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I know that $\mathcal{C}^k$ for $k\geq 1$ is dense in the space of Lipschitz funcions. My question in fact is: If $\{f_n\}_{n\geq 0}\subset \mathcal{C}^k$ such that $f_n \to f$ where $f$ is only Lipschitz. If such sequence $\{f_n\}_{n\geq 0}$ can be found in such a way that all derivatives are bounded and $\sup_n \|f_n^{(k)}\|_{\infty}<\infty$. So that I can upperbound "things" by the derivatives, and still get convergence when I let $n$ go to infinity.

Thanks a lot for your help! :)

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If $(f_n^{(k)})_n$ is bounded, by Arzela-Ascoli, $(f_n^{(k-1)})_n$ has a locally uniform convergent subsequence, so $f$ is $C^{k-1}$. –  martini Dec 12 '12 at 10:06
    
Does this then mean that I can not approximate an $f$ which is only Lipschitz by $f_n$ with bounded derivatives? mm.. what about $f(x)=|x|$? I can approximate it by $f_n\in \mathcal{C}^2$ with $f_n'$ bounded but then $f_n''$ is not bounded? Do I understand it correctly? Thank you very much for your help! –  Daniel Dec 12 '12 at 10:37
    
You can approximate an Lipschitz $f$ with functions having bounded first derivatives, as the above argument gives a $C^0$-convergent subsequence, no problem. But the second derivatives cannot be bounded. –  martini Dec 12 '12 at 13:19

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