# Lipschitz implies bounded derivative?

Is there any function which is Lipschitz but has an unbounded derivative ? I guess there isn't but I have not found one.

## 1 Answer

By definition of Lipschitz, $f'$ is necessarily bounded since $|f(x) -f(y)|\le L |x - y|$ which implies that $$|g'(x)| = \left|\lim_{y\to x} \frac{f(y) - f(x)}{y-x} \right| \le L$$
for all $x$ (assuming $g'(x)$ exists).

• I believe this also shows that if $f'$ exists and is (uniformly) bounded then $f$ is Lipschitz, correct? – Darrin Nov 9 '14 at 17:55
• Yes the condition is sufficient too, in addition to being necessary. – O_huck Nov 10 '14 at 12:48
• To be clear, we're assuming that $f$ has codomain $\mathbb{R}$, then? I ask because the notation $d_y$ suggests a more general context. – Jesse Madnick Apr 20 '15 at 11:47
• It should be noted that a Lipschitz function need not be differentiable. Example |x|. If a Lipschitz function is differentiable then the derivative is bounded. – Kavi Rama Murthy Dec 30 '16 at 8:21