Proof of Couplings of Finite Noise Cost: the Support Bound, Swap, Displacement Couplings and Lower Semicontinuity of the Noise Cost

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

Throughout, for $z\in X\times X$ we write $x=\pi_{1}(z)$, $y=\pi_{2}(z)$ and $\delta(z)=y-x$, so that $D_{a}=\delta^{-1}(X^{a})$ and $c_{a}=n_{a}\circ\delta$ by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §pairs; in particular $c_{a}(z)=|y-x|_{a}^{2}$ for $z\in D_{a}$ and $c_{a}(z)=0$ for $z\notin D_{a}$, 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. For $N\in\mathbb{N}$, $S_{N}:X\to\mathbb{R}$ is the function $S_{N}(h)=\sum_{k=1}^{N}a_{k}^{-1}h_{k}^{2}$ of The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability. For a coupling $\pi$, the complement $(X\times X)\setminus D_{a}$ is written $D_{a}^{c}$; it is Borel, and if $\pi(D_{a})=1$ then $\pi(D_{a}^{c})=1-1=0$ by claim 3 of Basic Properties of a Measure, so that $D_{a}^{c}$ is $\pi$-null.

Claim (support-bound). Let $\pi\in\Pi^{a}(\mu,\nu)$, so $\pi(D_{a})=1$ and $I^{a}(\pi)=\int c_{a}\,d\pi<\infty$ by Couplings of Finite Noise Cost and Their Noise Cost §finite and Couplings of Finite Noise Cost and Their Noise Cost §cost. For $z\in D_{a}$ the vector $y-x$ lies in $X^{a}$, so The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §embedding gives $|\pi_{1}(z)-\pi_{2}(z)|^{2}=|y-x|^{2}\le\bar{a}\,|y-x|_{a}^{2}=\bar{a}\,c_{a}(z)$, where $|x-y|=|y-x|$ by the symmetry of the metric (The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §metric). Both sides are nonnegative Borel functions of $z$ (Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost and The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §pairs, the multiple $\bar{a}\,c_{a}$ of the Borel function $c_{a}$ being Borel by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions), and the inequality holds outside the $\pi$-null set $D_{a}^{c}$, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison and the scalar rule of Linearity and Monotonicity of the Lebesgue Integral §nonnegative,

$$I(\pi)=\int_{X\times X}|\pi_{1}(z)-\pi_{2}(z)|^{2}\,\pi(dz)\le\int_{X\times X}\bar{a}\,c_{a}\,d\pi=\bar{a}\,I^{a}(\pi)<\infty .$$

Now suppose $\pi\in\Pi^{a}(\mu,\nu)$ exists. If $\mu\in\mathcal{P}_{2}(X)$, then $I(\pi)<\infty$ and the converse part of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §cost-finite give $\nu\in\mathcal{P}_{2}(X)$. If $\nu\in\mathcal{P}_{2}(X)$, then by Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §swap the swapped measure $\sigma_{\#}\pi$ lies in $\Pi(\nu,\mu)$ with $I(\sigma_{\#}\pi)=I(\pi)<\infty$, and the converse part of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §cost-finite, applied to the pair $(\nu,\mu)$ and the coupling $\sigma_{\#}\pi$, gives $\mu\in\mathcal{P}_{2}(X)$.

Claim (swap). Let $\pi\in\Pi^{a}(\mu,\nu)$. By Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §swap, $\sigma$ is Borel and $\sigma_{\#}\pi\in\Pi(\nu,\mu)$. For $z\in X\times X$ one has $\delta(\sigma(z))=x-y=-\delta(z)$. Since $X^{a}$ is a linear subspace of $X$ (The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §hilbert), $\delta(z)\in X^{a}$ if and only if $-\delta(z)\in X^{a}$; hence $\sigma^{-1}(D_{a})=D_{a}$. Moreover $c_{a}\circ\sigma=c_{a}$: off $D_{a}$ both sides vanish, and for $z\in D_{a}$, homogeneity and symmetry of $\langle\cdot,\cdot\rangle_{a}$ give $|-\delta(z)|_{a}^{2}=(-1)^{2}|\delta(z)|_{a}^{2}=|\delta(z)|_{a}^{2}$. By Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pushforward, that is by claims 1 and 2 of Image Measures, Measures with Densities, and Change of Variables,

$$\sigma_{\#}\pi(D_{a})=\pi\bigl(\sigma^{-1}(D_{a})\bigr)=\pi(D_{a})=1,\qquad\int_{X\times X}c_{a}\,d(\sigma_{\#}\pi)=\int_{X\times X}c_{a}\circ\sigma\,d\pi=\int_{X\times X}c_{a}\,d\pi=I^{a}(\pi)<\infty .$$

So $\sigma_{\#}\pi$ has finite noise cost by Couplings of Finite Noise Cost and Their Noise Cost §finite, whence $\sigma_{\#}\pi\in\Pi^{a}(\nu,\mu)$ by Couplings of Finite Noise Cost and Their Noise Cost §couplings, with $I^{a}(\sigma_{\#}\pi)=I^{a}(\pi)$ by Couplings of Finite Noise Cost and Their Noise Cost §cost. In particular, if $(\mu,\nu)$ is noise-connected then so is $(\nu,\mu)$, and applying this to the pair $(\nu,\mu)$ gives the converse (Couplings of Finite Noise Cost and Their Noise Cost §connected).

Claim (displacement). Let $T=(\mathrm{id}_{X},S)$, Borel as stated. By Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §pushforward, applied with its maps $S$ and $T$ taken to be $\mathrm{id}_{X}$ and our $S$, the measure $T_{\#}\mu$ lies in $\Pi((\mathrm{id}_{X})_{\#}\mu,S_{\#}\mu)$, and $(\mathrm{id}_{X})_{\#}\mu=\mu$ because $\mathrm{id}_{X}^{-1}(B)=B$ for every Borel $B$. Since $\delta(T(x))=S(x)-x\in X^{a}$ for every $x$, one has $T^{-1}(D_{a})=X$ and $c_{a}\circ T(x)=n_{a}(S(x)-x)$. By claims 1 and 2 of Image Measures, Measures with Densities, and Change of Variables,

$$T_{\#}\mu(D_{a})=\mu(X)=1,\qquad\int_{X\times X}c_{a}\,d(T_{\#}\mu)=\int_{X}n_{a}(S(x)-x)\,\mu(dx),$$

and the right-hand side is finite by hypothesis. Hence $T_{\#}\mu$ has finite noise cost by Couplings of Finite Noise Cost and Their Noise Cost §finite, so $T_{\#}\mu\in\Pi^{a}(\mu,S_{\#}\mu)$ by Couplings of Finite Noise Cost and Their Noise Cost §couplings, with the displayed noise cost by Couplings of Finite Noise Cost and Their Noise Cost §cost. For $S=\mathrm{id}_{X}$, which is Borel and satisfies $S(x)-x=0_{X}\in X^{a}$, the integrand is $n_{a}(0_{X})=\langle0_{X},0_{X}\rangle_{a}=0$ (homogeneity of $\langle\cdot,\cdot\rangle_{a}$ with the factor $0$), so its integral is $0$ by the scalar rule of Linearity and Monotonicity of the Lebesgue Integral §nonnegative with $c=0$; thus, by the case just proved (that is, by Couplings of Finite Noise Cost and Their Noise Cost §finite and Couplings of Finite Noise Cost and Their Noise Cost §couplings, with the value from Couplings of Finite Noise Cost and Their Noise Cost §cost), $(\mathrm{id}_{X},\mathrm{id}_{X})_{\#}\mu\in\Pi^{a}(\mu,\mu)$ has noise cost $0$, and $(\mu,\mu)$ is noise-connected.

Claim (lsc). Since $\Pi^{a}(\mu_{j},\nu_{j})\subseteq\Pi(\mu_{j},\nu_{j})$, Couplings on a Hilbert Space: Tightness, Closedness under Weak Convergence, and Lower Semicontinuity of the Quadratic Cost §closed gives $\pi\in\Pi(\mu,\nu)$. Let $L=\liminf_{j}I^{a}(\pi_{j})$, a real number by Limit Inferior of a Bounded Sequence of Real Numbers.

Step 1 (bounded continuous minorants). For $N,M\in\mathbb{N}$ let $g_{N,M}(z)=\min\{S_{N}(\delta(z)),M\}$. Then $0\le g_{N,M}\le M$, as $S_{N}\ge0$. It is continuous on $X\times X$: if $z_{m}\to z$, then $\pi_{i}(z_{m})\to\pi_{i}(z)$ for $i=1,2$ by Properties of the Product of Two Real Inner Product Spaces §componentwise, so $\delta(z_{m})\to\delta(z)$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits, so $S_{N}(\delta(z_{m}))\to S_{N}(\delta(z))$ by the continuity of $S_{N}$ (The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §partial-sums) and Continuity Between Metric Spaces is Equivalent to Sequential Continuity §sequential; and $|\min\{s,M\}-\min\{t,M\}|\le|s-t|$ for real $s,t$ (check the cases $s,t\le M$; $s,t\ge M$; $s\le M\le t$ and its mirror), so $g_{N,M}(z_{m})\to g_{N,M}(z)$, and Continuity Between Metric Spaces is Equivalent to Sequential Continuity §on-subset applies. Hence $g_{N,M}$ is Borel by claims 2 and 3 of Borel Measurability and Bounded Integration on a Metric Space, and its integral as a bounded integrable function equals its integral as a nonnegative function by claim 6 of that lemma. For $z\in D_{a}$, The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §partial-sums gives $g_{N,M}(z)\le S_{N}(\delta(z))\le\sup_{N'}S_{N'}(\delta(z))=|\delta(z)|_{a}^{2}=c_{a}(z)$.

Step 2 (passage to the limit in $j$). Fix $N,M$. Since $\pi_{j}(D_{a})=1$, the inequality $g_{N,M}\le c_{a}$ holds outside the $\pi_{j}$-null set $D_{a}^{c}$, so The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison gives $\alpha_{j}:=\int g_{N,M}\,d\pi_{j}\le\int c_{a}\,d\pi_{j}=I^{a}(\pi_{j})$. By Weak Convergence of Finite Borel Measures on a Metric Space, $\alpha_{j}\to\alpha:=\int g_{N,M}\,d\pi$. We show $\alpha\le L$. Let $\varepsilon>0$ and choose $K$ with $\alpha_{m}>\alpha-\varepsilon$ for all $m\ge K$. Then $I^{a}(\pi_{m})\ge\alpha_{m}>\alpha-\varepsilon$ for $m\ge K$, so $\alpha-\varepsilon$ is a lower bound of $A_{K}=\{I^{a}(\pi_{m}):m\ge K\}$, whence $\alpha-\varepsilon\le\inf A_{K}\le L$ by Limit Inferior of a Bounded Sequence of Real Numbers. As $\varepsilon$ was arbitrary, $\alpha\le L$. Thus $\int g_{N,M}\,d\pi\le L$ for all $N,M\in\mathbb{N}$.

Step 3 (limit in $N$). Fix $M$. By The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §partial-sums, $S_{N}\le S_{N+1}$, so $g_{N,M}\le g_{N+1,M}$ on $X\times X$. Let $G_{M}(z)=\sup_{N}g_{N,M}(z)$. By Monotone Convergence Theorem, $G_{M}$ is measurable and $\int G_{M}\,d\pi=\sup_{N}\int g_{N,M}\,d\pi\le L$. If $z\in D_{a}^{c}$, then $\delta(z)\notin X^{a}$, so by The Noise Space is a Real Hilbert Space: Orthonormal Basis, Continuous Embedding, Partial Sums, Closed Balls and Borel Measurability §partial-sums the sequence $(S_{N}(\delta(z)))_{N}$ is not bounded above, and some $N$ has $S_{N}(\delta(z))\ge M$, i.e. $g_{N,M}(z)=M$; hence $G_{M}(z)=M$. If $z\in D_{a}$ and $c_{a}(z)\le M$, then $g_{N,M}(z)=S_{N}(\delta(z))$ for all $N$ by Step 1, so $G_{M}(z)=\sup_{N}S_{N}(\delta(z))=c_{a}(z)$; if $z\in D_{a}$ and $c_{a}(z)>M$, then some $N$ has $S_{N}(\delta(z))>M$ (approximation of the supremum $c_{a}(z)$), so $G_{M}(z)=M$. In all cases $M\,\mathbf{1}_{D_{a}^{c}}\le G_{M}$, and $\min\{c_{a}(z),M\}\le G_{M}(z)$ for $z\in D_{a}$.

Step 4 ($\pi(D_{a})=1$). By The Integral of an Indicator Function is the Measure of the Set and Linearity and Monotonicity of the Lebesgue Integral §nonnegative, $M\,\pi(D_{a}^{c})=\int M\,\mathbf{1}_{D_{a}^{c}}\,d\pi\le\int G_{M}\,d\pi\le L$ for every $M\in\mathbb{N}$. If $p=\pi(D_{a}^{c})$ were positive, the Archimedean property would give $M\in\mathbb{N}$ with $Mp>L$, a contradiction. So $p=0$ and $\pi(D_{a})=1-p=1$ by claim 3 of Basic Properties of a Measure.

Step 5 (limit in $M$). Since $g_{N,M}\le g_{N,M+1}$, also $G_{M}\le G_{M+1}$. Let $H(z)=\sup_{M}G_{M}(z)\in[0,\infty]$. By Monotone Convergence Theorem, $H$ is measurable and $\int H\,d\pi=\sup_{M}\int G_{M}\,d\pi\le L$. Moreover $c_{a}\le H$ on $X\times X$: off $D_{a}$, $c_{a}=0\le H$; on $D_{a}$, choosing $M\ge c_{a}(z)$ (Archimedean property) gives $c_{a}(z)=\min\{c_{a}(z),M\}\le G_{M}(z)\le H(z)$ by Step 3. Hence, by the monotonicity part of Linearity and Monotonicity of the Lebesgue Integral §nonnegative,

$$\int_{X\times X}c_{a}\,d\pi\le\int_{X\times X}H\,d\pi\le L<\infty .$$

Together with Step 4 and $\pi\in\Pi(\mu,\nu)$, this shows that $\pi$ has finite noise cost (Couplings of Finite Noise Cost and Their Noise Cost §finite), so $\pi\in\Pi^{a}(\mu,\nu)$ (Couplings of Finite Noise Cost and Their Noise Cost §couplings), and, by Couplings of Finite Noise Cost and Their Noise Cost §cost, $I^{a}(\pi)=\int c_{a}\,d\pi\le\liminf_{j}I^{a}(\pi_{j})$.