The Cross Pairing of Two Square-Integrable Vector Fields Along a Coupling

In the setting of The Intrinsic Calculus on the Wasserstein Space: Standing Notation, let $\nu,\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$, let $\pi\in\Pi(\nu,\mu)$, and let $q\in L^{2}(\nu;\mathbb{R}^{d})$ and $\eta\in L^{2}(\mu;\mathbb{R}^{d})$; for $z\in\mathbb{R}^{d+d}$ we write $x=\mathrm{pr}_{1}(z)$ and $y=\mathrm{pr}_{2}(z)$, and the discrepancy of $q$ and $\eta$ along $\pi$ is the nonnegative real number $\int_{\mathbb{R}^{d+d}}\lVert q(x)-\eta(y)\rVert^{2}\,\pi(dz)$ fixed there. Then the following hold.

1. (The cross pairing)¶ For all representatives of $q$ and of $\eta$ the function $\mathbb{R}^{d+d}\to\mathbb{R}$ with value $q(x)\cdot\eta(y)$ at $z$ is Borel and integrable with respect to $\pi$, and its integral does not depend on the representatives chosen. That integral,

is called the cross pairing of $q$ and $\eta$ along $\pi$.

2. (Bound)¶

3. (Bilinearity)¶ For all $q'\in L^{2}(\nu;\mathbb{R}^{d})$, all $\eta'\in L^{2}(\mu;\mathbb{R}^{d})$ and all $a,b\in\mathbb{R}$,

4. (Polarisation)¶

5. (The diagonal coupling)¶ If $\mu=\nu$, so that $\eta\in L^{2}(\nu;\mathbb{R}^{d})$, and $\pi=(\mathrm{id},\mathrm{id})_{\#}\nu$, then