Proof of The Score of a Penalty Pair is Determined by the Penalty

Each result cited is universally quantified over the data in its own statement. Throughout, $L^{2}(\mu;\mathbb{R}^{d})$ is the real Hilbert space of Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii), with inner product $\langle\cdot,\cdot\rangle_{\mu}$ and norm $\lVert\cdot\rVert_{\mu}$, in particular a real inner product space, and $T_{\mu}$ is a subset of it by The Tangent Space of the Wasserstein Space at a Probability Measure (ii); and $\mu\in\mathcal{D}_{\Sigma}\subseteq\mathcal{D}\subseteq\mathcal{P}_{2}(\mathbb{R}^{d})$ by Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains (i).

Step 1: the functional $\psi\mapsto\langle\Sigma(\mu),\nabla\psi\rangle_{\mu}$ is linear and bounded. Define $\ell(\psi)=\langle\Sigma(\mu),\nabla\psi\rangle_{\mu}$ for $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$. Let $\psi,\phi\in C_{c}^{\infty}(\mathbb{R}^{d})$ and $a,b\in\mathbb{R}$. By The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (iii), $a\psi+b\phi\in C_{c}^{\infty}(\mathbb{R}^{d})$ and $\nabla(a\psi+b\phi)(x)=a\,\nabla\psi(x)+b\,\nabla\phi(x)$ for every $x\in\mathbb{R}^{d}$; since the operations on classes in $L^{2}(\mu;\mathbb{R}^{d})$ are formed from representatives (The Space of Square-Integrable Random Vectors (ii), applied on the probability space $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ as Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii) prescribes), the class of $\nabla(a\psi+b\phi)$ is $a\,\nabla\psi+b\,\nabla\phi$. Hence, by the bilinearity of the inner product, Elementary Identities in a Real Inner Product Space (i),

$$\ell(a\psi+b\phi)=\langle\Sigma(\mu),a\,\nabla\psi+b\,\nabla\phi\rangle_{\mu}=a\,\ell(\psi)+b\,\ell(\phi).$$

Moreover, by The Cauchy–Schwarz Inequality in a Real Inner Product Space, $|\ell(\psi)|\le\lVert\Sigma(\mu)\rVert_{\mu}\,\lVert\nabla\psi\rVert_{\mu}$ for every $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$, and $C=\lVert\Sigma(\mu)\rVert_{\mu}$ is a nonnegative real number (Real Inner Product Spaces (ii)).

Step 2: claim 1. By Step 1, $\ell$ satisfies the hypotheses of Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients (iv) with the constant $C$, so there is exactly one element of $T_{\mu}$ whose inner product with $\nabla\psi$ is $\ell(\psi)$ for every $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$; and $\Sigma(\mu)\in T_{\mu}$ is such an element by the definition of $\ell$, so it is the one. Since $\mu\in\mathcal{D}_{\Sigma}$, Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains (v) says precisely that $\Sigma(\mu)$ represents the first variation of $\mathcal{E}$ at $\mu$. Now let $\xi\in T_{\mu}$ represent the first variation of $\mathcal{E}$ at $\mu$, and fix $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$. There are then $t_{0}>0$ and $t_{0}'>0$ such that $f(t)=\mathcal{E}\bigl((\mathrm{id}+t\,\nabla\psi)_{\#}\mu\bigr)$ is defined for $t\in(-t_{0},t_{0})\cup(-t_{0}',t_{0}')$, that $f$ restricted to $(-t_{0},t_{0})$ is differentiable at $0$ with derivative $\langle\Sigma(\mu),\nabla\psi\rangle_{\mu}$, and that $f$ restricted to $(-t_{0}',t_{0}')$ is differentiable at $0$ with derivative $\langle\xi,\nabla\psi\rangle_{\mu}$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $s\in\{t_{0},t_{0}'\}$ with $s\le t_{0}$ and $s\le t_{0}'$; then $0<s$ in either case, $-t_{0}\le-s$ and $-t_{0}'\le-s$ by claim 4 of that lemma, and so, by the definition of the open interval and the mixed transitivity of claim 2 of that lemma, $(-s,s)\subseteq(-t_{0},t_{0})$ and $(-s,s)\subseteq(-t_{0}',t_{0}')$. Also $-s<-0=0<s$ by claim 4 of that lemma, so $(-s,s)$ is an interval of which $0$ is an interior point by Basic Facts about Intervals of the Real Line (iv). In Derivative of a Real Function on an Interval the condition defining a derivative $L$ at $0$ of a function on an interval $I$ quantifies over the increments $h$ with $0<|h|<\delta$ and $0+h=h\in I$; given $\varepsilon>0$, the $\delta$ furnished on the larger interval works on $(-s,s)$ as well, since every $h$ with $0<|h|<\delta$ and $h\in(-s,s)$ lies in the larger interval, and the difference quotients $(f(h)-f(0))/h$ are the same numbers, $0$ and $h$ lying in both intervals. Hence the restriction of $f$ to $(-s,s)$ is differentiable at $0$ with derivative $\langle\Sigma(\mu),\nabla\psi\rangle_{\mu}$ and also with derivative $\langle\xi,\nabla\psi\rangle_{\mu}$. The interval $(-s,s)$ is order-convex, the two conditions being the same, and $0$ lies strictly between two of its points by Interior Point of an Interval; hence Uniqueness of the Derivative of a Real Function on an Interval gives $\langle\xi,\nabla\psi\rangle_{\mu}=\langle\Sigma(\mu),\nabla\psi\rangle_{\mu}=\ell(\psi)$. As $\psi$ was arbitrary, $\xi$ is an element of $T_{\mu}$ whose inner product with $\nabla\psi$ is $\ell(\psi)$ for every $\psi$, hence $\xi=\Sigma(\mu)$ by the uniqueness established above. This proves claim 1.

Step 3: claim 2. Let $(\mathcal{D},\mathcal{D}_{\Sigma},\mathcal{E},\Sigma')$ be a penalty pair. Since $\mu\in\mathcal{D}_{\Sigma}$, Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains (v) for this pair says that $\Sigma'(\mu)\in T_{\mu}$ represents the first variation of $\mathcal{E}$ at $\mu$, the property being formulated in terms of $\mathcal{D}$, $\mathcal{E}$ and $\mu$ only. By claim 1, $\Sigma'(\mu)=\Sigma(\mu)$. $\blacksquare$