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Young's theorem integrates a Hölder function of unbounded variation against itself

Example

There is a 3/43/4-Hölder function f:[0,1]Rf:[0,1]\to\mathbb R of unbounded variation for which the Young integral 01fdf\int_0^1 f\,df nevertheless exists.

Facts & Assumptions

Given: Let S=n=1n4/3S=\sum_{n=1}^\infty n^{-4/3}, put wn=S1n4/3w_n=S^{-1}n^{-4/3}, and tile (0,1](0,1] by consecutive intervals InI_n of lengths wnw_n accumulating at zero. On InI_n, let ff be the symmetric triangular tent of height wn3/4w_n^{3/4}, and set f(0)=0f(0)=0.

[L1]

The pp-series converges for p>1p>1 and diverges for p=1p=1 (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1).

[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<0<S<\infty and nwn=1\sum_nw_n=1, so the intervals tile (0,1](0,1]. On one tent, the linear slope estimate and [L2] give f(x)f(y)2xy3/4|f(x)-f(y)|\le2|x-y|^{3/4}. If x,yx,y lie in different tents, let zxz_x be the endpoint of xx's tent toward yy and zyz_y the endpoint of yy's tent toward xx. Both have value zero and xzx,yzyxy|x-z_x|,|y-z_y|\le|x-y|, so the two one-tent estimates and the triangle inequality give f(x)f(y)4xy3/4|f(x)-f(y)|\le4|x-y|^{3/4}. Taking the endpoints of successively smaller tents gives the same estimate at zero. Thus ff is 3/43/4-Hölder.

L1L2
1.2

A partition through the endpoints and peaks of the first NN tents has variation at least [given] 2n=1Nwn3/4=2S3/4n=1N1n.2\sum_{n=1}^N w_n^{3/4}=2S^{-3/4}\sum_{n=1}^N\frac1n. This is unbounded by [L1], so ff is not BV.

2.1

Since 3/4+3/4>13/4+3/4>1, [L3] nonetheless gives existence of 01fdf\int_0^1 f\,df. This is genuinely outside the BV existence theorem.

L3

Depends on

Used by

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Dependency tree · next 3 levels

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