Differentiability of spherically symmetric functions

Question

Let $g\in C^{2n}(\mathbb R)$ be an even function, and suppose $f:\mathbb R^d\to \mathbb R$ is defined by $f(x) = g(|x|)$. Is it true that $f\in C^{2n}(\mathbb R^d)$?

My attempt: The norm function $|\cdot|:\mathbb R^d \to \mathbb R$ is smooth away from $0$, thus by the chain rule $f$ is $C^{2n}$ away from $0$. Also note that since $g$ is even, its existing odd derivatives at $0$ vanish there. Thus Taylor expanding $g$ about $0$ we get $$f(h) = g(0) +\frac{|h|^2}{2}g''(0) +\dots+ \frac{|h|^{2n}}{(2n)!}g^{(2n)}(0)+o(|h|^{2n}).$$ By using $|h|^{2k} = (h_1^2+\dots+h_d^2)^k$ for all $k$, we see that $f$ has a Taylor expansion of order $2n$ at $0$, or in other words that $f$ is $2n$-times Peano differentiable at $0$. However Peano differentiability is weaker that classical differentiability in general. I wanted to use Corollary 3 from https://www.ams.org/journals/proc/2013-141-07/S0002-9939-2013-11529-7/S0002-9939-2013-11529-7.pdf which says that if the Peano derivatives exist and are bounded in a neighbourhood, then the classical derivatives exist in that neighbourhood, however I don't know if/how to show boundedness of the higher order derivatives of $f$ around $0$, since the higher order derivatives of $|\cdot|$ are not bounded around $0$ in general (e.g. $\frac{\partial^2}{\partial x_1\partial x_2} |\cdot|$ is unbounded around $0$).

Any help is appreciated. I suspect there is a different way to look at this problem which makes things clear, but I am not able to see how.

Answer 1

After more thought, I have come up with the following solution. It is inspired from the answer to Smoothness at the origin of a radial function obtained by rotating an even function. I have adapted the proof method from that answer, as its direct application yields something like $g\in C^{2n}(\mathbb R) \Rightarrow f\in C^n(\mathbb R^d)$ as opposed to $g\in C^{2n}(\mathbb R) \Rightarrow f\in C^{2n}(\mathbb R^d)$.

Preliminary result: First we suppose $g\in C^n(\mathbb R\setminus\{0\})$ and that the $j$-th order partial derivatives of $g$ are $o(x^{k-j})$ as $x\to 0$, for some $k\in\mathbb Z$ and all $0\leq j\leq n$. We show by induction on $n$ that $f:= g\circ |\cdot|:\mathbb R^d\setminus\{0\}\to \mathbb R$ has $j$-th order partial derivatives which are $o(|x|^{k-j})$ as $x\to 0$, for all $0\leq j\leq n$.

$n=0$ is clear; the growth/decay of $f$ at $0$ is the same as that of $g$. For $n>0$, note that \begin{align}\label{eq: partial f} \frac{\partial f}{\partial x_i}(x) = g'(|x|)\frac{x_i}{|x|}= h(|x|)x_i \end{align} for $x\in\mathbb R^d\setminus\{0\}$ and $1\leq i\leq d$, where $$h(x) := \frac{g'(x)}{x}$$ for $x\in \mathbb R\setminus \{0\}$. Clearly $h\in C^{n-1}(\mathbb R\setminus \{0\})$ with $$h^{(j)}(x) = \sum_{l=0}^{j}g^{(l+1)}(x)(-1)^{j-l}(j-l)!x^{-1-j+l}=o(x^{k-j-1})$$ as $x\to 0$, for all $0\leq j\leq n-1$. By induction hypothesis the $j$-th order partial derivatives of $h\circ |\cdot|$ are $o(|x|^{k-j-1})$ as $x\to 0$, thus by the expression for $\frac{\partial f}{\partial x_i}$ we see that the $j$-th order partial derivatives of $f$ are $o(|x|^{k-j})$ as $x\to 0$. This concludes the induction step.

Main result: Now we are interested in $g\in C^{2n}(\mathbb R)$ even. So the existing odd derivatives of $g$ vanish at $0$. So defining $\tilde g\in C^n(\mathbb R)$ by $$\tilde g(x) = g(x)-\sum_{j=0}^n\frac{x^{2j}}{j!}g^{(2j)}(0),$$ we have that the $j$-th order partial derivatives of $\tilde g$ are $o(x^{2n-j})$ as $x\to 0$. So $\tilde f:= \tilde g\circ |\cdot|:\mathbb R^d \to \mathbb R$ is $2n$-times continuously differentiable on $\mathbb R^d\setminus\{0\}$, and the $j$-th order partial derivatives are $o(|x|^{2n-j})$ as $x\to 0$. In particular the $2n$-th partial derivative of $\tilde f$ has the limit $0$ at $0$, and hence by the mean value theorem $\tilde f$ is $2n$-times differentiable at $0$. Now $$f(x) = \tilde f(x) +\sum_{j=0}^n\frac{|x|^{2j}}{j!}g^{(2j)}(0)$$ for all $x\in \mathbb R^d$. Since $|x|^{2j} =(x_1^2+\dots+x_d^2)^j$ for all $j$, we see that $f$ is the sum of two $C^{2n}(\mathbb R^d)$ functions. So $f\in C^{2n}(\mathbb R^d)$.