# Is there a function on a compact interval that is differentiable but not Lipschitz continuous?

Consider a function $f:[a,b]\rightarrow \mathbb{R}$, does there exist a differentiable function that is not Lipschitz continuous?

After discussing this with friends we have come to the conclusion that none exist. However there is every chance we are wrong. If it is true that none exist how could we go about proving that? It is true that if $f$ is continuously differentiable then $f$ is Lipschitz, but what if we don't assume the derivative is continuous?

## 2 Answers

The map $f : [0,1] \to \mathbb{R}$, $f(0) = 0$ and $f(x) = x^{3/2} \sin(1/x)$ is differentiable on $[0,1]$ (in particular $f'(0) = \lim_{x \to 0^+} f(x)/x = 0$), but it is not Lipschitz (the derivative $f'(x)$ is unbounded).

• You mean $x^{3/2}\,\sin(1/x)$? – Julián Aguirre Jan 12 '16 at 10:31
• Yes @JuliánAguirre, thanks for catching that. – Najib Idrissi Jan 12 '16 at 10:32
• Boundedness of the derivative is a sufficient condition for a function being Lipschitzian. I'm not sure it is necessary. – egreg Jan 12 '16 at 10:39
• @egreg It is necessary: if $|f(x) - f(y)| \le K |x-y|$ for all $x$, $y$, then $|f'(x)| = |\lim_{y \to x} (f(y) - f(x))/(x-y)| \le K$. (The converse is proved used the MVT.) – Najib Idrissi Jan 12 '16 at 10:42
• Thanks for the answer. I'm unsure why you're looking at $\lim_{x \rightarrow 0^{+}} f(x)/x$ and not $\lim_{x\rightarrow 0^{+}} f^{\prime}(x)$. – mark Jan 12 '16 at 12:42

If you require $f'$ to be continious then there can be no $f$ since $f'([0,1])$ is compact and indeed bounded.

• No, Michael Albanese had alrady suggested that, but $f'(0)$ does not exist. – Hagen von Eitzen Jan 12 '16 at 10:37
• thanks! I would like to delete it but I can't find a button to do that on my phone app! – Maffred Jan 12 '16 at 10:41
• since I can't delete at least I write something uselfull i hope! :) – Maffred Jan 12 '16 at 10:58