TITLE: Is a function Lipschitz if and only if its derivative is bounded?
QUESTION [39 upvotes]: Is the following statement true?

Let $f: \mathbb{R}\to\mathbb{R}$ be continuous and differentiable.
$f$ Lipschitz $\leftrightarrow \exists M:\forall x\in\mathbb{R}\ |f'(x)|\leq M$

If $f'$ is bounded, it is Lipschitz, that's obvious.
Does that work the other way around?
Let $f$ be $M$-Lipschitz, that is to say $\forall x_1, x_2\in\mathbb{R},\ |f(x_1) - f(x_2)| \leq M|x_1 - x_2|$, where $M$ is independent of $x_1, x_2$.
Let $x_1 <x_2$ be arbitrary.
$f$ is continuous, so by the mean value theorem,
there exists $c\in\mathbb{R}, x_1 < c < x_2$, such that $f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \Rightarrow |f'(c)| = |\frac{f(x_2) - f(x_1)}{x_2-x_1}| \leq M$.
Does this hold? Can you, using the mean value theorem, "reach" every point in the derivative?
Also, another question: If $f$ is Lipschitz, is it necessarily differentiable?
Thanks!

REPLY [3 votes]: On request, here is a multidimensional version of Three.One.Four's answer.
\begin{align*}
\|\nabla f(x)\| 
&= \sup_{v} \frac{\|\nabla f(x)v\|}{\|v\|}
=\sup_{v} \lim_{\lambda\to0}\frac{|\lambda|\|\nabla f(x)v\|}{|\lambda|\|v\|}
=\sup_{v} \lim_{\lambda \to0}\frac{\|\nabla f(x)(\lambda v)\|}{\|\lambda v\|}\\
&\le \sup_{v} \lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}
+ \frac{ \|f(x+\lambda v) - f(x) \|}{\|\lambda v\|}\\
&\le \sup_{v} \lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}
+ \frac{ M\|(x+\lambda v) - (x) \|}{\|\lambda v\|}\\
&= \sup_{v} \underbrace{\lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}}_{=0}
+ M\\
&\le M
\end{align*}
Note that we simply used the definition of the multidimensional derivative here.