DERIVATIONS — QOT Localization Bounds & Affine Case (kcnuX4xEpL)
Notation: P = µ⊗ν, c(x,y)=½‖x−y‖², T Monge map, r = dist(sptπ_ε; grT) (directed Hausdorff), h = dπ_ε/dP, ω_d = π^(d/2)/Γ(d/2+1) = vol B(0,1), ∆_ε = QOT_ε − OT ≥ 0. Eckstein–Nutz rate (used as an input, eq. 3.2): C_val⁻¹ ε^(2/(d+2)) ≤ ∆_ε ≤ C_val ε^(2/(d+2)), ε∈(0,1].
Each derivation step below has an executable numerical check (CHK). Checks are cheap (seconds), CPU-only, use numpy only. They VERIFY the algebra/lemmas, not paper outcomes.
Claim 1 — Theorem 3.3: lower bound r ≥ c_sm ε^(1/(d+2))
D1. Lemma 3.1 (distance to a Lipschitz graph)
For T L-Lipschitz and any (x,y): dist((x,y);grT) ≥ ‖y−T(x)‖ / (1+L²)^(1/2). Proof: for x' with t=‖x−x'‖, reverse triangle + Lipschitz give ‖y−T(x')‖ ≥ max{0, a−Lt}, a=‖y−T(x)‖. Then ‖(x,y)−(x',T(x'))‖² ≥ t² + max{0,a−Lt}². Minimizing over t≥0: if a>Lt the min over t of (1+L²)t²−2Lat+a² is at t*=La/(1+L²) with value a²/(1+L²); if a≤Lt then t²≥a²/L²≥a²/(1+L²). Both cases ≥ a²/(1+L²). ∎ Remark (bias equivalence): dist ≤ ‖y−T(x)‖ ≤ √(1+L²)·dist. So directed Hausdorff and vertical bias sup‖y−T(x)‖ are equivalent up to √(1+L²). This is why exp01's dbias proxy decides a directed-Hausdorff (Claim 1) statement.
- CHK-D1: pick A=diag(λ_1..λ_d) (L=max λ_i), a random. For 200 random (x,y), numerically minimize g(x')=‖x−x'‖²+‖y−Ax'−a‖² over x' (closed form: x'=(I+A²)⁻¹(x+A(y−a))). Assert min g = ‖y−T(x)‖²/(1+L²) up to 1e-8 only when the vertical direction aligns with the top eigenvector; in general assert min g ≥ ‖y−T(x)‖²/(1+L²) − 1e-9 (inequality D1) and min g ≤ ‖y−T(x)‖² (upper remark). PASS if both hold for all samples.
D2. Lemma 3.2 (fiberwise L² lower bound)
(i) marginal ∫h(x,y)µ(dx)=1 ⇒ by Cauchy–Schwarz 1=(∫_{X_y}h dµ)² ≤ µ(X_y)∫h²dµ, so ∫h²dµ ≥ 1/µ(X_y). (ii) X_y ⊆ T⁻¹(B(y,(1+L²)^½ r)) (from D1), and Assumption 2 (ν ≤ λ_ν·Leb) ⇒ µ(X_y)=ν(image) ≤ λ_ν (1+L²)^(d/2) ω_d r^d. Integrating (i) over ν: ‖h‖²_{L²(P)} ≥ ∫ 1/µ(X_y) ν(dy) ≥ [λ_ν(1+L²)^(d/2)ω_d]⁻¹ r⁻d ≕ C_0⁻¹ r⁻d. (eq 3.1)
- CHK-D2: discretize µ,ν uniform on a grid; take an arbitrary nonneg h with row-marginal ∫h dµ=1 enforced (normalize each fiber). Numerically verify ∫h²dµ ≥ 1/µ(X_y) per fiber (X_y = support of that fiber) to 1e-10. Also verify the Cauchy–Schwarz constant chain: for random support masses m∈(0,1], 1/m ≥ 1 (monotone) and equality of C_0 formula C_0=λ_ν(1+L²)^(d/2)ω_d numerically vs its definition. PASS on exact algebra.
D3. Assembling Theorem 3.3 + value-gap rate (eq 3.2)
Quadratic penalty contributes (ε/2)‖h‖²_{L²(P)} to QOT_ε, and OT ≤ ∫c dπ_ε, so ∆ε ≥ (ε/2)‖h‖²{L²(P)} ≥ (ε/(2C_0)) r⁻d (eq A.1). Case ε∈(0,1]: combine with ∆_ε ≤ C_val ε^(2/(d+2)) ⇒ (ε/2C_0) r⁻d ≤ C_val ε^(2/(d+2)) ⇒ r ≥ (2C_0 C_val)^(−1/d) ε^(1/(d+2)) = c_sm ε^(1/(d+2)). (Case ε≥1 gives r ≥ const.) The exponent is 1/(d+2) because r⁻d ≲ ε^(2/(d+2)−1)=ε^(−d/(d+2)) ⇒ r ≳ ε^(1/(d+2)).
- CHK-D3 (rate-scaling fit, decisive low-d surrogate for the regime): the whole lower bound hinges on ∆_ε = Θ(ε^(2/(d+2))). Verify this rate directly at d=1 where the regime is accessible: build µ=ν=Unif[0,1] discretized N=M=150; solve discrete QOT (B.4/B.6, GS solver from exp01 core) for ε on a log grid ε∈{1e-3..1e-1}×c_med; fit slope of log∆_ε vs logε. Assert slope ∈ [2/(1+2)−0.15, 2/(1+2)+0.15] = [0.517, 0.817]. (∆_ε = ∫cdπ_ε−OT_LP, OT_LP from linprog/emd.) PASS if slope within band ⇒ the Θ(ε^(2/(d+2))) input rate holds, hence the r ≳ ε^(1/(d+2)) exponent is correct. FALSIFIES C1's rate if slope ≠ ~0.667.
Claim 2 — Theorem 3.7: affine sharp pointwise bound b ≲ √λ_max(A) ε^(1/(d+2))
D4. Exact reduction to self-transport (Step 1, Lemma A.2)
F(x,v)=(x,T(v)) ⇒ F_#(µ⊗µ)=µ⊗ν; for ẽπ∈Π(µ,µ), π=F_#ẽπ∈Π(µ,ν) with equal L²(P) density norm. For affine T=Ax+a, φ(x)=½⟨x,Ax⟩+⟨a,x⟩, so Fenchel–Young slack D_φ(x,T(v)) = ½⟨v−x, A(v−x)⟩. Hence ∫c dπ − OT = ½∫⟨v−x,A(v−x)⟩ dẽπ, and QOT(µ,ν) reduces (up to const OT) to self-transport functional (A.6) over Π(µ,µ).
- CHK-D4: sample x_i,v_i ~ µ (truncated Gaussian, d=5). With φ quadratic as above, assert D_φ(x,T(v)) := φ(x)+φ*(T(v))−⟨x,T(v)⟩ equals ½⟨v−x,A(v−x)⟩ to 1e-9 elementwise, where φ*(y)=½⟨y−a,A⁻¹(y−a)⟩. Also assert pushforward-norm invariance numerically on a small discrete coupling: ‖dπ/dP‖ unchanged under F. PASS on algebra.
D5. Whitening + tube transfer (Step 2)
u=A^(1/2)x, w=A^(1/2)v ⇒ self-transport QOT for µ_A=(A^(1/2))_#µ with cost ½‖u−w‖². For (x,y)∈sptπ_ε, v=T⁻¹(y): y−T(x)=A(v−x)=A^(1/2)(w−u), so ‖y−T(x)‖ ≤ ‖A^(1/2)‖_op‖w−u‖ = √λ_max(A)‖w−u‖. Thus a diagonal tube bound for bπ_ε transfers to a grT tube bound for π_ε.
- CHK-D5: random A∈S^d_++ (d=6), random x,v; set y=T(v)=Av+a. Assert ‖y−T(x)‖ = ‖A^(1/2)(w−u)‖ exactly (≤1e-10) and ≤ √λ_max(A)·‖w−u‖. PASS on identity.
D6. Wiesel–Xu self-transport bound + ε_0 boundary
sup‖u−w‖ ≤ 8 δ_{µ_A}(ε)^(1/2), δ_η(ε)=inf{r>0: r ρ_η(√r)>ε}, ρ_η(s)=inf_z η(B(z,s)). Density lower bound: ρ_{µ_A}(s) ≥ λ_{µ_A} κ_A ω_d s^d ⇒ for r≤r_A², r ρ(√r) ≥ λ_{µ_A}κ_A ω_d r^(1+d/2). If ε ≤ ε_0 := λ_{µ_A} κ_A ω_d r_A^(d+2), then r* = (ε/(λ_{µ_A}κ_A ω_d))^(2/(d+2)) ≤ r_A² satisfies r* ρ(√r*) ≥ ε, so δ(ε) ≤ r*, δ(ε)^(1/2) ≤ (ε/(λ_{µ_A}κ_A ω_d))^(1/(d+2)). Combine with D5: sup‖y−T(x)‖ ≤ 8√λ_max(A) (ε/(λ_{µ_A}κ_A ω_d))^(1/(d+2)). ∎ (eq 3.6)
- CHK-D6a (algebraic boundary): for random λ,κ∈(0,1], ω_d, r_A, and ε=ε_0, assert r*=r_A² exactly and rρ(√r)=ε_0 (the inf is tight at the boundary), to 1e-10.
- CHK-D6b (rate arithmetic): verify the map ε ↦ 8√λ_max (ε/(λκω_d))^(1/(d+2)) has log-log slope exactly 1/(d+2) (finite-difference of log over ε grid) for d∈{1,10,100}. PASS ⇒ the claimed exponent is algebraically 1/(d+2).
- CHK-D6c (regime boundary magnitude — the in-regime reality): compute ε_0 for the experiment's family (d=100: ω_100=π^50/Γ(51), r_A=0.08) via log-gamma; assert ε_0 < 1e-100. This DOCUMENTS that no accessible ε is in the strict regime (see STATE in-regime note); it is a fact-check, not a pass/fail on the theorem.
Summary of executable checks
CHK-D1, D2, D4, D5, D6a/b/c: pure-numpy algebra/lemma assertions (< 5 s total). CHK-D3: one small QOT rate-fit (d=1, N=M=150) reusing exp01's GS solver (~seconds); this is the decisive IN-REGIME numerical confirmation that ∆_ε=Θ(ε^(2/(d+2))) — the rate that both claims inherit. If CHK-D3 fails, both C1 and C2 rate exponents are in doubt regardless of exp01.