Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Tagged sums approximate a contour integral within oscillation times length

Statement

Let γ:[a,b]C be a rectifiable contour with a<b, let f be continuous on its trace γ, let a=t0<t1<<tr=b be a partition of [a,b], and choose a tag ξi[ti,ti+1] for each i<r. Write γi for the restriction γ[ti,ti+1] and

ωi:=sup{f(u)f(v) : u,vγ([ti,ti+1])},

which is a nonnegative real number. Then

γf(z)dzi<rf(γ(ξi))(γ(ti+1)γ(ti))  i<rωiL(γi).

In particular, if ω0 satisfies f(u)f(v)ω for all u,vγ, then the left-hand side is at most ωL(γ).

The bound is stated with the oscillations themselves and not as a limit, so a modulus of continuity for f on γ converts directly into an error estimate. For a singleton parameter interval [a,a] there is no partition, and both the integral and the empty tagged sum are 0.

Facts & Assumptions

Given: A rectifiable contour γ:[a,b]C with a<b, a continuous f on γ, a partition a=t0<t1<<tr=b, and tags ξi[ti,ti+1] for i<r.

[L1]

A complex contour is a rectifiable path γ:[a,b]C; if α,β:[0,1]C satisfy α(1)=β(0), their concatenation is (αβ)(s)=α(2s) for 0s12 and β(2s1) for 12s1 (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L2]

For a rectifiable γ:[a,b]C and f continuous on its trace, the complex line integral γfdz of The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L3]

If ϕ:[c,d][a,b] is a strictly increasing continuous bijection, γ:[a,b]C is rectifiable and f is continuous on the trace of γ, then γϕfdz=γfdz (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L4]

For composable rectifiable contours α,β, αβfdz=αfdz+βfdz (Complex line integrals change sign under reversal and add under concatenation).

[L5]

For continuous f,g on the trace of a rectifiable contour γ and α,βC, γ(αf+βg)dz=αγfdz+βγgdz (Complex line integrals are linear in the integrand).

[L6]

For cC and a rectifiable contour γ:[a,b]C, γcdz=c(γ(b)γ(a)) (The contour integral of a constant c is c times the endpoint displacement).

[L7]

If f(z)M on the trace of a rectifiable contour γ, with M0, then γf(z)dzML(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length).

[L8]

For a path γ:[a,b]Rn with n1 and c[a,b], L[a,b](γ)=L[a,c](γ[a,c])+L[c,b](γ[c,b]) in the nonnegative extended reals, and γ is rectifiable on [a,b] if and only if both restrictions are rectifiable (Arc length is additive across every subdivision point and decreases under restriction).

[L9]

A partition of [a,b] with a<b consists of a=t0<t1<<tr=b with r1, its subintervals [ti,ti+1] being indexed from i=0 (Partition of [a,b] as a finite strictly increasing list a=t0<t1<<tn=b, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L11]

If akbk for all k<n then k<nakk<nbk (Laws of finite sums and finite products).

[L12]

If a property holds at 0 and passes from n to n+1, it holds for every nN (The principle of mathematical induction).

[L15]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

Proof

technique · direct
1.1

By [L8] applied at t1, then to γ[t1,b] at t2, and so on, an induction on the number of partition points ([L12]) shows that each γi is rectifiable and that L(γ)=i<rL(γi).

givenL8L9L12
1.2

Each ωi is a nonnegative real: [ti,ti+1] is a closed bounded interval, hence compact by [L13]; fγ is continuous on it, so its image is compact by [L14] and bounded by [L15]; hence {f(u)f(v):u,vγ([ti,ti+1])} is a nonempty set of reals bounded above, and it has a supremum, which is 0 because u=v is allowed.

givenL13L14L15
1.3

For au<v<wb put α(s)=γ(u+s(vu)) and β(s)=γ(v+s(wv)) on [0,1]; then α(1)=γ(v)=β(0), so αβ is defined by [L1], and αβ=γ[u,w]ϕ where ϕ:[0,1][u,w] is the strictly increasing continuous bijection that is affine on [0,12] and on [12,1] with ϕ(12)=v. Since α and β are increasing reparametrisations of γ[u,v] and γ[v,w], [L3] and [L4] give γ[u,w]fdz=γ[u,v]fdz+γ[v,w]fdz.

givenL1L2L3L4
1.4

For each i<r, [L6] applied to the constant f(γ(ξi)) on γi gives γif(γ(ξi))dz=f(γ(ξi))(γ(ti+1)γ(ti)).

L6
2.1

Applying step 1.3 at t1, then to γ[t1,b] at t2, and so on, an induction on the number of partition points ([L12]) gives γfdz=i<rγifdz.

step 1.3L12
2.2

Fix i<r. The tag value γ(ξi) lies on the trace of γi, so f(z)f(γ(ξi))ωi for every z on that trace by the definition of ωi in step 1.2; by [L5] the difference γifdzγif(γ(ξi))dz equals γi(f(z)f(γ(ξi)))dz, and [L7] bounds its modulus by ωiL(γi).

step 1.2step 1.4L5L7
3.1

Subtracting the identity of step 1.4 from that of step 2.1 termwise, the quantity to be estimated is i<r(γifdzf(γ(ξi))(γ(ti+1)γ(ti))); the finite triangle inequality, obtained from [L10] by induction ([L12]), and then [L11] with the bounds of step 2.2, give the stated estimate i<rωiL(γi).

step 2.1step 2.2L10L11L12
4.1

If f(u)f(v)ω for all u,vγ then ωiω for every i<r, so step 3.1 and [L11] bound the error by ωi<rL(γi), which is ωL(γ) by step 1.1; and on a singleton interval [a,a] the integral is 0 and there is no partition, so the assertion made there is the stated one about the empty sum.

step 1.1step 3.1L11

Depends on

Used by

Dependency tree · two levels

85 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources