TITLE: Are radially symmetric maps smooth at the origin?
QUESTION [1 upvotes]: Let $U \subseteq \mathbb{R}^2$ be an open set containing the origin, and let $h:[0,1] \to [0,1]$ be a smooth strictly monotone function satisfying $h(0)=0,h'(x)>0$ for every $x \in [0,1]$.
Define $f:U \to \mathbb R^2$ by 
$$ f(x,y)=\frac{h(r(x,y))}{r(x,y)}(x,y), \, \,\text{where} \, \, r(x,y)=\sqrt{x^2+y^2}.$$
In polar coordinates $f$ is given by
$(r,\theta)\to  (h(r),\theta)$.

Is $f$ smooth in a neighbourhood of the origin?

The only possible problem might be at the origin itself. A direct computation shows that
$$ df_p(v)=\langle v,\frac{p}{|p|}\rangle (h'(|p|)-\frac{h(|p|)}{|p|}) \frac{p}{|p|}+\frac{h(|p|)}{|p|}v.$$
Since $\lim_{p \to 0}h'(|p|)-\frac{h(|p|)}{|p|}=h'(0)-h'(0)=0$,
it follows that $\lim_{p \to 0} df_p(v)=h'(0)v=df_0(v)$, so $f \in C^1$.
Is there a reasonable way to proceed?
This question is similar to this one, but here we have a vector-valued map.
Edit:
I wonder what are sufficient or necessary conditions on $h$ that are related to the smoothness of $f$.

The first observation is that $f(x,0)=\frac{h(|x|)}{|x|}(x,0)=\big((h(|x|)\text{sgn}(x),0\big)$.

Claim:
The function $x \to h(|x|)\text{sgn}(x)$ is infinitely differentiable at zero if and only if $h^{2k}(0)=0$ for every $k \in \mathbb N$. (In particular if $h$ is real-analytic, then it is odd).
Does this condition on $h$ imply that $f$ is smooth around the origin?

REPLY [1 votes]: No, $f$ need not be smooth. Take $h(t) = \frac{t(t+1)}{2}$. Then $f$ sends $(x, 0)$ to $(\frac{x^2\operatorname{sign}(x) + x}2, 0)$, so $f$ is at most once differentiable.
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