On radially invariant smooth $f : \mathbb R^n \to \mathbb R$ and smooth dependence on the radius

Question

I'm working with radially symmetric functions on isotropic manifolds, and ran into a smoothness issue which I reduced to the following equivalent problem:

Let $f : \mathbb R^n \to \mathbb R$ be a smooth function which only depends on the squared radius $r^2(x) = \sum x_i^2$ of its argument. That is, there exists a (unique) function $g : \mathbb R_{\geq0} \to \mathbb R$ such that $f(x)=g(r^2(x))$.

Is $g$ smooth?

Here is what I can prove:

• $g$ is smooth on $\mathbb R_{>0}$. This is because $r^2$ is a submersion outside of the origin, hence locally it has smooth sections $\sigma$, i.e. $g(t) = f(\sigma(t))$ is smooth.
• $g$ is continuous. By the same argument, because $r^2$ has a (global) continuous section. (E.g. the inclusion $\mathbb R^+\subseteq \mathbb R \times \{(0,\ldots, 0)\}$)
• $g$ is $C^1$ everywhere. The proof is the calculation below:

For all $i$ we have $$ \frac{\partial f}{\partial x_i}(x) = g'(r^2(x)) \frac{\partial r^2}{\partial x_i}(x) $$ for $x \neq 0$. Multiplying by $x_i$ and summing gives $$ \sum_i \frac{\partial f}{\partial x_i}(x) x_i = 2r^2 \cdot g'(r^2(x)) $$ Note that the symmetry of $f$ implies $\frac{\partial f}{\partial x_i}(0) = 0$, $\frac{\partial^2 f}{\partial x_ix_j}(0) = 0$ and $\frac{\partial^2 f}{\partial x_i^2}(0) = \frac{\partial^2 f}{\partial x_j^2}(0)$ for all $i,j$. Thus the LHS is $$\begin{align*} \sum_i x_i \left(D\frac{\partial f}{\partial x_i}(0)(x) + o(r)\right) &= \sum_{i,j} \frac{\partial^2 f}{\partial x_ix_j}(0) x_ix_j + o(r^2) \\ &= \frac{\partial^2 f}{\partial (x^1)^2}(0) r^2 + o(r^2) \end{align*}$$ where I used that $\sum |x_i| \ll r$ by Cauchy-Schwarz. Hence $g'(t)$ has limit $\frac12\frac{\partial^2 f}{\partial x_1^2}(0)$ for $t \to 0$.

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I wonder if there's a conceptual way to understand this proof and possibly generalize it to higher derivatives of $g$.

Answer 1

Yes, $g$ is smooth. As a start, note that $g(t^2)$ is smooth in $t \in \mathbb R$. This is because we can recover $g(t^2)$ from $f$ by composing $f$ with the inclusion of the positive real line.

The problem is now equivalent to the following:

If $h : \mathbb R \to \mathbb R$ is even, then $h = g(t^2)$ for some smooth $g$.

That is exactly the question Composing a smooth even function and square root which is answered on Mathoverflow: https://mathoverflow.net/questions/77120