Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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A continuous function on [0,1] can have unbounded variation

Example

Let d(t,Z)=inf⁡k∈Z∣t−k∣ and define f(0)=0,f(x)=x d(1/x,Z)(0<x≤1). Then f is continuous on [0,1] but has unbounded variation.

Facts & Assumptions

Given: The displayed function f.

[L2]

The integer-part property gives an integer n with n≤t<n+1; one of t−n and n+1−t is at most 1/2, so 0≤d(t,Z)≤1/2. The definition gives value zero at integers, and the two nearest integers to n+1/2 are both at distance 1/2 (Integer part: for every real x there is exactly one integer m with m≤x<m+1, The integers as equivalence classes of pairs of naturals).

[L3]

The harmonic series diverges (For rational p>0, ∑1/kp converges iff p>1 at exponent 1).

Verification

technique · construction
1.1

By [L1] and the algebra of limits, f is continuous for x>0. By [L2], ∣f(x)∣≤x/2, so f(x)→0=f(0) as x↓0. Hence f is continuous on [0,1].

L1L2
1.2

For every integer n≥1, f(1/n)=0, while [given] f(1n+1/2)=12n+1. The ordered partition containing these alternating zeros and peaks for 1≤n≤N therefore has variation sum at least 2∑n=1N(2n+1)−1, apart from at most one endpoint term.

L2
2.1

Since (2n+1)−1≥(3n)−1 and the harmonic partial sums are unbounded by [L3], the variation sums in step 1.2 are unbounded. Thus the continuous function f is not BV.

L3step 1.2∎

Depends on

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