# Prob. 3, Chap. 5 in Baby Rudin: Any positive real number and any real function with bounded derivative together give rise to an injective function

Here is Prob. 3, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $g$ is a real function on $\mathbb{R}^1$, with bounded derivative (say $| g^\prime | \leq M$). Fix $\varepsilon > 0$, and define $f(x) = x + \varepsilon g(x)$. Prove that $f$ is one-to-one if $\varepsilon$ is small enough. (A set of admissible values of $\varepsilon$ can be determined which depends only on $M$.)

My Attempt:

Suppose $x, y \in \mathbb{R}$ such that $x < y$. Then $g$ is continuous on $[x, y]$ and differentiable on $(x, y)$. So, by the Mean Value Theorem, there is some point $p \in (x, y)$ such that $$g(y) - g(x) = (y-x) g^\prime(p).$$ And so, $$f(y) - f(x) = y-x + \varepsilon \left( g(y) - g(x) \right) = (y-x)\left( 1 + \varepsilon g^\prime(p) \right).$$ Since $-M \leq g^\prime(t) \leq M$ for all $t \in \mathbb{R}$ and since $y-x > 0$, therefore, $$(y-x) \left( 1 - M \varepsilon \right) \leq f(y)-f(x) \leq (y-x) \left( 1 + M \varepsilon \right).$$ So if we choose $\varepsilon$ so that $$0 < \varepsilon < \frac{1}{M},$$ then we note that $$0 < (y-x) \left( 1 - M \varepsilon \right) \leq f(y)-f(x) \leq (y-x) \left( 1 + M \varepsilon \right)$$ for any real numbers $x$ and $y$ such that $x < y$; that is, $f(x) < f(y)$ for any real numbers $x$ and $y$ such that $x < y$, which implies that $f$ is strictly increasing and thus one-to-one.

Is this proof correct?

• That is the approach I would take. – Doug M May 4 '17 at 6:30
• More simple: Show that $g ' >0$ once $\epsilon$ is small enough. Then $g$ is strictly increasing and thus injective. – PhoemueX May 4 '17 at 7:55

A continuous function on $\mathbb{R}$ is injective if and only if it is strictly increasing or decreasing.
You have $f'(x)=1+\varepsilon g'(x)$, which is positive as soon as $$g'(x)>-\frac{1}{\varepsilon}$$ which is true when $0<\varepsilon<1/M$.