# Is there a counter example for this statement?

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 $$x the Mean-value theorem gives you a $$L \in (a,b)$$ such that $$f(y)−f(x)=f′(L)(y−x)$$ Since $$|f′(L)|, it follows that $$|f(y)−f(x)| Now for $$\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?

• What theorem? You stated some hypotheses but left off the conclusion... Nov 28 '21 at 14:09
• @lulu sorry, I edited my post Nov 28 '21 at 14:12
• 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. Nov 28 '21 at 17:49

$$\sqrt{x}$$ is uniformly continuous but $$\frac{1}{2\sqrt{x}}$$ is not bounded.
• 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. Nov 28 '21 at 23:28