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L2-normalised H1 atoms have uniformly bounded H1 norm

Statement

Assume Countable Choice. Fix an admissible kernel $φ$ with $\int φ \neq 0$ and an auxiliary integer order $\tilde{N} \geq \max {N_{0} (n, 1, φ), n + 1}$ for the grand-maximal characterisation. Let $a$ be a $(1, 2)$ -atom supported in a cube $Q$ : $a$ vanishes off $Q$ , $\int a = 0$ and $(∣ Q ∣^{- 1} \int_{Q} ∣ a ∣^{2})^{1 / 2} \leq ∣ Q ∣^{- 1}$ . Then $∥ a ∥_{H^{1}} = ∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ}$ , with the constant independent of $Q$ and the atom.

Facts & Assumptions

Given: Countable Choice, the fixed admissible kernel $φ$ and auxiliary integer order $\tilde{N}$ from the statement; also let $a$ be a $(1, 2)$ -atom supported in a cube $Q$ with centre $x_{Q}$ and side $ℓ (Q)$ , and the functionals and spaces of The real Hardy space $H^{p}$ defined by a radial maximal function and Grand maximal test class of order N and the grand maximal function.

[F1]

The test class is $F_{\tilde{N}} = {ψ \in S : P_{\tilde{N}} (ψ) \leq 1}$ with $P_{\tilde{N}} (ψ) = \sup_{x} (1 + ∣ x ∣)^{\tilde{N}} \max_{∣ α ∣ \leq \tilde{N} + 1} ∣ \partial^{α} ψ (x) ∣$ , $M_{\tilde{N}} f (x) = \sup_{ψ \in F_{\tilde{N}}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (f * ψ_{t}) (y) ∣$ with $ψ_{t} (u) = t^{- n} ψ (u / t)$ ; in particular $∣ ψ ∣ \leq 1$ and, from the componentwise derivative bounds in the seminorm, $∣ \nabla ψ (u) ∣ \leq \sqrt{n} (1 + ∣ u ∣)^{- \tilde{N}}$ for every $ψ \in F_{\tilde{N}}$ (Grand maximal test class of order N and the grand maximal function).

[F2]

Every $L^{2}$ class has a representative defining a tempered distribution, and for the regular distribution of a locally integrable compactly supported $a$ the convolution is $(a * ψ_{t}) (y) = \int a (z) ψ_{t} (y - z) d z$ (Polynomial growth functions define tempered distributions, Convolution of a tempered distribution with a schwartz function).

[F3]

The stated atom hypotheses give $supp a \subseteq Q$ , $\int a = 0$ and $∥ a ∥_{L^{2}} \leq ∣ Q ∣^{- 1 / 2}$ ; hence $∥ a ∥_{L^{1}} \leq ∣ Q ∣^{1 / 2} ∥ a ∥_{L^{2}} \leq 1$ by Cauchy-Schwarz on the finite-measure cube (Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

Once $M_{\tilde{N}} a \in L^{1}$ has been established, the maximal-function characterisation gives $a \in H^{1}$ and $∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ} ∥ M_{\tilde{N}} a ∥_{L^{1}}$ and $M_{\tilde{N}} a$ is Borel measurable (Maximal-function characterisations of real Hardy spaces, Measurability and lower semicontinuity of the smooth maximal functions).

[F5]

The centered Hardy-Littlewood maximal operator satisfies $∥ M f ∥_{L^{2}} \leq C_{n, 2} ∥ f ∥_{L^{2}}$ and $M f$ is Borel measurable for $f \in L_{l o c}^{1}$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered Hardy-Littlewood maximal function is Borel measurable, The centered and uncentered Hardy-Littlewood maximal functions).

[F6]

A Euclidean ball of radius $R$ is contained in the axis-parallel box of side $2 R$ with the same centre; under Countable Choice that box has Lebesgue measure $(2 R)^{n}$ , and Lebesgue measure is monotone (Axis-parallel rectangles in $R^{m}$ and their volume, 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, Measures are monotone).

[F7]

Under Countable Choice, $λ (B (x, r)) = c_{n} r^{n}$ for a finite positive constant $c_{n}$ depending only on dimension (Sphere and ball measures scale in Rn).

Proof

technique · direct

1.1F1F2F3F5F7algebra

Since $a \in L^{2}$ is supported in the finite-measure cube $Q$ , [F2] identifies its regular distribution with a tempered distribution; [F3] gives $∥ a ∥_{L^{1}} \leq 1$ . For every $ψ \in F_{\tilde{N}}$ , $∣ ψ (u) ∣ \leq (1 + ∣ u ∣)^{- \tilde{N}}$ . Fix $t > 0$ and $∣ y - x ∣ \leq t$ . Split the convolution integral into $∣ y - z ∣ < t$ and the annuli $2^{k - 1} t \leq ∣ y - z ∣ < 2^{k} t$ for $k \geq 1$ . The ball inclusions $B (y, t) \subseteq B (x, 2 t)$ and $B (y, 2^{k} t) \subseteq B (x, 2^{k + 1} t)$ , together with [F5] and [F7], give $∣ (a * ψ_{t}) (y) ∣ \leq t^{- n} \int_{B (y, t)} ∣ a (z) ∣ d z + \sum_{k \geq 1} 2^{- (k - 1) \tilde{N}} t^{- n} \int_{B (y, 2^{k} t)} ∣ a (z) ∣ d z$ $\leq c_{n} 2^{n} M (∣ a ∣) (x) + c_{n} \sum_{k \geq 1} 2^{- (k - 1) \tilde{N}} 2^{(k + 1) n} M (∣ a ∣) (x) \leq C_{n, \tilde{N}} M (∣ a ∣) (x),$ since $\tilde{N} \geq n + 1$ makes the geometric series converge. Taking suprema gives $M_{\tilde{N}} a \leq C_{n, \tilde{N}} M (∣ a ∣)$ pointwise.

2.1step 1.1F3F5F6

Near region. Put $ρ_{n} : = 4 \sqrt{n}$ and $Q^{*} : = {x : ∣ x - x_{Q} ∣ \leq ρ_{n} ℓ (Q)}$ . By [F6], $∣ Q^{*} ∣ \leq (2 ρ_{n} ℓ (Q))^{n} = C_{n}^{'} ∣ Q ∣$ , since the ball is contained in the corresponding axis-parallel cube. Cauchy-Schwarz together with the $L^{2}$ bound of [F5] and step 1.1 gives $\int_{Q^{*}} ∣ M_{\tilde{N}} a ∣ \leq ∣ Q^{*} ∣^{1 / 2} ∥ M_{\tilde{N}} a ∥_{2} \leq ∣ Q^{*} ∣^{1 / 2} C_{n, \tilde{N}} ∥ M (∣ a ∣) ∥_{2} \leq C_{n, 2}^{'} ∣ Q^{*} ∣^{1 / 2} ∥ a ∥_{2} \leq C_{n}$ .

2.2step 1.1F1F3F2

Far region, pointwise. Assume $r : = ∣ x - x_{Q} ∣ \geq ρ_{n} ℓ (Q)$ ; then every $z \in Q$ satisfies $∣ z - x_{Q} ∣ \leq \sqrt{n} ℓ (Q) \leq r / 4$ , so for $∣ y - x ∣ \leq t$ and $w$ on the segment between $y - z$ and $y - x_{Q}$ one has $∣ w ∣ \geq r - t - r / 4$ . If $t \leq r / 2$ , then $∣ w ∣ \geq r / 4$ and $∣ ψ_{t} (y - z) - ψ_{t} (y - x_{Q}) ∣ \leq \sqrt{n} ℓ (Q) \sup ∣ \nabla ψ_{t} ∣$ with $∣ \nabla ψ_{t} (w) ∣ = t^{- n - 1} ∣ \nabla ψ (w / t) ∣ \leq \sqrt{n} t^{- n - 1} (t / ∣ w ∣)^{\tilde{N}} \leq C_{\tilde{N}, n} r^{- n - 1}$ because $\tilde{N} \geq n + 1$ and $t \leq r / 2$ ; if $t > r / 2$ , the same difference is at most $\sqrt{n} ℓ (Q) \cdot \sqrt{n} (r / 2)^{- n - 1} \leq C_{n} ℓ (Q) r^{- n - 1}$ by [F1]. Using $\int a = 0$ to write $(a * ψ_{t}) (y) = \int a (z) [ψ_{t} (y - z) - ψ_{t} (y - x_{Q})] d z$ and $∥ a ∥_{L^{1}} \leq 1$ from [F3], both cases give $∣ (a * ψ_{t}) (y) ∣ \leq C_{\tilde{N}, n} ℓ (Q) r^{- n - 1}$ , and taking suprema over $ψ \in F_{\tilde{N}}$ , $t > 0$ and $∣ y - x ∣ \leq t$ yields $M_{\tilde{N}} a (x) \leq C_{\tilde{N}, n} ℓ (Q) r^{- n - 1}$ .

3.1step 2.2algebra

Far region, integration. Covering ${r \geq ρ_{n} ℓ (Q)}$ by the shells ${2^{j} ρ_{n} ℓ (Q) \leq r < 2^{j + 1} ρ_{n} ℓ (Q)}$ , each contained in a cube of side $4 \cdot 2^{j} ρ_{n} ℓ (Q)$ , and using step 2.2 gives $\int_{r \geq ρ_{n} ℓ (Q)} M_{\tilde{N}} a \leq C_{\tilde{N}, n} ℓ (Q) \sum_{j \geq 0} (2^{j} ρ_{n} ℓ (Q))^{- n - 1} (4 \cdot 2^{j} ρ_{n} ℓ (Q))^{n} = C_{\tilde{N}, n}^{'}$ .

4.1step 2.1step 3.1F4∎

Combining steps 2.1 and 3.1, $∥ M_{\tilde{N}} a ∥_{L^{1}} = \int_{Q^{*}} ∣ M_{\tilde{N}} a ∣ + \int_{R^{n} ∖ Q^{*}} ∣ M_{\tilde{N}} a ∣ \leq C_{n, \tilde{N}} < \infty$ , so [F4] gives $∥ a ∥_{H^{1}} = ∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ} ∥ M_{\tilde{N}} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ}$ , independent of $Q$ and of the atom. Countable Choice is inherited from the maximal-function and measure suppliers [F4]-[F7].

Depends on

Used by

Mean-zero L2 functions on a cube embed continuously into H1 Lemma

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

Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)