac{1}{n!} L|x|^{n} \\text { for }-a \\leq x \\leq a,\n\\]\nand obtain the proof for positive and negative arguments simultaneously.\nSecond Solution. Suppose that \\( \\boldsymbol{g}(\\boldsymbol{x}) \\) does not vanish identically. Then from the continuity of \\( g \\) it follows that there exist points \\( a \\) and \\( b \\) such that \\( g(a)=0, g(b) \\neq 0,|a-b|<1 \\), and \\( |g(b)| \\) is the maximum value of \\( |g(x)| \\) on the closed interval with endpoints \\( a \\) and \\( b \\). (For example, if \\( g \\) does not vanish on \\( [0, \\infty) \\), we can let \\( a \\) be the least non-negative halfinteger such that \\( g \\) does not vanish on \\( \\left[a, a+\\frac{1}{2}\\right] \\), and let \\( b \\) be the point at which \\( |g(x)| \\) attains its maximum for \\( a \\leq x \\leq a+\\frac{1}{2} \\).)\nFrom the mean value theorem it follows that there is a number \\( c \\) between \\( a \\) and \\( b \\) such that\n\\[\n\\left|g^{\\prime}(c)\\right|=\\left|\\frac{g(b)-g(a)}{b-a}\\right|=\\left|\\frac{g(b)}{b-a}\\right|>|g(b)| .\n\\]\n\nBut our choice of \\( b \\) shows that \\( |g(b)| \\geq|g(c)| \\), hence \\( \\left|g^{\\prime}(c)\\right|>|g(c)| \\), contrary to (ii). This contradiction proves that \\( g \\) vanishes identically.\nThis proof does not require the continuity of \\( g^{\\prime} \\).\nRemark. This is a standard result that arises in proving the uniqueness of the solutions of differential equations.",
  "vars": [
    "g",
    "h",
    "x",
    "t"
  ],
  "params": [
    "a",
    "b",
    "c",
    "n",
    "p",
    "K",
    "L"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "g": "anyfunc",
        "h": "reflect",
        "x": "varreal",
        "t": "vartemp",
        "a": "pointa",
        "b": "pointb",
        "c": "pointc",
        "n": "indexn",
        "p": "indexp",
        "K": "boundk",
        "L": "boundl"
      },
      "question": "4. Let \\( anyfunc(varreal) \\) be a function that has a continuous first derivative \\( anyfunc^{\\prime}(varreal) \\) for all values of \\( varreal \\). Suppose that the following conditions hold for every \\( varreal \\) : (i) \\( anyfunc(0)= 0 ; \\) (ii) \\( \\left|anyfunc^{\\prime}(varreal)\\right| \\leq|anyfunc(varreal)| \\). Prove that \\( anyfunc(varreal) \\) vanishes identically.",
      "solution": "First Solution. We first convert the differential inequality into an integral inequality. Suppose \\( varreal \\geq 0 \\). Using (i), (ii), and the continuity of \\( anyfunc^{\\prime} \\), we have\n\\[\n|anyfunc(varreal)|=\\left|\\int_{0}^{varreal} anyfunc^{\\prime}(vartemp) d vartemp\\right| \\leq \\int_{0}^{varreal}\\left|anyfunc^{\\prime}(vartemp)\\right| d vartemp \\leq \\int_{0}^{varreal}|anyfunc(vartemp)| d vartemp .\n\\]\n\nThus\n\\[\n|anyfunc(varreal)| \\leq \\int_{0}^{varreal}|anyfunc(vartemp)| d vartemp, \\quad \\text { for } varreal \\geq 0\n\\]\n\nLet \\( pointa \\geq 0 \\) be chosen arbitrarily. Since \\( anyfunc \\) is differentiable, it is continuous and therefore bounded on any finite interval. So there is a number \\( boundk \\) such that\n\\[\n|anyfunc(varreal)| \\leq boundk, \\text { for } 0 \\leq varreal \\leq pointa .\n\\]\n\nIf \\( 0 \\leq vartemp \\leq varreal \\leq pointa \\), we have \\( |anyfunc(vartemp)| \\leq boundk \\), so (1) gives\n\\[\n|anyfunc(varreal)| \\leq \\int_{0}^{varreal} boundk d vartemp=boundk varreal, \\quad \\text { for } \\quad 0 \\leq varreal \\leq pointa .\n\\]\n\nNow if \\( 0 \\leq vartemp \\leq varreal \\leq pointa \\), we have \\( |anyfunc(vartemp)| \\leq boundk vartemp \\), and (1) gives\n\\[\n|anyfunc(varreal)| \\leq \\int_{0}^{varreal} boundk vartemp d vartemp=\\frac{1}{2} boundk varreal^{2}, \\quad \\text { for } \\quad 0 \\leq varreal \\leq pointa .\n\\]\n\nContinuing in this way we find\n\\[\n|anyfunc(varreal)| \\leq \\frac{1}{indexn!} boundk varreal^{indexn}, \\text { for } 0 \\leq varreal \\leq pointa\n\\]\nand all positive integers \\( indexn \\).\nTo prove (2) formally, we use mathematical induction. We have shown that (2) is true for \\( indexn=1 \\). Suppose it is true for \\( indexn=indexp \\). Then for \\( 0 \\leq vartemp \\leq  varreal \\leq pointa \\), we have \\( |anyfunc(vartemp)| \\leq boundk vartemp^{indexp} / indexp! \\). Using this in (1), we find\n\\[\n|anyfunc(varreal)| \\leq \\int_{0}^{varreal} \\frac{1}{indexp!} boundk vartemp^{indexp} d vartemp=\\frac{1}{(indexp+1)!} boundk varreal^{indexp+1}, \\text { for } 0 \\leq varreal \\leq pointa .\n\\]\n\nThus (2) is true for \\( indexn=indexp+1 \\). We conclude it is true for all \\( indexn \\).\nSetting \\( varreal=pointa \\) in (2) and letting \\( indexn \\rightarrow \\infty \\), we have\n\\[\n|anyfunc(pointa)| \\leq \\lim _{indexn \\rightarrow \\infty} \\frac{1}{indexn!} boundk pointa^{indexn}=0 .\n\\]\n\nTherefore \\( anyfunc(pointa)=0 \\). But \\( pointa \\) was arbitrary, so \\( anyfunc \\) vanishes on all of \\( [0, \\infty) \\).\nConsider the function \\( reflect \\) defined by \\( reflect(varreal)=anyfunc(-varreal) \\). Since \\( reflect \\) has a continuous derivative and satisfies (i) and (ii), \\( reflect \\) vanishes on \\( [0, \\infty ) \\), by what we have already proved. Therefore \\( anyfunc \\) vanishes on \\( (-\\infty, 0 ] \\) as well.\n\nBy a slight variation of this reasoning, one can prove directly inequalities of the form\n\\[\n|anyfunc(varreal)| \\leq \\frac{1}{indexn!} boundl|varreal|^{indexn} \\text { for }-pointa \\leq varreal \\leq pointa,\n\\]\nand obtain the proof for positive and negative arguments simultaneously.\nSecond Solution. Suppose that \\( \\boldsymbol{anyfunc}(\\boldsymbol{varreal}) \\) does not vanish identically. Then from the continuity of \\( anyfunc \\) it follows that there exist points \\( pointa \\) and \\( pointb \\) such that \\( anyfunc(pointa)=0, anyfunc(pointb) \\neq 0,|pointa-pointb|<1 \\), and \\( |anyfunc(pointb)| \\) is the maximum value of \\( |anyfunc(varreal)| \\) on the closed interval with endpoints \\( pointa \\) and \\( pointb \\). (For example, if \\( anyfunc \\) does not vanish on \\( [0, \\infty) \\), we can let \\( pointa \\) be the least non-negative halfinteger such that \\( anyfunc \\) does not vanish on \\( \\left[pointa, pointa+\\frac{1}{2}\\right] \\), and let \\( pointb \\) be the point at which \\( |anyfunc(varreal)| \\) attains its maximum for \\( pointa \\leq varreal \\leq pointa+\\frac{1}{2} \\).)\nFrom the mean value theorem it follows that there is a number \\( pointc \\) between \\( pointa \\) and \\( pointb \\) such that\n\\[\n\\left|anyfunc^{\\prime}(pointc)\\right|=\\left|\\frac{anyfunc(pointb)-anyfunc(pointa)}{pointb-pointa}\\right|=\\left|\\frac{anyfunc(pointb)}{pointb-pointa}\\right|>|anyfunc(pointb)| .\n\\]\n\nBut our choice of \\( pointb \\) shows that \\( |anyfunc(pointb)| \\geq|anyfunc(pointc)| \\), hence \\( \\left|anyfunc^{\\prime}(pointc)\\right|>|anyfunc(pointc)| \\), contrary to (ii). This contradiction proves that \\( anyfunc \\) vanishes identically.\nThis proof does not require the continuity of \\( anyfunc^{\\prime} \\).\nRemark. This is a standard result that arises in proving the uniqueness of the solutions of differential equations."
    },
    "descriptive_long_confusing": {
      "map": {
        "g": "pinecones",
        "h": "marshland",
        "x": "tortoises",
        "t": "skylights",
        "a": "paintbrush",
        "b": "copperwire",
        "c": "sandcastle",
        "n": "dragonfly",
        "p": "quarterback",
        "K": "blueberry",
        "L": "snowflake"
      },
      "question": "4. Let \\( pinecones(tortoises) \\) be a function that has a continuous first derivative \\( pinecones^{\\prime}(tortoises) \\) for all values of \\( tortoises \\). Suppose that the following conditions hold for every \\( tortoises \\) : (i) \\( pinecones(0)= 0 ;\\) (ii) \\( \\left|pinecones^{\\prime}(tortoises)\\right| \\leq|pinecones(tortoises)| \\). Prove that \\( pinecones(tortoises) \\) vanishes identically.",
      "solution": "First Solution. We first convert the differential inequality into an integral inequality. Suppose \\( tortoises \\geq 0 \\). Using (i), (ii), and the continuity of \\( pinecones^{\\prime} \\), we have\n\\[\n|pinecones(tortoises)|=\\left|\\int_{0}^{tortoises} pinecones^{\\prime}(skylights) \\, d skylights\\right| \\leq \\int_{0}^{tortoises}\\left|pinecones^{\\prime}(skylights)\\right| \\, d skylights \\leq \\int_{0}^{tortoises}|pinecones(skylights)| \\, d skylights .\n\\]\nThus\n\\[\n|pinecones(tortoises)| \\leq \\int_{0}^{tortoises}|pinecones(skylights)| \\, d skylights, \\quad \\text { for } tortoises \\geq 0\n\\]\nLet \\( paintbrush \\geq 0 \\) be chosen arbitrarily. Since \\( pinecones \\) is differentiable, it is continuous and therefore bounded on any finite interval. So there is a number \\( blueberry \\) such that\n\\[\n|pinecones(tortoises)| \\leq blueberry, \\text { for } 0 \\leq tortoises \\leq paintbrush .\n\\]\nIf \\( 0 \\leq skylights \\leq tortoises \\leq paintbrush \\), we have \\( |pinecones(skylights)| \\leq blueberry \\), so (1) gives\n\\[\n|pinecones(tortoises)| \\leq \\int_{0}^{tortoises} blueberry \\, d skylights=blueberry\\,tortoises, \\quad \\text { for } \\quad 0 \\leq tortoises \\leq paintbrush .\n\\]\nNow if \\( 0 \\leq skylights \\leq tortoises \\leq paintbrush \\), we have \\( |pinecones(skylights)| \\leq blueberry\\,skylights \\), and (1) gives\n\\[\n|pinecones(tortoises)| \\leq \\int_{0}^{tortoises} blueberry\\,skylights \\, d skylights=\\frac{1}{2} blueberry\\,tortoises^{2}, \\quad \\text { for } \\quad 0 \\leq tortoises \\leq paintbrush .\n\\]\nContinuing in this way we find\n\\[\n|pinecones(tortoises)| \\leq \\frac{1}{dragonfly!} \\, blueberry\\,tortoises^{dragonfly}, \\text { for } 0 \\leq tortoises \\leq paintbrush\n\\]\nand all positive integers \\( dragonfly \\).\nTo prove (2) formally, we use mathematical induction. We have shown that (2) is true for \\( dragonfly=1 \\). Suppose it is true for \\( dragonfly=quarterback \\). Then for \\( 0 \\leq skylights \\leq tortoises \\leq paintbrush \\), we have \\( |pinecones(skylights)| \\leq blueberry\\,skylights^{quarterback} / quarterback! \\). Using this in (1), we find\n\\[\n|pinecones(tortoises)| \\leq \\int_{0}^{tortoises} \\frac{1}{quarterback!} \\, blueberry\\,skylights^{quarterback} \\, d skylights=\\frac{1}{(quarterback+1)!} \\, blueberry\\,tortoises^{quarterback+1}, \\text { for } 0 \\leq tortoises \\leq paintbrush .\n\\]\nThus (2) is true for \\( dragonfly=quarterback+1 \\). We conclude it is true for all \\( dragonfly \\).\nSetting \\( tortoises=paintbrush \\) in (2) and letting \\( dragonfly \\rightarrow \\infty \\), we have\n\\[\n|pinecones(paintbrush)| \\leq \\lim _{dragonfly \\rightarrow \\infty} \\frac{1}{dragonfly!} \\, blueberry\\,paintbrush^{dragonfly}=0 .\n\\]\nTherefore \\( pinecones(paintbrush)=0 \\). But \\( paintbrush \\) was arbitrary, so \\( pinecones \\) vanishes on all of \\( [0, \\infty) \\).\nConsider the function \\( marshland \\) defined by \\( marshland(tortoises)=pinecones(-tortoises) \\). Since \\( marshland \\) has a continuous derivative and satisfies (i) and (ii), \\( marshland \\) vanishes on \\( [0, \\infty) \\), by what we have already proved. Therefore \\( pinecones \\) vanishes on \\( (-\\infty, 0 ] \\) as well.\n\nBy a slight variation of this reasoning, one can prove directly inequalities of the form\n\\[\n|pinecones(tortoises)| \\leq \\frac{1}{dragonfly!} \\, snowflake|tortoises|^{dragonfly} \\text { for }-paintbrush \\leq tortoises \\leq paintbrush,\n\\]\nand obtain the proof for positive and negative arguments simultaneously.\n\nSecond Solution. Suppose that \\( \\boldsymbol{pinecones}(\\boldsymbol{tortoises}) \\) does not vanish identically. Then from the continuity of \\( pinecones \\) it follows that there exist points \\( paintbrush \\) and \\( copperwire \\) such that \\( pinecones(paintbrush)=0,\\; pinecones(copperwire) \\neq 0,\\; |paintbrush-copperwire|<1 \\), and \\( |pinecones(copperwire)| \\) is the maximum value of \\( |pinecones(tortoises)| \\) on the closed interval with endpoints \\( paintbrush \\) and \\( copperwire \\). (For example, if \\( pinecones \\) does not vanish on \\( [0, \\infty) \\), we can let \\( paintbrush \\) be the least non-negative half-integer such that \\( pinecones \\) does not vanish on \\( [paintbrush, paintbrush+\\tfrac{1}{2}] \\), and let \\( copperwire \\) be the point at which \\( |pinecones(tortoises)| \\) attains its maximum for \\( paintbrush \\leq tortoises \\leq paintbrush+\\tfrac{1}{2} \\).)\nFrom the mean value theorem it follows that there is a number \\( sandcastle \\) between \\( paintbrush \\) and \\( copperwire \\) such that\n\\[\n\\left|pinecones^{\\prime}(sandcastle)\\right|=\\left|\\frac{pinecones(copperwire)-pinecones(paintbrush)}{copperwire-paintbrush}\\right|=\\left|\\frac{pinecones(copperwire)}{copperwire-paintbrush}\\right|>|pinecones(copperwire)| .\n\\]\nBut our choice of \\( copperwire \\) shows that \\( |pinecones(copperwire)| \\geq|pinecones(sandcastle)| \\), hence \\( \\left|pinecones^{\\prime}(sandcastle)\\right|>|pinecones(sandcastle)| \\), contrary to (ii). This contradiction proves that \\( pinecones \\) vanishes identically.\nThis proof does not require the continuity of \\( pinecones^{\\prime} \\).\nRemark. This is a standard result that arises in proving the uniqueness of the solutions of differential equations."
    },
    "descriptive_long_misleading": {
      "map": {
        "g": "nonvarying",
        "h": "inertvalue",
        "x": "constant",
        "t": "fixedtime",
        "a": "limitless",
        "b": "boundless",
        "c": "restricted",
        "n": "continuum",
        "p": "fluidity",
        "K": "unbounded",
        "L": "variable"
      },
      "question": "4. Let \\( nonvarying(constant) \\) be a function that has a continuous first derivative \\( nonvarying^{\\prime}(constant) \\) for all values of \\( constant \\). Suppose that the following conditions hold for every \\( constant \\) : (i) \\( nonvarying(0)=0 ;\\) (ii) \\( \\left|nonvarying^{\\prime}(constant)\\right| \\leq|nonvarying(constant)| \\). Prove that \\( nonvarying(constant) \\) vanishes identically.",
      "solution": "First Solution. We first convert the differential inequality into an integral inequality. Suppose \\( constant \\geq 0 \\). Using (i), (ii), and the continuity of \\( nonvarying^{\\prime} \\), we have\n\\[\n|nonvarying(constant)|=\\left|\\int_{0}^{constant} nonvarying^{\\prime}(fixedtime)\\, d fixedtime\\right| \\leq \\int_{0}^{constant}\\left|nonvarying^{\\prime}(fixedtime)\\right|\\, d fixedtime \\leq \\int_{0}^{constant}|nonvarying(fixedtime)|\\, d fixedtime .\n\\]\n\nThus\n\\[\n|nonvarying(constant)| \\leq \\int_{0}^{constant}|nonvarying(fixedtime)|\\, d fixedtime, \\quad \\text { for } constant \\geq 0\n\\]\n\nLet \\( limitless \\geq 0 \\) be chosen arbitrarily. Since \\( nonvarying \\) is differentiable, it is continuous and therefore bounded on any finite interval. So there is a number \\( unbounded \\) such that\n\\[\n|nonvarying(constant)| \\leq unbounded, \\text { for } 0 \\leq constant \\leq limitless .\n\\]\n\nIf \\( 0 \\leq fixedtime \\leq constant \\leq limitless \\), we have \\( |nonvarying(fixedtime)| \\leq unbounded \\), so (1) gives\n\\[\n|nonvarying(constant)| \\leq \\int_{0}^{constant} unbounded\\, d fixedtime = unbounded\\, constant, \\quad \\text { for } \\quad 0 \\leq constant \\leq limitless .\n\\]\n\nNow if \\( 0 \\leq fixedtime \\leq constant \\leq limitless \\), we have \\( |nonvarying(fixedtime)| \\leq unbounded\\, fixedtime \\), and (1) gives\n\\[\n|nonvarying(constant)| \\leq \\int_{0}^{constant} unbounded\\, fixedtime\\, d fixedtime =\\frac{1}{2}\\, unbounded\\, constant^{2}, \\quad \\text { for } \\quad 0 \\leq constant \\leq limitless .\n\\]\n\nContinuing in this way we find\n\\[\n|nonvarying(constant)| \\leq \\frac{1}{continuum!}\\, unbounded\\, constant^{continuum}, \\text { for } 0 \\leq constant \\leq limitless\n\\]\nand all positive integers \\( continuum \\).\n\nTo prove (2) formally, we use mathematical induction. We have shown that (2) is true for \\( continuum=1 \\). Suppose it is true for \\( continuum=fluidity \\). Then for \\( 0 \\leq fixedtime \\leq constant \\leq limitless \\), we have \\( |nonvarying(fixedtime)| \\leq unbounded\\, fixedtime^{fluidity} / fluidity! \\). Using this in (1), we find\n\\[\n|nonvarying(constant)| \\leq \\int_{0}^{constant} \\frac{1}{fluidity!}\\, unbounded\\, fixedtime^{fluidity}\\, d fixedtime = \\frac{1}{(fluidity+1)!}\\, unbounded\\, constant^{fluidity+1}, \\text { for } 0 \\leq constant \\leq limitless .\n\\]\n\nThus (2) is true for \\( continuum=fluidity+1 \\). We conclude it is true for all \\( continuum \\).\n\nSetting \\( constant=limitless \\) in (2) and letting \\( continuum \\rightarrow \\infty \\), we have\n\\[\n|nonvarying(limitless)| \\leq \\lim _{continuum \\rightarrow \\infty} \\frac{1}{continuum!}\\, unbounded\\, limitless^{continuum}=0 .\n\\]\n\nTherefore \\( nonvarying(limitless)=0 \\). But \\( limitless \\) was arbitrary, so \\( nonvarying \\) vanishes on all of \\( [0, \\infty) \\).\n\nConsider the function \\( inertvalue \\) defined by \\( inertvalue(constant)=nonvarying(-constant) \\). Since \\( inertvalue \\) has a continuous derivative and satisfies (i) and (ii), \\( inertvalue \\) vanishes on \\( [0, \\infty) \\), by what we have already proved. Therefore \\( nonvarying \\) vanishes on \\( (-\\infty, 0] \\) as well.\n\nBy a slight variation of this reasoning, one can prove directly inequalities of the form\n\\[\n|nonvarying(constant)| \\leq \\frac{1}{continuum!}\\, variable\\, |constant|^{continuum} \\text { for } -limitless \\leq constant \\leq limitless,\n\\]\nand obtain the proof for positive and negative arguments simultaneously.\n\nSecond Solution. Suppose that \\( \\boldsymbol{nonvarying}(\\boldsymbol{constant}) \\) does not vanish identically. Then from the continuity of \\( nonvarying \\) it follows that there exist points \\( limitless \\) and \\( boundless \\) such that \\( nonvarying(limitless)=0,\\, nonvarying(boundless) \\neq 0,\\, |limitless-boundless|<1 \\), and \\( |nonvarying(boundless)| \\) is the maximum value of \\( |nonvarying(constant)| \\) on the closed interval with endpoints \\( limitless \\) and \\( boundless \\). (For example, if \\( nonvarying \\) does not vanish on \\( [0, \\infty) \\), we can let \\( limitless \\) be the least non-negative half-integer such that \\( nonvarying \\) does not vanish on \\( \\left[limitless,\\, limitless+\\frac{1}{2}\\right] \\), and let \\( boundless \\) be the point at which \\( |nonvarying(constant)| \\) attains its maximum for \\( limitless \\leq constant \\leq limitless+\\frac{1}{2} \\).)\n\nFrom the mean value theorem it follows that there is a number \\( restricted \\) between \\( limitless \\) and \\( boundless \\) such that\n\\[\n\\left|nonvarying^{\\prime}(restricted)\\right|=\\left|\\frac{nonvarying(boundless)-nonvarying(limitless)}{boundless-limitless}\\right|=\\left|\\frac{nonvarying(boundless)}{boundless-limitless}\\right|>|nonvarying(boundless)| .\n\\]\n\nBut our choice of \\( boundless \\) shows that \\( |nonvarying(boundless)| \\geq |nonvarying(restricted)| \\), hence \\( \\left|nonvarying^{\\prime}(restricted)\\right| > |nonvarying(restricted)| \\), contrary to (ii). This contradiction proves that \\( nonvarying \\) vanishes identically.\n\nThis proof does not require the continuity of \\( nonvarying^{\\prime} \\).\n\nRemark. This is a standard result that arises in proving the uniqueness of the solutions of differential equations."
    },
    "garbled_string": {
      "map": {
        "g": "qzxwvtnp",
        "h": "hjgrksla",
        "x": "lsfdkjwe",
        "t": "prxvence",
        "a": "vbgncmlq",
        "b": "wrethspo",
        "c": "mxnvjkal",
        "n": "zpqoridv",
        "p": "ugtiwcsr",
        "K": "dfghjklo",
        "L": "cvmnxzas"
      },
      "question": "4. Let \\( qzxwvtnp(lsfdkjwe) \\) be a function that has a continuous first derivative \\( qzxwvtnp^{\\prime}(lsfdkjwe) \\) for all values of \\( lsfdkjwe \\). Suppose that the following conditions hold for every \\( lsfdkjwe \\) : (i) \\( qzxwvtnp(0)= \\) 0 ; (ii) \\( \\left|qzxwvtnp^{\\prime}(lsfdkjwe)\\right| \\leq |qzxwvtnp(lsfdkjwe)| \\). Prove that \\( qzxwvtnp(lsfdkjwe) \\) vanishes identically.",
      "solution": "First Solution. We first convert the differential inequality into an integral inequality. Suppose \\( lsfdkjwe \\geq 0 \\). Using (i), (ii), and the continuity of \\( qzxwvtnp^{\\prime} \\), we have\n\\[\n|qzxwvtnp(lsfdkjwe)|=\\left|\\int_{0}^{lsfdkjwe} qzxwvtnp^{\\prime}(prxvence) \\, dprxvence\\right| \\leq \\int_{0}^{lsfdkjwe}\\left|qzxwvtnp^{\\prime}(prxvence)\\right| \\, dprxvence \\leq \\int_{0}^{lsfdkjwe}|qzxwvtnp(prxvence)| \\, dprxvence .\n\\]\nThus\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\int_{0}^{lsfdkjwe}|qzxwvtnp(prxvence)| \\, dprxvence, \\quad \\text{for } lsfdkjwe \\geq 0. \\tag{1}\n\\]\nLet \\( vbgncmlq \\geq 0 \\) be chosen arbitrarily. Since \\( qzxwvtnp \\) is differentiable, it is continuous and therefore bounded on any finite interval. So there is a number \\( dfghjklo \\) such that\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq dfghjklo, \\qquad 0 \\leq lsfdkjwe \\leq vbgncmlq .\n\\]\nIf \\( 0 \\leq prxvence \\leq lsfdkjwe \\leq vbgncmlq \\), we have \\( |qzxwvtnp(prxvence)| \\leq dfghjklo \\); using (1) gives\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\int_{0}^{lsfdkjwe} dfghjklo \\, dprxvence = dfghjklo\\, lsfdkjwe, \\qquad 0 \\leq lsfdkjwe \\leq vbgncmlq .\n\\]\nNow if \\( 0 \\leq prxvence \\leq lsfdkjwe \\leq vbgncmlq \\), we have \\( |qzxwvtnp(prxvence)| \\leq dfghjklo\\, prxvence \\), and (1) gives\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\int_{0}^{lsfdkjwe} dfghjklo\\, prxvence \\, dprxvence = \\frac{1}{2}dfghjklo\\, lsfdkjwe^{2}, \\qquad 0 \\leq lsfdkjwe \\leq vbgncmlq .\n\\]\nContinuing in this way we find\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\frac{1}{zpqoridv!}dfghjklo\\, lsfdkjwe^{zpqoridv}, \\qquad 0 \\leq lsfdkjwe \\leq vbgncmlq, \\tag{2}\n\\]\nfor all positive integers \\( zpqoridv \\).\n\nTo prove (2) formally, we use mathematical induction. We have shown that (2) is true for \\( zpqoridv=1 \\). Suppose it is true for \\( zpqoridv=ugtiwcsr \\). Then for \\( 0 \\leq prxvence \\leq lsfdkjwe \\leq vbgncmlq \\) we have \\( |qzxwvtnp(prxvence)| \\leq dfghjklo\\, prxvence^{ugtiwcsr}/ugtiwcsr! \\). Using this in (1), we find\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\int_{0}^{lsfdkjwe} \\frac{dfghjklo}{ugtiwcsr!} prxvence^{ugtiwcsr} \\, dprxvence = \\frac{dfghjklo}{(ugtiwcsr+1)!} lsfdkjwe^{ugtiwcsr+1}, \\qquad 0 \\leq lsfdkjwe \\leq vbgncmlq .\n\\]\nThus (2) holds for \\( zpqoridv=ugtiwcsr+1 \\), and hence for all \\( zpqoridv \\).\nSetting \\( lsfdkjwe = vbgncmlq \\) in (2) and letting \\( zpqoridv \\to \\infty \\), we obtain\n\\[\n|qzxwvtnp(vbgncmlq)| \\leq \\lim_{zpqoridv\\to\\infty} \\frac{dfghjklo\\, vbgncmlq^{zpqoridv}}{zpqoridv!}=0.\n\\]\nTherefore \\( qzxwvtnp(vbgncmlq)=0 \\). Since \\( vbgncmlq \\) was arbitrary, \\( qzxwvtnp \\) vanishes on all of \\([0,\\infty)\\).\n\nConsider the function \\( hjgrksla \\) defined by \\( hjgrksla(lsfdkjwe)=qzxwvtnp(-lsfdkjwe) \\). Because \\( hjgrksla \\) has a continuous derivative and satisfies (i) and (ii), \\( hjgrksla \\) vanishes on \\([0,\\infty)\\). Hence \\( qzxwvtnp \\) vanishes on \\(( -\\infty,0] \\) as well.\n\nBy a slight variation of this reasoning one can prove directly inequalities of the form\n\\[\n|qzxwvtnp(lsfdkjwe)| \\leq \\frac{1}{zpqoridv!} cvmnxzas |lsfdkjwe|^{zpqoridv}, \\qquad -vbgncmlq \\leq lsfdkjwe \\leq vbgncmlq,\n\\]\nand obtain the proof for positive and negative arguments simultaneously.\n\nSecond Solution. Suppose that \\( \\boldsymbol{qzxwvtnp}(\\boldsymbol{lsfdkjwe}) \\) does not vanish identically. By continuity there exist points \\( vbgncmlq \\) and \\( wrethspo \\) such that \\( qzxwvtnp(vbgncmlq)=0, \\; qzxwvtnp(wrethspo)\\neq0, \\; |vbgncmlq-wrethspo|<1, \\) and \\( |qzxwvtnp(wrethspo)| \\) is the maximum of \\(|qzxwvtnp(lsfdkjwe)|\\) on the closed interval with endpoints \\( vbgncmlq \\) and \\( wrethspo \\). (For instance, if \\( qzxwvtnp \\) does not vanish on \\([0,\\infty)\\), choose \\( vbgncmlq \\) to be the least non-negative half-integer such that \\( qzxwvtnp \\) does not vanish on \\([vbgncmlq, vbgncmlq+\\tfrac12]\\), and let \\( wrethspo \\) be the point where \\(|qzxwvtnp(lsfdkjwe)|\\) attains its maximum on this interval.)\n\nBy the mean-value theorem there exists \\( mxnvjkal \\) between \\( vbgncmlq \\) and \\( wrethspo \\) such that\n\\[\n|qzxwvtnp^{\\prime}(mxnvjkal)| = \\left|\\frac{qzxwvtnp(wrethspo)-qzxwvtnp(vbgncmlq)}{wrethspo-vbgncmlq}\\right| = \\left|\\frac{qzxwvtnp(wrethspo)}{wrethspo-vbgncmlq}\\right| > |qzxwvtnp(wrethspo)|.\n\\]\nBecause \\( |qzxwvtnp(wrethspo)| \\ge |qzxwvtnp(mxnvjkal)| \\), we obtain \\( |qzxwvtnp^{\\prime}(mxnvjkal)| > |qzxwvtnp(mxnvjkal)| \\), contradicting (ii). Hence \\( qzxwvtnp \\) must vanish identically.\n\nThis proof does not require the continuity of \\( qzxwvtnp^{\\prime} \\).\n\nRemark. This is a standard result that arises in proving the uniqueness of solutions of differential equations."
    },
    "kernel_variant": {
      "question": "Let H be a (possibly infinite–dimensional) real separable Hilbert space with inner product ⟨·,·⟩ and induced norm ‖·‖.  \n\nLet  \n\n  g : H → H  \n\nbe Fréchet–differentiable at every point and assume that its derivative  \n\n  Dg : H → B(H), x ↦ Dg(x)  \n\nis continuous, where B(H) denotes the Banach space of bounded linear operators on H endowed with the operator norm ‖·‖op.  \nSuppose  \n\n(1) g(0)=0,  \n\n(2) ‖Dg(x)‖op ≤ ‖g(x)‖ for every x ∈ H.  \n\nProve that g vanishes identically on H; that is, g(x)=0 for every x∈H.",
      "solution": "Throughout the proof we repeatedly use the following basic facts.\n\n• If γ:[0,1]→H is C¹, then by the fundamental theorem of calculus for Bochner-valued functions  \n  γ(1)−γ(0)=∫₀¹γ′(t) dt  (Bochner integral).  \n• The norm ‖·‖ on a Hilbert space is 1-Lipschitz and therefore absolutely continuous along absolutely continuous curves; in particular, if u:[0,1]→H is absolutely continuous, then t↦‖u(t)‖ is also absolutely continuous and differentiable almost everywhere (a.e.).\n\nStep 1.  Integral representation of g(x).  \nFix x∈H and consider the C¹ line γ(t)=tx (0≤t≤1).  By the chain rule\n\n  d/dt g(γ(t)) = Dg(tx)[x] for every t∈[0,1].\n\nUsing g(0)=0 and applying the Bochner fundamental theorem of calculus we obtain\n\n  g(x)=∫₀¹Dg(tx)[x] dt.  (3)\n\nStep 2.  An integral inequality for ‖g(x)‖.  \nTaking norms in (3) and using (2) gives\n\n  ‖g(x)‖ ≤ ∫₀¹‖Dg(tx)‖op‖x‖ dt ≤ ‖x‖∫₀¹‖g(tx)‖ dt.  (4)\n\nFor the fixed x set  \n\n  f(t):=‖g(tx)‖ (0≤t≤1).\n\nThen (4) rewrites as\n\n  f(1) ≤ ‖x‖∫₀¹f(t) dt.  (5)\n\nStep 3.  Absolute continuity of f and a differential inequality.  \nBecause t↦g(tx) is C¹, the mapping f(t)=‖g(tx)‖ is the composition of a C¹ map with the Lipschitz norm. Hence f is absolutely continuous on [0,1] and therefore differentiable a.e.\n\nFor a.e. t with g(tx)≠0 we have by the chain rule for a norm,\n\n  f ′(t)=⟨Dg(tx)[x], g(tx)/‖g(tx)‖⟩.\n\nIf g(tx)=0 we simply keep the convention f′(t)=0; this causes no harm because such t belong to a set on which f vanishes identically and hence f is flat there.  Estimate (2) then yields, for a.e. t∈[0,1],\n\n  |f ′(t)| ≤ ‖Dg(tx)‖op‖x‖ ≤ ‖g(tx)‖‖x‖ = f(t)‖x‖.  (6)\n\nCombining with sign information we may write\n\n  f ′(t) ≤ ‖x‖f(t)  a.e. on [0,1],   f(0)=0.  (7)\n\nStep 4.  Grönwall-type argument.  \nDefine h(t):=e^{−‖x‖t}f(t).  Equation (7) implies, for a.e. t,\n\n  h ′(t)=e^{−‖x‖t}(f ′(t)−‖x‖f(t)) ≤ 0.  (8)\n\nThus h is absolutely continuous and non-increasing.  Since h(0)=f(0)=0 and h≥0, we have h(t)=0 for every t∈[0,1].  Consequently f(t)=0 for all t, and in particular f(1)=‖g(x)‖=0.\n\nStep 5.  Conclusion.  \nBecause the choice of x∈H was arbitrary, we conclude that ‖g(x)‖=0 for every x∈H; hence g(x)=0 for every x∈H, as required.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.403238",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher‐dimensional / infinite‐dimensional setting  \n   • The problem is posed on an arbitrary separable Hilbert space, not on ℝ; the derivative is the full Fréchet derivative, a bounded linear operator, and norms are operator norms.  \n2. Additional layers of structure  \n   • The solution must manipulate operator norms, Fréchet differentiability, and path integrals in Banach spaces, none of which appear in the original problem.  \n3. Advanced techniques required  \n   • One has to convert the operator inequality into a scalar inequality along arbitrary rays in H, justify differentiability of composed maps, integrate operator‐valued functions, and employ the Grönwall lemma in an abstract setting.  \n4. Non-elementary tools  \n   • Knowledge of the Fundamental Theorem of Calculus in Banach spaces, continuity of the Fréchet derivative, and the differential form of Grönwall’s lemma are indispensable.  \n5. More steps and subtler reasoning  \n   • The argument now involves: (a) rewriting g via an integral of its Fréchet derivative, (b) turning an operator inequality into a scalar differential inequality, (c) transforming that inequality with an exponential integrating factor, and (d) concluding with Grönwall.  \nThese additions make the enhanced variant significantly tougher than the original single-variable, first-derivative inequality."
      }
    },
    "original_kernel_variant": {
      "question": "Let H be a (possibly infinite–dimensional) real separable Hilbert space with inner product ⟨·,·⟩ and induced norm ‖·‖.  \n\nLet  \n\n  g : H → H  \n\nbe Fréchet–differentiable at every point and assume that its derivative  \n\n  Dg : H → B(H), x ↦ Dg(x)  \n\nis continuous, where B(H) denotes the Banach space of bounded linear operators on H endowed with the operator norm ‖·‖op.  \nSuppose  \n\n(1) g(0)=0,  \n\n(2) ‖Dg(x)‖op ≤ ‖g(x)‖ for every x ∈ H.  \n\nProve that g vanishes identically on H; that is, g(x)=0 for every x∈H.",
      "solution": "Throughout the proof we repeatedly use the following basic facts.\n\n• If γ:[0,1]→H is C¹, then by the fundamental theorem of calculus for Bochner-valued functions  \n  γ(1)−γ(0)=∫₀¹γ′(t) dt  (Bochner integral).  \n• The norm ‖·‖ on a Hilbert space is 1-Lipschitz and therefore absolutely continuous along absolutely continuous curves; in particular, if u:[0,1]→H is absolutely continuous, then t↦‖u(t)‖ is also absolutely continuous and differentiable almost everywhere (a.e.).\n\nStep 1.  Integral representation of g(x).  \nFix x∈H and consider the C¹ line γ(t)=tx (0≤t≤1).  By the chain rule\n\n  d/dt g(γ(t)) = Dg(tx)[x] for every t∈[0,1].\n\nUsing g(0)=0 and applying the Bochner fundamental theorem of calculus we obtain\n\n  g(x)=∫₀¹Dg(tx)[x] dt.  (3)\n\nStep 2.  An integral inequality for ‖g(x)‖.  \nTaking norms in (3) and using (2) gives\n\n  ‖g(x)‖ ≤ ∫₀¹‖Dg(tx)‖op‖x‖ dt ≤ ‖x‖∫₀¹‖g(tx)‖ dt.  (4)\n\nFor the fixed x set  \n\n  f(t):=‖g(tx)‖ (0≤t≤1).\n\nThen (4) rewrites as\n\n  f(1) ≤ ‖x‖∫₀¹f(t) dt.  (5)\n\nStep 3.  Absolute continuity of f and a differential inequality.  \nBecause t↦g(tx) is C¹, the mapping f(t)=‖g(tx)‖ is the composition of a C¹ map with the Lipschitz norm. Hence f is absolutely continuous on [0,1] and therefore differentiable a.e.\n\nFor a.e. t with g(tx)≠0 we have by the chain rule for a norm,\n\n  f ′(t)=⟨Dg(tx)[x], g(tx)/‖g(tx)‖⟩.\n\nIf g(tx)=0 we simply keep the convention f′(t)=0; this causes no harm because such t belong to a set on which f vanishes identically and hence f is flat there.  Estimate (2) then yields, for a.e. t∈[0,1],\n\n  |f ′(t)| ≤ ‖Dg(tx)‖op‖x‖ ≤ ‖g(tx)‖‖x‖ = f(t)‖x‖.  (6)\n\nCombining with sign information we may write\n\n  f ′(t) ≤ ‖x‖f(t)  a.e. on [0,1],   f(0)=0.  (7)\n\nStep 4.  Grönwall-type argument.  \nDefine h(t):=e^{−‖x‖t}f(t).  Equation (7) implies, for a.e. t,\n\n  h ′(t)=e^{−‖x‖t}(f ′(t)−‖x‖f(t)) ≤ 0.  (8)\n\nThus h is absolutely continuous and non-increasing.  Since h(0)=f(0)=0 and h≥0, we have h(t)=0 for every t∈[0,1].  Consequently f(t)=0 for all t, and in particular f(1)=‖g(x)‖=0.\n\nStep 5.  Conclusion.  \nBecause the choice of x∈H was arbitrary, we conclude that ‖g(x)‖=0 for every x∈H; hence g(x)=0 for every x∈H, as required.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.344530",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher‐dimensional / infinite‐dimensional setting  \n   • The problem is posed on an arbitrary separable Hilbert space, not on ℝ; the derivative is the full Fréchet derivative, a bounded linear operator, and norms are operator norms.  \n2. Additional layers of structure  \n   • The solution must manipulate operator norms, Fréchet differentiability, and path integrals in Banach spaces, none of which appear in the original problem.  \n3. Advanced techniques required  \n   • One has to convert the operator inequality into a scalar inequality along arbitrary rays in H, justify differentiability of composed maps, integrate operator‐valued functions, and employ the Grönwall lemma in an abstract setting.  \n4. Non-elementary tools  \n   • Knowledge of the Fundamental Theorem of Calculus in Banach spaces, continuity of the Fréchet derivative, and the differential form of Grönwall’s lemma are indispensable.  \n5. More steps and subtler reasoning  \n   • The argument now involves: (a) rewriting g via an integral of its Fréchet derivative, (b) turning an operator inequality into a scalar differential inequality, (c) transforming that inequality with an exponential integrating factor, and (d) concluding with Grönwall.  \nThese additions make the enhanced variant significantly tougher than the original single-variable, first-derivative inequality."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}