# Lipschitz function where $f ^2$ is not Lipschitz [closed]

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.

## closed as off-topic by T. Bongers, MisterRiemann, Strants, YiFan, ShaileshMar 20 at 2:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – T. Bongers, MisterRiemann, Strants, YiFan, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• Try $f(x)=x$ :) – badatmath Mar 19 at 17:33
• @orange simple as that? really? – Tigran Minasyan Mar 19 at 17:34
• 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. – Thomas Andrews Mar 19 at 17:42

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