Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. Throughout, ∣s∣ is the absolute value of s∈R, 2α is the quotient of α by 2=1+1, α−1 is the multiplicative inverse of a positive α, 2k and (21)k are natural powers, and W2(μ,ν)2=W2(μ,ν)W2(μ,ν).
Suppose the conclusion fails. Since the order of R is total (an axiom of Ordered Field), there is then μ0∈D with v(μ0)<u(μ0); put θ0=u(μ0)−v(μ0), a positive real number by claim 1 of Elementary Order Arithmetic in an Ordered Field applied to v(μ0)<u(μ0) with the summand −v(μ0).
Step 1. The least penalty and the penalised suprema. Claim 1 of The Doubled Difference on the Lift of a Wasserstein-Coercive Penalty Pair: Bounds, Closed Superlevel Sets and the Least Penalty, read with δ=21, with the present u, v, b, b′ and with any lower bound for E on D as provided by Basic Properties of a Wasserstein-Coercive Penalty Pair §bounded-below, gives μmin∈D with E(μmin)≤E(σ) for every σ∈D. Put e0=E(μmin); this is a real number satisfying e0≤E(σ) for every σ∈D, and it is with this choice that The Structure Estimate at a Maximiser of the Wasserstein-Doubled Difference on the Lift and Existence, Penalty Bounds and Optimal Realisation at a Maximiser of the Wasserstein-Doubled Difference on the Lift are read below.
Let Z1=D×D, a nonempty set because D contains the nonempty DΣ by Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair and Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §nonempty, and let D:Z1→R have the value D(μ,ν)=E(μ)+E(ν)−2e0; then 0≤D(μ,ν) for every (μ,ν)∈Z1, and D(μmin,μmin)=0. For positive α∈R let ψα:Z1→R have the value u(μ)−v(ν)−2αW2(μ,ν)2; it is bounded above by b−b′.
For positive δ and α let Ψδ,α:Z1→R and M(δ,α) be as in The Structure Estimate at a Maximiser of the Wasserstein-Doubled Difference on the Lift. Since uδ−=u−δE and vδ+=v+δE on D by Basic Properties of a Wasserstein-Coercive Penalty Pair §envelopes,
Ψδ,α(μ,ν)=ψα(μ,ν)−δD(μ,ν)−2δe0((μ,ν)∈Z1).
Let g(α,δ) be the supremum of the values of ψα−δD on Z1, a real number by claim 1 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence read with Z=Z1, with ψ=ψα, with this D and with z0=(μmin,μmin). If (μ^,ν^)∈Z1 satisfies Ψδ,α(μ^,ν^)=M(δ,α), then adding the constant 2δe0 to the inequalities Ψδ,α(μ,ν)≤Ψδ,α(μ^,ν^) shows that (ψα−δD)(μ^,ν^) is an upper bound for the values of ψα−δD and is one of them, so that
g(α,δ)=(ψα−δD)(μ^,ν^),M(δ,α)=g(α,δ)−2δe0.
Let I={δ∈R:0<δ<1}, let Z2=Z1×I, and let ψ′,D′:Z2→R have the values ψ′((μ,ν),δ)=u(μ)−v(ν)−δD(μ,ν) and D′((μ,ν),δ)=21W2(μ,ν)2. Then ψ′ is bounded above by b−b′, the values of D′ are nonnegative, D′(z0′)=0 for z0′=((μmin,μmin),21)∈Z2, and
(ψ′−αD′)((μ,ν),δ)=(ψα−δD)(μ,ν)
for positive α. Let N(α) be the supremum of the values of ψ′−αD′ on Z2, a real number by claim 1 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence read with Z=Z2, ψ=ψ′, D=D′ and z0=z0′. Two consequences will be used. First, g(α,δ)≤N(α) for every δ∈I and every positive α: by the last display the values of ψα−δD on Z1 are among the values of ψ′−αD′ on Z2, for which N(α) is an upper bound. Secondly, by claim 2 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence, read once with Z1 and once with Z2, the function g(α,⋅) is nonincreasing on the positive reals for each fixed positive α, and N is nonincreasing on the positive reals.
Step 2. The level and the constants. Put
B=2(∣b∣+∣b′∣+∣e0∣+∣E(μ0)∣+θ0),R=2B,
positive because θ0 is. Let δ∈I and let α∈R satisfy 1<α, and let (μ^,ν^)∈Z1 satisfy Ψδ,α(μ^,ν^)=M(δ,α), as Existence, Penalty Bounds and Optimal Realisation at a Maximiser of the Wasserstein-Doubled Difference on the Lift §maximiser provides. Since W2(μ0,μ0)=0,
θ0−2δE(μ0)=Ψδ,α(μ0,μ0) ≤ M(δ,α)=Ψδ,α(μ^,ν^),
so claim 2 of Existence, Penalty Bounds and Optimal Realisation at a Maximiser of the Wasserstein-Doubled Difference on the Lift, read with this μ0, this e0 and with the nonnegative number there taken to be 0, gives E(μ^)≤c0 and E(ν^)≤c0, where c0=δ−1(b−b′−θ0+2δE(μ0))−e0. Multiplying by the positive δ (claim 5 of Elementary Arithmetic in an Ordered Field),
δE(μ^) ≤ b−b′−θ0+2δE(μ0)−δe0 ≤ ∣b∣+∣b′∣+θ0+2∣E(μ0)∣+∣e0∣ ≤ B,
the middle step by claim 3 of Properties of the Absolute Value in an Ordered Field, which bounds each of b, −b′, −θ0, 2δE(μ0) and −δe0 by the corresponding absolute value, combined with 0<δ<1 and claim 5 of Elementary Arithmetic in an Ordered Field. Since e0≤E(μ^) we also have −∣e0∣≤δe0≤δE(μ^) and ∣e0∣≤B, so −B≤δE(μ^)≤B and hence ∣δE(μ^)∣≤B by claim 6 of Properties of the Absolute Value in an Ordered Field; as δ is positive, δ∣E(μ^)∣=∣δE(μ^)∣≤B by claim 4 of that lemma; the same argument gives δ∣E(ν^)∣≤B. Also ∣b∣+∣b′∣+∣e0∣≤B and 0<2B≤R. These bounds hold for every δ∈I, every α>1 and every maximiser.
Since F is locally strictly proper, fix a properness constant λ for F at R; since F satisfies the second-order structure condition, fix a second-order structure pair (ω1,ω2) for F at R. Put ς=θ0/4 and κ=λθ0/8, both positive.
Step 3. The choice of the doubling strength. As ω1 is a modulus of continuity, its clause 2 provides a positive τ1 with ω1(t)≤κ whenever 0≤t≤τ1. Put η1=τ1/16 and β0=1+2τ1−1, both positive, and βk=2kβ0 for k∈N.
First, 1≤2k for every k∈N. The set S={k∈N:1≤2k} contains 1, since 21=2 and 1≤2, the latter because 0<1 by claim 6 of Elementary Order Arithmetic in an Ordered Field and claim 1 of that lemma then gives 1<1+1=2; and if k∈S then 2k+1=2⋅2k and 1≤2=2⋅1≤2⋅2k by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier 2, so k+1∈S. By Principle of Induction for the Natural Numbers, S=N. Multiplying 1≤2k by the nonnegative β0 gives β0≤βk for every k∈N.
By claim 4 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence, read with Z2, ψ′, D′, z0′ and this β0, the sequence whose k-th term is N(βk)−N(βk+1) converges to 0. Fix k∈N with N(βk)−N(βk+1)≤η1 and put α=βk+1. Then α=2k+1β0=2⋅2kβ0=2βk, so βk=α/2; and 1<β0≤α, so 1<α. Moreover 2τ1−1≤β0≤α, and multiplying first by the nonnegative α−1 and then by the nonnegative τ1/2 gives α−1≤τ1/2.
By clause 2 of The Second-Order Structure Condition at Optimally Coupled Pairs on the Lift of the Wasserstein Space the function on the nonnegative reals with value ω2(t,α) at t is a modulus of continuity; fix a positive τ2 with ω2(t,α)≤κ whenever 0≤t≤τ2.
Step 4. The choice of the penalty weight. Since N(α) is the supremum of the values of ψ′−αD′ on Z2, claim 3 of Approximation Property of the Supremum and the Infimum in R provides z=((μ,ν),δ1)∈Z2 with N(α)−η1<(ψ′−αD′)(z). By Step 1, (ψ′−αD′)(z)=(ψα−δ1D)(μ,ν)≤g(α,δ1), so N(α)−η1≤g(α,δ1) with δ1∈I.
For j∈N put ρj=(21)jδ1; then 0<ρj≤δ1<1, so ρj∈I, and ρj+1=ρj/2 because (21)j+1=(21)j⋅21. By claim 4 of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series, read with r=21, the sequence whose j-th term is (21)j converges to 0, so by claim 3 of Arithmetic of Limits of Real Sequences the sequence whose j-th term is ρj converges to 0.
Because g(α,⋅) is nonincreasing and ρj+1≤ρj, the sequence whose j-th term is g(α,ρj) is nondecreasing, and it is bounded above by N(α); by claim 1 of A Bounded Monotone Sequence of Real Numbers Converges it converges. The sequence whose j-th term is g(α,ρj+1) is a subsequence of it, hence converges to the same limit by A Subsequence of a Convergent Sequence Has the Same Limit, so by claim 3 of Arithmetic of Limits of Real Sequences the sequence whose j-th term is g(α,ρj+1)−g(α,ρj) converges to 0.
For each j∈N let (μ^j,ν^j)∈Z1 satisfy Ψρj,α(μ^j,ν^j)=M(ρj,α), as Existence, Penalty Bounds and Optimal Realisation at a Maximiser of the Wasserstein-Doubled Difference on the Lift §maximiser provides. By Step 1 it satisfies g(α,ρj)=(ψα−ρjD)(μ^j,ν^j), so claim 3 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence, read with Z1, ψα, D, with ρj in the role of the positive number written β there and with an arbitrary positive η, gives
2ρjD(μ^j,ν^j) ≤ g(α,ρj+1)−g(α,ρj)+η
for every positive η, whence ρjD(μ^j,ν^j)≤2(g(α,ρj+1)−g(α,ρj)) by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above. The right-hand side converges to 0 and the left-hand side is nonnegative, so the sequence whose j-th term is ρjD(μ^j,ν^j) converges to 0.
Since e0≤E(μ^j) the number E(μ^j)−e0 is nonnegative, so ∣E(μ^j)−e0∣=E(μ^j)−e0 by claim 1 of Properties of the Absolute Value in an Ordered Field, and the triangle inequality, claim 5 of that lemma, applied to E(μ^j)=(E(μ^j)−e0)+e0 gives ∣E(μ^j)∣≤E(μ^j)−e0+∣e0∣; likewise for ν^j. Adding the two and multiplying by the positive ρj,
ρj(∣E(μ^j)∣+∣E(ν^j)∣+1) ≤ ρjD(μ^j,ν^j)+ρj(2∣e0∣+1).
Both terms on the right converge to 0, and so does the sequence whose j-th term is 2ρj∣E(μ0)∣, by claim 3 of Arithmetic of Limits of Real Sequences. Choosing j beyond the three thresholds these convergences supply, fix j∈N with
ρjD(μ^j,ν^j)+ρj(2∣e0∣+1)≤τ2,2ρj∣E(μ0)∣≤ς,
and put δ=ρj, μ^=μ^j, ν^=ν^j. Then δ∈I and δ≤δ1, so N(α)−η1≤g(α,δ1)≤g(α,δ) because g(α,⋅) is nonincreasing; and
δ(∣E(μ^)∣+∣E(ν^)∣+1)≤τ2,soω2(δ(∣E(μ^)∣+∣E(ν^)∣+1),α)≤κ.
Step 5. The contradiction. By Step 2 and the choice of j,
θ0−ς ≤ θ0−2δE(μ0) ≤ M(δ,α),
using 2δE(μ0)≤2δ∣E(μ0)∣≤ς; and θ0−ς=3θ0/4 is positive, so 0≤M(δ,α).
Apply claim 3 of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence with Z2, ψ′, D′, z0′, with α in the role of β, with η=η1 and with the point z=((μ^,ν^),δ): its hypothesis N(α)−η1≤(ψ′−αD′)(z)=g(α,δ) holds by Step 4, so
2α⋅21W2(μ^,ν^)2 ≤ N(α/2)−N(α)+η1 ≤ η1+η1,
the last step because α/2=βk, α=βk+1 and N(βk)−N(βk+1)≤η1. Hence αW2(μ^,ν^)2≤8η1=τ1/2, and with α−1≤τ1/2 from Step 3,
αW2(μ^,ν^)2+α−1 ≤ τ1,soω1(αW2(μ^,ν^)2+α−1)≤κ.
All hypotheses of The Structure Estimate at a Maximiser of the Wasserstein-Doubled Difference on the Lift now hold: the probability space is rich, the penalty pair is Wasserstein-coercive with closed score, F satisfies the shift-coercivity and shift-semicontinuity conditions, u and v are as required with the bounds b and b′ and are respectively a viscosity subsolution and a viscosity supersolution on the lift, e0 is a lower bound for E on D, 0<δ<1 and 1<α, the pair (μ^,ν^) maximises Ψδ,α and 0≤M(δ,α), a pair X^,Y^ as required exists by Existence, Penalty Bounds and Optimal Realisation at a Maximiser of the Wasserstein-Doubled Difference on the Lift §optimal-pair, the numbers B and R satisfy the four inequalities by Step 2, and λ and (ω1,ω2) are a properness constant and a structure pair at R. Its clause The Structure Estimate at a Maximiser of the Wasserstein-Doubled Difference on the Lift §estimate therefore gives
λ(θ0−ς) ≤ λM(δ,α) ≤ κ+κ=4λθ0,
the first inequality by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier λ. As θ0−ς=3θ0/4, this reads 43λθ0≤41λθ0, that is 21λθ0≤0 by claim 3 of Elementary Arithmetic in an Ordered Field. But λ and θ0 are positive, so 0<21λθ0 by claim 5 of Elementary Order Arithmetic in an Ordered Field applied twice, a contradiction.
Therefore no such μ0 exists, and u(μ)≤v(μ) for every μ∈D.