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I would like to consider the following simple problem. I want to find two functions $f,g : \mathbb R \to \mathbb R$ such that, being given a collection $(h_v)_{v\in V}$ of real functions indexed by unit vectors $v=(v_1,v_2) \in \mathbb S^1$, the equation \begin{align*} f(v_1s)g(v_2s) = h_v(s) \end{align*} is satisfied for all $s\in \mathbb R$.

My question is: if $V$ is infinite is it possible to recover $f$ and $g$?

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    $\begingroup$ What are $v_1$ and $v_2$? $\endgroup$ – Matthias Ludewig Apr 29 '14 at 9:11
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    $\begingroup$ I guess $v=(v_1,v_2)$ and the equation is satisfied for all $s$ in $\mathbb R$. Are the functions assumed continuous? $\endgroup$ – Loïc Teyssier Apr 29 '14 at 9:34
  • $\begingroup$ Also, $f$ and $g$ should probably be normalized? $\endgroup$ – Joni Teräväinen Apr 29 '14 at 10:12
  • $\begingroup$ Yes, $v = (v_1 v_2)$. The functions can be assumed to be $C^\infty$ for the moment. $\endgroup$ – user45183 Apr 29 '14 at 10:42
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Fact: if you can recover $(f,g)$ from the collection $(h_v)_{v\in V}$ then it must be the unique solution to your problem.

The case $h_v=0$ leads to a lot of non-uniqueness. Otherwise, there is only one source of non-uniqueness: the existence of functions $\alpha,\beta$ solving $$ \alpha(v_1s)\beta(v_2s)=1 .$$

If $\alpha,\beta$ are continuous then they must be constant as long as $V$ contains two non-colinear vectors.

Answer: Yes under some additionnal hypothesis. An important one is $f(0)g(0)\neq0$, so let us assume that. Assume also that $f,g$ are differentiable at $0$ and that $V$ contains two non-colinear vectors $v, \tilde v$ (this hypothesis is sufficient), and define $$\tau:=\frac{v_1\tilde v_2}{v_2\tilde v_1}.$$ We can label $v,\tilde v$ in such a way that $0\leq|\tau|<1$ and $v_2\neq0$.

Fiddling around with the functional equation gives you, for $t\in \mathbb R$ with $f(\frac{v_1 }{ v_2}t)\neq0$: $$ g(t)=g(\tau t)\phi(t) ,$$ where $$\phi(t):=\frac{h_v(\frac{1}{v_2}t)}{h_{\tilde v}(\frac{v_1 }{\tilde v_1 v_2}t)}.$$ Now you can iterate: $$g(t)=g(0)\prod_{n=0}^{\infty}\phi(\tau^n t).$$ Because $\phi(0)=1$ and is differentiable at $0$, there exists a constant $C$ such that, for $n$ big enough, $$|\phi(\tau ^n t)-1|\leq C|\tau^n t|.$$ The infinite product converges absolutely (and locally uniformly in $t$), so that $g$ is computed in terms of $h_v$, $h_{\tilde v}$ and $g(0)$ only. You can recover $f$ easily from the functional equation.

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There is a big difference between the situations where $f$ and $g$ are allowed to have zeroes or not. In the latter case, we have $$f(s)=\frac{h_{(1,0)}(s)}{g(0)}, \quad g(s)=\frac{h_{(0,1)}(s)}{f(0)}.$$ Also, one can clearly only get the functions up to a multiplicative constant since replacing $f$ by $af$ and $g$ by $\dfrac g a$ doesn't change anything. The former case is a different kettle of fish entirely.

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  • $\begingroup$ How does having infinitely many $v \in \mathbb S^1$ imply that you have $v = (1,0)$ and $v=(0,1)$? Answer: it does not. Also, the answer that you posted is a trivial solution which I tried to ignore. $\endgroup$ – user45183 Apr 29 '14 at 10:43
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    $\begingroup$ You should try to be a bit less terse if you wish people to help you solve your problem. If you don't give enough details in your question, then how can people figure out what are the "trivial cases" you wish to ignore? $\endgroup$ – Loïc Teyssier Apr 29 '14 at 10:49

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