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Question: Suppose $f$ is differentiable in an interval $E \subset R$. Prove that $f'$ is bounded on $E$ if and only if a constant $M$ exists such that $|f(x) − f(y)| \le M|x − y|$ for all $x, y \in E$.

My approach: I'll skip the part about the $\Leftarrow$ direction since it is quite straightforward.

Assume that $f'$ is bounded. Set $\phi(t)$ as $f(t)−f(x)/t−x$. Then
$$|\lim_{t\to x} \phi(t)| = |f'(x)| \le M$$ according to the characteristic of boundedness.

The next step is the part where I'm unsure whether it is legit.

multiply $|t-x|$ on both sides of inequality then it will derive $|f(t) − f(x)| \le M|t − x|$.

Is this approach appropriate? Or is there more legitimate way to prove this condition?

Arctic Char
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cycla
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  • I wanted to ask whether I could rule out $lim$ from the first approach that I did. Because in order to derive the last inequality, the $lim$ should be removed. – cycla Sep 20 '20 at 14:23
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    That's why you should invoke some form of the mean value theorem; it allows you to equate the derivative with a legitimate slope without any limiting process. – Alex Provost Sep 20 '20 at 14:27

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Try to apply Mean Value Theorem, keeping in mind that $|f’(x)|<M$ for some $M>0$.