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The Slobodeckij seminorm bounds the dyadic level-set sum

Statement

Assume the Axiom of Countable Choice. Let $d \geq 1$ , $0 < θ < 1$ , $1 \leq p < \infty$ with $p θ < d$ , let $f \in L^{\infty} (R^{d})$ have compact support, and put $A_{k} : = {∣ f ∣ > 2^{k}}$ , $a_{k} : = ∣ A_{k} ∣$ . Then $[f]_{θ, p}^{p} \geq c (d, p, θ) \sum_{k \in Z : a_{k} \neq 0} a_{k + 1} a_{k}^{- p θ / d} 2^{p k},$ where $[\cdot]_{θ, p}$ is the Slobodeckij seminorm of The Gagliardo--Slobodeckij space on Euclidean space.

Facts & Assumptions

Given: the Axiom of Countable Choice, $d \geq 1$ , $0 < θ < 1$ , $1 \leq p < \infty$ with $p θ < d$ , a compactly supported $f \in L^{\infty} (R^{d})$ , and the sets $A_{k} = {∣ f ∣ > 2^{k}}$ with $a_{k} = ∣ A_{k} ∣$ . Write $α : = p θ / d \in (0, 1)$ , $T : = 2^{p} > 1$ , and $D_{k} : = A_{k} ∖ A_{k + 1}$ , $d_{k} : = ∣ D_{k} ∣$ .

[F1]

Level sets and annuli. $A_{k + 1} \subseteq A_{k}$ , so $a_{k + 1} \leq a_{k}$ and $d_{k} = a_{k} - a_{k + 1}$ ; the $D_{k}$ are pairwise disjoint, $A_{k} = ⋃_{ℓ \geq k} D_{ℓ}$ up to a null set, $a_{k} = \sum_{ℓ \geq k} d_{ℓ}$ , and $a_{k} = 0$ for all large $k$ because $f$ is bounded with compact support. With $Z : = {f = 0}$ , for every $i$ we have $A_{i - 1}^{c} = Z \cup ⋃_{j \leq i - 2} D_{j}$ up to a null set. All these sets are measurable. (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , Finite and countable subadditivity of measures, Measure of a set difference when the smaller set has finite measure)

[F2]

Kernel estimate. If $E$ is measurable with $0 < ∣ E ∣ < \infty$ and $x \in R^{d}$ , then $\int_{R^{d} ∖ E} ∣ x - y ∣^{- d - p θ} d y \geq c_{1} ∣ E ∣^{- p θ / d}$ with $c_{1} = c_{1} (d, p, θ) > 0$ independent of $x$ and $E$ . (The level-set kernel measure estimate for the Slobodeckij kernel)

[F3]

Slobodeckij seminorm on disjoint blocks. For measurable $B \subseteq R^{d} \times R^{d}$ , $\iint_{B} ∣ f (x) - f (y) ∣^{p} ∣ x - y ∣^{- d - p θ} d x d y \leq [f]_{θ, p}^{p}$ ; sums over pairwise disjoint such blocks of a nonnegative integrand are bounded by the total integral. (The Gagliardo--Slobodeckij space on Euclidean space, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

Proof

technique · Pair the annulus $D_i$ with the complement of $A_{i-1}$, where the level gap is at least $2^{i-1}$; sum the resulting block estimates against the geometric weights, control the overlap of the nested tails by a geometric series, and relabel

1.1F1given

By [F1] the annuli $D_{k}$ are pairwise disjoint measurable sets with $d_{k} = a_{k} - a_{k + 1}$ and $a_{k} = \sum_{ℓ \geq k} d_{ℓ}$ , and $a_{k} = 0$ for all $k$ large. Let $Z : = {f = 0}$ . If $x \in D_{i}$ and $y \in D_{j}$ with $j \leq i - 2$ , then $∣ f (x) ∣ > 2^{i}$ and $∣ f (y) ∣ \leq 2^{j + 1} \leq 2^{i - 1}$ , so $∣ f (x) - f (y) ∣ \geq 2^{i - 1}$ . The same bound holds for $y \in Z$ , since $∣ f (y) ∣ = 0$ and $∣ f (x) ∣ > 2^{i}$ . Thus the low positive bands together with $Z$ cover $A_{i - 1}^{c}$ up to a null set.

2.1F2F3step 1.1algebra

If $[f]_{θ, p} = \infty$ , the conclusion is immediate. Otherwise the disjoint-block sum $I$ below is finite by [F3]. The sums $X, S, Y$ are finite: $a_{i} a_{i - 1}^{- α} \leq a_{i - 1}^{1 - α}$ , the $a_{i}$ are bounded and eventually zero, and $T > 1$ , so the negative tail is geometric. Fix $i$ with $a_{i - 1} \neq 0$ . By [F2] applied to $E = A_{i - 1}$ (which has $0 < a_{i - 1} < \infty$ ), for every $x \in D_{i}$ the integral of $∣ x - y ∣^{- d - p θ}$ over $A_{i - 1}^{c}$ is at least $c_{1} a_{i - 1}^{- α}$ . Since $A_{i - 1}^{c} = Z \cup ⋃_{j \leq i - 2} D_{j}$ up to a null set, the lower bound from step 1.1 gives $I_{i} : = \sum_{j \leq i - 2} \iint_{D_{i} \times D_{j}} ∣ f (x) - f (y) ∣^{p} ∣ x - y ∣^{- d - p θ} d x d y + \iint_{D_{i} \times Z} ∣ f (x) - f (y) ∣^{p} ∣ x - y ∣^{- d - p θ} d x d y \geq c_{0} 2^{p i} a_{i - 1}^{- α} d_{i},$ where $c_{0} : = 2^{- p} c_{1}$ . Writing $d_{i} = a_{i} - \sum_{ℓ \geq i + 1} d_{ℓ}$ and summing over $i$ with $a_{i - 1} \neq 0$ , $I : = \sum_{i} I_{i} \geq c_{0} X - c_{0} Y$ where $X : = \sum_{i : a_{i - 1} \neq 0} 2^{p i} a_{i - 1}^{- α} a_{i}$ and $Y : = \sum_{i : a_{i - 1} \neq 0} \sum_{ℓ \geq i + 1} 2^{p i} a_{i - 1}^{- α} d_{ℓ}$ . Swapping the order of summation in $Y$ and using $a_{i - 1} \geq a_{ℓ - 1}$ whenever $i \leq ℓ$ gives $Y \leq \sum_{ℓ : a_{ℓ - 1} \neq 0} d_{ℓ} a_{ℓ - 1}^{- α} \sum_{i \leq ℓ - 1} 2^{p i} = \frac{T^{- 1}}{1 - T^{- 1}} \sum_{ℓ : a_{ℓ - 1} \neq 0} 2^{p ℓ} a_{ℓ - 1}^{- α} d_{ℓ} = \frac{1}{T - 1} S$ , where $S : = \sum_{ℓ : a_{ℓ - 1} \neq 0} 2^{p ℓ} a_{ℓ - 1}^{- α} d_{ℓ}$ . On the other hand, the block bound itself gives $I \geq c_{0} S$ , so $S \leq I / c_{0}$ and therefore $I \geq c_{0} X - \frac{c_{0}}{T - 1} S \geq c_{0} X - \frac{1}{T - 1} I$ , that is $I \geq \frac{c_{0} (T - 1)}{T} X$ .

3.1F3step 2.1algebra∎

The blocks $D_{i} \times D_{j}$ with $j \leq i - 2$ and $D_{i} \times Z$ are pairwise disjoint: the $D_{i}$ are disjoint in the first coordinate, and for each $i$ the second-coordinate bands and $Z$ are disjoint. Hence [F3] gives $I \leq [f]_{θ, p}^{p}$ . Relabelling $k = i - 1$ in $X = \sum_{i : a_{i - 1} \neq 0} 2^{p i} a_{i - 1}^{- α} a_{i}$ yields $X = 2^{p} \sum_{k : a_{k} \neq 0} 2^{p k} a_{k}^{- α} a_{k + 1}$ , so $[f]_{θ, p}^{p} \geq \frac{c_{0} (T - 1) 2^{p}}{T} \sum_{k : a_{k} \neq 0} a_{k + 1} a_{k}^{- p θ / d} 2^{p k}$ , which is the assertion with $c = \frac{c_{0} (T - 1) 2^{p}}{T} > 0$ .

Depends on

Used by

The critical fractional Sobolev inequality on ℝᵈ Theorem

Dependency tree · two levels

33 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

Eleonora Di Nezza, Giampiero Palatucci and Enrico Valdinoci, Hitchhiker's guide to the fractional Sobolev spaces (arXiv:1104.4345, survey) (standard reference, not scraped)