Let $f: \mathbb R \to \mathbb R$ be a differentiable function such that $|f'(x)| \le 100$ for all $x \in \mathbb R$. Prove that $f$ is uniformly continuous.
I don't know how to start this proof. I know the proof is going to have the following reasoning somewhere in the proof but I am having trouble putting the pieces together:
If we use the mean value theorem, then we have $$\left| \frac{f(b)-f(a)}{b-a}\right|=\left| f'(c) \right| \le 100.$$
So, $$\left| \frac{f(b)-f(a)}{b-a}\right| \le 100.$$ Which implies $$|f(b)-f(a)|\le 100|b-a| < 100 \delta.$$
So, if we let $100\delta =\epsilon$, then we can use $\delta= \epsilon /100$ in the definition of uniform continuity.
However, I am having trouble in the beginning part of the proof. How do I use $|x-x_0| < \delta$ in the mean value theorem? Or maybe I need to use $|x-x_0| < \delta$ after I use mean value theorem?
I am having trouble because MVT is stated in terms of a closed interval. How do I work in this closed interval to invoke MVT in the proof?
Your reasoning is almost going through: $|f(x)-f(y)|\leq 100|x-y|<100\delta=\epsilon$ whenever $|x-y|<\delta=\epsilon/100$.
Here are the proper steps: Fix an $\epsilon>0$, for every $x,y$ with $|x-y|<\epsilon/100$, then there exists some $\eta_{x,y}$ lies in between $x$ and $y$ such that $|f(x)-f(y)|=|f'(\xi_{x,y})||x-y|$, so $|f(x)-f(y)|<\epsilon$.