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Suppose that a function $f$ defined in an open set $U \subseteq \mathbb{R}^m$ is smooth at a point $p \in U$. Then we have that there exists an open set $U_n \subseteq U$ $($ say $U_{n+1} \subseteq U_{n} \subseteq U$ $)$ contains $p$ such that $f$ is $n$- derivable in $U_n$, that is, all $n$th-order partial derivatives of $f$ exist in $U_n$. Hence, the smooth domain of $f$ is $$V\overset {\text{def}}{=} \bigcap^{\infty}_{n=1}U_n ~.$$

My question:

Does there exist a function $f$ defined in an open set $U$ such that it is smooth at a point $p \in U$ and not smooth in any nonempty open subset of $U$ ? To put it another way, if we do not choose those $U_n$ too small artificially, then, is $V^{\circ}$ always an nonempty open set ?

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Idea too long for a comment: take for each $n\in\Bbb N$ a function $f_n$ s.t.:

$$0\le f_n^{(k)}\le 1, {0\le k\le n},$$ smooth in $(-1/n,n)$ and $\in C^n\setminus C^{n+1}$ in $(-2,2)\setminus[-1/n,1/n]$.

Define $$f=\sum_{n=1}^\infty\frac{f_n}{2^n}.$$

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