# Differentiable at a point with positive derivative implies increasing in neighborhood of point?

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?

• This doesn't seem differentiable at irrational $x.$ [Or actually at any $x \neq 0] May 6, 2018 at 11:25 • @coffeemath what about at 0? May 6, 2018 at 11:28 • Yes, diff at$0.$[use definition of derivative on the two separate formulas\ May 6, 2018 at 11:30 ## 3 Answers 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$f'(x_n)=-1$. • 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. 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 $$f(x_0+t) - f(x_0)\geq t f'(x_0)/2 > 0$$ UPDATE: 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. • 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)$? Feb 22, 2021 at 10:28 • +1 for neat subresult. At first, I thought this couldn't be true, because I didn't read carefully and assumed this proved the neighborhood was increasing, but that's not the case; it's just that all the values are higher than$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
– azi
Nov 6, 2021 at 18:16
• ...which is precisely what Hrit's answer below says, +1 for them too :)
– azi
Nov 6, 2021 at 19:24
• @Mr.GandalfSauron: the answer is no; see counterexample in Hagen von Eitzen's answer and my updated answer. 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.