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I encountered this theorem:

Suppose that $f$ is a differentiable function on $(a,b)$ such that there's $M \ge 0$ and for $x \in (a,b)$, $f'(x)$ is bounded by $M$, then $f$ is uniformly continuous.

So the theorem itself wasn't difficult to prove. I just used the MVT: For any $x,y \in R$ with $x<y$ the Mean-value theorem gives you a $L \in (a,b)$ such that $f(y)−f(x)=f′(L)(y−x)$ Since $|f′(L)|<M$, it follows that $|f(y)−f(x)|<M|y−x|$ Now for $\delta>0$, choose $\delta=\epsilon/M$, then $|x−y|<\delta$ then $|f(x)−f(y)|<\epsilon$

My question is about the other way of the theorem.

If the function was uniformly continuous then do we conclude that its derivative is bounded?

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  • $\begingroup$ What theorem? You stated some hypotheses but left off the conclusion... $\endgroup$ Nov 28 '21 at 14:09
  • $\begingroup$ @lulu sorry, I edited my post $\endgroup$
    – Hoda Bibo
    Nov 28 '21 at 14:12
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    $\begingroup$ The theorem you want is this: Suppose that $f:[a,b]\to \mathbb R$ is everywhere differentiable. Then the derivative $f'$ is bounded if and only if $f$ is Lipschitz. $\endgroup$ Nov 28 '21 at 17:49
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$\sqrt{x}$ is uniformly continuous but $\frac{1}{2\sqrt{x}}$ is not bounded.

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  • $\begingroup$ Ohhhhh! You know what I thought I needed to find a function that is uniformly continuous but not differentiable and its differentiation is not bounded. Thank you so much! $\endgroup$
    – Hoda Bibo
    Nov 28 '21 at 14:14
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    $\begingroup$ @HodaBibo Well it will be difficult to find the derivative of a function that is not differentiable $\endgroup$ Nov 28 '21 at 14:26
  • $\begingroup$ LOL! My brain just stopped working I was so confused, I'm sorry I bothered you with a dumb post. $\endgroup$
    – Hoda Bibo
    Nov 28 '21 at 14:30
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    $\begingroup$ @HodaBibo no problem at all. sometimes you can’t see the forrest for the trees. happens to me too a lot. $\endgroup$ Nov 28 '21 at 14:36
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    $\begingroup$ Depending on the conditions, you could still go with $\sqrt x$ on $(0,1)$. It's still uniformly continuous, it's differentiable on the entire interval, but the derivative is not bounded. $\endgroup$
    – Teepeemm
    Nov 28 '21 at 23:28

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