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My intuition says that the statement is false. Anyone out there know of counterexamples?

Suppose $f: R\to R$ and $c\in R$ such that $f'(c) > 0$. So, $\exists \varepsilon> 0 $ such that $f\mid_{(c-\varepsilon, c+\varepsilon)}$ is increasing.

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user125303
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2 Answers2

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No. Take $$f(x)=\cases{ x+2x^2\sin(1/x), & $x\ne0$\cr 0, &$x=0$}.$$ $f$ is differentiable everywhere and $f'(0)=1$. But $f$ is not monotonic in any neighborhood of $x=0$ (since in any neighborhood of $0$, $f'$ takes both positive and negative values).

This is Example 3.5 in Gelbaum and Olmsted's Counterexamples in Analysis.

David Mitra
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  • Ok, I appreciate your attention! Do you have an easier example? – user125303 Feb 01 '14 at 15:22
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    @user125303 I don't think there is an easier example. The concept is simple, though: take $y=x$ and add a differentiable "infinite oscillatory" part to it. – David Mitra Feb 01 '14 at 15:25
  • Or, being naive (and possibly wrong) about things, take $g(x)=x^2\sin(1/x)$, $x\ne0$; $g(0)=0$ (which has derivative $0$ at $x=0$) and tilt it's graph. – David Mitra Feb 01 '14 at 15:29
  • Now I got your point. Thank you!! The comments from the others were also very useful. – user125303 Feb 01 '14 at 20:34
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$f(x):=\begin{cases}x,&\text{if $x\in\boldsymbol {Q}$}\\ \tan(x),&\text{if $x\in\boldsymbol{R}\setminus\boldsymbol{Q}$}\end{cases}$, for $c=0$ we have $f'(c)=1$. Here we have a punctal monotony: there exists $\epsilon>0$ such that for all $x_1\in ]-\epsilon,0[$ and $x_2\in ]0,\epsilon[$ we have $f(x_1)<f(x_2)$, but $f$ is not increasing on $]-\epsilon,\epsilon[$ for any $\epsilon>0$: chose an irrational $a\in]0,\epsilon[$ and a rational $b\in]0,\epsilon[$ with $a<b$, then $f(a)>f(b)$.

Michael Hoppe
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