I found this problem when going through Lipschitz functions, I wanted to solve it but could not come up with any such function. Help is appreciated.

Give an example of a Lipschitz function $f$ on $[0,∞)$ such that its square $f^2$ is not a Lipschitz function.


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  • 3
    $\begingroup$ Try $f(x)=x$ :) $\endgroup$ – badatmath Mar 19 at 17:33
  • $\begingroup$ @orange simple as that? really? $\endgroup$ – Tigran Minasyan Mar 19 at 17:34
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    $\begingroup$ I believe, Ii the domain of $f$ is bounded and $f$ is Lipshitz, then so $f^2.$ So counterexamples will have to be be on and unbounded domain. $\endgroup$ – Thomas Andrews Mar 19 at 17:42

$f(x)=x$. $g(x)=x^2$ is not Lipschitz.

  • $\begingroup$ are you sure about f(x) = x? it says $[0, ∞]$ $\endgroup$ – Tigran Minasyan Mar 19 at 17:43
  • $\begingroup$ Your question says $[0,\infty)$ $\endgroup$ – badatmath Mar 19 at 17:45
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    $\begingroup$ @TigranMinasyan $x \mapsto x$ is Lipschitz everywhere, because $|x-y| \leq |x-y|$ for all $x,y$... $\endgroup$ – TheSilverDoe Mar 19 at 17:45
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    $\begingroup$ And $x^2$ is not because $|(x+1)^2-x^2|=2x+1$ is not bounded. $\endgroup$ – Thomas Andrews Mar 19 at 17:50

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