Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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Young's theorem integrates a Hölder function of unbounded variation against itself

Example

There is a 3/4-Hölder function f:[0,1]→R of unbounded variation for which the Young integral ∫01f df nevertheless exists.

Facts & Assumptions

Given: Let S=∑n=1∞n−4/3, put wn=S−1n−4/3, and tile (0,1] by consecutive intervals In of lengths wn accumulating at zero. On In, let f be the symmetric triangular tent of height wn3/4, and set f(0)=0.

[L1]

The p-series converges for p>1 and diverges for p=1 (For rational p>0, ∑1/kp converges iff p>1).

[L2]

Rational powers are monotone and obey their exponent laws (Monotonicity of r↦ar and of a↦ar, Laws of rational exponents).

[L3]

Young's theorem applies when the two Hölder exponents have sum greater than one (Young's Riemann–Stieltjes existence theorem for rational Hölder exponents).

Verification

technique · construction
1.1

By [L1], 0<S<∞ and ∑nwn=1, so the intervals tile (0,1]. On one tent, the linear slope estimate and [L2] give ∣f(x)−f(y)∣≤2∣x−y∣3/4. If x,y lie in different tents, let zx be the endpoint of x's tent toward y and zy the endpoint of y's tent toward x. Both have value zero and ∣x−zx∣,∣y−zy∣≤∣x−y∣, so the two one-tent estimates and the triangle inequality give ∣f(x)−f(y)∣≤4∣x−y∣3/4. Taking the endpoints of successively smaller tents gives the same estimate at zero. Thus f is 3/4-Hölder.

L1L2
1.2

A partition through the endpoints and peaks of the first N tents has variation at least [given] 2∑n=1Nwn3/4=2S−3/4∑n=1N1n. This is unbounded by [L1], so f is not BV.

2.1

Since 3/4+3/4>1, [L3] nonetheless gives existence of ∫01f df. This is genuinely outside the BV existence theorem.

L3∎

Depends on

Used by

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Dependency tree · two levels

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Sources