# 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 $$x_1 be arbitrary. $$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 $|f'(c)|\leq M$.

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.

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.

• You want $|h|$ in the denominator for your calculations to be true. – JKEG Nov 14 '19 at 22:32
• Thanks, fixed now. – Three.OneFour Nov 15 '19 at 16:49