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Let $ f : I\rightarrow\mathbb{R}$, and suppose that $f'$ is bounded on $I$. Prove that $f$ is uniformly continuous on $I$ .

I can do this using the Mean Value Theorem and the definition of uniform continuity. However that requires $f'$ to be continuous on the interval? Or is this a mistake in the question? All the other questions related on the forum they have the claim that $f'$ exists.

Axion004
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Yisroel Green
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1 Answers1

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No continuity of $f'$ is required. Rather, you can prove something stronger, $f$ is actually Lipschitz continuous: $|f(x)-f(y)|=|f'(c)||x-y|\leq(\sup|f'|)|x-y|$.

user284331
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  • Can you expand on the difference between uniform and Lipschitz? Is it claiming that there is a maximum $\epsilon$? – Yisroel Green Dec 23 '19 at 04:35
  • Look at this, and also this. – user284331 Dec 23 '19 at 04:36
  • Do I still need to claim an arbitrary closed interval $[x,y]$ for Lipschitz continuity? Or I need to anyways for the mean value basis? – Yisroel Green Dec 23 '19 at 04:40
  • I don't quite get what you said. But in your question, those $x,y$ can run within the interval $I$, because $\sup_{I}|f'|<\infty$. – user284331 Dec 23 '19 at 04:42
  • My question is , for the mean value theorem we need $f $ to be continuous on$[a,b]$ and differentiable on $(a,b)$ how do we claim that in this instance? – Yisroel Green Dec 23 '19 at 04:43
  • I think in your question the so called $I$ must at least mean an interval. For suppose that $I=[a,b]$, the existence of the derivative $f'$ should include also the left-right derivatives too. For $I=(a,b)$, yes, maybe it is better to stress about the interval $[x,y]\subseteq(a,b)$ for $x<y$. – user284331 Dec 23 '19 at 04:47
  • You can claim that the derivative exists because$f'$ is bounded? Is it not possible for there to be discontinuity in the derivative if it is bounded? – Yisroel Green Dec 23 '19 at 04:58
  • The problem says $f'$... on $I$, we must therefore interpret it as $f'$ exists for every point in $I$. – user284331 Dec 23 '19 at 05:00