Given functions $f,g \in L^2([-1,1]),$ we can construct antisymmetric functions in $L^2([-1,1]\times [-1,1])$ by defining: $$F(x,y) = f(x)g(y)-f(y)g(x),$$ which is the antisymmerization of $f(x)g(y),$ or, equivalently,the determinant $$F(x,y) = \begin{vmatrix} f(x) & f(y) \\ g(x) & g(y) \notag \end{vmatrix}. $$
In general, I call "determinant" to an antisymmetric function $F \in L^2([-1,1],[-1,1])$ that can be expressed as above for certain $f,g \in L^2([-1,1]).$
Now, for a generic antisymmetric $ F \in L^2([-1,1]\times [-1,1])$, $F(x,y) = -F(y,x)$, what are the $f,g \in L^2([-1,1])$ such that $f(x)g(y)-f(y)g(x)$ is closest to $F$ in the $L^2$ norm. For example, consider the determinants: $$x y^2 - y x^2$$ and $$\sin(x) e^y - \sin(y) e^x,$$ and combine them linearly: $$\alpha \left( x y^2 - y x^2 \right) + \beta \left( \sin(x) e^y - \sin(y) e^x \right)$$
It is intuitive that this combination cannot be expressed as a determinant for arbitrary $\alpha,\beta,$ but I cannot prove it. And how to find the closest determinant?