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Let $f$ be an infinitely differentiable real function and let $Z(f)$ denote the set of points on which all derivatives of $f$ vanish. It is not hard to describe an $f$ such that $Z(f)$ is any specified finite set.

Must $Z(f)$ be always finite? If not, must it be discrete? If not, must it be countable?

I suppose the same questions apply to complex functions.

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    $\begingroup$ It shouldn't be too hard to show that, at least for functions $\mathbb{R}\to\mathbb{R}$, the set $Z(f)$ can be any prescribed closed set, no? e.g. by appealing to the fact that a given closed set is the complement of a countable union of disjoint open intervals, and building $f$ in some inductive way (specifying it first on all intervals longer than $1$, say) $\endgroup$ Aug 1, 2023 at 22:18

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I claim that whenever $M$ is a paracompact connected smooth manifold and $C$ is a closed subset of $M$, there is a smooth function $f:M\rightarrow[0,\infty)$ where $C=f^{-1}[\{0\}]$ and where all derivatives of $f$ are zero on $C$.

Let $M$ be a paracompact $C^\infty$-manifold with at most countably many components. Then there exists a sequence of compact sets $(K_n)_{n=0}^\infty$ where $M=\bigcup_{n=0}^\infty K_n$ and where $K_n\subseteq K_n^\circ$. Therefore, for each $n$, there is a smooth function $f_n:M\rightarrow[0,1]$ where $f_n[K_n]=\{0\}$ but where $f_n[K_{n+1}^c]=\{1\}$.

In this case, set $f=1+\sum_{n=0}^\infty f_n$. Then $f:M\rightarrow[1,\infty)$ is a smooth function where $f^{-1}[[1,r]]$ is compact for each $r<\infty$.

Now, if $M$ is a paracompact smooth manifold and $C$ is a closed subset of $M$, then set $U=M\setminus C$. Then $U$ is a paracompact smooth manifold with at most countably many components. Therefore, there is some function $f:U\rightarrow[1,\infty)$ where $f^{-1}[[1,r]]$ is compact whenever $r<\infty$. Therefore, extend $f$ to a continuous function $f_0:M\rightarrow[1,\infty]$ by setting $g(c)=\infty$ for $c\in C$. Let $g:M\rightarrow[0,1]$ be the function defined by letting $g(x)=1/f_0(x)$. Then $g$ is a continuous function that is smooth on the set $U$.

Theorem: Let $M$ be a paracompact connected manifold, and let $g:M\rightarrow\mathbb{R}$ is a continuous function that is smooth on the set $\{x\in M:g(x)\neq 0\}$. Then there is a smooth bijective function $H:\mathbb{R}\rightarrow\mathbb{R}$ where $H(0)=0$, $H\circ g$ is smooth everywhere, and where all higher order derivatives of $H\circ g$ are zero on the set $\{x\in M:g(x)=0\}$.

I gave a proof of the above result in my other answer here.

By applying the above result to the function $g$ and setting $h=H\circ g$, we obtain $C=h^{-1}[\{0\}]$ and we may conclude that all derivatives of all orders of $h$ are zero on the set $C$.

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  • $\begingroup$ Thanks for all the answers. I learned a lot today. $\endgroup$ Aug 2, 2023 at 11:50
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$\newcommand{\R}{\mathbb R}\newcommand{\J}{\mathcal J}$Let $f\in C^{\infty}(\R)$. Then $f$ is continuous and hence the set $Z(f):=f^{-1}(\{0\})$ is closed.

On the other hand, take any closed subset $C$ of $\R$. Then $D:=\R\setminus C$ is open and hence the union of an at most countable set $\J$ of nonempty pairwise disjoint open intervals.
Let \begin{equation} f:=\sum_{I\in\J}f_I, \end{equation} where, for $I=(a-h,a+h)$ and all real $x$, \begin{equation} f_I(x):=e^{-1/h}\psi((x-a)/h) \end{equation} and \begin{equation} \psi(u):=e^{1/(u^2-1)}\,1(|u|<1). \end{equation} Then $Z(f)=C$ and $f\in C^{\infty}(\R)$ (the latter fact follows because for any $t\in\R$, any nonnegative integers $k$ and $m$, and a varying nonempty interval $I=(a-h,a+h)$ we have (i) $f^{(k)}_I(x)=o(h^m(x-(a-h))^m)=o((x-t)^m)$ if $x\in I$, $a-h\ge t$, and $h\downarrow0$ and (ii) $f^{(k)}_I(x)=o(h^m(a+h-x)^m)=o((t-x)^m)$ if $x\in I$, $a+h\le t$, and $h\downarrow0$.)

So, a subset $C$ of $\R$ is the zero set of a function in $C^{\infty}(\R)$ iff $C$ is closed.

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  • $\begingroup$ Thanks for all the answers. I learned a lot today. $\endgroup$ Aug 2, 2023 at 11:50

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