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Is every function of class $C^∞$ also (locally) Lipschitz continuous? If so, how can this be proven?

user76284
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    Lipschitz implies a linear growth of the function, which is pretty strong. You can easily imagine a smooth function which is not, the exponential function for example. Smoothness only implies locally Lipschitz. – Troy Woo Oct 15 '14 at 07:37
  • Sorry, I meant locally Lipschitz continuous. I have edited the question to reflect this intent. – user76284 Oct 15 '14 at 07:44

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Yes, in fact, every $C^1$ function is locally Lipchitz. This is because there is an interval around every $x$ where $f'(x)$ exists and is bounded; the bound on $f'(x)$ will be the Lipchitz constant in this neighborhood.

Mike Earnest
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