Proof of Noise Displacement and Cross Pairings Along Couplings: Bounds, Linearity, Displacement Couplings, Polarisation and a Vanishing Criterion

Each result cited is universally quantified over the data in its own statement.

Throughout, $\mu\in\mathcal{P}(X)$ is the measure fixed in the statement. Maps into $X$ are Borel in the sense of Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §measures, and measurability of maps into $X^{a}$ is that of The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §measurable. Every $\pi\in\Pi^{a}(\mu,\nu)$ belongs to $\Pi(\mu,\nu)$ by Couplings of Finite Noise Cost and Their Noise Cost §couplings, so $(\pi_{1})_{\#}\pi=\mu$ and $(\pi_{2})_{\#}\pi=\nu$ by Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling, and $\pi(D_{a})=1$ by Couplings of Finite Noise Cost and Their Noise Cost §finite. We record five facts used repeatedly.

(F1) The coordinate maps $\pi_{1},\pi_{2}$ are Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma, a pair of Borel maps into $X$ is a Borel map into $X\times X$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §pairing, and a composite of measurable maps between spaces among $X$, $X\times X$, $X^{a}$ is measurable by claim 4 of Borel Measurability and Bounded Integration on a Metric Space.

(F2) Let $\Phi$ be a Borel map between two of the spaces $X$, $X\times X$ and let $\lambda$ be a Borel probability measure on its domain. Then $\Phi_{\#}\lambda$ is a probability measure by claim 1 of Image Measures, Measures with Densities, and Change of Variables, and by the change-of-variables formula of Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pushforward one has $\int g\,d(\Phi_{\#}\lambda)=\int g\circ\Phi\,d\lambda$ in $[0,\infty]$ for every nonnegative Borel $g$ on the target, while a real-valued Borel $g$ is integrable against $\Phi_{\#}\lambda$ if and only if $g\circ\Phi$ is integrable against $\lambda$, the same identity then holding in $\mathbb{R}$. Moreover $(\Psi\circ\Phi)_{\#}\lambda=\Psi_{\#}(\Phi_{\#}\lambda)$ for every Borel $\Psi$ defined on the target of $\Phi$, because $(\Psi\circ\Phi)^{-1}(B)=\Phi^{-1}(\Psi^{-1}(B))$ for every Borel set $B$.

(F3) $X^{a}$ is a linear subspace of $X$ and a real Hilbert space with the inner product $\langle\cdot,\cdot\rangle_{a}$ by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §hilbert. For $u,v\in X^{a}$, expanding with the symmetry and the linearity in the first argument of the inner product (Real Inner Product Space §inner-product) gives $|u\pm v|_{a}^{2}=|u|_{a}^{2}\pm2\langle u,v\rangle_{a}+|v|_{a}^{2}$; adding the two signs, $|u+v|_{a}^{2}+|u-v|_{a}^{2}=2|u|_{a}^{2}+2|v|_{a}^{2}$, hence $|u\pm v|_{a}^{2}\le2|u|_{a}^{2}+2|v|_{a}^{2}$.

(F4) By the definition of $n_{a}$ in The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §borel and of $c_{a}$ in The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §pairs, $c_{a}(z)=n_{a}(y-x)$, which equals $|y-x|_{a}^{2}$ for $z\in D_{a}$ and $0$ for $z\notin D_{a}$.

(F5) Let $i\in\{1,2\}$, let $\kappa\in\mathcal{P}(X)$ and $\lambda\in\mathcal{P}(X\times X)$ with $(\pi_{i})_{\#}\lambda=\kappa$, and let $v:X\to X^{a}$ be measurable with $\int_{X}|v|_{a}^{2}\,d\kappa<\infty$. Then $v\circ\pi_{i}$ is measurable into $X^{a}$ by (F1); the function $|v\circ\pi_{i}|_{a}^{2}=|v|_{a}^{2}\circ\pi_{i}$ is nonnegative and Borel by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations, so (F2) gives $\int_{X\times X}|v\circ\pi_{i}|_{a}^{2}\,d\lambda=\int_{X}|v|_{a}^{2}\,d\kappa<\infty$. If moreover $v':X\to X^{a}$ is measurable with $v'\sim_{\kappa}v$, then the set $N=\{x:v(x)\ne v'(x)\}$ is Borel with $\kappa(N)=0$ by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §almost-everywhere, and $\{z:v(\pi_{i}z)\ne v'(\pi_{i}z)\}=\pi_{i}^{-1}(N)$ has $\lambda$-measure $\kappa(N)=0$ by the definition of the image measure; thus $v'\circ\pi_{i}\sim_{\lambda}v\circ\pi_{i}$.

Step 1 (claim 1). Let $\nu\in\mathcal{P}(X)$ and $\pi\in\Pi^{a}(\mu,\nu)$, and let $\delta$ be the map of claim 1. The map $z\mapsto y-x$ from $X\times X$ to $X$ is continuous, being Lipschitz with constant $2$: it is the negative of the map $z\mapsto\pi_{1}(z)-\pi_{2}(z)$ recorded in Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost as Lipschitz with constant $2$, and the Lipschitz bound passes to the negative because $|(-u)-(-u')|=|u-u'|$ for $u,u'\in X$ by Elementary Identities in a Real Inner Product Space §homogeneity; hence it is Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space; call it $\Delta$. For a Borel set $B\subseteq X$, $\delta^{-1}(B)$ is $D_{a}\cap\Delta^{-1}(B)$ if $0_{X}\notin B$ and $(D_{a}\cap\Delta^{-1}(B))\cup((X\times X)\setminus D_{a})$ if $0_{X}\in B$; since $D_{a}$ is Borel by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §pairs, $\delta^{-1}(B)$ is Borel. Thus $\delta$ is Borel as a map into $X$, and it takes values in $X^{a}$ (for $z\in D_{a}$ by the definition of $D_{a}$, and $0_{X}\in X^{a}$ by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §hilbert); by the last sentence of The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §measurable, $\delta$ is measurable into $X^{a}$. By (F4), $|\delta(z)|_{a}^{2}=|y-x|_{a}^{2}=c_{a}(z)$ for $z\in D_{a}$ and $|\delta(z)|_{a}^{2}=|0_{X}|_{a}^{2}=0=c_{a}(z)$ for $z\notin D_{a}$, so $|\delta|_{a}^{2}=c_{a}$ on all of $X\times X$. Hence $\int_{X\times X}|\delta|_{a}^{2}\,d\pi=I^{a}(\pi)<\infty$ by Couplings of Finite Noise Cost and Their Noise Cost §cost: $\delta$ is square-integrable with respect to $\pi$ in the sense of The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §space, and $\lVert\delta\rVert_{\pi}^{2}=I^{a}(\pi)$ by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations.

Step 2 (claim 2). Let $\nu$, $\pi$, $\eta$ and $\delta$ be as in claim 2, and fix a representative $\eta:X\to X^{a}$, which is measurable and square-integrable with respect to $\mu$ by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §classes. By (F5) with $i=1$, $\kappa=\mu$ and $\lambda=\pi$, the map $\eta\circ\pi_{1}$ is measurable into $X^{a}$ with $\int_{X\times X}|\eta\circ\pi_{1}|_{a}^{2}\,d\pi=\lVert\eta\rVert_{\mu}^{2}<\infty$, and $\delta$ is measurable with $\int_{X\times X}|\delta|_{a}^{2}\,d\pi=I^{a}(\pi)<\infty$ by Step 1. By Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations, applied with $E=X^{a}$, the basis $(f_{k})_{k\in\mathbb{N}}$ of The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §basis and the measurable space $(X\times X,\mathcal{B}(X\times X))$, the function $z\mapsto\langle\eta(x),\delta(z)\rangle_{a}$ is Borel, and by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §square-integrable it is integrable with respect to $\pi$. If $\eta'$ is another representative, then $\eta'\sim_{\mu}\eta$ by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §classes, so $\eta'\circ\pi_{1}\sim_{\pi}\eta\circ\pi_{1}$ by (F5), and Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §almost-everywhere, applied with $v=\eta\circ\pi_{1}$, $v'=\eta'\circ\pi_{1}$ and $w=w'=\delta$, shows that the two integrals coincide. Finally, for $z\in D_{a}$ one has $\delta(z)=y-x$, so the integrand equals $\langle\eta(x),y-x\rangle_{a}$ there.

Step 3 (claim 3). With the notation of Step 2, the inequality of Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §square-integrable gives

$$\bigl|\mathcal{J}^{a}(\eta,\pi)\bigr|\le\Bigl(\int_{X\times X}|\eta\circ\pi_{1}|_{a}^{2}\,d\pi\Bigr)^{1/2}\Bigl(\int_{X\times X}|\delta|_{a}^{2}\,d\pi\Bigr)^{1/2}=\lVert\eta\rVert_{\mu}\sqrt{I^{a}(\pi)},$$

the equality by the two integrals computed in Step 2 and the definition of the norm in The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations.

Step 4 (claim 4). Fix representatives $\eta$ and $\xi$. By The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations, the pointwise map $s\eta+t\xi$ is a representative of the class $s\eta+t\xi$, and by Step 2 the pairing may be computed with it. For every $z$, the linearity of $\langle\cdot,\cdot\rangle_{a}$ in its first argument (Real Inner Product Space §inner-product) gives $\langle s\eta(x)+t\xi(x),\delta(z)\rangle_{a}=s\langle\eta(x),\delta(z)\rangle_{a}+t\langle\xi(x),\delta(z)\rangle_{a}$. Both functions on the right are integrable against $\pi$ by Step 2, so the linearity of the integral, Linearity and Monotonicity of the Lebesgue Integral §integrable, gives $\mathcal{J}^{a}(s\eta+t\xi,\pi)=s\mathcal{J}^{a}(\eta,\pi)+t\mathcal{J}^{a}(\xi,\pi)$.

Step 5 (claim 5, the displacement couplings). Let $h:X\to X^{a}$ be a representative of $h\in L^{2}(\mu;X^{a})$ and $t\in\mathbb{R}$. Let $\iota:X^{a}\to X$ be the inclusion. For $u,u'\in X^{a}$ one has $u-u'\in X^{a}$, the vector operations of $X^{a}$ being those of $X$ by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §hilbert, so $|\iota(u)-\iota(u')|\le\bar{a}^{1/2}|u-u'|_{a}$ by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §embedding, with $\bar{a}>0$ by Probability Measures on a Hilbert Space Transported in the Noise Norm: Standing Notation §weights. Thus $\iota$ is Lipschitz, hence continuous by A Lipschitz Map is Uniformly Continuous, hence measurable from the Borel $\sigma$-algebra of $(X^{a},|\cdot|_{a})$ to $\mathcal{B}(X)$ by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Therefore $\iota\circ h:X\to X$ is Borel by (F1). The identity of $X$ is continuous, hence Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space, and Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations, applied with $E=X$, the basis $(e_{k})_{k\in\mathbb{N}}$ and the measurable space $(X,\mathcal{B}(X))$, shows that $S_{t}=\mathrm{id}+t\,(\iota\circ h)$ is Borel.

For every $x\in X$, $S_{t}(x)-x=t\,h(x)\in X^{a}$ by (F3), and $n_{a}(S_{t}(x)-x)=|t\,h(x)|_{a}^{2}=t^{2}|h(x)|_{a}^{2}$ by the definition of $n_{a}$ in The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §borel. By Linearity and Monotonicity of the Lebesgue Integral §nonnegative, $\int_{X}n_{a}(S_{t}(x)-x)\,\mu(dx)=t^{2}\int_{X}|h|_{a}^{2}\,d\mu=t^{2}\lVert h\rVert_{\mu}^{2}<\infty$. Hence Couplings of Finite Noise Cost: the Support Bound, Swap, Displacement Couplings and Lower Semicontinuity of the Noise Cost §displacement, applied with $S=S_{t}$, gives $\pi_{t}=(\mathrm{id},S_{t})_{\#}\mu\in\Pi^{a}(\mu,(S_{t})_{\#}\mu)$ and $I^{a}(\pi_{t})=t^{2}\lVert h\rVert_{\mu}^{2}$.

Let $\eta\in L^{2}(\mu;X^{a})$ with a representative $\eta$, and let $\delta$ be the displacement field of Step 1 for $\pi_{t}$. The map $(\mathrm{id},S_{t})$ is Borel by (F1). For every $x\in X$ the pair $(x,S_{t}(x))$ lies in $D_{a}$, since $S_{t}(x)-x\in X^{a}$, so $\delta(x,S_{t}(x))=t\,h(x)$, and the integrand of Step 2 composed with $(\mathrm{id},S_{t})$ is $x\mapsto\langle\eta(x),t\,h(x)\rangle_{a}=t\langle\eta(x),h(x)\rangle_{a}$. The integrand is integrable against $\pi_{t}$ by Step 2, and the function $x\mapsto\langle\eta(x),h(x)\rangle_{a}$ is integrable against $\mu$ by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §square-integrable, with integral $\langle\eta,h\rangle_{\mu}$ by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations; so (F2) and then Linearity and Monotonicity of the Lebesgue Integral §integrable and The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations give

$$\mathcal{J}^{a}(\eta,\pi_{t})=\int_{X}t\langle\eta(x),h(x)\rangle_{a}\,\mu(dx)=t\langle\eta,h\rangle_{\mu}.$$

Step 6 (claim 5, noise-connectedness to $\rho$). Suppose $\mu\in\mathcal{P}^{a}_{\rho}$. By The Measures Noise-Connected to the Reference Measure §space and Couplings of Finite Noise Cost and Their Noise Cost §connected there is $\pi'\in\Pi^{a}(\mu,\rho)$. Let $\Phi:X\times X\to X\times X$ be $\Phi(z)=(S_{t}(x),y)$; it is Borel by (F1), $S_{t}$ being Borel by Step 5. Put $\pi''=\Phi_{\#}\pi'\in\mathcal{P}(X\times X)$, using (F2). Since $\pi_{1}\circ\Phi=S_{t}\circ\pi_{1}$ and $\pi_{2}\circ\Phi=\pi_{2}$, (F2) gives $(\pi_{1})_{\#}\pi''=(S_{t})_{\#}((\pi_{1})_{\#}\pi')=(S_{t})_{\#}\mu$ and $(\pi_{2})_{\#}\pi''=(\pi_{2})_{\#}\pi'=\rho$, so $\pi''\in\Pi((S_{t})_{\#}\mu,\rho)$ by Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling. For $z\in D_{a}$, the difference of the coordinates of $\Phi(z)$ is $y-S_{t}(x)=(y-x)-t\,h(x)\in X^{a}$ by (F3), so $D_{a}\subseteq\Phi^{-1}(D_{a})$ and $1\ge\pi''(D_{a})=\pi'(\Phi^{-1}(D_{a}))\ge\pi'(D_{a})=1$. Moreover, for $z\in D_{a}$, (F4) and (F3) give

$$c_{a}(\Phi(z))=\bigl|(y-x)-t\,h(x)\bigr|_{a}^{2}\le2|y-x|_{a}^{2}+2t^{2}|h(x)|_{a}^{2}=2c_{a}(z)+2t^{2}|h(\pi_{1}z)|_{a}^{2}.$$

Since $\pi'(D_{a})=1$, this holds for $\pi'$-almost every $z$; both sides are nonnegative Borel functions of $z$ (by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §pairs, (F1) and (F5)). By (F2), The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, Linearity and Monotonicity of the Lebesgue Integral §nonnegative and (F5) with $i=1$, $\kappa=\mu$, $\lambda=\pi'$,

$$\int_{X\times X}c_{a}\,d\pi''=\int_{X\times X}c_{a}\circ\Phi\,d\pi'\le2I^{a}(\pi')+2t^{2}\lVert h\rVert_{\mu}^{2}<\infty .$$

Hence $\pi''$ has finite noise cost by Couplings of Finite Noise Cost and Their Noise Cost §finite, so $\pi''\in\Pi^{a}((S_{t})_{\#}\mu,\rho)$, the ordered pair $((S_{t})_{\#}\mu,\rho)$ is noise-connected by Couplings of Finite Noise Cost and Their Noise Cost §connected, and $(S_{t})_{\#}\mu\in\mathcal{P}^{a}_{\rho}$ by The Measures Noise-Connected to the Reference Measure §space.

Step 7 (claim 6). Let $\nu$, $\pi\in\Pi(\nu,\mu)$, $q$ and $\eta$ be as in claim 6, with representatives $q$ and $\eta$. By Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling, $(\pi_{1})_{\#}\pi=\nu$ and $(\pi_{2})_{\#}\pi=\mu$. By (F5), $v=q\circ\pi_{1}$ and $w=\eta\circ\pi_{2}$ are measurable into $X^{a}$ with $\int|v|_{a}^{2}\,d\pi=\lVert q\rVert_{\nu}^{2}$ and $\int|w|_{a}^{2}\,d\pi=\lVert\eta\rVert_{\mu}^{2}$, both finite. By Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations the function $\langle v,w\rangle_{a}$, whose value at $z$ is $\langle q(x),\eta(y)\rangle_{a}$, is Borel, and by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §square-integrable it is integrable with

$$\bigl|\mathcal{K}^{a}(q,\eta,\pi)\bigr|\le\Bigl(\int|v|_{a}^{2}\,d\pi\Bigr)^{1/2}\Bigl(\int|w|_{a}^{2}\,d\pi\Bigr)^{1/2}=\lVert q\rVert_{\nu}\lVert\eta\rVert_{\mu},$$

which is the inequality of the cross-bound part. If $q'$ and $\eta'$ are other representatives, then $q'\circ\pi_{1}\sim_{\pi}v$ and $\eta'\circ\pi_{2}\sim_{\pi}w$ by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §classes and (F5), so Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §almost-everywhere shows that the integral does not change.

For the discrepancy, $v-w=v+(-1)w$ is measurable into $X^{a}$ and $|v-w|_{a}^{2}$, whose value at $z$ is $|q(x)-\eta(y)|_{a}^{2}$, is a nonnegative Borel function by Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality §operations. By (F3), $|v-w|_{a}^{2}=|v|_{a}^{2}-2\langle v,w\rangle_{a}+|w|_{a}^{2}$ at every $z$. The functions $|v|_{a}^{2}$ and $|w|_{a}^{2}$ are nonnegative with finite integrals, hence integrable by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral, and $\langle v,w\rangle_{a}$ is integrable; so Linearity and Monotonicity of the Lebesgue Integral §integrable shows that $|v-w|_{a}^{2}$ is integrable with

$$\int_{X\times X}|q(x)-\eta(y)|_{a}^{2}\,\pi(dz)=\lVert q\rVert_{\nu}^{2}-2\mathcal{K}^{a}(q,\eta,\pi)+\lVert\eta\rVert_{\mu}^{2}.$$

The right-hand side depends only on the classes of $q$ and $\eta$, by The Space of Square-Integrable Maps from a Measure Space into a Hilbert Space with an Orthonormal Basis §operations and the first paragraph of this step, so the discrepancy does not depend on the representatives chosen.

Step 8 (claim 7). Suppose $\mu\in\mathcal{P}^{a}_{\rho}$ and $\eta$ satisfies the hypothesis of claim 7, and suppose, for a contradiction, that $\lVert\eta\rVert_{\mu}\ne0$, so $\lVert\eta\rVert_{\mu}>0$. Apply Steps 5 and 6 with $h=\eta$ (any representative): for every $t\in\mathbb{R}$ the measure $\nu_{t}=(S_{t})_{\#}\mu$ belongs to $\mathcal{P}^{a}_{\rho}$, the coupling $\pi_{t}$ belongs to $\Pi^{a}(\mu,\nu_{t})$, $I^{a}(\pi_{t})=t^{2}\lVert\eta\rVert_{\mu}^{2}$ and $\mathcal{J}^{a}(\eta,\pi_{t})=t\langle\eta,\eta\rangle_{\mu}=t\lVert\eta\rVert_{\mu}^{2}$, the norm being that of the inner product by The Space of Square-Integrable Hilbert-Valued Maps is a Real Hilbert Space: Coordinates and Synthesis §hilbert. The order of choices is: first $\varepsilon$, then $\theta$, then $t$. Let $\varepsilon=\lVert\eta\rVert_{\mu}/2>0$; the hypothesis provides $\theta>0$ for this $\varepsilon$. Let $t=\theta/(2\lVert\eta\rVert_{\mu})>0$. Then $I^{a}(\pi_{t})=\theta^{2}/4<\theta^{2}$, so the hypothesis applies to $\nu_{t}$ and $\pi_{t}$, and since $t\lVert\eta\rVert_{\mu}\ge0$ is the square root of $t^{2}\lVert\eta\rVert_{\mu}^{2}$ by Existence and Uniqueness of the Nonnegative Square Root,

$$t\lVert\eta\rVert_{\mu}^{2}=\bigl|\mathcal{J}^{a}(\eta,\pi_{t})\bigr|\le\varepsilon\sqrt{I^{a}(\pi_{t})}=\varepsilon\,t\lVert\eta\rVert_{\mu}.$$

Dividing by $t\lVert\eta\rVert_{\mu}>0$ gives $\lVert\eta\rVert_{\mu}\le\lVert\eta\rVert_{\mu}/2$, contradicting $\lVert\eta\rVert_{\mu}>0$. Hence $\lVert\eta\rVert_{\mu}=0$, so $\langle\eta,\eta\rangle_{\mu}=0$, and the positive definiteness of the inner product of the real Hilbert space $L^{2}(\mu;X^{a})$ (The Space of Square-Integrable Hilbert-Valued Maps is a Real Hilbert Space: Coordinates and Synthesis §hilbert and Real Inner Product Space §inner-product) shows that $\eta$ is the zero element of $L^{2}(\mu;X^{a})$.