# Is a function Lipschitz if and only if its derivative is bounded?

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!

Assume that $f$ is a M-Lipschitz function. So $|f(x+h)-f(x)|\leq M|h|, \quad \forall x, h \in \mathbb{R}.$ It is equivalent to $|\frac{f(x+h)-f(x)}{h}|< M$. By taking que limit, $|f'(x)|\leq M$.
For the converse, use the mean value theorem. Let $x,y \in \mathbb{R}$, there exists $c\in \mathbb {R}$ so that $f(x)-f(y)=(x-y)f'(c)$ (you have to assume that $f$ is differentiable) and now use the fact that $|f'(c)|\leq M$.
You don't have to use the mean value theorem. Just use the definition of the derivative: $$f'(x) = \lim_{h \to 0} \frac{|f(x+h)-f(x)| }{h} \leq \lim_{h \to 0} \frac{M |x+h -x|}{h} = M.$$
And no, Lipschitz functions don't have to be differentiable, e.g. the absolute value $| \cdot |$ is Lipschitz. Lipschitz functions are differentiable almost everywhere though.