# Derivative Greater Than 0 Implies One-To-One Function In Neighborhood

Let $$f: \textrm{dom}(f) \rightarrow \mathbb{R}.$$

Let $$x_0 \in \mathbb{R}.$$

Assume $$f'(x_0) > 0$$.

i.e. $$~ \displaystyle\lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h} > 0$$

i.e. $$~ \exists l > 0 \textrm{ s.t. } \forall \varepsilon_1 > 0, \exists \delta_1 > 0 \textrm{ s.t. } \forall h \in \mathbb{R}, 0 < |h| < \delta_1 \Rightarrow \Bigg| \displaystyle\frac{f(x_0 + h) - f(x_0)}{h} - l \Bigg| < \varepsilon_1$$

This implies that $$f$$ is continuous at $$x_0$$, as I have proven before.

i.e. $$~ \forall \varepsilon_2 > 0, \exists \delta_2 > 0 \textrm{ s.t. } \forall x \in \mathbb{R}, |x - x_0| < \delta_2 \Rightarrow |f(x) - f(x_0)| < \varepsilon_2$$

Also,
$$\exists \delta_3 > 0 \textrm{ s.t. } \forall x \in \mathbb{R}, 0 < |x - x_0| < \delta_3 \Rightarrow \displaystyle\frac{f(x) - f(x_0)}{x - x_0} > 0$$

I would like to prove there exists an open interval containing $$x_0$$ where $$f(x)$$ does not have a value of $$f(x_0)$$ for any $$x$$ in that interval apart from $$x_0$$.
i.e. $$~ \exists a, b \in \mathbb{R} \textrm{ s.t. } a < x_0 < b \wedge \big( \forall x \in (a, b), x \neq x_0 \Rightarrow f(x) \neq f(x_0) \big)$$

Unfortunately, I cannot say anything about double differentiability of $$f$$.

Because of this, I cannot mention continuity of $$f$$ in a neighborhood of $$x_0$$ (or can I?)

Perhaps there is a counterexample and I shouldn't try to prove this statement.

Help needed.

It is not true that if $$f'(x_0)>0$$ then $$f$$ is one-to-one in some open neighborhood of $$x_0.$$

But it is true that if $$f'(x_0)>0$$ then there is some open neighborhood of $$x_0$$ within which $$f$$ takes the value $$f(x_0)$$ only at $$x_0$$ and nowhere else.

Let $$\displaystyle f(x) = \begin{cases} f(x_0) & \text{if }x=x_0, \\[8pt] f(x_0) + (x-x_0) + (x-x_0)^2 \sin(1/(x-x_0)) & \text{if } x\ne x_0. \end{cases}$$

Then $$f'(x_0)=1,$$ but in every open neighborhood of $$x_0$$ there are values of $$x$$ for which $$f'(x)$$ is positive and others for which it is negative. Thus $$f$$ is not one-to-one in that neighborhood.

However, there is some open neighborhood of $$x_0$$ within which every value of $$x$$ other than $$x_0$$ satisfies $$\frac{f(x) - f(x_0)}{x-x_0} > \frac 1 2.$$ That implies $$f(x)-f(x_0)>0$$ if $$x>x_0$$ and $$f(x)-f(x_0)<0$$ if $$x

Consider the function $$f(x) = \begin{cases} x + x^2 & {\rm if \ } x \in \{\tfrac 1 2, \tfrac 1 4, \tfrac 1 8 , \tfrac 1 {16}, \dots \} \\ x & {\rm otherwise \ }\end{cases}$$

$$f$$ is differentiable at $$x = 0$$, with derivative $$f'(0) = 1$$.

But for every $$\delta > 0$$, there exists an $$n$$ such that $$\left|\tfrac 1 {2^n} \right| < \delta$$ and $$\left| \frac 1 {2^n} + \frac 1 {4^n}\right| < \delta$$.

Since $$f(\tfrac 1 {2^n}) = f(\tfrac 1 {2^n} + \tfrac 1 {4^n}) = \tfrac 1 {2^n} + \tfrac 1 {4^n},$$ $$f$$ is not injective on $$(-\delta, \delta)$$.