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L-p norm equivalence for finite Hadamard-lacunary sums

Statement

Let $q > 1$ and $0 < p < \infty$ . There are constants $c_{p, q}, C_{p, q} > 0$ such that every finite $q$ -Hadamard-lacunary sum $f = \sum_{j \in J} a_{j} e_{λ_{j}}$ satisfies

$c_{p, q} {(\sum_{j \in J} ∣ a_{j} ∣^{2})}^{1 / 2} \leq ∥ f ∥_{L^{p} (T)} \leq C_{p, q} {(\sum_{j \in J} ∣ a_{j} ∣^{2})}^{1 / 2} .$

Here $∥ f ∥_{p} = (\int_{0}^{1} ∣ f ∣^{p})^{1 / p}$ is a quasi-norm when $0 < p < 1$ . The $L^{2}$ identity is Finite lacunary Fourier sums have their coefficient ell-two norm, the additive input is Hadamard gaps bound the additive representations used in even moments, and the usual $p \geq 1$ integral inequality is Holder's inequality for integrals, including the endpoint cases.

Facts & Assumptions

Given: $p, q, J, (a_{j})$ and $f$ as in the Statement; write $A = (\sum_{j \in J} ∣ a_{j} ∣^{2})^{1 / 2}$ .

Proof

technique · residue-class even moments, followed by interpolation or a distribution bound

1.1givenalgebra

For every integer $m \geq 1$ , split $J$ into $r = r (m, q)$ residue classes [given, algebra] with $q^{r} > m$ . Expanding the $2 m$ -th moment of each class, the additive lemma says that only equal index multisets survive integration. Their permutations give at most $(m!)^{2}$ copies, and $∣ \sum_{ν = 1}^{r} z_{ν} ∣^{2 m} \leq r^{2 m - 1} \sum_{ν} ∣ z_{ν} ∣^{2 m}$ . Consequently $∥ f ∥_{2 m} \leq B_{m, q} A$ for a constant independent of $J$ and the coefficients.

2.1step 1.1algebra

The cited $L^{2}$ identity gives $∥ f ∥_{2} = A$ . If $1 \leq p < 2$ , Holder [step 1.1, algebra] applied to $∣ f ∣^{2} = ∣ f ∣^{θ p} ∣ f ∣^{2 m (1 - θ)}$ , with $1 / 2 = θ / p + (1 - θ) / (2 m)$ , combines this identity and step 1.1 to give $A \leq ∥ f ∥_{p}^{θ} (B_{m, q} A)^{1 - θ}$ . If $p \geq 2$ , choose $m$ with $p \leq 2 m$ and apply the same interpolation identity with $1 / p$ between $1 / 2$ and $1 / (2 m)$ for the upper bound. For the lower bound, normalized Haar measure has mass one, so Holder gives $∥ f ∥_{p} \geq ∥ f ∥_{2} = A$ whenever $p \geq 2$ . Thus the stated two-sided estimate holds for every $p \geq 1$ .

3.1step 1.1step 2.1algebra∎

Let $0 < p < 1$ . Step 1.1 with $m = 2$ and the $L^{2}$ identity give [step 1.1, step 2.1, algebra] $\int ∣ f ∣^{2} = A^{2}$ and $\int ∣ f ∣^{4} \leq B_{2, q}^{4} A^{4}$ . Cauchy--Schwarz applied to $∣ f ∣^{2} 1_{{∣ f ∣ \geq A / \sqrt{2}}}$ shows that this set has measure at least $d_{q} > 0$ ; otherwise its complement contributes at most $A^{2} / 2$ . Hence $∥ f ∥_{p}^{p} \geq d_{q} (A / \sqrt{2})^{p}$ . Conversely $t^{p} \leq 1 + t^{2}$ for $t \geq 0$ , and applying this to $t = ∣ f ∣ / A$ shows $∥ f ∥_{p}^{p} \leq 2 A^{p}$ when $A > 0$ ; the case $A = 0$ is immediate. This proves both bounds for $0 < p < 1$ .

Depends on

Used by

L-p convergence of a lacunary series is equivalent to ell-two coefficients Corollary

Dependency tree · two levels

14 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., Theorem 3.6.4 (standard reference, not scraped)