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Problem: Suppose $g$ is a real function on $\mathbb{R}$ with bounded derivative (say $|g'|). Fix $\epsilon>0$, 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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    I was told before that they should go in the answers. Perhaps I should I should ma$k$e a meta thread to clarify these issues?2012-06-30