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Let $f$ be a $C^1$ function.

If $|f(x+t)-f(x)|\le |f'(x)t|+|t^n|$ for sufficiently small $t$, does it follow that $f$ is a $C^m$ function (say for $m$ strictly less than $n$, or whatever we can generalize it)?

If it does hold for all positive whole $n$, does it follow that $f$ is $C^\infty$ smooth?


Rewrite it (not quite equivalent but related):

$f(x) = f(x-c) + (x-c)f'(x-c)+o((x-c)^n)$

Then it (that $f$ is $C^n$) follows from answers to https://mathoverflow.net/q/88501

So my question is answered. But I want also a short proof for my special case.

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  • $\begingroup$ Does $t$ dependent on $x$? $\endgroup$ – user228113 Aug 7 '16 at 6:16
  • $\begingroup$ @G.Sassatelli Consider both cases if $t$ depends on $x$ or not. I think the case if $t$ depends on $x$ is more interesting $\endgroup$ – porton Aug 7 '16 at 6:18
  • $\begingroup$ Related questions: math.stackexchange.com/q/107460/4876 and mathoverflow.net/q/88501 $\endgroup$ – porton Aug 7 '16 at 6:42
  • $\begingroup$ Your second sentence: Does this mean there is $\delta > 0$ such that for all $x$ and all $t$ with $|t|<\delta,$ the inequality holds? $\endgroup$ – zhw. Aug 7 '16 at 18:43
  • $\begingroup$ @zhw. I'd rather make it mean that for every $x$ there exists a $\delta>0$ such that for all $|t|<\delta$ the inequality holds. However you may interpret it another way, the question is not quite precise $\endgroup$ – porton Aug 7 '16 at 20:03

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