If I have a function $f:\mathbb R\to\mathbb R^n$ I can say that it is globally Lipschitz in $t$ if its Jacobian is bounded in $t$.
However, does it work the other way around? If I find that the function Jacobian is not bounded, does this mean that the function is not globally Lipschitz?
Best Answer
A function with unbounded derivative is not Lipschitz. Indeed, if $f'$ is unbounded, then for every $M$ there is $x$ such that $\|f'(x)\|>M$. By the definition of derivative (used with $\epsilon = \|f'(x)\|-M$, there is $h\ne 0$ such that $$\left\|\frac{f(x+h)-f(x)}{h}\right\|>M $$ The inequality $\|f(x+h)-f(x)\|>M|h|$ shows that $f$ is not Lipschitz with constant $M$. And since $M$ is arbitrary, it's not Lipschitz.