Smoothness of radially defined functions. Suppose that for each $v \in \mathbb{R}^2$ with $|v| = 1$, there is a smooth ($C^{\infty}$) function $f_v : [0, 1] \rightarrow \mathbb{R}$ such that $f_v(0) = 0$.
Now, let $\bar{D}$ be the closed unit disk in $\mathbb{R}^2$ and define $\phi : \bar{D} \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ by setting $\phi(0) = 0$ and, for all other $u \in \bar{D}$,
$$\phi(u) := f_{u/|u|}(|u|).$$
QUESTION: Are there reasonable conditions to impose on the family $\{f_v\}_{v~\in~\mathbb{S}^1}$ that guarantee smoothness of $\phi$? Under these conditions, how would you prove it?
I apologize for the imprecise use of the word reasonable here; but I'd be interested in pretty much any conditions, other than absolutely trivial ones, like "take all $f_v$ identically zero". Thanks!
 A: Here is a first attempt:  
Assume that $$g:\quad (x,y)\mapsto g(x,y):=\sum_{j,k\geq 0} a_{jk}\, x^j y^k$$ is a real analytic function defined in the unit disk $D:=\{(x,y)\ |\ x^2+y^2<1\}$ and that
$$f(r,\phi):=g(r\cos\phi, r\sin\phi)\qquad(r\geq0,\ \phi\in{\mathbb R})\ .$$
What can we say about $f\,$?
Right away we can say that $f$ is a real analytic function of $r$ and $\phi$ defined for $-1<r<1$ and all $\phi\in{\mathbb R}$;  furthermore $f$ is $2\pi$-periodic in $\phi$, and $f(-r,\phi)\equiv f(r,\phi+\pi)$. But there is more to it.
Putting 
$$z:=x+i y=r e^{i\phi}\ ,\quad \bar z:=x-i y= r e^{-i\phi}$$
we can write $g$ in the form
$$g^*(z,\bar z)=\sum_{j,k} c_{jk}\, z^j \bar z^k=\sum_{l\geq 0} r^l \ T_l(\phi)$$ where 
$$T_l(\phi)=\sum_{j,k\geq 0;\ j+k=l} c_{jk}e^{(j-k)\phi}\ .$$
Therefore we can conclude the following: The polar representation $f$ of $g$ is necessarily of the form
$$f(r,\phi)=\sum_{l=0}^\infty r^l\ T_l(\phi)\ ,$$
where $T_l(\cdot)$ is a trigonometric polynomial of degree $\leq l$ that satisfies $T(\phi+\pi)=(-1)^lT(\phi)$. By going backwards it is easy to see that any such $f$ is the polar representation of a real analytic $g:\ (x,y)\mapsto g(x,y)$.
The $C^m$-case is not so easy. If we want $g$ to be in $C^m$ then $f=g\circ{\rm rect}\ $ will have to be in $C^m$ also. As $g$ is at at least $m$ times differentiable at $(0,0)$ one has $$g(x,y)=\sum_{j,k=0}^m a_{jk}\, x^j y^k + o(r^m)\qquad (r\to 0)\ .$$ It follows that  necessarily
$$f(r,\phi)=\sum_{l=0}^m r^l T_l(\phi)+o(r^m)\qquad (r\to 0)\ ,$$
where the $T_l$ are as before. But this is not enough to guarantee the continuity of the derivatives of $g$. In the case $m=1$ the complete answer looks as follows:
The function $f$ is the polar representation of a $C^1$ function $g$ defined in a neighborhood of the origin iff there are constants $a$, $b$, $c$ and a $C^1$-function $R$ such that
$$f(r,\phi)=c + a\, r \cos\phi+b\, r\sin\phi +R(r,\phi)\ ,$$
$$R(0,\phi)=0,\quad R_r(0,\phi)=0, \qquad R_\phi(r,\phi)=o(r)\quad(r\to0).$$
Proof. Let $g(0,0)=:c$, $g_x(0,0)=:a$, $g_y(0,0)=:b$, and put
$$h(x,y):=g(x,y)- ax - by\ , \quad R(r,\phi):=h(r\cos\phi,r\sin\phi)\ .$$
The facts $h(0,0)=h_x(0,0)=h_y(0,0)=0$ together with the equations
$$R_r=h_x\cos\phi + h_y\sin \phi\,\qquad R_\phi=r\ (-h_x\sin\phi + h_y\cos\phi)$$
show that the stated conditions on ($a$, $b$, $c$ and) $R$ are necessary, and the dual equations
$$h_x= R_r\cos\phi-{1\over r}R_\phi\sin\phi\ ,\qquad h_y=R_r\sin\phi +{1\over r}R_\phi\cos\phi$$
show that these conditions are also sufficient to guarantee the continuity of $h_x$, $h_y$, resp., $g_x$, $g_y$.
