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I am struggling to prove $$x+|x|$$ to be locally Lipschitz. Since it is not continuously differentiable - not differentiable at $x=0$ - it is hence not globally Lipschitz. But how do I proceed for the local case?

$$|f(x) - f(y)| \leq L|x-y|$$ $$\Rightarrow |x+|x|-y-|y||=...$$

mathsnovice
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2 Answers2

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$f(x)=x+|x|=\max\{0,2x\}$, so it's globally Lipschitz with constant 2.

max
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$|x+|x|-y-|y||=|(x-y)+(|x|-|y|)|\le |x-y|+||x|-|y|| \le |x-y|+|x-y| = 2|x-y|$. Thus $k = 2$. The key inequality is $xy \le |xy| = |x||y|$ which leads to $||x| - |y|| \le |x-y|$ which can be shown by simply squaring both sides.

  • I must have started completely wrong... what I did was: $f'(x) = 1 + \dfrac{x}{|x|}$ – mathsnovice Mar 08 '21 at 03:25
  • @mathsnovice: Yes, I think you forgot the triangle inequality... –  Mar 08 '21 at 03:26
  • I am still new to this and I have a question: For the case of $x^2$, what I did was $\text{sup}\dfrac{|x^2-y^2|}{|x-y|} = \text{sup}|x+y|=\infty$. I understand this completely. But I don't understand how can you add the term $||x|-|y||$ on the RHS of the inequality. Is that because we could try to create an inequality that will eventually lead to the Lipschitz form $|f(x)-f(y)|\leq L|x-y|$? – mathsnovice Mar 08 '21 at 04:09
  • @mathsnovice: That term comes from the triangle inequality, and yes you are right about the intention to create a situation that leads to the $L|x-y|$ –  Mar 08 '21 at 04:12
  • Thanks! I am not sure if I am doing it right if the first term is an $x^2$, i.e. $f = x^2 + |x|$: $|x^2+|x| - y^2 - |y|| = |x^2-y^2+(|x|-|y|) \leq |x-y|(|x+y|+1)$, $L$ here is then $L=(|x+y|+1)$ which is unbounded and is therefore locally Lipschitz. I can reason it from its sketch but not sure how to prove it. – mathsnovice Mar 08 '21 at 06:00
  • @mathsnovice: Ah, for the new one $f(x) = x^2+|x|$, you don't have global Lipschitz but you do have local Lipschitz on a bounded interval. –  Mar 08 '21 at 06:02
  • I tried to come up with some sort of inequality just like how you did when $x$ is linear.
    $|x^2+|x| - y^2 - |y|| = |x^2-y^2+(|x|-|y|) \leq |x-y|(|x+y|+1)$, $L$ here is then $L=(|x+y|+1)$ Does this look right?     – mathsnovice Mar 08 '21 at 06:06
  • @mathsnovice: $L$ must be constant. –  Mar 08 '21 at 18:49