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Let $f: \mathbb R \to \mathbb R$ be a continuous function, and there exists a $k>0$ such that for each $y \in \mathbb R$, there are at most $k$ distinct $x$ with $f(x)=y$. Prove that $f$ is differentiable a.e.

My approach is trying to show $f$ is absolutly continuous, but need a hint to make a start. Thanks.

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You’ll not be able to show that $f$ is absolutely continuous: $f(x)=x^3$ is injective but not even uniformly continuous on $\mathbb{R}$. – Brian M. Scott Dec 17 '11 at 4:43
In addition to Brian's point, there are strictly increasing continuous functions that are not locally absolutely continuous, like the Cantor function plus $x$. It seems you'd be better off trying to show that $f$ is locally of bounded variation. – Jonas Meyer Dec 17 '11 at 4:46
up vote 4 down vote accepted

It's enough to show $f$ is locally of bounded variation.

Let $I$ be a compact interval. Then $f$ achieves a maximum and minimum value on $I.$ Let $m^+,m^-$ be these values, respectively.

Try to show the variation of $f$ on $I$ is bounded by $k(m^+ - m^-).$

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