Let $\mathfrak g$ be a finite dimensional real Lie algebra. If a bilinear map $A:\mathfrak g\times\mathfrak g\to\mathbb R$ vanishes on commuting elements (i.e. $[U,V]=0\implies A(U,V)=0$), is there a linear map $\phi:\mathfrak g\to\mathbb R$ so that $A(U,V)=\phi([U,V])$? If this is not generally true, does it hold for some non-abelian Lie algebras?
I have tried to find a proof or a counterexample, but to no avail. The reason I think this has some hope of being true is this result for linear maps:
Let $E$ and $F$ be real vector spaces and $\alpha:E\to F$ a linear map. If a linear map $A:E\to\mathbb R$ satisfies $\ker\alpha\subset\ker A$, then there is a linear map $\phi:F\to\mathbb R$ so that $A=\phi\circ\alpha$.
Here $\alpha$ plays the role of the commutator. I do not see how to generalize the proof of this linear result to the bilinear realm, especially since the kernel (preimage of zero) of a bilinear map is not generally a vector space.
If two Lie algebras $\mathfrak g$ and $\mathfrak h$ satisfy the desired property, so does their direct sum $\mathfrak g\oplus\mathfrak h$. Abelian Lie algebras satisfy it trivially. Therefore if one shows that all simple Lie algebras satisfy it, the result follows for all reductive Lie algebras. This is such a large class of Lie algebras that I would be satisfied with an answer under the additional assumption that $\mathfrak g$ is simple.