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Let $f$ be a differentiable function on open interval $(a,b)$. Suppose $f'(x)$ is not identically zero. Show that there exists an subinterval $(c,d)$ such that $f(x)$ is strictly increasing or strictly decreasing on $(c,d)$.

How to prove this?

I think this statement is wrong...

XLDD
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    $f'(x)$ is strictly... or $f(x)$ is strictly...? – Mikasa Mar 26 '13 at 07:19
  • I mean $f'(x)$ is strictly increasing... – XLDD Mar 26 '13 at 07:32
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    You must mean $f(x)$ strictly increasing or decreasing. Because on $(0,1)$ the function $f(x)=x$ has $f'(x)=1$ at all $x$, which is not strictly increasing or decreasing on any subinterval $(c,d)$ of $(0,1)$. – coffeemath Mar 26 '13 at 09:17
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    In relation to the comments under the answer by Taro, I would like to ask whether the question specifically wants to not assume that $f'$ is continuous. Whether or not that is assumed makes a large difference for the level of difficulty of the question. – Marc van Leeuwen Mar 26 '13 at 13:17
  • I'm trying to prove this without assuming the derivative is continuous, but first I gotta ask: is it actually true? – Javier Apr 08 '13 at 12:35

3 Answers3

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I show that it's true in case $f$ is not only differentiable but also its derivative is continuous.

Since $f'$ is non-zero function, there is a point $\tilde{x} \in (a, b)$ such that $f'(\tilde{x}) > 0$ or $f'(\tilde{x}) < 0$. Suppose $f'(\tilde{x}) > 0$. Then take $c$ and $d$ as follows: $$ \begin{align*} c &:= \inf\{\, \tilde{c} \mid a \leq \tilde{c} \leq \tilde{x}, \quad f'(c') > 0 \quad \text{for all $c' \in (\tilde{c}, \tilde{x}]$} \,\} \\ d &:= \sup\{\, \tilde{d} \mid \tilde{x} \leq \tilde{d} \leq b, \quad f'(d') > 0 \quad \text{for all $d' \in [\tilde{x}, \tilde{d})$} \,\}. \\ \end{align*}$$ From assumption that $f'$ is continuous, $c \neq \tilde{x} \neq d$. The interval $(c, d)$ is the required one (indeed, the largest interval containing $\tilde{x}$).

Orat
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  • But this does not use that $f'$ is the derivative of something, the same would go for any continuous function in the place of $f'$. – Marc van Leeuwen Mar 26 '13 at 11:54
  • The required interval $(c, d)$ is the one which $f$ is strictly increasing for all $x \in (c, d)$ in the situation above. To show this, I showed $f'(x) > 0$ for all $x \in (c, d)$. This argument make sense since $f'$ is the derivative of $f$, doesn't it? – Orat Mar 26 '13 at 12:05
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    You are entirely correct. What I meant to say is that the question immediately translates into "if $f'$ not identically zero then either $f'(x)>0$ holds on some interval, or $f'(x)<0$ does"; since this statement is obviously true when replacing $f'$ by any continuous function, it is the case where $f'$ is not continuous that is the most interesting one. – Marc van Leeuwen Mar 26 '13 at 13:11
  • Wouldn't the fact that derivatives satisfy the intermediate value property suffice? – Javier Apr 07 '13 at 12:32
  • @JavierBadia I cannot see how to use it. Why not post it as an answer if the argument become much simpler? – Orat Apr 07 '13 at 21:00
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The statement is indeed wrong. You can construct for example a function $f:\mathbb{R} \rightarrow\mathbb{R}$ which is differentiable everywhere such that both $\{x \in \mathbb{R} : f'(x) > 0\}$ and $\{x \in \mathbb{R} : f'(x) < 0\}$ are dense in $\mathbb R$ and thus $f$ is monotone on no interval. You can find such a construction on page 80 of A Second Course on Real Functions by van Rooij and Schikhof. See also here.

Community
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Frunobulax
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I also think the statement is not true. There are examples of a function that has derivative positive but it is not monotone in any neighborhood of that point. Consider $\phi(0)=0$ and for $x\neq 0$ consider $$\phi(x)=x^2 \sin\left(\dfrac{1}{x}\right)+\dfrac{x}{2}$$ This function is continuous, $\phi^\prime(0)>0$ but not monotone near zero

Luiz
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