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Young's Riemann–Stieltjes existence theorem for rational Hölder exponents

Statement

Let $p, q \in Q \cap (0, 1]$ with $p + q > 1$ . If $f : [a, b] \to R$ is $p$ -Hölder and $g : [a, b] \to R$ is $q$ -Hölder, then both $\int_{a}^{b} f d g$ and $\int_{a}^{b} g d f$ exist. They satisfy

$\int_{a}^{b} f d g + \int_{a}^{b} g d f = f (b) g (b) - f (a) g (a) .$

Facts & Assumptions

Given: Hölder functions $f, g$ with rational exponents whose sum exceeds one.

[L1]

The Young partition estimate controls refinement errors by a constant times $∥ P ∥^{p + q - 1}$ (Young's partition estimate for rational Hölder exponents).

[L2]

Every Cauchy sequence of real numbers converges (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

[L3]

A Stieltjes integral is the common limit of all sufficiently fine tagged sums (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral).

[L4]

Such a limit is unique (The Riemann–Stieltjes integral is unique).

[L5]

Tails of a geometric series with ratio in $(0, 1)$ tend to zero (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Proof

technique · direct

1.1

For the dyadic left sums $L_{m}$ , the first estimate in [L1] and the geometric-tail fact [L5] make $(L_{m})$ a Cauchy sequence. It therefore converges to some $I$ by [L2].

L1L2L5

2.1

Given a partition $P$ , compare it and a sufficiently fine dyadic partition $D_{m}$ with their common refinement. The second estimate in [L1] bounds the two refinement errors by a constant times $∥ P ∥^{r - 1} + ∥ D_{m} ∥^{r - 1}$ . Together with $L_{m} \to I$ , this shows that every sufficiently fine left-endpoint sum is close to $I$ . Replacing a left endpoint $t_{i}$ by an arbitrary tag $ξ_{i}$ changes the $i$ th term by at most $K_{f} K_{g} ∣ t_{i + 1} - t_{i} ∣^{r}$ ; the total is at most $K_{f} K_{g} (b - a) ∥ P ∥^{r - 1}$ . Thus every fine tagged sum tends to $I$ , and [L3] gives $\int f d g$ . Interchanging $f$ and $g$ gives $\int g d f$ .

step 1.1L1L2L3L4

3.1

On every partition, the right-endpoint sum for $f d g$ plus the left-endpoint sum for $g d f$ telescopes exactly to $f (b) g (b) - f (a) g (a)$ . Passing to the two limits established in step 2.1 proves the formula.

step 2.1L3∎

Depends on

Used by

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Sources

Nourdin, Nualart, and Peccati, The Breuer–Major theorem in total variation: improved rates under minimal regularity, Section 2.2 (standard reference, not scraped)