Per-Term PV Vanishing: CPV Helpers and Assembly #
CPV helpers for continuous functions, holomorphic CPV vanishing on convex domains, scalar/sum lemmas for CPV integrands, integrability, and the assembly of per-term vanishing into the complete higher-order cancellation.
Main Results #
tendsto_cpv_of_continuousOn_zero_integral— CPV with zero contour integralholomorphic_cpv_tendsto_zero_on_convex— holomorphic CPV → 0cpvIntegrandOn_const_smul/cpvIntegrandOn_add/cpvIntegrandOn_finset_sumintervalIntegrable_cpvIntegrandOn_of_continuousOn_diffresidueAt_sub_residueSum_eq_zero— residue of f minus residue sum vanishescpv_tendsto_zero_of_add_decomposition— final assembly
Helper: CPV of a function continuous along γ with zero contour integral #
If g is continuous on γ's image and ∮_γ g dz = 0, then cpv(S0, g, ε) → 0. The CPV integrand converges a.e. to g(γ(t)) * γ'(t) as ε → 0 (the crossing set has measure zero), and is dominated by ‖g(γ(t))‖ * ‖γ'(t)‖. By DCT, the limit equals ∮_γ g dz = 0.
CPV integral of a function continuous along γ with zero ordinary contour integral tends to 0. This is the DCT core of the assembly proof, abstracting the zero-integral condition.
Sublemma 3: Holomorphic CPV integral → 0 on closed curve #
Sublemma 3: For a function holomorphic on a convex open U containing the
closed curve γ, the multi-point CPV integral tends to 0.
The CPV integrand 1_{∀s∈S0, ‖γ(t)-s‖>ε} · g(γ(t)) · γ'(t) converges a.e.
to g(γ(t)) · γ'(t) as ε → 0 (the cutout set shrinks to a null set), and
is dominated by ‖g(γ(t))‖ · ‖γ'(t)‖ (bounded since g is continuous on
the compact image of γ). By DCT, the CPV integral converges to the ordinary
integral ∮_γ g dz, which is 0 by Cauchy's integral theorem on convex U.
Helper: CPV integral of scalar multiple #
CPV integrand of c • f equals c • cpv(f) pointwise (the indicator set is the same).
CPV integrand of f + g equals cpv(f) + cpv(g) pointwise (same indicator).
CPV integrand of a finset sum decomposes.
Integrability of CPV integrand for functions continuous on U \ S0. For any g continuous on U \ S0 with γ mapping into U, the multi-point CPV integrand is interval integrable on [a,b] for fixed ε > 0. The key insight is that the CPV integrand is bounded (it's either 0 or g(γ(t)) * γ'(t) where γ(t) is far from all poles) and ae strongly measurable (continuous off a finite set within [a,b]).
Assembly helper: CPV of h = f - f_res tends to 0 #
This helper packages the assembly of Sublemmas 1-3 into the main bridge lemma.
It shows that ∫ cpv(S0, h, ε) → 0 where h z = f z - Σ res(f,s)/(z-s).
The proof decomposes h into:
- Polar terms at each crossed point:
a_k/(z-s)^{k+1}fork ≥ 1, each tending to 0 bymultipoint_pv_zpow_tendsto_zero(Sublemma 2). - Holomorphic remainder: continuous along γ with zero contour integral,
tending to 0 by
tendsto_cpv_of_continuousOn_zero_integral.
The identification a_0 = residueAt f s uses residueAt_eq_laurent_head_coeff
(Sublemma 1).
Helper: residueAt (f - Σ res(f,s')/(z-s')) s = 0 for s ∈ S0.
The function h z = f z - Σ_{s' ∈ S0} residueAt f s' / (z - s') has the same
higher-order poles as f at s but no simple pole (the s' = s term in the
sum cancels the residue). Hence residueAt h s = 0.
The proof uses circle integral decomposition: for small r,
∮_{C(s,r)} h = ∮ f - Σ residueAt f s' * ∮ 1/(z-s'). For s' ≠ s with
|s'-s| > r, ∮ 1/(z-s') = 0 (Cauchy); for s' = s, ∮ 1/(z-s) = 2πi.
So (2πi)⁻¹ ∮ h = (2πi)⁻¹ ∮ f - residueAt f s. Taking r → 0:
residueAt h s = residueAt f s - residueAt f s = 0.