Differentiability of a Function on the Wasserstein Space Along Couplings, and Its Gradient

In the setting of Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation, let $\varphi:\mathcal{P}_{2}(\mathbb{R}^{d})\to\mathbb{R}$ and 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 $\mu$, let $\Pi(\mu,\nu)$ and $I$ be the couplings and the quadratic cost, and let $\mathcal{J}(\eta,\pi)$ be the displacement pairing of $\eta\in L^{2}(\mu;\mathbb{R}^{d})$ along $\pi\in\Pi(\mu,\nu)$.

1. (Differentiability along couplings)¶ Let $\eta\in L^{2}(\mu;\mathbb{R}^{d})$. The function $\varphi$ is differentiable along couplings at $\mu$ with gradient $\eta$ if 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 $I(\pi)<\theta^{2}$. The function $\varphi$ is differentiable along couplings at $\mu$ if there is an $\eta\in L^{2}(\mu;\mathbb{R}^{d})$ with this property.

2. (The gradient along couplings)¶ Suppose that $\varphi$ is differentiable along couplings at $\mu$. Then exactly one $\eta\in L^{2}(\mu;\mathbb{R}^{d})$ has the property of clause 1. Indeed, one exists by that hypothesis. Suppose $\eta$ and $\eta'$ both have it and let $\varepsilon\in\mathbb{R}$ be positive, so that $\varepsilon\cdot2^{-1}$ is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field. Let $\theta$ and $\theta'$ be as in clause 1 for $\eta$ and for $\eta'$ respectively, with $\varepsilon\cdot2^{-1}$ there in place of $\varepsilon$, and let $\theta''=\min\{\theta,\theta'\}$ be their minimum, which is positive because it equals $\theta$ or $\theta'$ by claim 2 of Elementary Properties of the Minimum of Two Elements. Let $\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ and $\pi\in\Pi(\mu,\nu)$ satisfy $I(\pi)<(\theta'')^{2}$. Since $\theta''\le\theta$ and $\theta''\le\theta'$ by claim 1 of that lemma, and both sides of each of these inequalities are nonnegative, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $(\theta'')^{2}\le\theta^{2}$ and $(\theta'')^{2}\le(\theta')^{2}$, whence $I(\pi)<\theta^{2}$ and $I(\pi)<(\theta')^{2}$ by the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field. Both estimates of clause 1 are therefore available at $\pi$, and, by the linearity of the displacement pairing (The Displacement Pairing of a Square-Integrable Vector Field Along a Coupling §linear, applied with $a=1$ and $b=-1$),

so that the triangle inequality of claim 5 of Properties of the Absolute Value in an Ordered Field, together with the symmetry of the absolute value (claim 2 there), bounds $|\mathcal{J}(\eta-\eta',\pi)|$ by $\varepsilon\cdot2^{-1}\sqrt{I(\pi)}+\varepsilon\cdot2^{-1}\sqrt{I(\pi)}$, which equals $\varepsilon\sqrt{I(\pi)}$ by the distributive law of Field and the halving identity of claim 8 of Elementary Order Arithmetic in an Ordered Field. Thus

So $\eta-\eta'$ satisfies the hypothesis of A Square-Integrable Vector Field Whose Displacement Pairings Vanish to First Order is Zero, whence $\eta-\eta'$ is the zero element by A Square-Integrable Vector Field Whose Displacement Pairings Vanish to First Order is Zero §vanishing and $\eta=\eta'$. This unique $\eta$ is written $\nabla\varphi(\mu)$ and called the gradient along couplings of $\varphi$ at $\mu$.