10
$\begingroup$

Does there exist any function $f \in C^2[0,1]$; $f: [0,1] \mapsto [0,1]$, for which the derivative changes sign more than countably many times?

$\endgroup$
10
  • 11
    $\begingroup$ No. Since $f\in\mathcal C^2[0,1]$, $f^\prime$ is continuous, so $\left\{\,x\,\big|\,f^\prime(x)>0\,\right\}$ is an open set and is the union of countable open intervals. $\endgroup$
    – Yai0Phah
    Feb 20, 2013 at 14:22
  • 4
    $\begingroup$ @FrankScience Why hide that perfectly good answer as a comment? $\endgroup$
    – mrf
    Feb 20, 2013 at 14:25
  • 1
    $\begingroup$ I agree with @mrf, make it an answer Frank. $\endgroup$
    – Tomás
    Feb 20, 2013 at 14:27
  • 1
    $\begingroup$ @mrf I doubt it's a typo. We can consider a stronger one, say, if $f$ is only differentiable on $[0,1]$. $\endgroup$
    – Yai0Phah
    Feb 20, 2013 at 14:27
  • 2
    $\begingroup$ @FrankScience, I don't understand the "typo" comment. Was that really meant for me? $\endgroup$
    – mrf
    Feb 20, 2013 at 14:43

1 Answer 1

2
$\begingroup$

How about a function $f'$ whose zero set is the Cantor set. And on complementary intervals with length $3^{-k}$, $k$ odd, the function is positive, while on complementary intervals with $3^{-k}$, $k$ even, the function is negative. Then could we say $f'$ "changes sign" at each point of the Cantor set (since there are positive values and negative values clustering at that point)?

picture

$\endgroup$
6
  • $\begingroup$ Would this be differentiable? $\endgroup$
    – Pedro
    May 29, 2013 at 0:44
  • 1
    $\begingroup$ @Peter: Almost everywhere! $\endgroup$
    – Asaf Karagila
    May 29, 2013 at 0:45
  • 1
    $\begingroup$ @AsafKaragila ¡=)! But don't we want it to be everywhere differentiable? $\endgroup$
    – Pedro
    May 29, 2013 at 0:47
  • 2
    $\begingroup$ Each little bump is a $C^\infty$ bump... $\endgroup$
    – GEdgar
    May 29, 2013 at 0:47
  • 1
    $\begingroup$ @Peter: We? I don't mind a negligible set of mistakes... :-) $\endgroup$
    – Asaf Karagila
    May 29, 2013 at 0:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .