Proof of A Square-Integrable Vector Field Whose Displacement Pairings Vanish to First Order is Zero

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement above.

Step 1: the pairings along displacement couplings. Let $\xi\in L^{2}(\mu;\mathbb{R}^{d})$ and let $t\in\mathbb{R}$ be positive. Fix a representative of $\xi$, again written $\xi$: a Borel map $\mathbb{R}^{d}\to\mathbb{R}^{d}$ with $\int_{\mathbb{R}^{d}}\lVert\xi\rVert^{2}\,d\mu<\infty$, by Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation (ii). Let $S$ be the map $x\mapsto x+t\,\xi(x)$. Each component of $S$ is the sum of a coordinate map and a real scalar multiple of a Borel real function, hence Borel by claim 2 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and a map into $\mathbb{R}^{d}$ with Borel components is Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (ii); so $S$ is Borel.

The map $S-\mathrm{id}$ is $x\mapsto t\,\xi(x)$, which represents the class $t\,\xi$ of $L^{2}(\mu;\mathbb{R}^{d})$ by Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation (ii). By the absolute homogeneity of the Euclidean norm, claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and $|t|=t$ for the positive $t$ --- by claim 1 of Properties of the Absolute Value in an Ordered Field the value $|t|$ is $t$ or $-t$, and $|t|=-t$ would give $0\le-t$, hence $t\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, contradicting $0<t$ --- one has $\lVert t\,\xi(x)\rVert^{2}=t^{2}\lVert\xi(x)\rVert^{2}$ for every $x$, so by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion

$$\int_{\mathbb{R}^{d}}\lVert S-\mathrm{id}\rVert^{2}\,d\mu=t^{2}\int_{\mathbb{R}^{d}}\lVert\xi\rVert^{2}\,d\mu=t^{2}\,\lVert\xi\rVert_{\mu}^{2}<\infty .$$

Therefore The Displacement Pairing of a Square-Integrable Vector Field Along a Coupling (iv) applies to $S$: the push-forward $S_{\#}\mu$ belongs to $\mathcal{P}_{2}(\mathbb{R}^{d})$, the coupling $\pi_{S}=(\mathrm{id},S)_{\#}\mu$ belongs to $\Pi(\mu,S_{\#}\mu)$ with

$$I(\pi_{S})=\lVert S-\mathrm{id}\rVert_{\mu}^{2}=t^{2}\lVert\xi\rVert_{\mu}^{2}=\bigl(t\,\lVert\xi\rVert_{\mu}\bigr)^{2},$$

and, $\eta$ being the field of the statement,

$$\mathcal{J}(\eta,\pi_{S})=\langle\eta,\,S-\mathrm{id}\rangle_{\mu}=\langle\eta,\,t\,\xi\rangle_{\mu}=t\,\langle\eta,\xi\rangle_{\mu},$$

the last equality by the homogeneity of the inner product in its second argument, Elementary Identities in a Real Inner Product Space (i).

Step 2: the estimate on the inner products. We show that $\langle\eta,\xi\rangle_{\mu}=0$ for every $\xi\in L^{2}(\mu;\mathbb{R}^{d})$.

If $\lVert\xi\rVert_{\mu}=0$ then $\xi$ is the zero class by Elementary Identities in a Real Inner Product Space (iii) and $\langle\eta,\xi\rangle_{\mu}=0$ by Elementary Identities in a Real Inner Product Space (ii). So assume $\lVert\xi\rVert_{\mu}$ is positive, the norm being nonnegative.

Let $\varepsilon\in\mathbb{R}$ be positive. Then $\varepsilon\,\lVert\xi\rVert_{\mu}^{-1}$ is positive by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field; let $\theta$ be a positive real supplied by the hypothesis of the statement for that positive number in place of the $\varepsilon$ there. The number $\theta\,\lVert\xi\rVert_{\mu}^{-1}$ is positive by the same two claims, and $t=\theta\,\lVert\xi\rVert_{\mu}^{-1}\cdot2^{-1}$ is positive with $t<\theta\,\lVert\xi\rVert_{\mu}^{-1}$, by claim 8 of Elementary Order Arithmetic in an Ordered Field. Multiplying that strict inequality by the positive $\lVert\xi\rVert_{\mu}$, by claim 10 of Elementary Order Arithmetic in an Ordered Field, gives $t\,\lVert\xi\rVert_{\mu}<\theta$; both sides being nonnegative, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $(t\,\lVert\xi\rVert_{\mu})^{2}<\theta^{2}$, that is $I(\pi_{S})<\theta^{2}$ for the $S$ built from this $\xi$ and $t$ in step 1.

The hypothesis of the statement, applied to $\nu=S_{\#}\mu$ and $\pi=\pi_{S}$, therefore yields

$$\bigl|\mathcal{J}(\eta,\pi_{S})\bigr|\le\varepsilon\,\lVert\xi\rVert_{\mu}^{-1}\,\sqrt{I(\pi_{S})} .$$

Here $\sqrt{I(\pi_{S})}=t\,\lVert\xi\rVert_{\mu}$ by Existence and Uniqueness of the Nonnegative Square Root, that number being nonnegative with square $I(\pi_{S})$; and $|\mathcal{J}(\eta,\pi_{S})|=|t\,\langle\eta,\xi\rangle_{\mu}|=t\,|\langle\eta,\xi\rangle_{\mu}|$ by step 1, the multiplicativity of the absolute value (claim 4 of Properties of the Absolute Value in an Ordered Field) and $|t|=t$, established in step 1. So

$$t\,\bigl|\langle\eta,\xi\rangle_{\mu}\bigr|\le\varepsilon\,\lVert\xi\rVert_{\mu}^{-1}\,t\,\lVert\xi\rVert_{\mu}=\varepsilon\,t ,$$

using the associativity and commutativity of multiplication in the field of Field and $\lVert\xi\rVert_{\mu}^{-1}\lVert\xi\rVert_{\mu}=1$. Multiplying by the positive $t^{-1}$ (claim 7 of Elementary Order Arithmetic in an Ordered Field), by claim 5 of Elementary Arithmetic in an Ordered Field, gives

$$\bigl|\langle\eta,\xi\rangle_{\mu}\bigr|\le\varepsilon .$$

As $\varepsilon$ was an arbitrary positive real and $|\langle\eta,\xi\rangle_{\mu}|$ is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, Comparison of Real Numbers with Arbitrary Positive Slack (iii) gives $|\langle\eta,\xi\rangle_{\mu}|=0$, whence $\langle\eta,\xi\rangle_{\mu}=0$ by claim 1 of Properties of the Absolute Value in an Ordered Field again.

Step 3: conclusion. Taking $\xi=\eta$ in step 2 gives $\langle\eta,\eta\rangle_{\mu}=0$, that is $\lVert\eta\rVert_{\mu}=0$ by Real Inner Product Spaces (ii) and Existence and Uniqueness of the Nonnegative Square Root. Hence $\eta$ is the zero element of $L^{2}(\mu;\mathbb{R}^{d})$ by Elementary Identities in a Real Inner Product Space (iii).