A function is smooth at a point and not smooth in any neighbourhood of it, exist or not? 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 ?
 A: 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
$$f=\sum_{n=1}^\infty\frac{f_n}{2^n}.$$
A: For continuous $f$ on $U:=(-1,1)$, let $I(f)$ be the function $x\mapsto \int_0^xf(t)\,\mathrm dt$.
Let $$f_n=I^n(x\mapsto \max\{|x|-\tfrac1n,0\}).$$
Note that this makes $|f_n|<\frac1{n!}$ on $U$. Then $f:=\sum_n f_n$ converges uniformly. We see that $f^{(n)}$ exists on $(\frac1n,\frac1n)$, but $f^{(n+1)}(\pm\frac1n)$ does not. So all derivatives of $f$ exist (and are zero) at $p=0$, but $f$ is not smooth in any neighbourhood of $p$. This is only halfway an answer to your question though (as $V$ does have a larger open subset).
So we repeat the above construction with a twist:
Let
$$ g_n=I^n(x\mapsto \tfrac1n\lfloor n|x|\rfloor )$$
Again, we have uniform convergence and $g^{(n)}$ exists on $(-\frac1n,\frac1n)$, but now $g^{(n+1)}(\tfrac kn)$ does not exist for $-n<k<n$, $k\ne 0$.  This time, we find that $V\cap \Bbb Q=\{0\}$, so $V$ has empty interior, as desired.
