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Fejer means converge at Lebesgue points

Statement

Assume the Axiom of Countable Choice.

Let $f : R \to C$ be one-periodic with $f ∣_{[0, 1]} \in L^{1} ([0, 1])$ . If $x \in R$ is a Lebesgue point of $f$ , then

$σ_{N} f (x) ⟶ f (x) (N \to \infty) .$

In particular, $σ_{N} f (x) \to f (x)$ for almost every $x$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a one-periodic function $f$ with $f ∣_{[0, 1]} \in L^{1} ([0, 1])$ , and a Lebesgue point $x$ of $f$ .

[L1]

The Cesaro means satisfy $σ_{N} f = f * F_{N}$ (Cesaro and Abel means of a Fourier series).

[L2]

The Fejer kernels are nonnegative, have integral $1$ , and obey the square formula and tail estimate from The Fejer kernel is a positive approximate identity.

[L3]

At a Lebesgue point, $\lim_{r \to 0^{+}} \frac{1}{2 r} \int_{- r}^{r} ∣ f (x - t) - f (x) ∣ d t = 0$ in the one-dimensional case of Lebesgue points and the Lebesgue set of an $L_{l o c}^{1}$ class.

[L4]

Assuming the Axiom of Countable Choice, almost every point is a Lebesgue point (Almost every point is a Lebesgue point of a locally integrable function).

Proof

technique · direct

1.1L3chooseconstructalgebra

Let $ε > 0$ . By [L3], choose $δ \in (0, 1 / 2]$ so that $\int_{- h}^{h} ∣ f (x - t) - f (x) ∣ d t < \frac{ε}{4} h (0 < h \leq δ) .$ For $0 < h \leq δ$ , put $G (h) : = \int_{0}^{h} (∣ f (x + t) - f (x) ∣ + ∣ f (x - t) - f (x) ∣) d t .$ Then $G (h) < ε h / 4$ for $0 < h \leq δ$ .

2.1L1L2step 1.1algebra

Using [L1], pair the intervals $[0, 1 / 2]$ and $[1 / 2, 1]$ exactly as in the Dirichlet symmetric-difference formula. This gives $∣ σ_{N} f (x) - f (x) ∣ \leq \int_{0}^{1 / 2} (∣ f (x + t) - f (x) ∣ + ∣ f (x - t) - f (x) ∣) F_{N} (t) d t .$ Set $a_{N} : = \min (δ, (N + 1)^{- 1})$ . Since step 1.1 gives $G (a_{N}) < ε a_{N} / 4$ and the square formula in [L2] yields $F_{N} (t) \leq N + 1$ , the interval $(0, a_{N})$ contributes at most $ε / 4$ . If $a_{N} < (δ)$ , then for $t \in [a_{N}, δ]$ one has $\sin (π t) \geq 2 t$ , so [L2] gives $F_{N} (t) \leq \frac{1}{4 (N + 1) t^{2}} .$ Integration by parts with $G^{'} (t)$ equal almost everywhere to the displayed integrand therefore gives $\int_{a_{N}}^{δ} \frac{G^{'} (t)}{t^{2}} d t = \frac{G (δ)}{δ^{2}} - \frac{G (a_{N})}{a_{N}^{2}} + 2 \int_{a_{N}}^{δ} \frac{G (t)}{t^{3}} d t \leq \frac{3 ε}{4 a_{N}},$ so the interval $[a_{N}, δ]$ contributes at most $3 ε / 16$ . Consequently $\int_{0}^{δ} (∣ f (x + t) - f (x) ∣ + ∣ f (x - t) - f (x) ∣) F_{N} (t) d t \leq \frac{7 ε}{16}$ for every $N$ .

3.1L2step 2.1choosealgebra

On $[δ, 1 / 2]$ , the integrand is integrable and [L2] gives $\sup_{t \in [δ, 1 - δ]} F_{N} (t) \to 0.$ Hence $\int_{δ}^{1 / 2} (∣ f (x + t) - f (x) ∣ + ∣ f (x - t) - f (x) ∣) F_{N} (t) d t \to 0.$ Choose $N_{0}$ so large that this far contribution is $< 9 ε / 16$ for all $N \geq N_{0}$ . Then step 2.1 yields $∣ σ_{N} f (x) - f (x) ∣ < ε (N \geq N_{0}) .$ Thus $σ_{N} f (x) \to f (x)$ .

4.1L4step 3.1∎

The first claim holds at every Lebesgue point by step 3.1. Applying [L4] therefore gives $σ_{N} f (x) \to f (x)$ for almost every $x$ .

Depends on

Used by

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Sources

Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)
Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)