Multivariate differentiability verification I have tried an attempt on the following:
Let $F:U\in \Bbb{R}^n\to \Bbb{R}$ and $f:\Bbb{R}\to \Bbb{R}$ where $f$ is an even function. Now $F(\mathbf{x})=f(|\mathbf{x}|)$, where $|\ . |$ is the Euclidean norm.
Given $f$ is $C^r$, I have to show that $F$ is $C^r$.
My attempt was to argument by induction. I showed that $F$ is $C^1$, basically using univariate chain rule on $\frac{\partial f\left(\sqrt{x_1^2+\dots+x_n^2}\right)}{\partial x_1}$ and arguing that it would be continuous using the fact that $f$ is $C^r$. Since the argument works for all $x_i$, it establishes first-differntiability for $F$ and continuity of the operator $DF(\mathbf{x})$. Then I moved on to show $F$ is $C^r$ assuming $F$ is $C^{r-1}$ and $f$ is $C^r$. Now what I did was partially differentiate:
$$
{\partial \over \partial x_2} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right] = \underbrace{{\partial \over \partial |\mathbf{x}|} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right]}_{(1)} . \underbrace{{\partial |\mathbf{x}| \over \partial x_2}}_{(2)}
$$
My argument runs as follows: this partial derivative is continuous since $f$ is $C^r$ and hence $(1)$ is continuous. Continuity of $(2)$ is establishd in a similar manner, since $\sqrt{\mathbf{x}}$ is $C^r$ if $\mathbf{x} \neq \mathbf{0}$. Then the product would be continuous. I am uneasy on multivariate arguments and dont know if my approach is completely off track. I didnt use the even function property, so I am sure I might me missing something. Any hints and references would be welcome. Thanks!
 A: Problem: ${\bf x}\mapsto |{\bf x}|$ isn't differentiable at ${\bf x}=0$.
EDIT: Special reasoning is required at $|{\bf x}|=0$. Proof for $r=1$:
As $f$ is even, $f'(0)=0$. The presumed differential $DF({\bf 0}):\Bbb{R}^n\longrightarrow\Bbb{R}$ must be zero and using the MVT:
$$
{f(|{\bf x}|)-f(|{\bf 0}|)-0\over|{\bf x}|}=f'(\xi_{\bf x}),
$$
with $0<\xi_{\bf x}<|{\bf x}|$.
Then, $\lim_{{\bf x}\to 0}\xi_{\bf x}=0$ and using than $f'$ is continuous,
$$
\lim_{{\bf x}\to 0}{f(|{\bf x}|)-f(|{\bf 0}|)-0\over|{\bf x}|}=\lim_{{\bf x}\to 0}\,f'(\xi_{\bf x})=f'(0)=0,$$
and by definition $DF({\bf 0})=$ zero indeed.
EDIT 2: If we can assume ${\bf x}\ne{\bf 0}$, then all is trivial. As $N: {\bf x}\mapsto |{\bf x}|$ is $C^\infty$ in ${\Bbb R}^n\backslash\{{\bf 0}\}$ and $f$ is $C^r$, by the chain rule $F=f\circ N$ is $C^r$ without more hypothesis about $f$. But this problem is interesting only at ${\bf x}={\bf 0}$.
A: Since $F$ is rotation invariant, if $\partial_1 F$ exists and continuous at a point, then the same is true for $\partial_j F$ in the corresponding rotated point. If we want to prove the continuity of (mixed and pure) partial derivatives it is enough to consider $\partial_1^m F$ in the points $\mathbf{x}=(x_1,0,\ldots,0)$. Here $F(\mathbf{x})=f(|x_1|)$. For the sake of simplicity in what follows I write $x$ instead of $x_1$. Since $f$ is even we can write the Taylor formula in the following way
$$
f(x)=f(0)+\frac{f^{(2)}(0)}{2!}x^2+\ldots+\frac{f^{(2m)}(0)}{(2m)!}x^{2m}+\ldots+\int_0^x\frac{f^{(r)}(t)}{(r-1)!}(x-t)^{r-1}\,dt.
$$
Here we used Lagrange remainder in Taylor series expansion. Since $F(\mathbf{x})=f(|x|)$ we have to investigate the remainder term.
$$
g(x):=\int_0^{|x|}\frac{f^{(r)}(t)}{(r-1)!}(|x|-t)^{r-1}\,dt.
$$
Now we need the Leibniz integral rule. If $x>0$ and $0<k<r$ then 
$$
g^{(k)}(x)=\int_0^{x}\frac{f^{(r)}(t)}{(r-1)!}(r-1)\cdots(r-k)(x-t)^{r-1-k}\,dt,
$$
and
$$
g^{(r)}(x)=f^{(r)}(x).
$$
In a similar way we obtain $g^{(k)}(x)$ and $g^{(r)}(x)$ if $x<0$. 
Since $g^{(r-1)}(x)=\int_0^x f^{(r)}(t)\,dt$ for every $x$, $g^{(r)}(x)$ is continuous everywhere.
