This is just a quick question:

I'm a little confused as to whether or not globally Lipschitz continuous implies Locally Lipschitz?

I'm aware that if $f$ is globally lipschitz, it means there is a positve $K$ such that for all $x, y \in \mathbb{R}^n$ then:

\begin{equation} |f(x)-f(y)|\leq K|x-y| \end{equation}

If it's locally, then for every point $a$ contained in an open subset of $\mathbb{R}^n$, there exists a small neighborhood around $a$ and a positive constant $L$ such that for all $x,y \in N_{\delta}(a)$ then \begin{equation} |f(x)-f(y)|\leq L|x-y| \end{equation}

I would think it global implies local, but I'm not sure to be honest.

Thank you for your time.


closed as off-topic by Guy Fsone, kimchi lover, JonMark Perry, Parcly Taxel, Namaste Jan 30 '18 at 2:42

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Guy Fsone, kimchi lover, JonMark Perry, Parcly Taxel, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ By definition $L=K$ will work. Typically "global property" implies "local property". $\endgroup$ – charlestoncrabb Nov 13 '15 at 19:58

Yes, it does. Take $L=K$.

Dot dot dot.


Not the answer you're looking for? Browse other questions tagged or ask your own question.