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Young's partition estimate for rational Hölder exponents

Statement

Let $p, q \in Q \cap (0, 1]$ satisfy $p + q > 1$ . Suppose $∣ f (y) - f (x) ∣ \leq K_{f} ∣ y - x ∣^{p}$ and $∣ g (y) - g (x) ∣ \leq K_{g} ∣ y - x ∣^{q}$ on $[a, b]$ . If $D_{m}$ is the partition into $2^{m}$ equal intervals and $L_{m}$ is its left-endpoint Stieltjes sum, then

$∣ L_{m + 1} - L_{m} ∣ \leq K_{f} K_{g} (b - a)^{p + q} 2^{- m (p + q - 1)} .$

Put $r = p + q$ and $C_{r} : = \frac{2^{r}}{1 - 2^{1 - r}} .$ If $R$ refines an arbitrary partition $P$ , then their left-endpoint sums satisfy $∣ L (R) - L (P) ∣ \leq C_{r} K_{f} K_{g} (b - a) ∥ P ∥^{r - 1} .$

Facts & Assumptions

Given: Rational Hölder exponents $p, q$ with $p + q > 1$ , Hölder constants $K_{f}, K_{g}$ , and the stated partitions.

[L1]

Rational powers are monotone and satisfy the exponent laws (Rational powers $a^{r}$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L2]

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

[L3]

Finite sums obey the triangle inequality and may be regrouped (Laws of finite sums and finite products, The triangle inequality).

Proof

technique · direct

1.1

Insert a point $v$ between adjacent points $u < w$ . The change from the old left-endpoint term to the two new terms is [given] $(f (v) - f (u)) (g (w) - g (v))$ up to sign. Its absolute value is at most $K_{f} K_{g} (v - u)^{p} (w - v)^{q}$ , hence at most $K_{f} K_{g} (w - u)^{p + q}$ by [L1].

2.1

Passing from $D_{m}$ to $D_{m + 1}$ inserts one midpoint in each of $2^{m}$ intervals of length $(b - a) 2^{- m}$ . Summing step 1.1 gives the first displayed bound. More generally, if a partition of an interval $I$ has $k \geq 2$ subintervals, some interior point has two adjacent lengths whose sum is at most $2 ∣ I ∣ / (k - 1)$ : the sum of all such two-interval lengths is at most $2 ∣ I ∣$ . Removing that point therefore changes the left sum by at most $K_{f} K_{g} (2 ∣ I ∣ / (k - 1))^{r}$ .

step 1.1L1L3

3.1

Remove the extra points of $R$ inside a fixed interval $I$ of $P$ , one at a time, always using step 2.1. The total error is at most $2^{r} K_{f} K_{g} ∣ I ∣^{r} \sum_{j \geq 1} j^{- r}$ . Grouping the positive integers into $[2^{m}, 2^{m + 1})$ bounds this series by $\sum_{m \geq 0} 2^{- m (r - 1)} = (1 - 2^{1 - r})^{- 1}$ via [L2]. Thus the error on $I$ is at most $C_{r} K_{f} K_{g} ∣ I ∣^{r}$ . Summing over $I \in P$ and using $∣ I ∣^{r} \leq ∥ P ∥^{r - 1} ∣ I ∣$ proves the refinement estimate. If $K_{f} K_{g} = 0$ or $a = b$ , every error is zero.

step 2.1L1L2L3∎

Depends on

Used by

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

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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