Assume f(x1,...,xn) is twice differentiable from Rn to R. If for all (i, j) i ≠ j (dxi dxj)f = 0 then f is additively separable.
This seems intuitively obvious since if there exists an xjxi term in f the cross partial would not be 0. However, I am having a hard time articulating this. Any direction would be appreciated
I will write the proof for $n=2$ and the induction argument for general n will be quite obvious from this. We have $\frac {\partial f} {\partial x}$ has derivative 0 w.r.t. y which makes it a function of x alone. Call it $g(x)$. Now integrating the equation $\frac {\partial f} {\partial x}=g(x)$ we get $f(x,y)=h(x)+\phi (y)$ where h is an anti-derivative of g. [ Note that the constant of integration can be a function $\phi$ of y since we are integrating partially w.r.t. x]. For a general n use induction].