Does eventually all the successive derivatives of a bounded functions become bounded if one of them becomes bounded?(for entire number line case)

  • 2
    $\begingroup$ No, take for example $f(x) = \sin(\frac{1}{x})$ which is bounded, with unbounded derivative $\endgroup$
    – fGDu94
    Commented Dec 26, 2019 at 11:38

1 Answer 1


No, they don't. Consider $$ \sin(x^2) $$ which is clearly bounded, but whose derivatives all are very much unbounded.

However, if your function is defined on a closed and bounded interval, and all its derivatives exist everywhere (and are therefore all continuous), then they must be bounded.

  • $\begingroup$ @AnshulArgrawal None of the derivatives are bounded. So they aren't eventually bounded either. As opposed to a polynomial, whose derivatives are all eventually $0$ and therefore bounded. $\endgroup$
    – Arthur
    Commented Dec 26, 2019 at 11:42
  • $\begingroup$ The second example need a correction. $\endgroup$ Commented Dec 26, 2019 at 11:45
  • $\begingroup$ @KaviRamaMurthy It's usually better to point out exactly what the typo is, so I understand that that's what you mean. I first thought you meant the function itself needed a correction. $\endgroup$
    – Arthur
    Commented Dec 26, 2019 at 11:48
  • $\begingroup$ OK, sorry about that. $\endgroup$ Commented Dec 26, 2019 at 11:49
  • $\begingroup$ @Arthur thanks.Is it possible that a function's derivative oscillates between being bounded and unbounded? $\endgroup$ Commented Dec 26, 2019 at 12:26

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