If $f(x)$ is Lipschitz, then is $xf(x)$ Lipschitz?

Question

Suppose $f(x)$ is Lipschitz (globally or locally), what can we say about the product $xf(x)$. Is it also Lipschitz?

Answer 1

If the domain is bounded, yes. Otherwise, this need not be true: $f(x) = x\sin x$ is not Lipschitz because $f'(x) = \sin x + x \cos x$ is unbounded (or more simply, as stated the comment, $f(x) = x^2$ is not Lipschitz because $f'(x)$ is unbounded)