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Suppose a continuous function $f(u)<K(1+|u|)$ for some positive number $K$. How can we find a sequence of Lipschitz functions $f_{n}$ that converge to $f$ uniformly on $\mathbb{R}$. If we require $f(u)<K(1+|u|^{\gamma})$ for $\gamma>1$, then the uniform convergency would be changed to just convergency by point limit? My idea is to use the cutoff function to approximate it. But I do not know how to prove the uniformly convergency using the assumption on $f$.

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2 Answers 2

I would try the following. For $n>K$, consider the function $f_n(x)=\sup_{u\in\mathbb R} (f(u)-n|u-x|)$. Then check that

  • $f_n$ is well defined (supremum is attained and is finite)
  • $f_n$ is Lipschitz
  • $f_n\ge f$
  • $f_n\to f$ pointwise ... uniformity is not so clear.

In fact, I don't believe in uniform approximation. The function $f(u) = \sin u^2$ obviously satisfies the given bound, but for any Lipschitz function $g:\mathbb R\to\mathbb R$ we have $\sup|f-g|\ge 1$.

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The answer given above is proven in San Martin and Lepeltier paper.

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Please elaborate: In what publication? in what year? For your information, "link only" answers, as well as "citation only" answers are to be avoided. And here, you don't even have a complete citation. –  amWhy Dec 6 at 17:08

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