{
  "index": "1998-A-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $f$ be a real function on the real line with continuous third\nderivative.  Prove that there exists a point $a$ such that\n\\[f(a)\\cdot f'(a) \\cdot f''(a) \\cdot f'''(a)\\geq 0 .\\]",
  "solution": "If at least one of $f(a)$, $f'(a)$, $f''(a)$, or $f'''(a)$ vanishes\nat some point $a$, then we are done.  Hence we may assume each of\n$f(x)$, $f'(x)$, $f''(x)$, and $f'''(x)$ is either strictly positive\nor strictly negative on the real line.  By replacing $f(x)$ by $-f(x)$\nif necessary, we may assume $f''(x)>0$; by replacing $f(x)$\nby $f(-x)$ if necessary, we may assume $f'''(x)>0$.  (Notice that these\nsubstitutions do not change the sign of $f(x) f'(x) f''(x) f'''(x)$.)\nNow $f''(x)>0$ implies that $f'(x)$ is increasing, and $f'''(x)>0$\nimplies that $f'(x)$ is convex, so that $f'(x+a)>f'(x)+a f''(x)$\nfor all $x$ and $a$.  By\nletting $a$ increase in the latter inequality, we see that $f'(x+a)$\nmust be positive for sufficiently large $a$; it follows that\n$f'(x)>0$\nfor all $x$.  Similarly, $f'(x)>0$ and $f''(x)>0$ imply\nthat $f(x)>0$ for all $x$.  Therefore $f(x) f'(x) f''(x) f'''(x)>0$ for\nall $x$, and we are done.",
  "vars": [
    "x"
  ],
  "params": [
    "f",
    "a"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "f": "realfunc",
        "a": "specialpt"
      },
      "question": "Let $realfunc$ be a real function on the real line with continuous third\nderivative.  Prove that there exists a point $specialpt$ such that\n\\[\nrealfunc(specialpt)\\cdot realfunc'(specialpt) \\cdot realfunc''(specialpt) \\cdot realfunc'''(specialpt)\\geq 0 .\n\\]",
      "solution": "If at least one of $realfunc(specialpt)$, $realfunc'(specialpt)$, $realfunc''(specialpt)$, or $realfunc'''(specialpt)$ vanishes\nat some point $specialpt$, then we are done.  Hence we may assume each of\n$realfunc(realvar)$, $realfunc'(realvar)$, $realfunc''(realvar)$, and $realfunc'''(realvar)$ is either strictly positive\nor strictly negative on the real line.  By replacing $realfunc(realvar)$ by $-realfunc(realvar)$\nif necessary, we may assume $realfunc''(realvar)>0$; by replacing $realfunc(realvar)$\nby $realfunc(-realvar)$ if necessary, we may assume $realfunc'''(realvar)>0$.  (Notice that these\nsubstitutions do not change the sign of $realfunc(realvar) realfunc'(realvar) realfunc''(realvar) realfunc'''(realvar)$.)\nNow $realfunc''(realvar)>0$ implies that $realfunc'(realvar)$ is increasing, and $realfunc'''(realvar)>0$\nimplies that $realfunc'(realvar)$ is convex, so that $realfunc'(realvar+specialpt)>realfunc'(realvar)+specialpt realfunc''(realvar)$\nfor all $realvar$ and $specialpt$.  By\nletting $specialpt$ increase in the latter inequality, we see that $realfunc'(realvar+specialpt)$\nmust be positive for sufficiently large $specialpt$; it follows that\n$realfunc'(realvar)>0$\nfor all $realvar$.  Similarly, $realfunc'(realvar)>0$ and $realfunc''(realvar)>0$ imply\nthat $realfunc(realvar)>0$ for all $realvar$.  Therefore $realfunc(realvar) realfunc'(realvar) realfunc''(realvar) realfunc'''(realvar)>0$ for\nall $realvar$, and we are done."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "porcelain",
        "f": "watering",
        "a": "sunflower"
      },
      "question": "Let $watering$ be a real function on the real line with continuous third\nderivative.  Prove that there exists a point $sunflower$ such that\n\\[watering(sunflower)\\cdot watering'(sunflower) \\cdot watering''(sunflower) \\cdot watering'''(sunflower)\\geq 0 .\\]\n",
      "solution": "If at least one of $watering(sunflower)$, $watering'(sunflower)$, $watering''(sunflower)$, or $watering'''(sunflower)$ vanishes\nat some point $sunflower$, then we are done.  Hence we may assume each of\n$watering(porcelain)$, $watering'(porcelain)$, $watering''(porcelain)$, and $watering'''(porcelain)$ is either strictly positive\nor strictly negative on the real line.  By replacing $watering(porcelain)$ by $-watering(porcelain)$\nif necessary, we may assume $watering''(porcelain)>0$; by replacing $watering(porcelain)$\nby $watering(-porcelain)$ if necessary, we may assume $watering'''(porcelain)>0$.  (Notice that these\nsubstitutions do not change the sign of $watering(porcelain) watering'(porcelain) watering''(porcelain) watering'''(porcelain)$.)\nNow $watering''(porcelain)>0$ implies that $watering'(porcelain)$ is increasing, and $watering'''(porcelain)>0$\nimplies that $watering'(porcelain)$ is convex, so that $watering'(porcelain+sunflower)>watering'(porcelain)+sunflower watering''(porcelain)$\nfor all $porcelain$ and $sunflower$.  By\nletting $sunflower$ increase in the latter inequality, we see that $watering'(porcelain+sunflower)$\nmust be positive for sufficiently large $sunflower$; it follows that\n$watering'(porcelain)>0$\nfor all $porcelain$.  Similarly, $watering'(porcelain)>0$ and $watering''(porcelain)>0$ imply\nthat $watering(porcelain)>0$ for all $porcelain$.  Therefore $watering(porcelain) watering'(porcelain) watering''(porcelain) watering'''(porcelain)>0$ for\nall $porcelain$, and we are done.\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "staticvalue",
        "f": "nonvarying",
        "a": "dispersion"
      },
      "question": "Let $nonvarying$ be a real function on the real line with continuous third\nderivative.  Prove that there exists a point $dispersion$ such that\n\\[nonvarying(dispersion)\\cdot nonvarying'(dispersion) \\cdot nonvarying''(dispersion) \\cdot nonvarying'''(dispersion)\\geq 0 .\\]",
      "solution": "If at least one of $nonvarying(dispersion)$, $nonvarying'(dispersion)$, $nonvarying''(dispersion)$, or $nonvarying'''(dispersion)$ vanishes\nat some point $dispersion$, then we are done.  Hence we may assume each of\n$nonvarying(staticvalue)$, $nonvarying'(staticvalue)$, $nonvarying''(staticvalue)$, and $nonvarying'''(staticvalue)$ is either strictly positive\nor strictly negative on the real line.  By replacing $nonvarying(staticvalue)$ by $-nonvarying(staticvalue)$\nif necessary, we may assume $nonvarying''(staticvalue)>0$; by replacing $nonvarying(staticvalue)$\nby $nonvarying(-staticvalue)$ if necessary, we may assume $nonvarying'''(staticvalue)>0$.  (Notice that these\nsubstitutions do not change the sign of $nonvarying(staticvalue)\\, nonvarying'(staticvalue)\\, nonvarying''(staticvalue)\\, nonvarying'''(staticvalue)$.)\nNow $nonvarying''(staticvalue)>0$ implies that $nonvarying'(staticvalue)$ is increasing, and $nonvarying'''(staticvalue)>0$\nimplies that $nonvarying'(staticvalue)$ is convex, so that $nonvarying'(staticvalue+dispersion)>nonvarying'(staticvalue)+dispersion\\, nonvarying''(staticvalue)$\nfor all $staticvalue$ and $dispersion$.  By\nletting $dispersion$ increase in the latter inequality, we see that $nonvarying'(staticvalue+dispersion)$\nmust be positive for sufficiently large $dispersion$; it follows that\n$nonvarying'(staticvalue)>0$\nfor all $staticvalue$.  Similarly, $nonvarying'(staticvalue)>0$ and $nonvarying''(staticvalue)>0$ imply\nthat $nonvarying(staticvalue)>0$ for all $staticvalue$.  Therefore $nonvarying(staticvalue)\\, nonvarying'(staticvalue)\\, nonvarying''(staticvalue)\\, nonvarying'''(staticvalue)>0$ for\nall $staticvalue$, and we are done."
    },
    "garbled_string": {
      "map": {
        "x": "hjgrksla",
        "f": "qzxwvtnp",
        "a": "kmlpqrst"
      },
      "question": "Let $qzxwvtnp$ be a real function on the real line with continuous third\nderivative.  Prove that there exists a point $kmlpqrst$ such that\n\\[qzxwvtnp(kmlpqrst)\\cdot qzxwvtnp'(kmlpqrst) \\cdot qzxwvtnp''(kmlpqrst) \\cdot qzxwvtnp'''(kmlpqrst)\\geq 0 .\\]",
      "solution": "If at least one of $qzxwvtnp(kmlpqrst)$, $qzxwvtnp'(kmlpqrst)$, $qzxwvtnp''(kmlpqrst)$, or $qzxwvtnp'''(kmlpqrst)$ vanishes\nat some point $kmlpqrst$, then we are done.  Hence we may assume each of\n$qzxwvtnp(hjgrksla)$, $qzxwvtnp'(hjgrksla)$, $qzxwvtnp''(hjgrksla)$, and $qzxwvtnp'''(hjgrksla)$ is either strictly positive\nor strictly negative on the real line.  By replacing $qzxwvtnp(hjgrksla)$ by $-qzxwvtnp(hjgrksla)$\nif necessary, we may assume $qzxwvtnp''(hjgrksla)>0$; by replacing $qzxwvtnp(hjgrksla)$\nby $qzxwvtnp(-hjgrksla)$ if necessary, we may assume $qzxwvtnp'''(hjgrksla)>0$.  (Notice that these\nsubstitutions do not change the sign of $qzxwvtnp(hjgrksla) qzxwvtnp'(hjgrksla) qzxwvtnp''(hjgrksla) qzxwvtnp'''(hjgrksla)$.)\nNow $qzxwvtnp''(hjgrksla)>0$ implies that $qzxwvtnp'(hjgrksla)$ is increasing, and $qzxwvtnp'''(hjgrksla)>0$\nimplies that $qzxwvtnp'(hjgrksla)$ is convex, so that $qzxwvtnp'(hjgrksla+kmlpqrst)>qzxwvtnp'(hjgrksla)+kmlpqrst qzxwvtnp''(hjgrksla)$\nfor all $hjgrksla$ and $kmlpqrst$.  By\nletting $kmlpqrst$ increase in the latter inequality, we see that $qzxwvtnp'(hjgrksla+kmlpqrst)$\nmust be positive for sufficiently large $kmlpqrst$; it follows that\n$qzxwvtnp'(hjgrksla)>0$\nfor all $hjgrksla$.  Similarly, $qzxwvtnp'(hjgrksla)>0$ and $qzxwvtnp''(hjgrksla)>0$ imply\nthat $qzxwvtnp(hjgrksla)>0$ for all $hjgrksla$.  Therefore $qzxwvtnp(hjgrksla) qzxwvtnp'(hjgrksla) qzxwvtnp''(hjgrksla) qzxwvtnp'''(hjgrksla)>0$ for\nall $hjgrksla$, and we are done."
    },
    "kernel_variant": {
      "question": "Let f: ℝ → ℝ be three-times differentiable and assume that the third derivative f‴ is continuous on ℝ.  Prove that there exists a point a ∈ ℝ for which\n\n          f(a)·f′(a)·f″(a)·f‴(a) ≥ 0.",
      "solution": "Step 1.  If any of f, f′, f″, f‴ vanishes somewhere we are done.\n---------------------------------------------------------------\nSuppose there exists x with f(x)=0 or f′(x)=0 or f″(x)=0 or f‴(x)=0.  Then at this x the product f(x)f′(x)f″(x)f‴(x) is 0 ≥ 0, so the assertion holds with a = x.\n\nFrom now on we therefore assume that none of the four functions ever vanishes:\n(1) f(x), f′(x), f″(x), f‴(x) keep a fixed (strict) sign on ℝ.\n\nStep 2.  Normalising the signs of f″ and f‴ without touching the product.\n-----------------------------------------------------------------------\nTwo operations leave the sign of the product f f′ f″ f‴ unchanged:\n•  replacing f by −f, which multiplies each factor by −1;\n•  reflecting the graph, i.e. passing to g(x)=f(−x); then g, g″ keep the sign of f, f″ while g′, g‴ change sign, so again the product is unaltered.\n\nBy first multiplying by −1 we may achieve f″>0 everywhere, and if afterwards f‴ is negative we additionally reflect.  Consequently we may assume once and for all that\n(2) f″(x) > 0 and f‴(x) > 0 for every x∈ℝ.\n\nStep 3.  Consequences of (2).\n----------------------------\nBecause f″>0, the derivative f′ is strictly increasing; because f‴>0, the function f″ itself is increasing.\n\nFix x₀ ∈ ℝ and write m := f″(x₀) > 0.  For every x ≥ x₀ we have\n      f′(x) = f′(x₀) + ∫_{x₀}^{x} f″(t)\n             ≥ f′(x₀) + m(x − x₀).                                (3)\nThe right–hand side is positive once x > x₀ − f′(x₀)/m.  Thus there exists a real number N such that\n      f′(x) > 0  for all x ≥ N.                                    (4)\n(Note that no statement is made for x < N; f′ could be negative there.)\n\nIntegrating (3) once more, for x ≥ x₀ we obtain\n      f(x) = f(x₀) + ∫_{x₀}^{x} f′(t)\n           ≥ f(x₀) + ∫_{x₀}^{x} (f′(x₀) + m(t − x₀))dt\n           = f(x₀) + (x − x₀)f′(x₀) + m(x − x₀)^2/2.              (5)\nThe quadratic term dominates for large x; hence there exists R ≥ max{x₀,N} such that\n      f(x) > 0 for every x ≥ R.                                   (6)\n\nStep 4.  Choice of the required point.\n-------------------------------------\nPick any a ≥ R.  Then by (4) and (6) we have\n      f(a) > 0, f′(a) > 0, and, by (2), f″(a) > 0, f‴(a) > 0.\nConsequently\n      f(a)·f′(a)·f″(a)·f‴(a) > 0 ≥ 0.\nThus the demanded point a exists, completing the proof.\n\n∎",
      "_meta": {
        "core_steps": [
          "If any of f, f', f'', f''' vanishes somewhere, the desired sign is attained immediately",
          "Else make sign-normalizations f→±f and x→−x so that f''>0 and f''' >0 while leaving the product’s sign unchanged",
          "With f''>0, f' is increasing; with f''' >0, f' is convex ⇒ f'(x+a) > f'(x)+a f''(x)",
          "Choosing large a forces f'(x+a)>0; monotonicity then gives f'>0 everywhere, and f'>0 together with f''>0 yields f>0 everywhere",
          "Thus f·f'·f''·f''' is positive (hence non-negative) everywhere, guaranteeing a point a with the required inequality"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The domain only needs to be unbounded so that x+a is always admissible; any unbounded interval would suffice",
            "original": "the real line ℝ"
          },
          "slot2": {
            "description": "Because the argument can start with −f if desired, the final inequality may be written with ≤ instead of ≥ without altering the proof",
            "original": "f(a)·f'(a)·f''(a)·f'''(a) ≥ 0"
          },
          "slot3": {
            "description": "Continuity of the third derivative can be relaxed; the proof uses only the existence and constant sign of f'', f''' (which already give monotonicity/convexity)",
            "original": "\"continuous third derivative\" assumption"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}