Recently I've came across the following situation when studying Quantum Mechanics: suppose we have two operators $A,B$ on the space of functions $L^2(\mathbb{R}^3)$ and suppose they have the following property
If we use spherical coordinates in $\mathbb{R}^3$ then $A$ just acts on the $r$-dependence of functions and $B$ just acts on the $\theta,\phi$-dependence of functions.
If a function is separable, that is, if it has the form $f(r,\theta,\phi)=f_1(r)f_2(\theta,\phi)$ then we can precisely formulate that property by saying that $A$ is such that
$$Af(r,\theta,\phi)=[Af_1(r)]f_2(\theta,\phi)$$
where $Af_1(r)$ really means $A$ applied to $(r,\theta,\phi)\mapsto f_1(r)$ and computed at $(r,\theta,\phi)$. The same idea would be used for $B$.
Now, the point is that if $f$ is separable we know that it's $r$-dependence is $f_1$ and it's angular dependence is $f_2$.
If $f$ were not separable, it is not clear at all what "$r$-dependence of $f$" and "$\theta,\phi$-dependence of $f$" really means and I can't see how to precisely formulate this property.
The background here is the following: I want to prove that if $A$ and $B$ are operators like that then $[A,B]=0$. In general if we pick $L^2(\mathbb{R}^n)$ and if $A,B$ are operators such that $A$ acts just on the $x^i$ dependence of functions and $B$ acts just on the $x^j$ dependence of functions I want to show that $[A,B]=0$.
This is intuitively obvious. But formulating this property in a precise way for general (not separable) functions is being the real problem here for me.
The $L^2$ functions on the square $0 \le x,y \le 1$ with Lebesgue measure can be written as $$ f = \sum_{n}\sum_{k}a_{n,k}e^{2\pi inx}e^{2\pi iky}, $$ where $$ a_{n,k} = \int_{0}^{1}\int_{0}^{1}f(x',y')e^{-2\pi nx'}e^{-2\pi iky'}dx'dy' $$ I suspect you can make sense of what you're doing in this context.