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?