I have the following question from my notes, where $f$ and $g$ are Lipschitz functions on $A ⊂ \Bbb R$.

I'm able to show that the sum $f + g$ is also a Lipschitz function, however I'm stuck on trying to show that if $f,g$ are bounded on $A$, then the product $fg$ is Lipschitz on $A$.

Also, is there a valid example of a Lipschitz function $f$ on $[0,+∞)$ such that $f^2$ is not Lipschitz on $[0,+∞)$?

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    $\begingroup$ As for the valid example, the simplest possible works wonder. $\endgroup$ Commented Nov 19, 2014 at 9:57

1 Answer 1


Suppose $|f| \leq M$ and $|g| \leq M$.

Then $$\begin{aligned} |(fg)(x) - (fg)(y)| &\leq |f(x)g(x) - f(x)g(y)| + |f(x)g(y) - f(y)g(y)| \\&\leq M(|g(x) - g(y)|+|f(x)-f(y)|) \end{aligned}$$

Consider $f(x) = x \, \forall x \in [0,\infty)$ which is Lipchitz on $[0, \infty)$ but $f^2$ is not.

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    $\begingroup$ @ABIM I don't think that is true. If we assume $f,g$ to be $C^1$ then the statement is if $f', g', f, g$ are bounded, then $(fg)'=f'g + fg'$ is bounded. Clearly this is no longer true if only one of $f,g$ is bounded. A concrete counterexample would be $f(x)=x, g(x)=\sin(x)$. Pick $x_n=2\pi n +\pi, y_n=2\pi n$, then we have $$\vert f(x_n)g(x_n)-f(y_n)g(y_n)\vert = \vert x_n - 0 \vert = 2\pi n +\pi =(2n+1)\pi = (2n+1) \vert x_n -y_n\vert.$$ $\endgroup$ Commented Mar 25, 2023 at 16:48

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