TheoremBase

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

For a coupling of two measures and square-integrable vector fields against each, the integral of the dot product of the first field at the first coordinate with the second field at the second coordinate is well defined, bounded by the product of the norms, bilinear, related to the discrepancy by polarisation, and equal to the inner product along the diagonal coupling.

Statement

In the setting of The Intrinsic Calculus on the Wasserstein Space: Standing Notation, let ν,μ∈P2(Rd)\nu,\mu\in\mathcal{P}_{2}(\mathbb{R}^{d}), let π∈Π(ν,μ)\pi\in\Pi(\nu,\mu), and let q∈L2(ν;Rd)q\in L^{2}(\nu;\mathbb{R}^{d}) and η∈L2(μ;Rd)\eta\in L^{2}(\mu;\mathbb{R}^{d}); for z∈Rd+dz\in\mathbb{R}^{d+d} we write x=pr1(z)x=\mathrm{pr}_{1}(z) and y=pr2(z)y=\mathrm{pr}_{2}(z), and the discrepancy of qq and η\eta along π\pi is the nonnegative real number ∫Rd+d∥q(x)−η(y)∥2 π(dz)\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 qq and of η\eta the function Rd+d→R\mathbb{R}^{d+d}\to\mathbb{R} with value q(x)⋅η(y)q(x)\cdot\eta(y) at zz is Borel and integrable with respect to π\pi, and its integral does not depend on the representatives chosen. That integral,

K(q,η,π)=∫Rd+dq(x)⋅η(y) π(dz)∈R,\mathcal{K}(q,\eta,\pi)=\int_{\mathbb{R}^{d+d}}q(x)\cdot\eta(y)\,\pi(dz)\in\mathbb{R},

is called the cross pairing of qq and η\eta along π\pi.

2. (Bound)

∣K(q,η,π)∣≤∥q∥ν ∥η∥μ.\bigl|\mathcal{K}(q,\eta,\pi)\bigr|\le\lVert q\rVert_{\nu}\,\lVert\eta\rVert_{\mu}.

3. (Bilinearity) For all q′∈L2(ν;Rd)q'\in L^{2}(\nu;\mathbb{R}^{d}), all η′∈L2(μ;Rd)\eta'\in L^{2}(\mu;\mathbb{R}^{d}) and all a,b∈Ra,b\in\mathbb{R},

K(a q+b q′,η,π)=a K(q,η,π)+b K(q′,η,π),K(q,a η+b η′,π)=a K(q,η,π)+b K(q,η′,π).\mathcal{K}(a\,q+b\,q',\eta,\pi)=a\,\mathcal{K}(q,\eta,\pi)+b\,\mathcal{K}(q',\eta,\pi),\qquad\mathcal{K}(q,a\,\eta+b\,\eta',\pi)=a\,\mathcal{K}(q,\eta,\pi)+b\,\mathcal{K}(q,\eta',\pi).

4. (Polarisation)

∫Rd+d∥q(x)−η(y)∥2 π(dz)=∥q∥ν2−2 K(q,η,π)+∥η∥μ2.\int_{\mathbb{R}^{d+d}}\lVert q(x)-\eta(y)\rVert^{2}\,\pi(dz)=\lVert q\rVert_{\nu}^{2}-2\,\mathcal{K}(q,\eta,\pi)+\lVert\eta\rVert_{\mu}^{2}.

5. (The diagonal coupling) If μ=ν\mu=\nu, so that η∈L2(ν;Rd)\eta\in L^{2}(\nu;\mathbb{R}^{d}), and π=(id,id)#ν\pi=(\mathrm{id},\mathrm{id})_{\#}\nu, then

K(q,η,π)=⟨q,η⟩ν.\mathcal{K}(q,\eta,\pi)=\langle q,\eta\rangle_{\nu}.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…