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Dyadic annulus far-field estimates for the maximal function

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $f \in L_{l o c}^{1} (R^{n})$ , $z \in R^{n}$ , $r > 0$ and $δ > 0$ . Then $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq C_{n, δ} r^{- δ} M f (z),$ where $M$ is the centred Hardy-Littlewood maximal function (The centered and uncentered Hardy-Littlewood maximal functions) and $C_{n, δ} = 2^{2 n} (1 - 2^{- δ})^{- 1}$ depends only on $n$ and $δ$ .

Facts & Assumptions

Given: Countable Choice, $f \in L_{l o c}^{1} (R^{n})$ , $z \in R^{n}$ , $r > 0$ and $δ > 0$ .

[F1]

For every locally integrable $f$ one has $M f (z) = \sup_{ρ > 0} λ (B (z, ρ))^{- 1} \int_{B (z, ρ)} ∣ f ∣ d λ$ , and each average is finite because $f$ is integrable over balls (The centered and uncentered Hardy-Littlewood maximal functions, A locally integrable function on $R^{n}$ ).

[F2]

Every ball $B (x, ρ)$ is Lebesgue measurable with $0 < λ (B (x, ρ)) < \infty$ (Euclidean balls have positive finite Lebesgue measure), the dilates of the unit ball satisfy $λ (B (0, ρ)) = v_{n} ρ^{n}$ with $v_{n} = λ (B (0, 1)) \in (0, \infty)$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it), and a ball is contained in the axis-parallel cube $Q (z, ρ) = \prod_{i} (z_{i} - ρ, z_{i} + ρ)$ of measure $(2 ρ)^{n}$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F3]

For a sequence of nonnegative measurable functions increasing to $g$ , the integrals increase to $\int g$ (Monotone convergence for the integral).

[F4]

If $A \subseteq B$ are measurable then $\int_{A} ∣ f ∣ \leq \int_{B} ∣ f ∣$ (Measures are monotone).

Proof

technique · Decompose the far field into dyadic annuli, bound each annulus by the maximal function through its measure, and sum the resulting geometric series

1.1F1F2F3given

The sets $A_{k} = {t : 2^{k} r \leq ∣ t - z ∣ < 2^{k + 1} r}$ , $k \geq 0$ , are pairwise disjoint measurable sets whose union is ${∣ t - z ∣ \geq r}$ . Each partial sum $\sum_{k < K} ∣ f ∣ ∣ t - z ∣^{- n - δ} 1_{A_{k}}$ increases with $K$ to $∣ f (t) ∣ ∣ t - z ∣^{- n - δ} 1_{{∣ t - z ∣ \geq r}}$ , so monotone convergence [F3] gives $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t = \sum_{k \geq 0} \int_{A_{k}} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t$ , and every term is finite because $(2^{k} r)^{- n - δ} \int_{B (z, 2^{k + 1} r)} ∣ f ∣ < \infty$ by [F1] and [F2].

2.1F1F2F4step 1.1algebra

For $t \in A_{k}$ one has $∣ t - z ∣^{- n - δ} \leq (2^{k} r)^{- n - δ}$ , and $A_{k} \subseteq B (z, 2^{k + 1} r) \subseteq Q (z, 2^{k + 1} r)$ , so [F4] and [F2] give $\int_{A_{k}} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq (2^{k} r)^{- n - δ} \int_{B (z, 2^{k + 1} r)} ∣ f ∣ \leq (2^{k} r)^{- n - δ} λ (B (z, 2^{k + 1} r)) M f (z) \leq (2^{k} r)^{- n - δ} (2^{k + 2} r)^{n} M f (z) = 2^{2 n} 2^{- k δ} r^{- δ} M f (z)$ .

3.1step 1.1step 2.1algebra∎

Summing the geometric series in step 2.1 with ratio $2^{- δ} < 1$ gives $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq 2^{2 n} (1 - 2^{- δ})^{- 1} r^{- δ} M f (z)$ , which is the asserted inequality with $C_{n, δ} = 2^{2 n} (1 - 2^{- δ})^{- 1}$ .

Why the exponent range is $δ > 0$ . The endpoint $δ = 0$ is not available: for the locally integrable function $f = 1_{B (0, R)}$ , the point $z = 0$ and $r = 1$ one has $M f (0) = 1$ , while $\int_{∣ t ∣ \geq 1} ∣ f (t) ∣ ∣ t ∣^{- n} d t = \int_{1 \leq ∣ t ∣ \leq R} ∣ t ∣^{- n} d t = ∣ S^{n - 1} ∣ \log R$ grows without bound as $R \to \infty$ , by Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma and The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t. Hence no constant independent of $R$ can bound that integral by $M f (0)$ ; the divergence of $\sum_{k \geq 0} 2^{- k δ}$ at $δ = 0$ is not removable, and only exponents $δ > 0$ occur in the Hölder estimates for standard kernels used on this page.

Depends on

Used by

Unweighted local good-lambda estimate for maximal truncations Lemma

Dependency tree · two levels

69 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

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249, 2014) (standard reference, not scraped)
Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)