Fix a function $f:\mathbb{R}^2\rightarrow \mathbb{R}$. A decomposition of $f$ is a sum $f(x,y)=\sum_i u_i(x)v_i(y)$. A decomposition is called irreducible if all the $u_i$ are linearly independent, and all the $v_i$ are linearly independent.
I am attempting to establish under what circumstances the following conjecture is true:
Conjecture. If $f$ is everywhere positive and has a decomposition, then there exists an irreducible decomposition $f(x,y)=\sum_i u_i(x)v_i(y)$ such that one of the $u_k$ is everywhere positive.
In my applications, I am usually dealing with functions $f$ that are smooth and/or have partial derivatives that are never zero. I am not sure if that helps.
So far, I have established the following result:
If $\sum_i u_i(x)v_i(y)$ and $\sum_i p_i(x)q_i(y)$ are irreducible decompositions of $f$, then they have the same number of terms, and are related by an invertible linear transformation $T$ where $T\cdot [u_1\ldots u_n] = [p_1,\ldots,p_n]$ and $[v_1,\ldots,v_n] = T\cdot [q_1,\cdots,q_n]$. Conversely, any invertible transformation $T$ maps irreducible decompositions onto irreducible decompositions.
This allows you to restate the conjecture as:
If $f$ is everywhere positive and has an irreducible decomposition $\sum_i u_iv_i$, then there exists a linear combination of $u_1,\ldots,u_n$ which is strictly positive.
The case is true at least when $f=u(x)v(y)$, i.e. in a sum with one term. But it is not clear to me when it extends to larger sums.
This is easy when you think about the circumstances under which $f$ has a decomposition. For each $y\in\mathbb{R}$, let $f_y(x)=f(x,y)$. Then $f$ has a decomposition iff the span of the set of functions $f_y$ is finite-dimensional, in which case you can get an irreducible decomposition by taking any basis for this space of functions as the $u_i$ (and then $v_i(y)$ has to be the coefficient of $u_i$ when you write $f_y$ in terms of this basis). Now you can always find such a basis in which each $u_i$ is actually equal to $f_y$ for some $y$ (just take a maximal linearly independent subset of $\{f_y:y\in\mathbb{R}\}$). If $f$ is positive, then so is each $f_y$, and thus this gives an irreducible decomposition where every $u_i$ is positive (not just one!).