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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11
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The positive and negative variations are nondecreasing and give the Jordan identities

Statement

For a bounded-variation function f:[a,b]→R, the functions Pf and Nf are nondecreasing and

f(x)=f(a)+Pf(x)−Nf(x),Vf(x)=Pf(x)+Nf(x).

Both Pf(a) and Nf(a) are 0.

Facts & Assumptions

Given: A bounded-variation function f:[a,b]→R and its functions Vf,Pf,Nf.

[L1]

Vf,Pf,Nf are defined by the displayed formulas in Variation function and positive and negative variations.

[L2]

For x≤y, Vf(y)−Vf(x)=Var⁡[x,y](f) (Total variation is additive over adjacent subintervals and decreases under restriction).

[L3]

∣f(y)−f(x)∣≤Var⁡[x,y](f) (Total variation bounds increments; bounded-variation functions are bounded; zero variation means constant).

[L5]

∣u∣≥u and ∣u∣≥−u (Absolute value in an ordered field).

Proof

technique · direct
1.1

For x≤y, [L2] and [L3] give Vf(y)−Vf(x)≥∣f(y)−f(x)∣, hence this difference is at least both f(y)−f(x) and f(x)−f(y). Therefore Pf(y)−Pf(x)≥0 and Nf(y)−Nf(x)≥0, so both functions are nondecreasing.

L1L2L3L4L5
1.2

Adding and subtracting the defining formulas gives Pf(x)+Nf(x)=Vf(x) and Pf(x)−Nf(x)=f(x)−f(a). At x=a, Vf(a)=0, so Pf(a)=Nf(a)=0.

L1algebra
2.1

Rearranging the second identity in step 1.2 gives f=f(a)+Pf−Nf, while the first is the asserted variation identity.

step 1.2algebra∎

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