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The tangential maximal function is controlled by the aperture-one nontangential maximal function in $L^{p}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < q < p < \infty$ , $T = n / q$ and $φ \in S (R^{n})$ with $\int φ \neq 0$ . For $f \in S^{'} (R^{n})$ define the tangential maximal function $M_{φ, T} f (x) = \sup_{t > 0} \sup_{y \in R^{n}} \frac{∣ (f * φ_{t}) (x - y) ∣}{(1 + ∣ y ∣ / t)^{T}}, x \in R^{n} .$ For a nonnegative Borel function $g$ define the extended centered average $\tilde{M} g (x) : = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} g,$ where the nonnegative Lebesgue integral may be $+ \infty$ . If $g \in L_{l o c}^{1}$ , then $\tilde{M} g$ agrees with the centered Hardy-Littlewood maximal function $M g$ of The centered and uncentered Hardy-Littlewood maximal functions. The aperture-one nontangential maximal function $M_{φ}^{*, 1} f$ is the one of Radial and nontangential maximal functions of a tempered distribution. Then, pointwise, $M_{φ, T} f (x)^{q} \leq \tilde{M} ((M_{φ}^{*, 1} f)^{q}) (x) .$ If $∥ M_{φ}^{*, 1} f ∥_{L^{p}} < \infty$ , then $∥ M_{φ, T} f ∥_{L^{p}} \leq C_{n, p, q} ∥ M_{φ}^{*, 1} f ∥_{L^{p}},$ with $C_{n, p, q}$ depending only on $n, p, q$ . The norm inequality also holds in the extended sense when the right-hand side is infinite, in which case it is trivial.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < q < p < \infty$ , $T = n / q$ , $φ \in S$ with $\int φ \neq 0$ , $f \in S^{'}$ and $x \in R^{n}$ .

[L1]

For every $a \in R^{n}$ and $r > 0$ , $λ (B (a, r)) = c_{n} r^{n}$ with $c_{n} = ω_{n - 1} / n > 0$ : the centred-ball formula is Sphere and ball measures scale in Rn and translation invariance is Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation. Hence $B (x - y, t) \subseteq B (x, ∣ y ∣ + t)$ and the volume ratio is $(1 + ∣ y ∣ / t)^{n}$ .

[F1]

For every $z \in B (x - y, t)$ one has $∣ (f * φ_{t}) (x - y) ∣ \leq M_{φ}^{*, 1} f (z)$ , because $∣ z - (x - y) ∣ < t$ and the supremum defining $M_{φ}^{*, 1} f (z)$ runs over $t > 0$ and all points within distance $t$ of $z$ (Radial and nontangential maximal functions of a tempered distribution).

[F2]

The centered Hardy-Littlewood maximal operator satisfies the strong $L^{r}$ bound $∥ M g ∥_{r} \leq C_{n, r} ∥ g ∥_{r}$ for $1 < r < \infty$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered and uncentered Hardy-Littlewood maximal functions).

[F3]

The tangential maximal function is Borel because it is a supremum, over fixed $t, y$ , of continuous functions of $x$ . The aperture-one nontangential maximal function is Borel for $f \in S^{'}$ (Measurability and lower semicontinuity of the smooth maximal functions). Also, for every $r > 1$ , $L^{r} (R^{n}) \subset L_{l o c}^{1} (R^{n})$ : for compact $K$ , Holder gives $\int_{K} ∣ g ∣ \leq λ (K)^{1 - 1 / r} ∥ g ∥_{r}$ , and bounded sets have finite measure (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

Proof technique: local averaging over the ball of radius $t$ and the Hardy-Littlewood maximal bound.

Proof

technique · direct

1.1L1F1F3algebra

Pointwise bound. Fix $x, y, t$ and put $g = (M_{φ}^{*, 1} f)^{q}$ , which is nonnegative Borel by [F3]. By [F1], $∣ (f * φ_{t}) (x - y) ∣^{q} \leq g (z)$ for every $z \in B (x - y, t)$ . Averaging (allowing an infinite integral) and enlarging to $B (x, ∣ y ∣ + t)$ gives $∣ (f * φ_{t}) (x - y) ∣^{q} \leq \frac{1}{λ (B (x - y, t))} \int_{B (x - y, t)} g \leq \frac{λ (B (x, ∣ y ∣ + t))}{λ (B (x - y, t))} \cdot \frac{1}{λ (B (x, ∣ y ∣ + t))} \int_{B (x, ∣ y ∣ + t)} g \leq (1 + ∣ y ∣ / t)^{n} \tilde{M} g (x),$ by [L1]. Since $T q = n$ , division by $(1 + ∣ y ∣ / t)^{T q}$ and taking the supremum over $t, y$ proves the pointwise inequality.

2.1step 1.1F2F3algebra

$L^{p}$ bound. If $∥ M_{φ}^{*, 1} f ∥_{L^{p}} = \infty$ , the asserted norm inequality is trivial. Otherwise [F3] gives that $g = (M_{φ}^{*, 1} f)^{q}$ is Borel and belongs to $L^{p / q}$ . Since $p / q > 1$ , [F3] also gives $g \in L_{l o c}^{1}$ , so $\tilde{M} g = M g$ . Apply [F2] with $r = p / q$ and use step 1.1: $∥ M_{φ, T} f ∥_{L^{p}}^{q} = ∥ (M_{φ, T} f)^{q} ∥_{L^{p / q}} \leq ∥ M g ∥_{L^{p / q}} \leq C_{n, p, q} ∥ g ∥_{L^{p / q}} = C_{n, p, q} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{q} .$ Taking $q$ -th roots proves the estimate (with the constant renamed).

3.1step 1.1step 2.1∎

Conclusion. Step 1.1 gives pointwise domination by the extended centered average, which agrees with the ordinary maximal operator on the locally integrable input in step 2.1; the strong $L^{p / q}$ bound then proves the norm estimate.

Depends on

Used by

Maximal-function characterisations of real Hardy spaces Theorem

Dependency tree · two levels

73 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

David Cruz-Uribe SFO, Li-An Daniel Wang, Variable Hardy Spaces, arXiv:1211.6505 (2012) (standard reference, not scraped)
Marcin Bownik, Anisotropic Hardy Spaces and Wavelets, Memoirs of the American Mathematical Society 164 (2003), no. 781 (standard reference, not scraped)

The tangential maximal function is controlled by the aperture-one nontangential maximal function in LpLp