I have a function $f(x,y)$, where both $x$ and $y$ are $n$-dimensional vectors, $n\ge 2$. I know that this function
has the following property:
$$
 \frac{\partial}{\partial x\_j} \frac{\partial}{\partial y\_k} f = 
 a\_j(x,y) b\_k(x,y)
$$
This can be expressed saying that the $n\times n$ block of the Hessian, out of diagonal, is a diadyc matrix. Examples of such a $f$ are:
$$
 f(x,y) = A(x)B(y) + C(x) + D(y)
$$
Actually, this example is similar to the form used in the "separation of variables" method for solving partial differential equations.


The second equation implies the first. My question is: is it also true that the first equation implies the second? Or a counter-example can be found?



Relation between separation of variables and Hessian properties
The answer is 'no, the first equation does not imply the second when $n=2$'.


The reason is that when $n=2$, the equation is essentially equivalent to requiring that the off-diagonal $n$-by-$n$ block of the Hessian of $f$ have determinant equal to zero.


This is one (non-linear) second-order equation for $f$ as a function on $\mathbb{R}^4$, and it is easy to show that the local solutions depend on 3 functions of 4 variables in Cartan's sense. (Basically, one can prescribe $f$ and its first normal derivative along a generic hypersurface in $\mathbb{R}^4$ subject to some generic conditions.)


However, the second form depends only on 4 functions of 2 variables, so not every local solution can be put in the second form, even locally.


In higher dimensions, the condition is equivalent to requiring that that $n$-by-$n$ matrix have rank at most equal to $1$. This is $(n{-}1)^2$ second order equations for $f$, so it's overdetermined when $n>2$. One would need to do the Cartan-Kähler analysis of this system to determine the generality of its space of solutions, but, probably, the answer is still 'no'.


