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A weak maximal bound implies almost-everywhere Fourier convergence

Statement

Assume countable choice and fix $1 \leq p < \infty$ . On the period-one torus with normalized Haar measure, suppose a finite constant $A \geq 0$ satisfies

$m {C g > λ} \leq A λ^{- p} ∥ g ∥_{p}^{p} (g \in L^{p} (T), λ > 0) .$

Then for every $f \in L^{p} (T)$ , $S_{N} f (x) \to f (x)$ almost everywhere. The conclusion holds for any measurable representative of $f$ .

Facts & Assumptions

Given: Countable choice, $1 \leq p < \infty$ , a finite weak-bound constant $A \geq 0$ as in the statement, and $f \in L^{p} (T)$ .

[F1]

For $g \in L^{1} (T)$ , $C g = \sup_{N \geq 0} ∣ S_{N} g ∣$ is a measurable extended nonnegative function, defined from the continuous finite Fourier sums and independent of the representative (Carleson maximal partial-sum operator).

[F2]

Assuming countable choice, for each one-periodic complex $f \in L^{p} ([0, 1])$ , $1 \leq p < \infty$ , the Fejer means satisfy $∥ σ_{j} f - f ∥_{p} \to 0$ (Fejer means converge in L^p for 1 <= p < infinity).

[F3]

For a measurable nonnegative extended function $h$ and $t > 0$ , $m {h \geq t} \leq t^{- 1} \int h d m$ (Chebyshev-Markov inequality for the integral).

Proof

technique · polynomial approximation and level-set estimates

1.1

Use a finite-valued measurable representative of $f$ , changing it on a null set if needed. It is integrable since $∣ f ∣ \leq 1 + ∣ f ∣^{p}$ and the torus has mass one. Set $Q_{j} = σ_{j} f = (j + 1)^{- 1} \sum_{r = 0}^{j} S_{r} f$ . These are explicit polynomials and $∥ f - Q_{j} ∥_{p} \to 0$ . Integrating $e^{2 π i (k - ℓ) x}$ over $[0, 1]$ gives one for $k = ℓ$ and zero otherwise, so $S_{N} Q_{j} = Q_{j}$ whenever $N \geq j$ , including constants and the zero polynomial.

F1F2given

2.1

The extended function $H (x) = {lim sup}_{N \to \infty} ∣ S_{N} f (x) - f (x) ∣$ is measurable: a limsup is a countable infimum of countable suprema of measurable functions. For each $j$ and $N \geq j$ , linearity and the triangle inequality give $∣ S_{N} f - f ∣ \leq ∣ S_{N} (f - Q_{j}) ∣ + ∣ f - Q_{j} ∣$ . Hence $H \leq C (f - Q_{j}) + ∣ f - Q_{j} ∣$ .

F1step 1.1

3.1

For every $λ > 0$ , the set ${H > 2 λ}$ is contained in ${C (f - Q_{j}) > λ} \cup {∣ f - Q_{j} ∣ > λ}$ . Apply the assumed weak estimate to the first set and the integral inequality to $h = ∣ f - Q_{j} ∣^{p}$ , $t = λ^{p}$ , for the second. The latter strict superlevel set is contained in ${h \geq λ^{p}}$ . Subadditivity yields $m {H > 2 λ} \leq (A + 1) λ^{- p} ∥ f - Q_{j} ∥_{p}^{p}$ .

F3step 2.1given

4.1

Let $j \to \infty$ with $λ$ fixed. The right side tends to zero, so $m {H > 2 λ} = 0$ . Because $H \geq 0$ , ${H > 0} = ⋃_{r = 1}^{\infty} {H > 2 / r}$ has measure at most the sum of these zero measures, hence zero. Outside it the nonnegative errors have limsup zero and therefore tend to zero. Changing the representative alters the conclusion on only a null set. This proves convergence almost everywhere, including $p = 1$ whenever its hypothesized weak estimate holds.

step 1.1step 3.1algebra∎

Depends on

Used by

Almost-everywhere convergence from the Carleson–Hunt estimate Corollary
The L1 endpoint is excluded Remark

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

Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)