Is a nonconvex function not locally Lipschitz?

Question

A convex function is locally Lipschitz on a closed and bounded set. This implies that a nonconvex function may not be locally Lipschitz. Is every nonconvex function not locally Lipschitz?

Answer 1

No, that would be ridiculous. For instance, every $C^1$ function is locally Lipschitz, but it needs not be convex.

Answer 2

No, take for example $\sin(x)$.

Answer 3

If this were true, why would we need different terms "Lipschitz" and "convex"? A convex function is a function that is concave upward, so convexity is not preserved by vertical reflection. But Lipschitz continuity is preserved by relections, so given any function $x \rightarrow f(x)$ that is convex, $x \rightarrow -f(x)$ will be nonconvex but Lipschitz. (And if you expand the question to "Okay, will any Lipschitz function be either always concave up or always concave down?", that hypothesis can be falsified very easily by simply gluing two functions with different concavities together. For instance, a function that is $x^2$ for positive $x$ and $-x^2$ for negative $x$ will be neither always concave up nor always concave down, but will be Lipschitz.)

BTW "Is a locally Lipschitz function convex>" is equivalent to "Is a nonconvex function not locally Lipschitz?", and a bit more direct way of phrasing it.