Let $f,g:A\to \mathbb R $ be Lipschitz continuous functions. Suppose $f $ is bounded.
Is $f\cdot g $ Lipschitz continuous? (We can ask the question for several $A $, say $A=]0,1[$ or $A=\mathbb R$ if the answer depends on $A $)
It is easy to prove that if both function a are bounded, this is true. But what if only one is assumed to be bounded?
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$\begingroup$If $A$ is bounded then the result is trivially true, because both $f$ and $g$, being Lipschitz continuous, are necessarily bounded.
If $A$ is unbounded then the claim is in general not true. Consider, for example, $f(x) = \cos(x)$, $g(x) = x$, $x\in\mathbb{R}$. Then $h(x) := f(x) g(x) = x\, \cos(x)$ is not a Lipschitz function in $\mathbb{R}$.
To check this claim, it is enough to consider the points $x_k := 2k\pi$, $y_k := 2k\pi + \pi/2$, $k\in\mathbb{N}$, so that $$ |x_k - y_k| = \pi/2, \qquad |h(x_k) - h(y_k)| = 2k\pi. $$
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