Linking push-forward of measures and the Legendre transform

Question

Suppose that $\rho_1$ and $\rho_2$ are absolutely continuous w.r.t Lebesgue measure on $\mathbb{R}^n$, for which the second moment of both measures are finite. By Brenier's theorem, there exists a convex function $\phi$ such that $\nabla \phi_{\#} \rho_1 = \rho_2$, or equivalently, $$\int_{\mathbb{R}^n} g\, \mathrm{d}\rho_2 = \int_{\mathbb{R}^n} g(\nabla \phi)\, \mathrm{d}\rho_1.$$ I found in a paper that we also have $$\nabla \phi^*_{\#} \rho_2 = \rho_1,\tag{*}$$ in which $\phi^*$ is the Legendre transform (or convex dual) of $\phi$, defined by $$\phi^*(q) = \sup_{x\in \mathbb{R}^n} \{q\cdot x - \phi(q)\}.$$ May I know how can we justify (*)? Thank you very much!

Answer 1

I think that this basically follows from $(\nabla \phi)^{-1} = \nabla \phi^*$ which holds if both, $\phi$ and $\phi^*$ are differentiable. This formula can also be extended to the non-differentiable case via $x^* \in \partial \phi(x)$ if and only if $x \in \partial \phi(x^*)$, where "$\partial$" denotes the subdifferential.

Suppose that $\phi$ and $\phi^*$ are differentiable. We have $\rho_2(A) = \rho_1(\nabla \phi^{-1}(A))$ for all measurable sets $A$. If we plug in $A = (\nabla\phi^*)^{-1}(B)$, we get $\rho_2((\nabla\phi^*)^{-1}(B)) = \rho_1(B)$ due to $\nabla \phi^{-1}((\nabla\phi^*)^{-1}(B)) = B$.

In the general case, it might be possible to use the fact that $\phi$ and $\phi^*$ are a.e. differentiable (together with the absolute continuity of the measures).