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

In the setting of Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation, let $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$, let $L^{2}(\mu;\mathbb{R}^{d})$ be the space of square-integrable vector fields against $\mu$, with norm $\lVert\cdot\rVert_{\mu}$, and let $\mathcal{J}(\eta,\pi)$ denote the displacement pairing of $\eta\in L^{2}(\mu;\mathbb{R}^{d})$ along a coupling $\pi\in\Pi(\mu,\nu)$ with $\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$, with $I(\pi)$ the quadratic cost of $\pi$.

Let $\eta\in L^{2}(\mu;\mathbb{R}^{d})$ and suppose that for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\theta\in\mathbb{R}$ with $0<\theta$ such that

for every $\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ and every $\pi\in\Pi(\mu,\nu)$ with $I(\pi)<\theta^{2}$.

1. (The field vanishes)¶ $\eta$ is the zero element of $L^{2}(\mu;\mathbb{R}^{d})$.