Let $\,f: \mathbb{R} \rightarrow\mathbb{R}$ be some function st $f(0)=0$ and $f'(0) > 0$. Is it the case that $f$ must be increasing in some neightborhood of zero? If $f$ is differentiable in some neighborhood of $0$ then the answer is trivial with the MVT, however all we have is differentiability at a point. I don't think the premise holds, take for example $f(x) = \begin{cases} \sin(x) & \text{, $x\in\mathbb{Q}$} \\ x & \text{, $x \notin \mathbb{Q}$} \end{cases} $

The function seems to be differentiable near $0$ with derivative $1$ but is neither increasing nor decreasing near $0$.

Is this correct? Would you have anymore counterexamples?

  • $\begingroup$ This doesn't seem differentiable at irrational $x.$ [Or actually at any $x \neq 0] $\endgroup$
    – coffeemath
    May 6, 2018 at 11:25
  • $\begingroup$ @coffeemath what about at 0? $\endgroup$ May 6, 2018 at 11:28
  • $\begingroup$ Yes, diff at $0.$ [use definition of derivative on the two separate formulas\ $\endgroup$
    – coffeemath
    May 6, 2018 at 11:30

3 Answers 3


Let $$ f(x)=\begin{cases}x+2x^2\sin\frac1x&x\ne0\\0&x=0\end{cases}$$ This $f$ is continuous and has derivative $$ f'(x)=\begin{cases}1+4x\sin\frac1x-2\cos\frac1x&x\ne0\\1&x=0\end{cases}$$ So $f$ is differentiable on all of $\Bbb R$, $f(0)=0$, $f'(0)=1$, and yet it is not increasing in any neighbourhood of $0$ because at $x_n=\frac1{2n\pi}$ we have $f'(x_n)=-1$.

  • 2
    $\begingroup$ For more extreme examples, see Everywhere differentiable function that is nowhere monotonic. Among other things, an everywhere differentiable function can have a positive derivative on a set that is dense in $\mathbb R$ (indeed, it can even have a positive derivative on a set that is $c$-dense in ${\mathbb R})$ and still not be monotone on any interval. $\endgroup$ May 6, 2018 at 12:37

The original statement is false, as the previous answer already showed.

However, a weaker statement holds: if $f$ is differentiable at a limit point $x_0$ with positive derivative $f'(x_0) > 0$, then there exists a small neighborhood to the right of $x_0$ that all have higher function value than $f(x_0)$, i.e., $\exists \delta >0, \forall x \in (x_0, x_0 + \delta), f(x) > f(x_0)$.

Proof: by definition of differentiability:

$$\lim_{t\to 0} \frac{f(x_0+t) - f(x_0)}{t} = f'(x_0) > 0 $$ Take $\epsilon = f'(x_0)/2$, then there exists $\delta>0$ such that for all $t \in (0, \delta)$, $$ |\frac{f(x_0+t) - f(x_0)}{t} - f'(x_0)| \leq f'(x_0)/2 $$ so $$f(x_0+t) - f(x_0)\geq t f'(x_0)/2 > 0$$


Even if $f$ is differentiable in a neighborhood of $x_0$ (not just at $x_0$) and $f'(x_0)>0$, this is still not enough to guarantee $f$ to be increasing in a neighborhood of $x_0$; this is already shown by Hagen von Eitzen's answer.

However if we assume $f$ is, in addition, continuously differentiable in a neighborhood of $x_0$, then $f$ will be strictly increasing near $x_0$ (see Terrance Tao, Analysis II Section 6.7). The continuity of $f'$ guarantees that it will be strictly positive in a neighborhood of $x_0$, say $(a, b)$, which implies $f$ will be strictly increasing on $(a, b)$ by a simple application of MVT.

  • $\begingroup$ 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 $(a,b)$? $\endgroup$ Feb 22, 2021 at 10:28
  • $\begingroup$ +1 for neat subresult. At first, I thought this couldn't be true, because I didn't read carefully and assumed this proved the neighborhood was increasing, but that's not the case; it's just that all the values are higher than $f(x_0)$. If you did have a proof for increasing, you could probably argue the same thing symmetrically for a small neighborhood to the left of $x_0$, and then combining the left and right neighborhoods, you'd have an increasing neighborhood around $x_0$ -- contradicting the original result. Posting this misunderstanding for posterity $\endgroup$
    – azi
    Nov 6, 2021 at 18:16
  • $\begingroup$ ...which is precisely what Hrit's answer below says, +1 for them too :) $\endgroup$
    – azi
    Nov 6, 2021 at 19:24
  • $\begingroup$ @Mr.GandalfSauron: the answer is no; see counterexample in Hagen von Eitzen's answer and my updated answer. $\endgroup$
    – Yibo Yang
    Feb 26, 2022 at 7:34

If we start writing an $\epsilon-\delta$ argument we'll find a neighbourhood of $0$ such that points on the right side of $0$ give higher functional values and points on the left side of $0$ give lower functional values. We can't say anything about the relation among the functional values for points other than $0$ even in that neighbourhood as demonstrated by the example given above.


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