Exercise 5.3

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\def\eps{\varepsilon}
Exercise 3: Suppose $g$ is a real function on $\mathbf R$, with bounded derivative (say $|g'|\le M$). 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 $M$.)

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

The derivative of $f$ is $f'=1+\varepsilon g'$, which is positive if $\varepsilon<1/M$. In that case, by Exercise 2, $f$ is strictly increasing, hence one-to-one.

Exercise 5.3

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Exercise 5.3

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