Does global lipschitz imply locally lipschitz? [closed]

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:

$$|f(x)-f(y)|\leq K|x-y|$$

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 $$|f(x)-f(y)|\leq L|x-y|$$

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

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

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

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