4

Suppose $x\in \mathbb{R}^n$ and $\|.\|_2$ is the Euclidean norm. Is $e^{-\|x\|^2}$ Lipschitz? If so, what is the Lipschitz norm?

Here is my attempt:

$\big(e^{-\|x\|^2}-e^{-\|y\|^2}\big) \leq |\|x\|^2-\|y\|^2| = |\|x\|-\|y\||(\|x\|+\|y\|)\leq \|x-y\|(\|x\|+\|y\|)$

Do we need to define it on a compact set to make it Lipschitz?

mathlover
  • 121
  • 2
    Let's say that I wrote $$2= 1+1\le 5+5\le 10$$ Did I just prove that $2\le3$ is false? –  May 23 '17 at 07:59
  • The negation of "there exists $K \ge 0$ such that for all $x, y \in \mathbb{R}^n$ $|f(x) - f(y)| \le K||x - y||$" is "for all $K \ge 0$ there exists $x, y \in \mathbb{R}^n$ such that $|f(x) - f(y)| > K||x - y||$" which is not what you have proved. – Alex Vong May 23 '17 at 08:08
  • Thanks. I didn't really mean that. That was bad wording. I just meant I can't find that $K$. – mathlover May 23 '17 at 08:13
  • @mathlover : Have a look at https://math.stackexchange.com/questions/1929299/proving-e-x-is-lipschitz – pitchounet May 23 '17 at 08:13
  • @jibounet Thanks but you can't bound $||x|^2-|y|^2|$ with $K |x-y|$ – mathlover May 23 '17 at 08:23
  • @mathlover: Is the one-variable function $f(x) = e^{-x^{2}}$ Lipschitz? If so, what is the Lipschitz norm? And then, why does this solve the general problem? (It probably helps to think about the level sets of your function, and how, if $x$ is fixed, to move $y$ along a level set until $y$ is "as close as possible" to $x$.) – Andrew D. Hwang May 23 '17 at 11:02
  • 2
    @AndrewD.Hwang So since |$-2x e^{-x^2}|\leq \sqrt{2/e}$ and based on https://math.stackexchange.com/questions/1257553/a-multivariate-function-with-bounded-partial-derivatives-is-lipschitz , we can conclude that $e^{-|x|^2}$ is lipschitz with the same Lipschitz norm, i.e. $\sqrt{2/e}$. Right? – mathlover May 24 '17 at 18:05
  • 1
    Yes to your estimate for $e^{-x^{2}}$, and yes, it does appear you can use the linked post (which differs from the strategy I suggested, the linked posting being more general, but possibly requiring a bit more work). – Andrew D. Hwang May 24 '17 at 18:22
  • @AndrewD.Hwang Thanks a lot! – mathlover May 24 '17 at 18:43

0 Answers0