real analysis - Continuous differentiability implies Lipschitz continuity

Here's a statement in Zygmund's Measure and Integral on page 17:

If $f$ has a continuous derivative on $[a,b]$, then (by the mean-value theorem) $f$ satisfies a Lipschitz condition on $[a,b]$.

This does not seem obvious to me. How can I show it?

Also, what does a continuous derivative imply? Can we conclude the function is differentiable? If so, how can I prove it?

Answer

By the mean value theorem,

$$f(x) - f(y) = f'(\xi)(x-y)$$
for some $\xi \in (y,x)$. But since $f'$ is continuous and $[a,b]$ is compact, then $f'$ is bounded in that interval, say by $C$. Thus taking absolute values yields

$$\lvert f(x) - f(y)\rvert \le C \lvert x-y\rvert$$