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I am curious if the following statement holds.

Let $f:[a,b] \rightarrow \mathbb{R}$ be a continuous function differentiable on the open interval $(a,b)$. Then if $f'(c)>0$ for some $c \in (a,b)$, there exists a neighbourhood of $c$ in which $f$ is monotonically increasing.

An ideal answer to this question would include either a proof or a counterexample.

MathematicsStudent1122
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3 Answers3

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As noted in other answers the results holds if $f\in C^1$. Consider the following counterexample:

$$f(x)=\begin{cases} x+2x^2 \sin(\frac{1}{x})\quad \text{ if } x\neq 0 \\ 0\qquad \quad \qquad \quad \quad \text{ if } x=0 \end{cases}$$

with derivative

$$f^\prime(x)=\begin{cases} 1+4x \sin(\frac{1}{x})-2\cos(\frac{1}{x}) \quad \text{ if } x\neq 0 \\ 1\quad \qquad\qquad\qquad\quad \quad\qquad \text{ if } x=0 \end{cases}$$

Note that $f^\prime(0)>0$ but every neighbourhood of $0$ has negative values (and positive values).

Chazz
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    Why $f$ is differentiable in $0$? – Maffred Jan 08 '16 at 04:07
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    $x^2 \sin(1/x)$ is a standard example of a function that oscillates around 0, yet is still differentiable (if you define $f(0) = 0$). It's a consequence of the fact that $\lim_{x \rightarrow 0} x \sin(1/x) = 0.$ You can prove this with the squeeze theorem. – Philip Hoskins Jan 08 '16 at 04:13
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    @Maffred. Straight from the definition: $f^\prime(0)=\lim_{x\rightarrow 0} \frac{f(x)-f(0)}{x}=\lim_{x\rightarrow 0} 1+2x\sin(\frac{1}{x})=1$ – Chazz Jan 08 '16 at 04:16
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    Oh well, I was trying to take limit in $f'$ and I couldn't of course! Thanks! – Maffred Jan 08 '16 at 04:17
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If you know $f'$ is continous, this is trivially true of course.

If it is not, that could be false.

Maffred
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  • Can we adapt this http://math.stackexchange.com/a/526286/1284 to create a counterexample? – Dan Brumleve Jan 08 '16 at 03:42
  • I don't understand, since that function is definited in $[0,1]$ thus there is no sense to ask for differentiability in $0$. – Maffred Jan 08 '16 at 03:47
  • You can define it on [-1,1] and it will be differentiable in (-1,1). – Philip Hoskins Jan 08 '16 at 03:52
  • Yes that is why that example doesn't work, can it be fixed somehow? Could we extend its domain to $[-1,1]$, make its derivative at $0$ non-zero (add $x$ to it?) and still not have monotonicity in $(-\epsilon,\epsilon)$? – Dan Brumleve Jan 08 '16 at 03:52
  • I definitely don't know!!! I hope there is a professor walking around here that can help us, I've never seen an example of a function differentiable that is not $C^1$! – Maffred Jan 08 '16 at 03:59
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This is true if $f \in C^1((a,b))$, i.e. $f'$ is continuous on $(a,b)$. To see this:

Suppose $f'(c)>0$, where $c \in (a,b)$ and $f \in C^1((a,b))$. Then $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that $|x-c|<\delta$ $\Rightarrow$ $|f'(x)-f'(c)|<\epsilon$. Take $\epsilon= \frac{1}{2} f'(c)>0$. Then $\exists$ $\delta>0$ with $|x-c|<\delta$ $\Rightarrow$ $0<\frac{1}{2}f'(c)<f'(x)$, so $f$ is monotone increasing $\forall$ $x \in (a,b)$ with $|x-c|<\delta$.

If $f \notin C^1((a,b))$ I see no reason for the claim to hold.

M10687
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