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Is $I\subseteq\mathbb{R}$ an interval and $f: I\to\mathbb{R}$ a differentiable function with bounded derivative $f': I\to\mathbb{R}$, then $f$ is Lipschitz-continuous.

This is supposed to be an application of the mean-value theorem. What gets me is the use of unspecified intervals. So $I=[a,b], (a,b), (a,b], [a,b)$, as the mean-value theorem holds for differentiable functions defined on a compact interval [a,b].

Every resource I looked it up proofs this result for compact intervals, but I was unable to give a counterexample for say $I=(a,b)$, because of the bounded derivative.

But how does one relax the condition to $I=[a,b]$ to apply the mean-value theorem? I thought that one might can proof that for $I=(a,b)$ you are able to continuously extend to $[a,b]$.

Thanks in advance.

Cornman
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1 Answers1

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You apply the mean-value theorem not to $I$, but to intervals $[x, y] \subseteq I$: If $|f'|$ is bounded by $M$ and $x, y \in I$ with $x < y$ then $$ |f(x) - f(y)| = |f'(c) (x-y)| \le M |x-y| \, . $$

It does not matter which kind of interval $I$ is, only that it is a connected set, i.e. that $x, y \in I$ implies $[x, y] \subseteq I$.

Martin R
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  • Ok, thanks for the reply. Would there still be a way to give a continuous extension of $f$ to a compact interval when the derivative is bounded? – Cornman Apr 14 '21 at 19:39
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    @Cornman: Yes. A bounded derivative implies that the function is uniformly continuous, and that is equivalent to the fact that it can be extended continuously to the closed interval, see for example https://math.stackexchange.com/q/245237/42969. (The underlying reason is that a uniformly continuous function transforms Cauchy sequences into Cauchy sequences.) – Martin R Apr 14 '21 at 19:41
  • Nice. Thank you! – Cornman Apr 14 '21 at 19:48