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Let $K$ be a compact set of $R^n,$ and let $U$ be an open neighborhood of $K$.

I have the following question: Show that there exists a smooth, compactly supported function $f\in {\cal C}^\infty(R^n)$ supported in $U$ which equals $1$ on $K$ and its derivatives $f^{(k)}$ are such that $ |f^{(k)}(x)|\le 1,\; a.e. \;x\in U, k=0,1,2.$

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    @IgorBelegradek I don't see how to do the normalization that you mentioned. Suppose, for example, $n=1$, $K=[0,1]$, and $U=(-\epsilon,1+\epsilon)$. Then $f(x)$ has to smoothly increase from 0 to 1 as $x$ goes from $-\epsilon$ to $0$, which is incompatible with $|f'(x)|\leq1$ when $\epsilon<1$. (The "a.e." doesn't help, as $f'$ is supposed to be continuous too.) – Andreas Blass Oct 15 '16 at 22:58
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    @AndreasBlass: you are right of course. I overlooked the assumption that $f$ is $1$ on $K$ and thought it just has to be constant there. For the OP: $f$ does not exist because $1=|f(1+\epsilon)-f(1)|=|\int_1^{1+\epsilon} f^\prime|\le (1+\epsilon-1) max|f^\prime|\le \epsilon$. – Igor Belegradek Oct 15 '16 at 23:38

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