I would like to know how to prove that a $C^1$ Lipschitz function has linear growth. (Actually I don't even know if it is true, it is a question in my exam – it says prove that, so it means it is true).
Thanks in advance.
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By "linear growth" do you mean that $\tfrac{|f(x)|}{|x|}$ is bounded for $|x|>1$? – amsmath Mar 27 '19 at 20:40
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yes, more precisely |f(x)|<=k(1+|x|) – Minkowski Yaacov Mar 27 '19 at 20:45
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No need for $C^1$: $$\frac{|f(x)|}{1+|x|}\le\frac{|f(x)-f(0)| + |f(0)|}{1+|x|}\le\frac{L(1+|x|) + |f(0)|}{1+|x|}\le L+|f(0)|.$$
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thanks but i just realize that i wanted to proof the opposite: that a function with linear growth is C^1 and Lipschitz. I'm sorry, my bad. – Minkowski Yaacov Mar 27 '19 at 20:51
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But I can answer that question right away: no. For a very simple reason: $|f(x)|\le k(1+|x|)$ just means that the graph of $f$ is contained in some sector. But it can be anything there. Even completely non-continuous. – amsmath Mar 27 '19 at 20:59