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!
Best Answer
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.