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Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$

Statement

Let $(X, A, μ)$ be a measure space, let $T$ be a sublinear operator on measurable functions, and suppose:

$T$ is of weak type $(1, 1)$ with constant $A$ ;
$T$ is of strong type $(\infty, \infty)$ with constant $B$ .

Then for every $1 < p < \infty$ and every $f \in L^{p} (μ)$ , $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} B^{1 - 1 / p} ∥ f ∥_{p} .$ In particular, $T$ is of strong type $(p, p)$ for every $1 < p < \infty$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , a sublinear operator $T$ , constants $A, B \geq 0$ , an exponent $1 < p < \infty$ , and a function $f \in L^{p} (μ)$ .

[L1]

Sublinearity, weak type $(1, 1)$ , and strong type $(\infty, \infty)$ are as defined in Sublinear operators and weak or strong type $(p, q)$ bounds.

[L2]

The distribution function of a measurable function $g$ is $A_{g} (t) = μ ({∣ g ∣ > t}) .$ (The distribution function of absolute value)

[L3]

For $0 < p < \infty$ , $\int ∣ g ∣^{p} d μ = p \int_{0}^{\infty} t^{p - 1} μ ({∣ g ∣ > t}) d t$ for every measurable $g$ . (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function)

Proof

technique · direct

1.1

Fix $t > 0$ and $η > 0$ , and put $C : = B + η > 0$ . Since the strong [L1, given, construct, algebra] $(\infty, \infty)$ bound with constant $B$ also holds with the larger constant $C$ , split $f = f_{t}^{>} + f_{t}^{\leq}, f_{t}^{>} : = f 1_{{∣ f ∣ > t / (2 C)}}, f_{t}^{\leq} : = f 1_{{∣ f ∣ \leq t / (2 C)}} .$ Sublinearity gives $∣ T f ∣ \leq ∣ T f_{t}^{>} ∣ + ∣ T f_{t}^{\leq} ∣ .$ Since $∥ f_{t}^{\leq} ∥_{\infty} \leq t / (2 C)$ , the strong $(\infty, \infty)$ bound with constant $C$ yields $∣ T f_{t}^{\leq} ∣ \leq C ∥ f_{t}^{\leq} ∥_{\infty} \leq t / 2.$ Therefore ${∣ T f ∣ > t} \subseteq {∣ T f_{t}^{>} ∣ > t / 2} .$

L1givenconstructalgebra

2.1

Apply the weak $(1, 1)$ bound to $f_{t}^{>}$ : [L1, step 1.1, algebra] $μ ({∣ T f ∣ > t}) \leq μ ({∣ T f_{t}^{>} ∣ > t / 2}) \leq \frac{2 A}{t} ∥ f_{t}^{>} ∥_{1} = \frac{2 A}{t} \int_{{∣ f ∣ > t / (2 C)}} ∣ f ∣ d μ .$

L1step 1.1algebra

3.1

Using [L3] with $g = T f$ and then step 2.1, [L2, L3, step 2.1, algebra] $∥ T f ∥_{p}^{p} = p \int_{0}^{\infty} t^{p - 1} μ ({∣ T f ∣ > t}) d t \leq 2 A p \int_{0}^{\infty} t^{p - 2} (\int_{{∣ f ∣ > t / (2 C)}} ∣ f ∣ d μ) d t .$

L2L3step 2.1algebra

4.1

The integrand in step 3.1 is nonnegative, so Tonelli's theorem for [step 3.1, algebra] nonnegative integrals lets us swap the order: $∥ T f ∥_{p}^{p} \leq 2 A p \int_{X} ∣ f (x) ∣ (\int_{0}^{2 C ∣ f (x) ∣} t^{p - 2} d t) d μ (x) .$ Because $p > 1$ , $\int_{0}^{2 C ∣ f (x) ∣} t^{p - 2} d t = \frac{(2 C ∣ f (x) ∣)^{p - 1}}{p - 1},$ so $∥ T f ∥_{p}^{p} \leq \frac{2^{p} A p}{p - 1} C^{p - 1} \int_{X} ∣ f ∣^{p} d μ .$

step 3.1algebra

5.1

Taking $p$ th roots in step 4.1 gives [step 4.1, algebra] $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} C^{1 - 1 / p} ∥ f ∥_{p} .$ Because $η > 0$ was arbitrary, letting $η ↓ 0$ yields $∥ T f ∥_{p} \leq 2 {(\frac{A p}{p - 1})}^{1 / p} B^{1 - 1 / p} ∥ f ∥_{p} .$ Thus $T$ is of strong type $(p, p)$ for every $1 < p < \infty$ .

step 4.1algebra∎

Depends on

Used by

The centered maximal operator is bounded on Lᵖ(ℝⁿ) for 1<p<∞ Corollary

Dependency tree · two levels

11 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

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Theorem 6.28 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Chapter 24.1 (standard reference, not scraped)

Marcinkiewicz interpolation from weak (1,1)(1,1) and strong (∞,∞)(∞,∞)