{
  "index": "1962-A-4",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "4. Assume that \\( |f(x)| \\leq 1 \\) and \\( \\left|f^{\\prime \\prime}(x)\\right| \\leq 1 \\) for all \\( x \\) on an interval of length at least 2 . Show that \\( \\left|f^{\\prime}(x)\\right| \\leq 2 \\) on the interval.",
  "solution": "First Solution. We may suppose without loss of generality that the interval in question is \\( [-1,+1] \\). Using Taylor's formula to expand about the point \\( x \\in[-1,1] \\), we find\n\\[\n\\begin{array}{c}\nf(1)=f(x)+(1-x) f^{\\prime}(x)+\\frac{1}{2}(1-x)^{2} f^{\\prime \\prime}(\\xi) \\\\\nf(-1)=f(x)+(-1-x) f^{\\prime}(x)+\\frac{1}{2}(-1-x)^{2} f^{\\prime \\prime}(\\eta)\n\\end{array}\n\\]\nwhere \\( \\xi \\in(x, 1) \\) and \\( \\eta \\in(-1, x) \\). Hence\n\\[\nf(1)-f(-1)=2 f^{\\prime}(x)+\\frac{1}{2}(1-x)^{2} f^{\\prime \\prime}(\\xi)-\\frac{1}{2}(1+x)^{2} f^{\\prime \\prime}(\\eta)\n\\]\n\nUsing the given bounds for \\( f \\) and \\( f^{\\prime \\prime} \\), we get\n\\[\n\\begin{aligned}\n2\\left|f^{\\prime}(x)\\right| & \\leq|f(1)|+|f(-1)|+\\frac{1}{2}(1-x)^{2}\\left|f^{\\prime \\prime}(\\xi)\\right|+\\frac{1}{2}(1+x)^{2}\\left|f^{\\prime \\prime}(\\eta)\\right| \\\\\n& \\leq 2+\\frac{1}{2}(1-x)^{2}+\\frac{1}{2}(1+x)^{2}=3+x^{2} \\leq 4\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\left|f^{\\prime}(x)\\right| \\leq 2\n\\]\n\nSecond Solution. Essentially the same argument, but less elegantly phrased, is the following.\n\nSuppose there is an \\( x \\) in \\( [-1,1] \\) such that\n\\[\nf^{\\prime}(x)=2+\\epsilon, \\quad \\text { where } \\epsilon>0\n\\]\n\nSince \\( \\left|f^{\\prime \\prime}(t)\\right| \\leq 1 \\) for all \\( t \\in[-1,1] \\), we have\n\\[\n\\begin{aligned}\nf^{\\prime}(t) & \\geq 2+\\epsilon+t-x \\quad \\text { for }-1 \\leq t \\leq x \\\\\n& \\geq 2+\\epsilon-t+x \\text { for } x \\leq t \\leq 1 .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nf(1)-f(-1) & =\\int_{-1}^{+1} f^{\\prime}(t) d t \\\\\n& \\geq 4+2 \\epsilon-\\frac{1}{2}(1+x)^{2}-\\frac{1}{2}(1-x)^{2}=3+2 \\epsilon-x^{2} \\\\\n& \\geq 2+2 \\epsilon\n\\end{aligned}\n\\]\n\nBut \\( |f(1)-f(-1)| \\leq|f(1)|+|f(-1)| \\leq 2 \\). Hence there can be no such \\( x \\).\n\nSimilarly we cannot have \\( f^{\\prime}(x)<-2 \\).\nRemarks. The inequality is the best possible, for, if \\( f(x)=\\frac{1}{2}(x+1)^{2} \\) -1 , the hypothesis is satisfied and \\( f^{\\prime}(1)=2 \\). The first solution shows that \\( \\left|f^{\\prime}\\right| \\) can attain the value 2 only for \\( x= \\pm 1 \\), and then it follows easily that the function just described and three others obtained from it by reflection are the only extremal functions.\n\nThe result was first established by Landau in \"Einige Ungleichungen fur zweimal differentzierbaren Funktionen,\" Proceedings of the London Mathematical Society (2), vol. 13 (1914), pages 43-49. For a treatment of several similar inequalities, see I. J. Schoenberg, \"The Elementary Cases of Landau's Problem of Inequalities between Derivatives,\" American Mathematical Monthly, vol. 80 (1973), pages 121-158.",
  "vars": [
    "x",
    "t",
    "\\\\xi",
    "\\\\eta"
  ],
  "params": [
    "f",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "t": "auxiliary",
        "\\xi": "greekxi",
        "\\eta": "greeketa",
        "f": "functionf",
        "\\epsilon": "smalleps"
      },
      "question": "Assume that \\( |functionf(variablex)| \\leq 1 \\) and \\( \\left|functionf^{\\prime \\prime}(variablex)\\right| \\leq 1 \\) for all \\( variablex \\) on an interval of length at least 2 . Show that \\( \\left|functionf^{\\prime}(variablex)\\right| \\leq 2 \\) on the interval.",
      "solution": "First Solution. We may suppose without loss of generality that the interval in question is \\( [-1,+1] \\). Using Taylor's formula to expand about the point \\( variablex \\in[-1,1] \\), we find\n\\[\n\\begin{array}{c}\nfunctionf(1)=functionf(variablex)+(1-variablex) functionf^{\\prime}(variablex)+\\frac{1}{2}(1-variablex)^{2} functionf^{\\prime \\prime}(greekxi) \\\\\nfunctionf(-1)=functionf(variablex)+(-1-variablex) functionf^{\\prime}(variablex)+\\frac{1}{2}(-1-variablex)^{2} functionf^{\\prime \\prime}(greeketa)\n\\end{array}\n\\]\nwhere \\( greekxi \\in(variablex, 1) \\) and \\( greeketa \\in(-1, variablex) \\). Hence\n\\[\nfunctionf(1)-functionf(-1)=2 functionf^{\\prime}(variablex)+\\frac{1}{2}(1-variablex)^{2} functionf^{\\prime \\prime}(greekxi)-\\frac{1}{2}(1+variablex)^{2} functionf^{\\prime \\prime}(greeketa)\n\\]\n\nUsing the given bounds for \\( functionf \\) and \\( functionf^{\\prime \\prime} \\), we get\n\\[\n\\begin{aligned}\n2\\left|functionf^{\\prime}(variablex)\\right| & \\leq|functionf(1)|+|functionf(-1)|+\\frac{1}{2}(1-variablex)^{2}\\left|functionf^{\\prime \\prime}(greekxi)\\right|+\\frac{1}{2}(1+variablex)^{2}\\left|functionf^{\\prime \\prime}(greeketa)\\right| \\\\\n& \\leq 2+\\frac{1}{2}(1-variablex)^{2}+\\frac{1}{2}(1+variablex)^{2}=3+variablex^{2} \\leq 4\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\left|functionf^{\\prime}(variablex)\\right| \\leq 2\n\\]\n\nSecond Solution. Essentially the same argument, but less elegantly phrased, is the following.\n\nSuppose there is an \\( variablex \\) in \\( [-1,1] \\) such that\n\\[\nfunctionf^{\\prime}(variablex)=2+smalleps, \\quad \\text { where } smalleps>0\n\\]\n\nSince \\( \\left|functionf^{\\prime \\prime}(auxiliary)\\right| \\leq 1 \\) for all \\( auxiliary \\in[-1,1] \\), we have\n\\[\n\\begin{aligned}\nfunctionf^{\\prime}(auxiliary) & \\geq 2+smalleps+auxiliary-variablex \\quad \\text { for }-1 \\leq auxiliary \\leq variablex \\\\\n& \\geq 2+smalleps-auxiliary+variablex \\text { for } variablex \\leq auxiliary \\leq 1 .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nfunctionf(1)-functionf(-1) & =\\int_{-1}^{+1} functionf^{\\prime}(auxiliary) d auxiliary \\\\\n& \\geq 4+2 smalleps-\\frac{1}{2}(1+variablex)^{2}-\\frac{1}{2}(1-variablex)^{2}=3+2 smalleps-variablex^{2} \\\\\n& \\geq 2+2 smalleps\n\\end{aligned}\n\\]\n\nBut \\( |functionf(1)-functionf(-1)| \\leq|functionf(1)|+|functionf(-1)| \\leq 2 \\). Hence there can be no such \\( variablex \\).\n\nSimilarly we cannot have \\( functionf^{\\prime}(variablex)<-2 \\).\n\nRemarks. The inequality is the best possible, for, if \\( functionf(variablex)=\\frac{1}{2}(variablex+1)^{2}-1 \\), the hypothesis is satisfied and \\( functionf^{\\prime}(1)=2 \\). The first solution shows that \\( \\left|functionf^{\\prime}\\right| \\) can attain the value 2 only for \\( variablex= \\pm 1 \\), and then it follows easily that the function just described and three others obtained from it by reflection are the only extremal functions.\n\nThe result was first established by Landau in \"Einige Ungleichungen fur zweimal differentzierbaren Funktionen,\" Proceedings of the London Mathematical Society (2), vol. 13 (1914), pages 43-49. For a treatment of several similar inequalities, see I. J. Schoenberg, \"The Elementary Cases of Landau's Problem of Inequalities between Derivatives,\" American Mathematical Monthly, vol. 80 (1973), pages 121-158."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "labyrinth",
        "x": "waterfall",
        "t": "persimmon",
        "\\\\xi": "driftwood",
        "\\\\eta": "moonlight",
        "\\\\epsilon": "sunflower"
      },
      "question": "4. Assume that \\( |labyrinth(waterfall)| \\leq 1 \\) and \\( \\left|labyrinth^{\\prime \\prime}(waterfall)\\right| \\leq 1 \\) for all \\( waterfall \\) on an interval of length at least 2 . Show that \\( \\left|labyrinth^{\\prime}(waterfall)\\right| \\leq 2 \\) on the interval.",
      "solution": "First Solution. We may suppose without loss of generality that the interval in question is \\( [-1,+1] \\). Using Taylor's formula to expand about the point \\( waterfall \\in[-1,1] \\), we find\n\\[\n\\begin{array}{c}\nlabyrinth(1)=labyrinth(waterfall)+(1-waterfall) labyrinth^{\\prime}(waterfall)+\\frac{1}{2}(1-waterfall)^{2} labyrinth^{\\prime \\prime}(driftwood) \\\\\nlabyrinth(-1)=labyrinth(waterfall)+(-1-waterfall) labyrinth^{\\prime}(waterfall)+\\frac{1}{2}(-1-waterfall)^{2} labyrinth^{\\prime \\prime}(moonlight)\n\\end{array}\n\\]\nwhere \\( driftwood \\in(waterfall, 1) \\) and \\( moonlight \\in(-1, waterfall) \\). Hence\n\\[\nlabyrinth(1)-labyrinth(-1)=2 labyrinth^{\\prime}(waterfall)+\\frac{1}{2}(1-waterfall)^{2} labyrinth^{\\prime \\prime}(driftwood)-\\frac{1}{2}(1+waterfall)^{2} labyrinth^{\\prime \\prime}(moonlight)\n\\]\n\nUsing the given bounds for \\( labyrinth \\) and \\( labyrinth^{\\prime \\prime} \\), we get\n\\[\n\\begin{aligned}\n2\\left|labyrinth^{\\prime}(waterfall)\\right| & \\leq|labyrinth(1)|+|labyrinth(-1)|+\\frac{1}{2}(1-waterfall)^{2}\\left|labyrinth^{\\prime \\prime}(driftwood)\\right|+\\frac{1}{2}(1+waterfall)^{2}\\left|labyrinth^{\\prime \\prime}(moonlight)\\right| \\\\\n& \\leq 2+\\frac{1}{2}(1-waterfall)^{2}+\\frac{1}{2}(1+waterfall)^{2}=3+waterfall^{2} \\leq 4\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\left|labyrinth^{\\prime}(waterfall)\\right| \\leq 2\n\\]\n\nSecond Solution. Essentially the same argument, but less elegantly phrased, is the following.\n\nSuppose there is an \\( waterfall \\) in \\( [-1,1] \\) such that\n\\[\nlabyrinth^{\\prime}(waterfall)=2+sunflower, \\quad \\text { where } sunflower>0\n\\]\n\nSince \\( \\left|labyrinth^{\\prime \\prime}(persimmon)\\right| \\leq 1 \\) for all \\( persimmon \\in[-1,1] \\), we have\n\\[\n\\begin{aligned}\nlabyrinth^{\\prime}(persimmon) & \\geq 2+sunflower+persimmon-waterfall \\quad \\text { for }-1 \\leq persimmon \\leq waterfall \\\\\n& \\geq 2+sunflower-persimmon+waterfall \\text { for } waterfall \\leq persimmon \\leq 1 .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nlabyrinth(1)-labyrinth(-1) & =\\int_{-1}^{+1} labyrinth^{\\prime}(persimmon) \\, d persimmon \\\\\n& \\geq 4+2 sunflower-\\frac{1}{2}(1+waterfall)^{2}-\\frac{1}{2}(1-waterfall)^{2}=3+2 sunflower-waterfall^{2} \\\\\n& \\geq 2+2 sunflower\n\\end{aligned}\n\\]\n\nBut \\( |labyrinth(1)-labyrinth(-1)| \\leq|labyrinth(1)|+|labyrinth(-1)| \\leq 2 \\). Hence there can be no such \\( waterfall \\).\n\nSimilarly we cannot have \\( labyrinth^{\\prime}(waterfall)<-2 \\).\n\nRemarks. The inequality is the best possible, for, if \\( labyrinth(waterfall)=\\frac{1}{2}(waterfall+1)^{2} -1 \\), the hypothesis is satisfied and \\( labyrinth^{\\prime}(1)=2 \\). The first solution shows that \\( \\left|labyrinth^{\\prime}\\right| \\) can attain the value 2 only for \\( waterfall= \\pm 1 \\), and then it follows easily that the function just described and three others obtained from it by reflection are the only extremal functions.\n\nThe result was first established by Landau in \"Einige Ungleichungen fur zweimal differentzierbaren Funktionen,\" Proceedings of the London Mathematical Society (2), vol. 13 (1914), pages 43-49. For a treatment of several similar inequalities, see I. J. Schoenberg, \"The Elementary Cases of Landau's Problem of Inequalities between Derivatives,\" American Mathematical Monthly, vol. 80 (1973), pages 121-158."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "timelessness",
        "\\xi": "outsider",
        "\\eta": "interior",
        "f": "nonfunction",
        "\\epsilon": "largeneg"
      },
      "question": "4. Assume that \\( |nonfunction(fixedvalue)| \\leq 1 \\) and \\( \\left|nonfunction^{\\prime \\prime}(fixedvalue)\\right| \\leq 1 \\) for all \\( fixedvalue \\) on an interval of length at least 2 . Show that \\( \\left|nonfunction^{\\prime}(fixedvalue)\\right| \\leq 2 \\) on the interval.",
      "solution": "First Solution. We may suppose without loss of generality that the interval in question is \\( [-1,+1] \\). Using Taylor's formula to expand about the point \\( fixedvalue \\in[-1,1] \\), we find\n\\[\n\\begin{array}{c}\nnonfunction(1)=nonfunction(fixedvalue)+(1-fixedvalue) nonfunction^{\\prime}(fixedvalue)+\\frac{1}{2}(1-fixedvalue)^{2} nonfunction^{\\prime \\prime}(outsider) \\\\\nnonfunction(-1)=nonfunction(fixedvalue)+(-1-fixedvalue) nonfunction^{\\prime}(fixedvalue)+\\frac{1}{2}(-1-fixedvalue)^{2} nonfunction^{\\prime \\prime}(interior)\n\\end{array}\n\\]\nwhere \\( outsider \\in(fixedvalue, 1) \\) and \\( interior \\in(-1, fixedvalue) \\). Hence\n\\[\nnonfunction(1)-nonfunction(-1)=2 nonfunction^{\\prime}(fixedvalue)+\\frac{1}{2}(1-fixedvalue)^{2} nonfunction^{\\prime \\prime}(outsider)-\\frac{1}{2}(1+fixedvalue)^{2} nonfunction^{\\prime \\prime}(interior)\n\\]\n\nUsing the given bounds for \\( nonfunction \\) and \\( nonfunction^{\\prime \\prime} \\), we get\n\\[\n\\begin{aligned}\n2\\left|nonfunction^{\\prime}(fixedvalue)\\right| & \\leq|nonfunction(1)|+|nonfunction(-1)|+\\frac{1}{2}(1-fixedvalue)^{2}\\left|nonfunction^{\\prime \\prime}(outsider)\\right|+\\frac{1}{2}(1+fixedvalue)^{2}\\left|nonfunction^{\\prime \\prime}(interior)\\right| \\\\\n& \\leq 2+\\frac{1}{2}(1-fixedvalue)^{2}+\\frac{1}{2}(1+fixedvalue)^{2}=3+fixedvalue^{2} \\leq 4\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\left|nonfunction^{\\prime}(fixedvalue)\\right| \\leq 2\n\\]\n\nSecond Solution. Essentially the same argument, but less elegantly phrased, is the following.\n\nSuppose there is an \\( fixedvalue \\) in \\( [-1,1] \\) such that\n\\[\nnonfunction^{\\prime}(fixedvalue)=2+largeneg, \\quad \\text { where } largeneg>0\n\\]\n\nSince \\( \\left|nonfunction^{\\prime \\prime}(timelessness)\\right| \\leq 1 \\) for all \\( timelessness \\in[-1,1] \\), we have\n\\[\n\\begin{aligned}\nnonfunction^{\\prime}(timelessness) & \\geq 2+largeneg+timelessness-fixedvalue \\quad \\text { for }-1 \\leq timelessness \\leq fixedvalue \\\\\n& \\geq 2+largeneg-timelessness+fixedvalue \\text { for } fixedvalue \\leq timelessness \\leq 1 .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nnonfunction(1)-nonfunction(-1) & =\\int_{-1}^{+1} nonfunction^{\\prime}(timelessness) d timelessness \\\\\n& \\geq 4+2 largeneg-\\frac{1}{2}(1+fixedvalue)^{2}-\\frac{1}{2}(1-fixedvalue)^{2}=3+2 largeneg-fixedvalue^{2} \\\\\n& \\geq 2+2 largeneg\n\\end{aligned}\n\\]\n\nBut \\( |nonfunction(1)-nonfunction(-1)| \\leq|nonfunction(1)|+|nonfunction(-1)| \\leq 2 \\). Hence there can be no such \\( fixedvalue \\).\n\nSimilarly we cannot have \\( nonfunction^{\\prime}(fixedvalue)<-2 \\).\nRemarks. The inequality is the best possible, for, if \\( nonfunction(fixedvalue)=\\frac{1}{2}(fixedvalue+1)^{2}-1 \\), the hypothesis is satisfied and \\( nonfunction^{\\prime}(1)=2 \\). The first solution shows that \\( \\left|nonfunction^{\\prime}\\right| \\) can attain the value 2 only for \\( fixedvalue= \\pm 1 \\), and then it follows easily that the function just described and three others obtained from it by reflection are the only extremal functions.\n\nThe result was first established by Landau in \"Einige Ungleichungen fur zweimal differentzierbaren Funktionen,\" Proceedings of the London Mathematical Society (2), vol. 13 (1914), pages 43-49. For a treatment of several similar inequalities, see I. J. Schoenberg, \"The Elementary Cases of Landau's Problem of Inequalities between Derivatives,\" American Mathematical Monthly, vol. 80 (1973), pages 121-158."
    },
    "garbled_string": {
      "map": {
        "x": "pavxmrqz",
        "t": "zodjhkly",
        "\\xi": "rcxfkpte",
        "\\eta": "gwylsnav",
        "f": "uvrblmeq",
        "\\epsilon": "ckqsdvha"
      },
      "question": "Assume that \\( |uvrblmeq(pavxmrqz)| \\leq 1 \\) and \\( \\left|uvrblmeq^{\\prime \\prime}(pavxmrqz)\\right| \\leq 1 \\) for all \\( pavxmrqz \\) on an interval of length at least 2 . Show that \\( \\left|uvrblmeq^{\\prime}(pavxmrqz)\\right| \\leq 2 \\) on the interval.",
      "solution": "First Solution. We may suppose without loss of generality that the interval in question is \\( [-1,+1] \\). Using Taylor's formula to expand about the point \\( pavxmrqz \\in[-1,1] \\), we find\n\\[\n\\begin{array}{c}\nuvrblmeq(1)=uvrblmeq(pavxmrqz)+(1-pavxmrqz) uvrblmeq^{\\prime}(pavxmrqz)+\\frac{1}{2}(1-pavxmrqz)^{2} uvrblmeq^{\\prime \\prime}(rcxfkpte) \\\\\nuvrblmeq(-1)=uvrblmeq(pavxmrqz)+(-1-pavxmrqz) uvrblmeq^{\\prime}(pavxmrqz)+\\frac{1}{2}(-1-pavxmrqz)^{2} uvrblmeq^{\\prime \\prime}(gwylsnav)\n\\end{array}\n\\]\nwhere \\( rcxfkpte \\in(pavxmrqz, 1) \\) and \\( gwylsnav \\in(-1, pavxmrqz) \\). Hence\n\\[\nuvrblmeq(1)-uvrblmeq(-1)=2 uvrblmeq^{\\prime}(pavxmrqz)+\\frac{1}{2}(1-pavxmrqz)^{2} uvrblmeq^{\\prime \\prime}(rcxfkpte)-\\frac{1}{2}(1+pavxmrqz)^{2} uvrblmeq^{\\prime \\prime}(gwylsnav)\n\\]\n\nUsing the given bounds for \\( uvrblmeq \\) and \\( uvrblmeq^{\\prime \\prime} \\), we get\n\\[\n\\begin{aligned}\n2\\left|uvrblmeq^{\\prime}(pavxmrqz)\\right| & \\leq|uvrblmeq(1)|+|uvrblmeq(-1)|+\\frac{1}{2}(1-pavxmrqz)^{2}\\left|uvrblmeq^{\\prime \\prime}(rcxfkpte)\\right|+\\frac{1}{2}(1+pavxmrqz)^{2}\\left|uvrblmeq^{\\prime \\prime}(gwylsnav)\\right| \\\\\n& \\leq 2+\\frac{1}{2}(1-pavxmrqz)^{2}+\\frac{1}{2}(1+pavxmrqz)^{2}=3+pavxmrqz^{2} \\leq 4\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\left|uvrblmeq^{\\prime}(pavxmrqz)\\right| \\leq 2\n\\]\n\nSecond Solution. Essentially the same argument, but less elegantly phrased, is the following.\n\nSuppose there is an \\( pavxmrqz \\) in \\( [-1,1] \\) such that\n\\[\nuvrblmeq^{\\prime}(pavxmrqz)=2+ckqsdvha, \\quad \\text { where } ckqsdvha>0\n\\]\n\nSince \\( \\left|uvrblmeq^{\\prime \\prime}(zodjhkly)\\right| \\leq 1 \\) for all \\( zodjhkly \\in[-1,1] \\), we have\n\\[\n\\begin{aligned}\nuvrblmeq^{\\prime}(zodjhkly) & \\geq 2+ckqsdvha+zodjhkly-pavxmrqz \\quad \\text { for }-1 \\leq zodjhkly \\leq pavxmrqz \\\\\n& \\geq 2+ckqsdvha-zodjhkly+pavxmrqz \\text { for } pavxmrqz \\leq zodjhkly \\leq 1 .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nuvrblmeq(1)-uvrblmeq(-1) & =\\int_{-1}^{+1} uvrblmeq^{\\prime}(zodjhkly) d zodjhkly \\\\\n& \\geq 4+2 ckqsdvha-\\frac{1}{2}(1+pavxmrqz)^{2}-\\frac{1}{2}(1-pavxmrqz)^{2}=3+2 ckqsdvha-pavxmrqz^{2} \\\\\n& \\geq 2+2 ckqsdvha\n\\end{aligned}\n\\]\n\nBut \\( |uvrblmeq(1)-uvrblmeq(-1)| \\leq|uvrblmeq(1)|+|uvrblmeq(-1)| \\leq 2 \\). Hence there can be no such \\( pavxmrqz \\).\n\nSimilarly we cannot have \\( uvrblmeq^{\\prime}(pavxmrqz)<-2 \\).\n\nRemarks. The inequality is the best possible, for, if \\( uvrblmeq(pavxmrqz)=\\frac{1}{2}(pavxmrqz+1)^{2} \\) -1 , the hypothesis is satisfied and \\( uvrblmeq^{\\prime}(1)=2 \\). The first solution shows that \\( \\left|uvrblmeq^{\\prime}\\right| \\) can attain the value 2 only for \\( pavxmrqz= \\pm 1 \\), and then it follows easily that the function just described and three others obtained from it by reflection are the only extremal functions.\n\nThe result was first established by Landau in \"Einige Ungleichungen fur zweimal differentzierbaren Funktionen,\" Proceedings of the London Mathematical Society (2), vol. 13 (1914), pages 43-49. For a treatment of several similar inequalities, see I. J. Schoenberg, \"The Elementary Cases of Landau's Problem of Inequalities between Derivatives,\" American Mathematical Monthly, vol. 80 (1973), pages 121-158."
    },
    "kernel_variant": {
      "question": "Let \\((M,g)\\) be a complete \\(n\\)-dimensional Riemannian manifold and fix a point \\(p\\in M\\).\nAssume that the closed geodesic ball  \n\n  \\(B_{5}(p):=\\{x\\in M : d(p,x)\\le 5\\}\\)\n\nis strongly geodesically convex; equivalently  \n\n  \\(\\mathrm{inj}(x)\\ge 5\\qquad\\forall x\\in B_{5}(p).\\)\n\nLet \\(f\\in C^{3}(B_{5}(p))\\) satisfy the uniform bounds  \n\n (1) \\(|f(x)|\\le 1,\\)  \n\n (2) \\(\\displaystyle\\|\\nabla^{3}f(x)\\|_{\\mathrm{op}}\\le 1,\\qquad\\forall x\\in B_{5}(p),\\)\n\nwhere \\(\\|T\\|_{\\mathrm{op}}\\!:=\\!\\sup_{|v_{1}|=\\dots=|v_{k}|=1}|T[v_{1},\\dots ,v_{k}]|\\) for a symmetric \\(k\\)-tensor \\(T\\).\n\nShow that for every point \\(y\\in B_{1}(p)\\) one has simultaneously  \n\n (a) \\(|\\nabla f(y)|\\le \\dfrac76,\\)  \n\n (b) \\(\\displaystyle\\|\\nabla^{2}f(y)\\|_{\\mathrm{op}}\\le\\dfrac53.\\)\n\n(The numerical constants \\(7/6\\approx1.17\\) and \\(5/3\\approx1.67\\) are still not sharp; improving them is a separate question.  Both inequalities remain valid, of course, with the weaker bound “\\(\\le 2\\)’’.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout, norms and inner products are taken with respect to the metric \\(g\\); the letter \\(C\\) denotes an absolute constant that may change from line to line.\n\nStep 0 – Preparing geodesics.  \nFix \\(y\\in B_{1}(p)\\) and a unit vector \\(v\\in T_{y}M\\).  \nBecause \\(\\mathrm{inj}(x)\\ge5\\) on \\(B_{5}(p)\\), the geodesic  \n\n\\[\n\\gamma_{v}(t):=\\exp_{y}(tv),\\qquad |t|\\le 2,\n\\]\n\nis well defined, length–minimising and contained in \\(B_{3}(p)\\subset B_{5}(p)\\); hence the bounds (1)–(2) hold along \\(\\gamma_{v}\\bigl([-2,2]\\bigr)\\).\n\nStep 1 – Reduction to one variable.  \nDefine  \n\n\\[\n\\varphi(t):=f\\bigl(\\gamma_{v}(t)\\bigr),\\qquad |t|\\le 2.\n\\]\n\nBecause \\(\\gamma_{v}\\) is a geodesic,\n\n\\[\n\\varphi'(t)=\\bigl\\langle\\nabla f,\\dot\\gamma_{v}\\bigr\\rangle,\\quad\n\\varphi''(t)=\\nabla^{2}f[\\dot\\gamma_{v},\\dot\\gamma_{v}],\\quad\n\\varphi'''(t)=\\nabla^{3}f[\\dot\\gamma_{v},\\dot\\gamma_{v},\\dot\\gamma_{v}].\n\\]\n\nSince \\(|\\dot\\gamma_{v}(t)|\\equiv1\\), (1)–(2) give  \n\n\\[\n|\\varphi(t)|\\le 1,\\qquad |\\varphi'''(t)|\\le 1\\qquad\\forall t\\in[-2,2].\\tag{3}\n\\]\n\nStep 2 – Landau-type estimate for \\(\\varphi'(0)\\).  \nLet \\(h\\in(0,2]\\) (later we take \\(h=2\\)).  \nTaylor’s theorem with Lagrange remainder yields\n\n\\[\n\\begin{aligned}\n\\varphi(h)&=\\varphi(0)+h\\varphi'(0)+\\frac{h^{2}}{2}\\varphi''(0)+\\frac{h^{3}}{6}\\varphi'''(\\xi_{h}),\\\\\n\\varphi(-h)&=\\varphi(0)-h\\varphi'(0)+\\frac{h^{2}}{2}\\varphi''(0)-\\frac{h^{3}}{6}\\varphi'''(\\xi_{-h}),\n\\end{aligned}\n\\]\nwith some \\(\\xi_{\\pm h}\\in(0,\\pm h)\\).  \nSubtracting,\n\n\\[\n2h\\,\\varphi'(0)=\\varphi(h)-\\varphi(-h)-\\frac{h^{3}}{6}\\!\\bigl[\\varphi'''(\\xi_{h})+\\varphi'''(\\xi_{-h})\\bigr].\n\\]\n\nUsing (3) we get  \n\n\\[\n2h\\,|\\varphi'(0)|\\le 2+\\frac{h^{3}}{3},\n\\qquad\\Longrightarrow\\qquad\n|\\varphi'(0)|\\le\\frac{1}{2h}\\Bigl(2+\\frac{h^{3}}{3}\\Bigr).\\tag{4}\n\\]\n\nFor \\(h=2\\) this gives  \n\n\\[\n|\\varphi'(0)|\\le\\frac{1}{4}\\Bigl(2+\\frac{8}{3}\\Bigr)=\\frac76.\\tag{5}\n\\]\n\n(Choosing \\(h=3^{1/3}\\) yields the slightly better constant \\(\\tfrac49 3^{2/3}\\approx1.04<7/6\\); any bound \\(\\,<2\\) suffices below.)\n\nStep 3 – Landau-type estimate for \\(\\varphi''(0)\\).  \nAdding the two Taylor expansions instead gives\n\n\\[\n\\varphi(h)+\\varphi(-h)=2\\varphi(0)+h^{2}\\varphi''(0)+\\frac{h^{3}}{6}\\bigl[\\varphi'''(\\xi_{h})-\\varphi'''(\\xi_{-h})\\bigr].\n\\]\n\nHence  \n\n\\[\nh^{2}\\,|\\varphi''(0)|\\le 4+\\frac{h^{3}}{3},\\qquad\\text{so}\\qquad\n|\\varphi''(0)|\\le\\frac{4+\\frac{h^{3}}{3}}{h^{2}}.\\tag{6}\n\\]\n\nWithin the admissible range \\(0<h\\le2\\) the right–hand side is decreasing (its derivative is \\(-\\tfrac{8}{h^{3}}+\\tfrac13<0\\) because \\(h^{3}\\le8\\)).  \nTherefore the minimum on \\((0,2]\\) is attained at \\(h=2\\):\n\n\\[\n|\\varphi''(0)|\\le\\frac{4+\\frac{8}{3}}{4}=\\frac53.\\tag{7}\n\\]\n\nStep 4 – Returning to \\(\\nabla f\\) and \\(\\nabla^{2}f\\).  \nBecause the unit vector \\(v\\in T_{y}M\\) was arbitrary, (5) implies\n\n\\[\n|\\langle\\nabla f(y),v\\rangle|\\le\\frac76\\qquad\\forall v,\\ |v|=1,\n\\]\nand therefore  \n\n\\[\n|\\nabla f(y)|\\le\\frac76.\\tag{8}\n\\]\n\nSimilarly, (7) gives  \n\n\\[\n|\\nabla^{2}f(y)[v,v]|\\le\\frac53\\qquad\\forall v,\\ |v|=1.\n\\]\n\nFor a symmetric bilinear form \\(A\\) one has  \n\n\\[\n\\|A\\|_{\\mathrm{op}}=\\sup_{|v|=|w|=1}|A[v,w]|=\\sup_{|v|=1}|A[v,v]|;\n\\]\ndiagonalising \\(A\\) in an orthonormal eigenbasis shows both suprema equal the spectral radius.  Consequently  \n\n\\[\n\\|\\nabla^{2}f(y)\\|_{\\mathrm{op}}\\le\\frac53.\\tag{9}\n\\]\n\nInequalities (8)–(9) establish assertions (a) and (b).\n\nStep 5 – Discussion of the geometric hypotheses.  \nThe only geometric ingredient used is that every unit-speed geodesic segment of length \\(2\\) starting at a point of \\(B_{1}(p)\\) stays inside \\(B_{5}(p)\\), where the bounds (1)–(2) are known.  No curvature bound is required; the injectivity-radius assumption is therefore sufficient.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.536135",
        "was_fixed": false,
        "difficulty_analysis": "1. Ambient curvature – The problem is no longer set in flat $\\mathbb R^{n}$ but on a manifold with non-zero sectional curvature.  One must deal with geodesics, parallel transport and normal coordinates, introducing geometric analysis tools that have no analogue in the original statement.\n\n2. Higher-order data – A bound on the third covariant derivative is added.  Controlling remainders therefore requires an understanding of how successive covariant derivatives behave along geodesics.\n\n3. Two simultaneous conclusions – The task is to estimate both the gradient and the Hessian, not merely the first derivative.  This forces the solver to manipulate second-order central differences in addition to first-order ones.\n\n4. Tensorial notation – Operator norms of covariant $k$-tensors must be handled.  The solver has to keep track of how these norms behave under contraction with unit vectors.\n\n5. Remainder terms and curvature corrections – Because the manifold is curved, Euclidean Taylor expansions are unavailable verbatim; geodesic versions with curvature error terms must be employed and shown to be sub-dominant.\n\n6. Non-trivial scale separation – The hypotheses are imposed on $B_{4}$ but the desired estimates hold only on the smaller $B_{1}$; geometric convexity must be invoked to guarantee that the geodesics used in the proof stay inside the larger ball.\n\nAltogether these additions oblige the solver to combine techniques from elementary real analysis, multivariable calculus, and Riemannian geometry—well beyond what is required for either the original problem or the simpler kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let \\((M,g)\\) be a complete \\(n\\)-dimensional Riemannian manifold and fix a point \\(p\\in M\\).\nAssume that the closed geodesic ball  \n\n  \\(B_{5}(p):=\\{x\\in M : d(p,x)\\le 5\\}\\)\n\nis strongly geodesically convex; equivalently  \n\n  \\(\\mathrm{inj}(x)\\ge 5\\qquad\\forall x\\in B_{5}(p).\\)\n\nLet \\(f\\in C^{3}(B_{5}(p))\\) satisfy the uniform bounds  \n\n (1) \\(|f(x)|\\le 1,\\)  \n\n (2) \\(\\displaystyle\\|\\nabla^{3}f(x)\\|_{\\mathrm{op}}\\le 1,\\qquad\\forall x\\in B_{5}(p),\\)\n\nwhere \\(\\|T\\|_{\\mathrm{op}}\\!:=\\!\\sup_{|v_{1}|=\\dots=|v_{k}|=1}|T[v_{1},\\dots ,v_{k}]|\\) for a symmetric \\(k\\)-tensor \\(T\\).\n\nShow that for every point \\(y\\in B_{1}(p)\\) one has simultaneously  \n\n (a) \\(|\\nabla f(y)|\\le \\dfrac76,\\)  \n\n (b) \\(\\displaystyle\\|\\nabla^{2}f(y)\\|_{\\mathrm{op}}\\le\\dfrac53.\\)\n\n(The numerical constants \\(7/6\\approx1.17\\) and \\(5/3\\approx1.67\\) are still not sharp; improving them is a separate question.  Both inequalities remain valid, of course, with the weaker bound “\\(\\le 2\\)’’.)\n\n--------------------------------------------------------------------",
      "solution": "Throughout, norms and inner products are taken with respect to the metric \\(g\\); the letter \\(C\\) denotes an absolute constant that may change from line to line.\n\nStep 0 – Preparing geodesics.  \nFix \\(y\\in B_{1}(p)\\) and a unit vector \\(v\\in T_{y}M\\).  \nBecause \\(\\mathrm{inj}(x)\\ge5\\) on \\(B_{5}(p)\\), the geodesic  \n\n\\[\n\\gamma_{v}(t):=\\exp_{y}(tv),\\qquad |t|\\le 2,\n\\]\n\nis well defined, length–minimising and contained in \\(B_{3}(p)\\subset B_{5}(p)\\); hence the bounds (1)–(2) hold along \\(\\gamma_{v}\\bigl([-2,2]\\bigr)\\).\n\nStep 1 – Reduction to one variable.  \nDefine  \n\n\\[\n\\varphi(t):=f\\bigl(\\gamma_{v}(t)\\bigr),\\qquad |t|\\le 2.\n\\]\n\nBecause \\(\\gamma_{v}\\) is a geodesic,\n\n\\[\n\\varphi'(t)=\\bigl\\langle\\nabla f,\\dot\\gamma_{v}\\bigr\\rangle,\\quad\n\\varphi''(t)=\\nabla^{2}f[\\dot\\gamma_{v},\\dot\\gamma_{v}],\\quad\n\\varphi'''(t)=\\nabla^{3}f[\\dot\\gamma_{v},\\dot\\gamma_{v},\\dot\\gamma_{v}].\n\\]\n\nSince \\(|\\dot\\gamma_{v}(t)|\\equiv1\\), (1)–(2) give  \n\n\\[\n|\\varphi(t)|\\le 1,\\qquad |\\varphi'''(t)|\\le 1\\qquad\\forall t\\in[-2,2].\\tag{3}\n\\]\n\nStep 2 – Landau-type estimate for \\(\\varphi'(0)\\).  \nLet \\(h\\in(0,2]\\) (later we take \\(h=2\\)).  \nTaylor’s theorem with Lagrange remainder yields\n\n\\[\n\\begin{aligned}\n\\varphi(h)&=\\varphi(0)+h\\varphi'(0)+\\frac{h^{2}}{2}\\varphi''(0)+\\frac{h^{3}}{6}\\varphi'''(\\xi_{h}),\\\\\n\\varphi(-h)&=\\varphi(0)-h\\varphi'(0)+\\frac{h^{2}}{2}\\varphi''(0)-\\frac{h^{3}}{6}\\varphi'''(\\xi_{-h}),\n\\end{aligned}\n\\]\nwith some \\(\\xi_{\\pm h}\\in(0,\\pm h)\\).  \nSubtracting,\n\n\\[\n2h\\,\\varphi'(0)=\\varphi(h)-\\varphi(-h)-\\frac{h^{3}}{6}\\!\\bigl[\\varphi'''(\\xi_{h})+\\varphi'''(\\xi_{-h})\\bigr].\n\\]\n\nUsing (3) we get  \n\n\\[\n2h\\,|\\varphi'(0)|\\le 2+\\frac{h^{3}}{3},\n\\qquad\\Longrightarrow\\qquad\n|\\varphi'(0)|\\le\\frac{1}{2h}\\Bigl(2+\\frac{h^{3}}{3}\\Bigr).\\tag{4}\n\\]\n\nFor \\(h=2\\) this gives  \n\n\\[\n|\\varphi'(0)|\\le\\frac{1}{4}\\Bigl(2+\\frac{8}{3}\\Bigr)=\\frac76.\\tag{5}\n\\]\n\n(Choosing \\(h=3^{1/3}\\) yields the slightly better constant \\(\\tfrac49 3^{2/3}\\approx1.04<7/6\\); any bound \\(\\,<2\\) suffices below.)\n\nStep 3 – Landau-type estimate for \\(\\varphi''(0)\\).  \nAdding the two Taylor expansions instead gives\n\n\\[\n\\varphi(h)+\\varphi(-h)=2\\varphi(0)+h^{2}\\varphi''(0)+\\frac{h^{3}}{6}\\bigl[\\varphi'''(\\xi_{h})-\\varphi'''(\\xi_{-h})\\bigr].\n\\]\n\nHence  \n\n\\[\nh^{2}\\,|\\varphi''(0)|\\le 4+\\frac{h^{3}}{3},\\qquad\\text{so}\\qquad\n|\\varphi''(0)|\\le\\frac{4+\\frac{h^{3}}{3}}{h^{2}}.\\tag{6}\n\\]\n\nWithin the admissible range \\(0<h\\le2\\) the right–hand side is decreasing (its derivative is \\(-\\tfrac{8}{h^{3}}+\\tfrac13<0\\) because \\(h^{3}\\le8\\)).  \nTherefore the minimum on \\((0,2]\\) is attained at \\(h=2\\):\n\n\\[\n|\\varphi''(0)|\\le\\frac{4+\\frac{8}{3}}{4}=\\frac53.\\tag{7}\n\\]\n\nStep 4 – Returning to \\(\\nabla f\\) and \\(\\nabla^{2}f\\).  \nBecause the unit vector \\(v\\in T_{y}M\\) was arbitrary, (5) implies\n\n\\[\n|\\langle\\nabla f(y),v\\rangle|\\le\\frac76\\qquad\\forall v,\\ |v|=1,\n\\]\nand therefore  \n\n\\[\n|\\nabla f(y)|\\le\\frac76.\\tag{8}\n\\]\n\nSimilarly, (7) gives  \n\n\\[\n|\\nabla^{2}f(y)[v,v]|\\le\\frac53\\qquad\\forall v,\\ |v|=1.\n\\]\n\nFor a symmetric bilinear form \\(A\\) one has  \n\n\\[\n\\|A\\|_{\\mathrm{op}}=\\sup_{|v|=|w|=1}|A[v,w]|=\\sup_{|v|=1}|A[v,v]|;\n\\]\ndiagonalising \\(A\\) in an orthonormal eigenbasis shows both suprema equal the spectral radius.  Consequently  \n\n\\[\n\\|\\nabla^{2}f(y)\\|_{\\mathrm{op}}\\le\\frac53.\\tag{9}\n\\]\n\nInequalities (8)–(9) establish assertions (a) and (b).\n\nStep 5 – Discussion of the geometric hypotheses.  \nThe only geometric ingredient used is that every unit-speed geodesic segment of length \\(2\\) starting at a point of \\(B_{1}(p)\\) stays inside \\(B_{5}(p)\\), where the bounds (1)–(2) are known.  No curvature bound is required; the injectivity-radius assumption is therefore sufficient.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.446696",
        "was_fixed": false,
        "difficulty_analysis": "1. Ambient curvature – The problem is no longer set in flat $\\mathbb R^{n}$ but on a manifold with non-zero sectional curvature.  One must deal with geodesics, parallel transport and normal coordinates, introducing geometric analysis tools that have no analogue in the original statement.\n\n2. Higher-order data – A bound on the third covariant derivative is added.  Controlling remainders therefore requires an understanding of how successive covariant derivatives behave along geodesics.\n\n3. Two simultaneous conclusions – The task is to estimate both the gradient and the Hessian, not merely the first derivative.  This forces the solver to manipulate second-order central differences in addition to first-order ones.\n\n4. Tensorial notation – Operator norms of covariant $k$-tensors must be handled.  The solver has to keep track of how these norms behave under contraction with unit vectors.\n\n5. Remainder terms and curvature corrections – Because the manifold is curved, Euclidean Taylor expansions are unavailable verbatim; geodesic versions with curvature error terms must be employed and shown to be sub-dominant.\n\n6. Non-trivial scale separation – The hypotheses are imposed on $B_{4}$ but the desired estimates hold only on the smaller $B_{1}$; geometric convexity must be invoked to guarantee that the geodesics used in the proof stay inside the larger ball.\n\nAltogether these additions oblige the solver to combine techniques from elementary real analysis, multivariable calculus, and Riemannian geometry—well beyond what is required for either the original problem or the simpler kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}