Suppose $f:(a,b)\rightarrow (a,b)$ is differentiable on $(a,b)$ and for an $x_{0}$ such that $a<x_{0}<b$ , $f'(x_{0}) >0 $ then is $f$ increasing in some neighborhood of $x_{0}$?

I have seen examples on this site on disproving this for the interval $(0,1)$ by taking the function $x+2x^{2}\sin(\frac{1}{x})$ when $x\neq 0$ and $0$ if $x=0$. But I have a doubt whether this would be true for $x_{0}$ being an interior point of the open interval $(a,b)$ . Can someone please clarify . I have tried to prove it using LMVT but since nothing is said about continuity on $[a,b]$ I am unable to proceed.

  • $\begingroup$ "in some neighborhood of $(a,b)$" : what do you mean ? $\endgroup$ – TheSilverDoe Feb 22 at 10:39
  • $\begingroup$ Oh no...typo...I meant in some neighborhood of $x_{0}$ . Thanks for noticing $\endgroup$ – Arghyadeep Chatterjee Feb 22 at 10:56
  • 2
    $\begingroup$ Take the same counterexample $x+2x^{2}\sin(x)$ on $(-1,1)$. $\endgroup$ – TheSilverDoe Feb 22 at 10:59
  • $\begingroup$ I think the result holds true when $f'$ is continuous at $x_0$. $\endgroup$ – cxh007 Feb 22 at 11:08
  • $\begingroup$ @cxh007 math.stackexchange.com/questions/1603958/… see this . $\endgroup$ – Arghyadeep Chatterjee Feb 22 at 11:21

If we can assume that $f'$ is continuous at $x_0$, then by $f'(x_0)>0$ we know that there exists a closed interval $[c,d]$ satisfying $x_0 \in [c,d]\subset (a,b)$ and s.t. $\forall t \in [c,d], f'(t)>0$. Moreover $f$ is differentiable on $(c,d)$, $f$ is continuous at $[c,d]$. Then we apply LMVT: if $x,y\in[c,d], x<y$, then $f(y)-f(x)=f'(\xi)(y-x)$ where $\xi \in [c,d]$, and thus $f'(\xi)>0, (y-x) >0$, finally for $x,y\in [c,d]$, $f(y)>f(x)$ whenever $y>x$.
As noted, for counterexample just take $(a,b)=(-1,1), x_0=0,f(x)=x+2x^2 \sin(\frac{1}{x})$ when $x \neq 0$, $f(x)=0$ when $x=0$.


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