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The grand maximal function dominates every admissible radial and nontangential maximal function

Statement

Let $n \geq 1$ , $f \in S^{'} (R^{n})$ , $φ \in S (R^{n})$ with $\int_{R^{n}} φ \neq 0$ , $a \geq 1$ , and an integer $N \geq 1$ . Then, with $P_{N}$ and $F_{N}$ as in Grand maximal test class of order N and the grand maximal function and the maximal functions of Radial and nontangential maximal functions of a tempered distribution, $M_{φ}^{*, a} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x) for every x \in R^{n},$ and in particular $M_{φ}^{0} f (x) \leq 2^{N} P_{N} (φ) M_{N} f (x), ∥ M_{φ}^{0} f ∥_{L^{p}} \leq 2^{N} P_{N} (φ) ∥ M_{N} f ∥_{L^{p}}$ whenever $0 < p \leq \infty$ and the right-hand side is finite. The constants $(1 + a)^{N}$ differ from the source's sharper $a^{N}$ but are equivalent for fixed $a, N$ and are the ones produced by the elementary translate estimate below. Consequently every admissible radial maximal function is pointwise dominated by the grand maximal function of every sufficiently large order, and the space defined by the radial maximal function of one kernel contains the space defined by $M_{N}$ .

Facts & Assumptions

Given: $n \geq 1$ , $φ \in S$ with $\int φ \neq 0$ , $a \geq 1$ , an integer $N \geq 1$ , $f \in S^{'}$ , and a point $x \in R^{n}$ .

[F1]

$(f * Ψ_{t}) (y) = ⟨ f_{z}, Ψ_{t} (y - z) ⟩$ for $Ψ \in S$ and $t > 0$ , and $M_{N} f (x) = \sup_{Ψ \in F_{N}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (f * Ψ_{t}) (y) ∣$ (Convolution of a tempered distribution with a schwartz function, Grand maximal test class of order N and the grand maximal function).

[F2]

The test seminorm satisfies $P_{N} (Ψ) \leq 1$ exactly for $Ψ \in F_{N}$ , and for every nonzero $Ψ$ , $P_{N} (Ψ / P_{N} (Ψ)) = 1$ ; $P_{N} (Ψ) = \sup_{w} (1 + ∣ w ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} Ψ (w) ∣$ (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

[F3]

If $∣ y - x ∣ \leq a t$ then $y = x + t γ$ with $∣ γ ∣ \leq a$ ; the translation identity $φ_{t} (y - z) = G_{t} (x - z)$ for $G (w) = φ (w + γ)$ holds for every $z$ , as both sides equal $t^{- n} φ ((x - z) / t + γ)$ (Dilations and their normalisations preserve Schwartz space, with scaling identities).

[F4]

Maximal functions are Borel measurable, so the $L^{p}$ statement is meaningful (Measurability and lower semicontinuity of the smooth maximal functions).

Proof technique: translate the kernel into the aperture-one cone at the base point, using the translate bound for $P_{N}$ .

Proof

technique · direct

1.1F2F3givenalgebra

The translate bound. Fix $γ \in R^{n}$ and put $G (w) = φ (w + γ)$ . Then $G \in S (R^{n})$ and $P_{N} (G) \leq (1 + ∣ γ ∣)^{N} P_{N} (φ)$ : indeed $\partial^{α} G (w) = (\partial^{α} φ) (w + γ)$ and $(1 + ∣ w ∣) \leq (1 + ∣ γ ∣) (1 + ∣ w + γ ∣)$ , so $(1 + ∣ w ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} φ (w + γ) ∣ \leq (1 + ∣ γ ∣)^{N} P_{N} (φ)$ pointwise in $w$ , and taking the supremum proves the claim. If $P_{N} (G) = 0$ then $G = 0$ and $f * G_{t} = 0$ ; otherwise $G / P_{N} (G) \in F_{N}$ by [F2], and $t^{- n}$ normalisation is the same for $G$ .

2.1step 1.1F1F3algebra

Pointwise domination. Fix $t > 0$ and $y$ with $∣ y - x ∣ \leq a t$ , and write $y = x + t γ$ with $∣ γ ∣ \leq a$ . By [F3], $(f * φ_{t}) (y) = (f * G_{t}) (x)$ for $G = φ (\cdot + γ)$ . Taking absolute values and applying the definition of $M_{N}$ through [F1] and step 1.1, $∣ (f * φ_{t}) (y) ∣ = ∣ (f * G_{t}) (x) ∣ \leq P_{N} (G) M_{N} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x) .$ Taking the supremum over all such $t, y$ gives $M_{φ}^{*, a} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x)$ .

3.1step 2.1F1F4∎

Radial case and $L^{p}$ consequence. Since $M_{φ}^{0} f (x) = \sup_{t > 0} ∣ (f * φ_{t}) (x) ∣$ is the diagonal $y = x$ instance of the aperture-one supremum, $M_{φ}^{0} f (x) \leq M_{φ}^{*, 1} f (x) \leq 2^{N} P_{N} (φ) M_{N} f (x)$ by step 2.1 with $a = 1$ . If $∥ M_{N} f ∥_{p} < \infty$ , the pointwise inequality and the Borel measurability of [F4] give $∥ M_{φ}^{0} f ∥_{p} \leq 2^{N} P_{N} (φ) ∥ M_{N} f ∥_{p}$ for every $0 < p \leq \infty$ by monotonicity of the integral. This proves the lemma.

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16 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

Shai Dekel, Gerard Kerkyacharian, George Kyriazis, Pencho Petrushev, A New Proof of the Atomic Decomposition of Hardy Spaces, Constructive Theory of Functions (Sozopol 2016), pp. 59-73 (standard reference, not scraped)
Li-An Daniel Wang, Multiplier Theorems on Anisotropic Hardy Spaces (PhD dissertation, University of Oregon, 2012) (standard reference, not scraped)