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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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A continuous function on [0,1][0,1] can have unbounded variation

Example

Let d(t,Z)=infkZtkd(t,\mathbb Z)=\inf_{k\in\mathbb Z}|t-k| and define f(0)=0,f(x)=xd(1/x,Z)(0<x1).f(0)=0,\qquad f(x)=x\,d(1/x,\mathbb Z)\quad(0<x\le1). Then ff is continuous on [0,1][0,1] but has unbounded variation.

Facts & Assumptions

Given: The displayed function ff.

[L2]

The integer-part property gives an integer nn with nt<n+1n\le t<n+1; one of tnt-n and n+1tn+1-t is at most 1/21/2, so 0d(t,Z)1/20\le d(t,\mathbb Z)\le1/2. The definition gives value zero at integers, and the two nearest integers to n+1/2n+1/2 are both at distance 1/21/2 (Integer part: for every real xx there is exactly one integer mm with mx<m+1m \le x < m + 1, The integers as equivalence classes of pairs of naturals).

[L3]

Verification

technique · construction
1.1

By [L1] and the algebra of limits, ff is continuous for x>0x>0. By [L2], f(x)x/2|f(x)|\le x/2, so f(x)0=f(0)f(x)\to0=f(0) as x0x\downarrow0. Hence ff is continuous on [0,1][0,1].

L1L2
1.2

For every integer n1n\ge1, f(1/n)=0f(1/n)=0, while [given] f(1n+1/2)=12n+1.f\left(\frac1{n+1/2}\right)=\frac1{2n+1}. The ordered partition containing these alternating zeros and peaks for 1nN1\le n\le N therefore has variation sum at least 2n=1N(2n+1)12\sum_{n=1}^N(2n+1)^{-1}, apart from at most one endpoint term.

L2
2.1

Since (2n+1)1(3n)1(2n+1)^{-1}\ge(3n)^{-1} and the harmonic partial sums are unbounded by [L3], the variation sums in step 1.2 are unbounded. Thus the continuous function ff is not BV.

L3step 1.2

Depends on

Used by

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Sources