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Intro

All functions are smooth real-valued functions of real variables.

Suppose you can decompose a function $f(x,y)$ as a product as one-variable functions $f(x,y) = \sum_{i=1}^n u_i(x)\cdot v_i(y)$. The rank of $f$, if it exists, is the smallest $n$ for which this decomposition is possible. You can prove that a decomposition is minimal if and only if the $u_i$ and $v_i$ are linearly independent, and that all minimal decompositions have the same number of terms.

Similarly for a three-variable function $f(x,y,z)$ you can define the rank with respect to $x$ as the smallest $n$ for which the following kind of decomposition exists: $f(x,y,z)=\sum_{i=1}^n u_i(x)\cdot v_i(y,z)$.

Problem

Let us say that a function $f(x,y,z)$ is strongly decomposable if its ranks with respect to $x$, $y$, and $z$ are each finite; in particular, equal to 2 or 3. I'm exploring the idea of weakly decomposable functions: a function $f(x,y,z)$ is weakly decomposable if there is a strictly positive function $g(x,y,z)$ such that the product $f\cdot g$ is strongly decomposable. (Note that $g$ does not need to be decomposable at all.)

The idea is that even if $f$ is not strongly decomposable, there may be another function $f\cdot g$ which has the same zeroes as $f$ and which is strongly decomposable. My question is: can this happen?

Question: Are there weakly decomposable functions that are not strongly decomposable?

So far I have established that:

  1. Every minimal decomposition has the same rank, and a decomposition is minimal if and only if the $u_i$ are linearly independent as are the $v_i$.
  2. It is possible for a product $f\cdot g$ to be decomposable even if $g$ is not decomposable.
  3. For a fixed value of $x$, define the function $f_x(y,z) = f(x,y,z)$. Then we can alternatively characterize the rank of $f$ with respect to $x$ as the dimension of the span of $\{ f_x : x\in \mathbb{R}\}$.
  4. Moreover, if $f$ is decomposable, we can always find a decomposition where each $u_i$ is equal to $f_{x_i}$ for some $x_i$.

But I do not know how to proceed further. If the answer is no, it implies that "weakly decomposable" is an empty concept, and that the rank of a function is determined (or constrained) by its zeroes (kernel).

user326210
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    Isn't this answered by my example in one of your previous questions? Just take $f(x,y,z)=\frac{1}{1+x^2+y^2}$ (I gave that example in a 2-variable context but you can basically just ignore the third variable). – Eric Wofsey Jul 10 '21 at 20:22
  • Or, more indirectly: every strictly positive function is weakly decomposable. But every function can be written as a difference of two strictly positive functions, and a linear combination of strongly decomposable functions is strongly decomposable. So if every weakly decomposable function were strongly decomposable, then every function would be strongly decomposable! – Eric Wofsey Jul 10 '21 at 20:39
  • @EricWofsey Thanks; this is helpful. I see that the argument works when "strongly" means "finite rank" and fns are not required to be smooth. I'm still figuring out if we can extend it when strongly means "either rank 2 or 3, nothing else", when you lose closure under linear combinations. Do you know if that difference-of-positive-functions property holds when all fns are required to be smooth? I can see why it's true in the unrestricted case. – user326210 Jul 13 '21 at 21:06
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    For instance, you can just note that for any $f$, $f^2+1>f$ and also $f^2+1>0$ everywhere. So you can write $f$ as the difference of the positive functions $f^2+1$ and $f^2+1-f$. – Eric Wofsey Jul 13 '21 at 21:09
  • Perfect. Do you know a way to adapt the proof that weakly-but-not-strongly decomposable fns exist when "strongly decomposable" means "the rank w.r.t. x, y, z is equal to 2 or 3 in each case" instead of "the rank wrt x,y,z is finite"? – user326210 Jul 13 '21 at 21:15
  • I wouldn't expect to be able to prove many general statements from that definition, which seems rather ad hoc and very poorly behaved. – Eric Wofsey Jul 13 '21 at 21:19
  • But if you just want a single example of a weakly but not strongly decomposable function, then $\frac{1}{1+x^2+y^2}$ should still work. – Eric Wofsey Jul 13 '21 at 21:22
  • This is a nice result. If a function has no zeroes, then it is "weakly equivalent to" functions of each positive rank, e.g. using exp(k(x+y)). Even if there aren't many general statements we can prove, I wonder if it's possible to find an example of a weakly-but-not-strongly decomposable function which is zero somewhere. This feels like it would answer the underlying question, 'Do the zeroes of a function determine its rank?' – user326210 Jul 13 '21 at 22:20

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