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Problem: Suppose $g$ is a real function on $\mathbb{R}$ with bounded derivative (say $|g'|0$, and define $f(x)=x+\epsilon g(x)$. Prove that $f$ is one-to-one if $\epsilon$ is small enough.

(A set of admissible values of $\epsilon$ can be determined which depends only on $M$.)

Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 3.

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Suppose not. Then for every $\epsilon>0$, there exists $a,b\in\mathbb{R}$, $a\neq b$, with $f(a)=f(b)$. Using the mean value theorem, we see there exists $x\in(a,b)$ with $f'(x)=0$. Note that $f$ is differentiable, as it is the sum of two differentiable functions. For such an $x$, we have

$$f'(x)=1+\epsilon g'(x)=0 \Rightarrow g'(x)=-\frac{1}{\epsilon}.$$

Taking $\epsilon$ small enough, we can force $g'(x)$ to be arbitrarily large. This is a contradiction, as $g'$ is bounded.

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    Don't worry. I've done weirder things, I assure you.2012-06-30
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    You didn't quite say it: $\varepsilon<\dfrac{1}{M}$ suffices.2012-06-30
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    @avatar: Potato is asking for feedback on his solutions to the problems. He posts he question he is trying to work out, and his proposed solution as an answer. Look at his history.2012-06-30
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    Precisely. If the Powers That Be disapprove, I will stop.2012-06-30
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    If that's the case, then I think his solutions should be in the body of the question.2012-06-30
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    I was told before that they should go in the answers. Perhaps I should I should make a meta thread to clarify these issues?2012-06-30