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

Question

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?

Answer 1

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 $f'(x_n)=-1$.

Answer 2

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.

Answer 3

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.