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?

$f(x)=x$ is Lipschitz on $\mathbb{R}$, but $xf(x)=x^2$ is not. — carmichael561, Jul 05 '16 at 21:54

score 4 · Accepted Answer · answered Jul 05 '16 at 21:57

4

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)

answered Jul 05 '16 at 21:57

You're right; removing comments. – egreg Jul 06 '16 at 05:40

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

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