I've been reviewing some analysis for an exam and have gotten stuck on this problem (Strichartz 10.2.4 #20):
$$\text{If }f:\mathbb{R} \rightarrow\mathbb{R} \in \mathcal{C}^k,f(-x)=f(x)\text{, show that }F:\mathbb{R}^n \rightarrow \mathbb{R}\text{ defined by }F(x)=f(||x||_2)\text{ is }\mathcal{C}^k$$
I started out by checking the special case where $k=1$, which came down to just using the chain rule: $\partial_{x_i}F(x)=f'(|x|)\dfrac{x_i}{|x|}$ and we get that it is $0$ when $x=0$ because of the even property, so it is continuous at all points for any arbitrary partial derivative. I feel like there's a way to make an induction argument out of this by taking a derivative and showing that the partial is $\mathcal{C}^k$ by the induction hypothesis, but I can't get it to work. So far, I have:
Define $g_i(t)=x_i\dfrac{f'(t)}{t}$ for $t \geq 0$. It suffices to show that $g_i \in \mathcal{C^k}$ because then $\partial_{x_i}F(x)=g_i(|x|)$ and the problem claim holds by the induction hypothesis. We have that $g_i^{(k)}(t)=x_i\sum\limits_{j=0}^k\binom{k}{j}f^{(k-j+1)}(t)\cdot(-1)^{j}\dfrac{j!}{t^{j+1}}$ but I don't see how to proceed from here, or even why this is necessarily continuous. I would appreciate any suggestions.
We have $f(|x|)=f(x)$ for $x\geq0$ while $f(|x|)=f(-x)=f(x)$ for $x<0$ where the second equality follows by the evenness of $f(x)$. So $f(|x|)=f(x)$ for all $x\in\Bbb R$. Thus, if $f(x)$ is $k$-times differentiable, so is $f(|x|)$ when $f$ is an even function.