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Refinement and tag-change estimates for Stieltjes sums

Statement

Let α have bounded variation on [a,b], let P=(n,t) be a partition, and let Q refine P. If the oscillation of f on [ti,ti+1] is at most ωi, then any tagged sum on P and any tagged sum on Q satisfy

∣SQ−SP∣≤∑i<nωiVar⁡[ti,ti+1](α).

In particular, if every ωi≤ω, the bound is ωVar⁡[a,b](α). Two tagged sums on arbitrary partitions whose intervals all have oscillation at most ω differ by at most 2ωVar⁡[a,b](α).

Facts & Assumptions

Given: Functions f,α:[a,b]→R, a partition P, a refinement Q, and tags on both.

[L1]

Stieltjes sums are weighted finite sums of integrator increments (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L2]

Total variation bounds every sum of absolute increments and is additive on adjacent subintervals (Bounded variation and total variation on an interval, Total variation is additive over adjacent subintervals and decreases under restriction).

[L5]

The absolute value of a finite sum is at most the sum of absolute values (The triangle inequality).

Proof

technique · direct
1.1

Inside one coarse interval [ti,ti+1], the refined integrator increments telescope to α(ti+1)−α(ti). Subtract the coarse term by assigning its tag value to every refined increment. Each coefficient difference has absolute value at most ωi, so the absolute difference contributed by that block is at most ωi times the sum of the absolute refined increments, hence at most ωiVar⁡[ti,ti+1](α).

L1L2L3L4L5
2.1

Summing step 1.1 over the coarse blocks proves the first estimate. If ωi≤ω, additivity of variation gives the uniform bound. The conclusions remain 0 when the variation or the interval is 0.

step 1.1L2L4L5
3.1

For two arbitrary partitions, pass to their common refinement and apply the uniform estimate once from each original sum to the refined sum. The triangle inequality gives the factor 2.

step 2.1L3L5∎

Depends on

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Dependency tree · two levels

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Sources