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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.

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

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

Dot dot dot.

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