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

Question

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!

Answer 1

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 $|f'(c)|\leq M$.

As the previous poster said, Rademacher's theorem says that every Lipschitz function is almost everywhere differentiable.

A simple example of non differentiable Lipschitz function is the absolute value.

Answer 2

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.

Answer 3

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.