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Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $g'(x) > 0$ for all $x \neq 0$. Prove that $g$ is one-to-one.

Proof: Case 1: Consider $x_{2} > x_{1}>0$. Then by the Mean Value Theorem, there exists a point $x \in \mathbb{R}$ such that $g'(x) = \frac{g(x_{2}) - g(x_{1})}{x_{2}- x_{1}}$ which means that $g(x_{2}) > g_(x_{1})$ since $g'(x)> 0$ . Thus $g$ is $1-1$ in this case.

Case 2: Similarily if $x_{1}<x_{2}<0$, then $g(x_{1})<g(x_{2})$.

Case 3: The only case left to consider is that if $x_{1} \leq 0< x_{2}$

I am not sure how to approach case 3. Do I use the MVT again or something else. Also, am I missing any circumstances?

Jack
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  • The inequality $x_2 - x_1>0$ holds all the same in case 3. – Thomas Apr 07 '15 at 04:44
  • But what about if my function crosses through 0? – Jack Apr 07 '15 at 04:45
  • @Tim When you say $x_2>x_1>0$, are you really using the fact that $x_1$ and $x_2$ are greater than zero? Simply assume $x_1,x_2\neq 0$, the case $x=0$ can be handled using continuity of $g$. – Daniel Apr 07 '15 at 04:48
  • I was because I am trying to avoid my function crossing zero as we do not know what the derivative is there – Jack Apr 07 '15 at 04:50
  • @Tim Read my comment again. – Daniel Apr 07 '15 at 04:51
  • @Tim Ok, I've got your point. You need the function to be differentiable over the interval $(x_1,x_2)$, that's why you're avoiding $x_1\leq 0 < x_2$. However, as I've said, the only thing you need to complete the proof if consider continuity of $g$ – Daniel Apr 07 '15 at 04:54
  • Yes I need to show that my function is 1-1 even in the third case. – Jack Apr 07 '15 at 04:56

2 Answers2

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You need the function to be differentiable over the interval $(x_1,x_2)$ and that's why you're considering the case $x_1\leq 0 < x_2$ apart. However, as I've said, the only thing you need to complete the proof is considering continuity of $g$ at $0$. This is a necessary condition (and you have it from differentiability condition). For example, you could have the function

$$g(x)=\left\{ \begin{align} 2x &\ \ \mbox{ if } x>0\\ x &\ \ \mbox{ if } x<0\\ 2 &\ \ \mbox{ if } x=0\end{align}\right.$$

This function satisfy your hypothesis (except for differentiability at $0$) but it's not continuous at $0$, you can see it's not $1-1$ since $g(0)=g(1)$.

Daniel
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  • I know that $g'(0)$ cannot be negative or else it would not be continuous. Still not quit understanding what you are saying here. I will keep reading and respond when I come to a better understanding. Thanks! – Jack Apr 07 '15 at 05:07
  • @Tim The function $g'$ need not be continuous, there are few cases in which this doesn't hold but they exist! http://math.stackexchange.com/questions/292275/discontinuous-derivative. You have case $1$ and $2$, just use continuity of $g$ at $0$ to conclude that $g$ is one to one on the whole domain (don't try to prove $g'(0)>0$, $g(x)=x^3$ is a counter example). – Daniel Apr 07 '15 at 05:09
  • So if I mention that $g$ is continuous at $0$ since, it is differentiable there, and that in every $\epsilon >0$ neighborhood around $g(0)$, $g$ is strictly increasing, thus $g$ is $1-1$. – Jack Apr 07 '15 at 20:28
  • It's the reason, yes, but it's not the proof. I prefer zhw's answer, this comes directly from the Mean Value Theorem. – Daniel Apr 07 '15 at 23:13
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The MVT shows $f$ is strictly increasing on both $(-\infty,0]$ and $[0,\infty).$ Therefore $f$ is strictly increasing on $\mathbb {R}.$

zhw.
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