# Show that $\exists \epsilon$ such that $f_{\epsilon}:=x+\epsilon g(x)$ is one-to-one where $g$ has bounded derivative

Consider a differentiable function $g : \mathbb{R} \to \mathbb{R}$ with bounded derivative $g'$, i.e. $\exists M>0$ such that $|g'(x)|\leq M$ for all $x\in \mathbb{R}$. Prove that for sufficiently small $\epsilon$ the function $f_{\epsilon}:\mathbb{R}\to \mathbb{R}$ defined by: $$f_{\epsilon}:=x+\epsilon g(x),$$ is one-to-one.

My workings so far: Because $|g'(x)|$ is bounded by $M$ on the entire real line, we know that $$f_{\epsilon}'(x)=1+\epsilon g'(x)$$ And thus: $$1-\epsilon M \leq f_{\epsilon}'(x) \leq 1+\epsilon M$$ If we choose $\epsilon$ such that $\epsilon=\frac{1}{2M}$ (which is always possible as $M>0$) then $$\frac{1}{2} \leq f'_{\epsilon}(x) \leq \frac{3}{2}$$ Therefore $f$ is increasing on the entire real line.

Now comes my problem, I want to use the fact that $f_{\epsilon}$ is monotonically increasing to show that it is one-to-one. To that end, let's argue by contradiction. Suppose $f_{\epsilon}$ is not one-to-one and therefore $\exists x_1,x_2$ such that $f_{\epsilon}(x_1)=f_{\epsilon}(x_2)$ where $x_1 \neq x_2$. Without loss of generality, let's assume $x_1<x_2$.

However, as $f_{\epsilon}$ is increasing we know that $f_{\epsilon}$ satisfies the strict inequality $$f_{\epsilon}(x_1) < f_{\epsilon}(x_2)$$ which contradicts our assumption. Therefore $f_{\epsilon}$ is one-to-one.

Is this enough to show $f_{\epsilon}$ is one-to-one? Thanks

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Yes ${}{}{}{}{}$ – Amr Feb 1 '13 at 15:07
Well I showed that $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$, but dont I need to also show the converse? namely that $f(x_1)=f(x_2)$ implies $x_1=x_2$? – Slugger Feb 1 '13 at 15:09
This is just the contrapositive – Amr Feb 1 '13 at 15:11

What you say is all correct. If $g$ is also bounded you could prove that such $f_\epsilon$ is a bijection.