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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