# Is there a restriction on the structure of the set of points where all derivatives of a $C^\infty$ real function are 0? [duplicate]

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.

• 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) Aug 1, 2023 at 22:18

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 $$C$$.

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

$$\newcommand{\R}{\mathbb R}\newcommand{\J}{\mathcal J}$$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 $$$$f:=\sum_{I\in\J}f_I,$$$$ where, for $$I=(a-h,a+h)$$ and all real $$x$$, $$$$f_I(x):=e^{-1/h}\psi((x-a)/h)$$$$ and $$$$\psi(u):=e^{1/(u^2-1)}\,1(|u|<1).$$$$ Then $$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.

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