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If $f:\mathbb{R}\rightarrow \mathbb{R}$ is locally Lipschitz, that is to say, on any compact set of $\mathbb{R}$, $f$ is Lipschitz on it. I am thinking about if in addition, $f$ is also bounded, can we conclude that $f$ is globally Lipschitz on $\mathbb{R}$?

Since I cannot imagine an example of bounded and locally Lipschitz function is globally Lipschitz, I tend to believe the answer is yes. Can anyone help with a counterexample or proof?

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The answer is no. As a counterexample, consider $f:\Bbb R \to \Bbb R$ given by $$ f(x) = \cos(x^2) $$ Notably, any $C^1$ function with an unbounded derivative will fail to be globally Lipschitz.

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A locally Lipschitz function is globally Lipschitz if it's quasiconvex (see http://www.math.jyu.fi/research/reports/rep100.pdf Lemma 2.2). Obviously a globally Lipschitz function is locally Lipschitz without needing the quasiconvexity requirement.

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    $\begingroup$ Not the function, but its domain is assumed quasiconvex there. $\endgroup$ – user357151 Jul 1 '18 at 9:52

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