If $f\cdot g$ is decomposable and $g$ is decomposable, is $f$ decomposable? A decomposition of a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a sum of products of single-value functions: $f(x,y)=\sum_{i=1}^n u_i(x)\cdot v_i(y)$. If a function has at least one decomposition, it is called decomposable; otherwise, it is undecomposable.
It is straightforward to show that sums and products of decomposable functions are likewise decomposable. I do not know of specific undecomposable functions, but my intuition is that they exist and that most functions are undecomposable.
Question. I'd like to know: given $f\cdot g$ is decomposable and $g$ is decomposable, [when] is $f$ decomposable?
If that's hard, I'd appreciate any additional results about when products and sums involving undecomposable functions can result in decomposable functions.

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*So far, my intuition is that decomposable functions and undecomposable functions will have similar arithmetic closure properties to rational and irrational numbers.


*I have been looking for a provably undecompsable function—besides functions like $x^y$, I've thought about using an infinite sequence of linearly-independent functions, because I've shown that if a function has a decomposition, it has a minimal decomposition where all the $u_i$ are linearly independent and all the $v_i$ are linearly independent. If a function were somehow an  irreducibly infinite sum of pairs $u_iv_i$ where the $u_i$ are all linearly independent as are the $v_i$, it would not be decomposable.


*I've also been looking into function spaces where these functions and their decompositions live. Based on what I've learned, I believe the space of decomposable functions is the tensor space $\mathscr{F}_X \otimes \mathscr{F}_Y$, consisting of linear combinations of pairs of functions of $x$ and of $y$, and this can be embedded in the vector space $\mathscr{F}_{X\times Y}$ of two-variable functions. But I have not yet found a way to prove my desired result by arguing generally about vector/tensor spaces.
 A: I wouldn't expect any simple criterion.  Certainly, given that $f\cdot g$ and $g$ are decomposable, $f$ will usually not be decomposable.  For instance, let's consider the case $fg=1$, so you are asking: if $g$ is decomposable and nowhere $0$, is $1/g$ decomposable? Writing $g_x(y)=g(x,y)$, then $g$ is decomposable iff the span of the functions $g_x$ is finite-dimensional.  So you are asking, if you have some finite collection of functions $v_1,\dots,v_n$, and you consider a set of functions formed by taking reciprocals of different linear combinations of the $v_i$, will that set of functions have finite-dimensional span?  There is no reason to expect it to.  For instance, the functions $$y\mapsto\frac{1}{a+y}$$ are all linearly independent as $a\in\mathbb{R}$ varies.  This example doesn't literally work in this context because these functions are not defined on all of $\mathbb{R}$, but for instance you could similarly consider functions of the form $$y\mapsto\frac{1}{a+y^2}$$ where $a>0$.  These infinitely many functions are all linearly independent (for instance, by considering their poles when you analytically continue them to $\mathbb{C}$), even though they are reciprocals of functions which are all spanned by just two different functions.
Explicitly, following these ideas, you could take $g(x,y)=1+x^2+y^2$  and $f(x,y)=\frac{1}{1+x^2+y^2}$.  Then $g$ is decomposable and nowhere $0$ and so is $fg=1$, but $f$ is not decomposable.
A better analogy than rational and irrational numbers would be that decomposable and undecomposable functions are like integers and non-integer real numbers.  Sometimes a quotient of two integers is an integer, but usually it isn't, and there isn't going to really be any simpler way to say when it is.  More abstractly, the decomposable functions form a subring of the ring of all functions, and I wouldn't expect them to have any particularly nicer closure properties than that.
