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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!

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At least to be differentiable at $0$ you need the condition that for any three $u, v, w$ with $|u|=|v|=|w|=1$ the $3 \times 3$ matrix $$ \begin{pmatrix} u & v & w \\ f_u'(0) & f_v'(0) & f_w'(0) \end{pmatrix} $$ has the same rank a $(u \; v \; w)$. – WimC Feb 28 '12 at 17:12
... or in other words: $\delta: \{u \in \mathbb{R}^2 \mid |u|=1\} \rightarrow \mathbb{R}$ such that $\delta: u \mapsto f_u'(0)$ extends to a linear function on $\mathbb{R}^2$. – WimC Feb 28 '12 at 17:21
This is an important problem, and it is not dealt with in analysis texts. I would formulate it differently: Under which conditions on $(r,\phi)\mapsto f(r,\phi)$ is there a $C^m$-function $(x,y)\mapsto g(x,y)$ such that $g(r\cos\phi, r\sin\phi)\equiv f(r,\phi)\ $? – Christian Blatter Feb 28 '12 at 19:29
What do you mean by "some interesting cases" (in the bounty text)? – joriki Mar 1 '12 at 19:58
up vote 1 down vote accepted

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., $g_x$, $g_y$.

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