Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 3

Problem

Suppose $g$ is a real function on $\mathbb{R}^1$, with bounded derivative (say $|g'| \leq M$). 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 $M$.)

Answer

It is enough to show that $f'>0$ for a small $\epsilon>0$. Note that

$$f'(x) = 1+ \epsilon g'(x) \geq 1 - \epsilon M.$$

Hence, if we choose $\epsilon < 1/M$, $f$ is one-to-one.

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