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?
 A: Since $f'$ is continuous it is bounded on $[a,b]$. Choose $M\in \Bbb R$ such that $|f'(x)|\le M$ for all $x\in [a,b]$. If $x,y\in [a,b]$, then by the MVT there is $\xi\in [a,b]$ with $f(x)-f(y)=f'(\xi)(x-y)$, thus $|f(x)-f(y)|\le M|x-y|$. In fact you can show that a differentiable function on an open interval (not necessarily a bounded interval) is Lipschitz continuous if and only if it has a bounded derivative. This is because any Lipschitz constant gives a bound on the derivative and conversely any bound on the derivative gives a Lipschitz constant.
To your other question: Continuous derivative does not imply twice differentiable. In fact, we know that there are functions $f$ (such as the Weierstrass-function) which are continuous but not differentiable. Taking the antiderivative $F$ gives us a differentiable function with derivative $f$ (by the Fundamental Theorem of Calculus), so it has a continuous derivative, but as $f$ is not differentiable, $F$ is not twice differentiable.
A: 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 
$$\lvert f(x) - f(y)\rvert \le C \lvert x-y\rvert$$ 
