One-Dimensional Fourier Series

Abstract

In this chapter, we present some theorems for one-dimensional Fourier series and for the Hardy-Littlewood maximal function. In Sect.  1.1, we introduce the $L_p({\mathbb T})$ spaces and prove some basic inequalities. In Sect. 1.2, we prove that the partial sums of the Fourier series are uniformly bounded on the $L_p({\mathbb T})$ spaces when $1<p<\infty$. As a consequence, we obtain the norm convergence of the partial sums.

In this chapter, we present some theorems for one-dimensional Fourier series and for the Hardy-Littlewood maximal function. In Sect. 1.1, we introduce the $L_p({\mathbb T})$ spaces and prove some basic inequalities. In Sect. 1.2, we prove that the partial sums of the Fourier series are uniformly bounded on the $L_p({\mathbb T})$ spaces when $1<p<\infty$. As a consequence, we obtain the norm convergence of the partial sums. We do not give the proof of the almost everywhere convergence because it can be found at several places, e.g., in Carleson [51], Grafakos [143], Arias de Reyna [9], Muscalu and Schlag [242], Lacey [192], or Demeter [80].

In the next section, the Hardy-Littlewood maximal function is considered and we prove that it is bounded on the $L_p({\mathbb T})$ spaces $(1<p\le \infty )$ and is of weak type (1, 1). Lebesgue’s differentiation theorem is also proved. We introduce the Lebesgue points and show that almost every point is a Lebesgue point.

It was proved by Fejér [107] that the Fejér means of the one-dimensional Fourier series of a continuous function converge uniformly to the function. A similar problem for integrable functions was investigated by Lebesgue [197]. He proved that for every integrable function f,

$$ \frac{1}{n} \sum _{k=0}^{n-1} s_kf(x) \rightarrow f(x) \quad \text{ as } \quad n\rightarrow \infty $$

at each Lebesgue point of f, thus almost everywhere, where $s_kf$ denotes the kth partial sum of the Fourier series of f. Later, Riesz [260], Butzer and Nessel [47], Stein and Weiss [293], and Torchinsky [310] proved the same convergence result for the Riesz, Weierstrass, Picard, Bessel, and de La Vallée-Poussin summations. In Sects. 1.4 and 1.5, we will generalize these results to Cesàro summability.

1.1 The $L_p$ Spaces

Let us denote the set of complex numbers, the set of real numbers, the set of rational numbers, the set of integers, the set of non-negative integers, and the set of positive integers by ${\mathbb C}$ , ${\mathbb R}$ , ${\mathbb Q}$ , ${\mathbb Z}$ , ${\mathbb N}$ , and ${\mathbb P}$ , respectively. The subsets of ${\mathbb R}$ and ${\mathbb Q}$ containing only positive numbers are denoted by ${\mathbb R}_+$ and ${\mathbb Q}_+$ , respectively. ${\mathbb T}$ denotes the torus, which can be identified naturally with the interval $[-\pi ,\pi )$.

In this book, the constants C are absolute constants and the constants $C_p$ are depending only on p and may denote different constants in different contexts.

Definition 1.1.1

The space $L_p({\mathbb X})$ is consisting of all Lebesgue measurable functions $f:{\mathbb X}\rightarrow {\mathbb C}$, for which

$$ \Vert f\Vert _{p}:=\left( \int _{{\mathbb X}}|f|^p \, d\lambda \right) ^{1/p}, \qquad \text{ if } 0<p<\infty $$

and

$$ \Vert f\Vert _{\infty }:= \sup _{{\mathbb X}} |f|, \qquad \text{ if } p=\infty , $$

where ${\mathbb X}\subset {\mathbb R}$ is an arbitrary Lebesgue measurable set and $\lambda$ denotes the Lebesgue measure.

Two functions in $L_p({\mathbb X})$ will be considered equal if they are equal $\lambda$-almost everywhere. It is known that $L_p({\mathbb X})$ is a Banach space if $1\le p \le \infty$ and a complete quasi-normed space if $0<p<1$. We also use the notation |I| for the Lebesgue measure of the set I. Most often we will use the notation ${\mathbb X}={\mathbb R}$ or ${\mathbb X}={\mathbb T}$. The functions from the $L_p({\mathbb T})$ space can be extended to ${\mathbb R}$ such that they are periodic with respect to $2\pi$. In case of ${\mathbb X}={\mathbb Z}$, the corresponding space will be denoted by $\ell _p({\mathbb Z})$ and it is consisting of all complex sequences $c=(c_k,k\in {\mathbb Z})$, for which

$$ \Vert c\Vert _{\ell _p}:=\left( \sum _{k\in {\mathbb Z}} \left| c_k\right| ^p \right) ^{1/p}, \qquad \text{ if } 0<p<\infty $$

and

$$ \Vert c\Vert _{\ell _\infty }:= \sup _{k\in {\mathbb Z}} \left| c_k\right| , \qquad \text{ if } p=\infty . $$

The space of continuous functions with the supremum norm is denoted by $C({\mathbb X})$ and $C_c({\mathbb R})$ denotes the space of continuous functions having compact support. We will use the notation $C_0({\mathbb R})$ for the space of continuous functions vanishing at infinity, i.e.,

$$ C_{0}(\mathbb {R}) := \left\{ f: \mathbb {R} \rightarrow \mathbb {C}: f \in C(\mathbb {R}), \lim _{|x| \rightarrow \infty }f(x) = 0\right\} . $$

We also introduce the notion of weak $L_p({\mathbb T})$ spaces.

Definition 1.1.2

A measurable function f is in the weak $L_p({\mathbb T})$ space , or, in other words, in the $L_{p,\infty }({\mathbb T})$ $(0<p<\infty )$ space if

$$ \Vert f\Vert _{p,\infty } := \sup _{\rho> 0} \rho \, \lambda (|f| > \rho )^{1/p}<\infty . $$

In case of $p=\infty$, let $L_{p,\infty }({\mathbb T}):=L_\infty ({\mathbb T})$.

The weak $L_p({\mathbb T})$ spaces are quasi-norm spaces because

$$ \Vert f\Vert _{p,\infty }=0 \ \Longleftrightarrow \ f=0 \qquad \text{ a.e. } $$

$$ \Vert cf\Vert _{p,\infty } = |c| \Vert f\Vert _{p,\infty } \qquad (c\in {\mathbb C}), $$

$$ \Vert f+g\Vert _{p,\infty } \le c_p (\Vert f\Vert _{p,\infty } + \Vert g\Vert _{p,\infty }), $$

where $c_p=\max (2,2^{1/p})$.

We show that the weak $L_p({\mathbb T})$ spaces are larger than the $L_p({\mathbb T})$ spaces.

Proposition 1.1.3

If $0<p<\infty$, then $L_p({\mathbb T})\subset L_{p,\infty }({\mathbb T})$ and

$$ \Vert f \Vert _{p,\infty } \le \Vert f \Vert _{p}. $$

Proof

It is easy to see that

$$ \int _{\mathbb {T}} |f(x)|^{p} \, dx \ge \int _{\{ x: |f(x)|> \rho \}} |f(x)|^{p} \,dx \ge \rho ^{p} \lambda \left( |f| > \rho \right) , $$

which proves the proposition.    $\blacksquare$

If $h(x):=|x|^{-1/p}$, then obviously $h\not \in L_p({\mathbb R})$, but $h\in L_{p,\infty }({\mathbb R})$ because

$$ \rho ^{p} \lambda \left( \left\{ x : |x|^{-1/p} > \rho \right\} \right) = 2\rho ^{p}\rho ^{-p}=2. $$

Thus, the inclusion $L_p({\mathbb R})\subset L_{p,\infty }({\mathbb R})$ is proper if $0<p<\infty$. Recall that the weak space $L_{p,\infty }({\mathbb R})$ is also complete for each p.

1.2 Convergence of Fourier Series

We introduce the trigonometric Fourier series and show that the partial sums of a function $f \in L_p({\mathbb T})$ $(1<p<\infty )$ converge almost everywhere as well in the $L_p({\mathbb T})$-norm to the function f.

Definition 1.2.1

For an integrable function $f\in L_1({\mathbb T})$, its kth Fourier coefficient is defined by

$$ \widehat{f}(k) = \frac{1}{2\pi } \int _{{\mathbb T}} f(x) e^{-\imath kx} \, dx \qquad (k\in {\mathbb Z}). $$

The formal trigonometric series

$$ \sum _{k\in {\mathbb Z}} \widehat{f}(k) e^{\imath kx}\qquad (x\in {\mathbb T}) $$

is called the Fourier series of f.

Definition 1.2.2

For $f\in L_1({\mathbb T})$ and $n \in {\mathbb N}$, the nth partial sum $s_nf$ of the Fourier series of f and the nth Dirichlet kernel $D_n$ are introduced by

$$ s_{n} f(x) := \sum _{k=-n}^n \widehat{f}(k) e^{\imath kx} $$

and

$$ D_{n}(t) := \sum _{k=-n}^{n} e^{\imath kt}, $$

respectively.

We get immediately that

$$\begin{aligned} s_{n} f(x)&= \sum _{k=-n}^{n} \frac{1}{2\pi }\int _{{\mathbb T}} f(t) e^{\imath k(x-t)} \, dt \nonumber \\&= \frac{1}{2\pi }\int _{{\mathbb T}} f(x-t) D_n(t) \, dt \qquad (n \in {\mathbb N}) \end{aligned}$$

(1.2.1)

(see Fig. 1.1).

Fig. 1.1

Dirichlet kernel $D_n$ for $n=5$

Lemma 1.2.3

For all $n \in {\mathbb N}$ and $t \in {\mathbb T}$, $t\ne 0$,

$$\begin{aligned} D_{n}(t) = \frac{\sin ((n+1/2)t)}{\sin (t/2)}. \end{aligned}$$

(1.2.2)

Proof

Using some simple trigonometric identities, we obtain

$$\begin{aligned} D_{n}(t)&= 1+ 2\sum _{k=1}^n \cos (kt) \nonumber \\&= \frac{1}{\sin (t/2)} \left( \sin (t/2)+ 2\sum _{k=1}^n \cos (kt)\sin (t/2) \right) \nonumber \\&= \frac{1}{\sin (t/2)} \left( \sin (t/2)+ \sum _{k=1}^n \left( \sin ((k+1/2)t) - \sin ((k-1/2)t) \right) \right) , \end{aligned}$$

which shows the lemma.    $\blacksquare$

The next lemma follows easily from this.

Lemma 1.2.4

For all $n \in {\mathbb N}$ and $t \in {\mathbb T}$, $t\ne 0$, we have

$$ |D_{n}|\le 2n+1 \qquad \text{ and } \qquad |D_{n}(t|\le C/t. $$

It is easy to see that the $L_1({\mathbb T})$-norms of $D_n$ are not uniformly bounded, more exactly $\Vert D_{n}\Vert _1 \sim \log n$.

Before proving the norm convergence of the partial sums, we need some other definitions and results. We follow the proof of Grafakos [143].

Definition 1.2.5

For some $n \in {\mathbb N}$, the function

$$ \sum _{k=-n}^{n} c_k e^{\imath kx} \qquad (x \in {\mathbb R}) $$

is said to be a trigonometric polynomial.

It is a well-known result that the trigonometric polynomials are dense in $L_p({\mathbb T})$ for any $1\le p<\infty$.

Definition 1.2.6

For a trigonometric polynomial f define the conjugate function $\widetilde{f}$ by

$$ \widetilde{f}(x) := -\imath \sum _{k\in {\mathbb Z}} \mathrm{sign \, }(k) \widehat{f}(k) e^{\imath kx}. $$

Now we show that $\widetilde{f}$ is bounded on $L_p({\mathbb T})$ $(1<p<\infty )$ (see Riesz [261, 262]).

Theorem 1.2.7

If $1<p<\infty$, then

$$ \left\| \widetilde{f}\right\| _p\le C_p \left\| f\right\| _p \qquad (f\in L_p({\mathbb T})). $$

Proof

First suppose that f is a real trigonometric polynomial and $\widehat{f}(0)=0$. It is easy to see that $\widetilde{f}$ is also real valued and $f+\imath \widetilde{f}$ contains only positive frequencies. Since $\int _{{\mathbb T}} e^{\imath kx}\,dx=0$ $(k\ne 0)$, we have

$$ \int _{{\mathbb T}} \left( f(x)+\imath \widetilde{f}(x) \right) ^{2k} \, dx =0, $$

where k is a positive natural number. Taking the real part of the integral and using that f and $\widetilde{f}$ are real valued, we obtain

$$\begin{aligned} 0&= \sum _{j=0}^{k} (-1)^{k-j} \left( {\begin{array}{c}2k\\ 2j\end{array}}\right) \int _{{\mathbb T}} f(x)^{2j} \widetilde{f}(x)^{2k-2j}\, dx \\&= (-1)^{k} \int _{{\mathbb T}} \widetilde{f}(x)^{2k}\, dx + \sum _{j=1}^{k} (-1)^{k-j} \left( {\begin{array}{c}2k\\ 2j\end{array}}\right) \int _{{\mathbb T}} f(x)^{2j} \widetilde{f}(x)^{2k-2j}\, dx . \end{aligned}$$

This and Hölder’s inequality imply that

$$\begin{aligned} \left\| \widetilde{f}\right\| _{2k}^{2k}\le & {} \sum _{j=1}^{k} \left( {\begin{array}{c}2k\\ 2j\end{array}}\right) \int _{{\mathbb T}} f(x)^{2j} \widetilde{f}(x)^{2k-2j}\, dx \\\le & {} \sum _{j=1}^{k} \left( {\begin{array}{c}2k\\ 2j\end{array}}\right) \left\| f\right\| _{2k}^{2j} \left\| \widetilde{f}\right\| _{2k}^{2k-2j}. \end{aligned}$$

Let $R=\left\| \widetilde{f}\right\| _{2k}/\left\| f\right\| _{2k}$ and divide by $\left\| f\right\| _{2k}^{2k}$ to obtain

$$ R^{2k} - \sum _{j=1}^{k} \left( {\begin{array}{c}2k\\ 2j\end{array}}\right) R^{2k-2j} \le 0. $$

Then R is smaller than the largest root in absolute value of the polynomial on the left-hand side, say $R\le C_{2k}$, in other words

$$\begin{aligned} \left\| \widetilde{f}\right\| _p\le C_p \left\| f\right\| _p \qquad \text{ for } p=2k. \end{aligned}$$

(1.2.3)

If $\widehat{f}(0)\ne 0$, then apply this inequality to $f-\widehat{f}(0)$. Since $|\widehat{f}(0)|\le \left\| f\right\| _p$, we get the preceding inequality with $2C_p$. Every general trigonometric polynomial can be written as the sum of two real-valued trigonometric polynomials. Therefore, (1.2.3) holds for every trigonometric polynomials and by density for all $f\in L_p({\mathbb T})$, $p=2k$. By interpolation (see, e.g., Berg and Löfström [33] or Weisz [346]), (1.2.3) holds for all $2\le p<\infty$. Finally, observe that the adjoint operator of $f\mapsto \widetilde{f}$ is $f\mapsto -\widetilde{f}$, which implies by duality that (1.2.3) holds also for $1<p\le 2$.    $\blacksquare$

Definition 1.2.8

For a trigonometric polynomial f, the Riesz projections $P^+$ and $P^-$ are defined by

$$ P^+f(x) \sim \sum _{k=1}^\infty \widehat{f}(k) e^{\imath kx} $$

and

$$ P^-f(x) \sim \sum _{k=-\infty }^{-1} \widehat{f}(k) e^{\imath kx}. $$

Observe that $f=P^+f+P^-f+\widehat{f}(0)$ and $\widetilde{f}=-\imath P^+f+\imath P^-f$.

Theorem 1.2.9

If $1<p<\infty$ and $f\in L_p({\mathbb T})$, then

$$ \left\| P^+f\right\| _p\le C_p \left\| f\right\| _p $$

and

$$ \left\| P^-f\right\| _p\le C_p \left\| f\right\| _p. $$

Proof

Since

$$ P^+f = \frac{1}{2}(f+\imath \widetilde{f}) -\frac{1}{2} \widehat{f}(0), $$

and $|\widehat{f}(0)|\le \left\| f\right\| _p$, the first inequality follows from Theorem 1.2.7. The second one can be proved similarly.    $\blacksquare$

The following theorem is a fundamental result and it can be found in most books about trigonometric Fourier series (e.g., Zygmund [367], Bary [19], Torchinsky [310], or Grafakos [143]). It is due to Riesz [260].

Theorem 1.2.10

If $f\in L_p({\mathbb T})$ for some $1<p< \infty$, then

$$\begin{aligned} \sup _{n \in {\mathbb N}} \left\| s_{n}f\right\| _p \le C_p \left\| f\right\| _p \end{aligned}$$

(1.2.4)

and

$$\begin{aligned} \lim _{n\rightarrow \infty } s_{n}f=f \qquad \text{ in } \text{ the } L_p({\mathbb T})\text {-norm}. \end{aligned}$$

(1.2.5)

Proof

Define

$$ P_n^+ g(x)= \sum _{k=0}^{2n} \widehat{g}(k) e^{\imath kx}. $$

It is easy to see that

$$ \sum _{k=-n}^{n} \widehat{f}(k) e^{\imath kx} = e^{-\imath nx} \sum _{k=0}^{2n} \widehat{(f(\cdot )e^{\imath n (\cdot )})}(k) e^{\imath kx}. $$

This implies that the norm of $s_n:L_p({\mathbb T}) \rightarrow L_p({\mathbb T})$ is equal to the norm of $P_n^+:L_p({\mathbb T}) \rightarrow L_p({\mathbb T})$.

We have

$$\begin{aligned} P_n^+f= & {} \sum _{k=0}^{\infty } \widehat{f}(k) e^{\imath kx} - \sum _{k=2n+1}^{\infty } \widehat{f}(k) e^{\imath kx} \\= & {} \sum _{k=0}^{\infty } \widehat{f}(k) e^{\imath kx} - e^{\imath (2n+1)x}\sum _{k=0}^{\infty } \widehat{f}(k+2n+1) e^{\imath kx} \\= & {} P^+f(x) - e^{\imath (2n+1)x} P^+(e^{-\imath (2n+1)(\cdot )}f) - \widehat{f}(0)(1-e^{\imath (2n+1)x}) \end{aligned}$$

for all trigonometric polynomials. By density this yields that

$$ \left\| P_n^+f\right\| _p \le \left( 2 \left\| P^+\right\| +2\right) \left\| f\right\| _p $$

for all $f\in L_p({\mathbb R})$ and $n\in {\mathbb N}$, which proves (1.2.4). The convergence (1.2.5) is clearly valid for all trigonometric polynomials and so the convergence follows for all $f\in L_p({\mathbb T})$ $(1<p<\infty )$ by density.    $\blacksquare$

Since the $L_1$-norms of $D_n$ are not uniformly bounded, Theorem 1.2.10 is not true for $p=1$ and $p=\infty$.

One of the deepest  results in harmonic analysis is Carleson’s theorem that the partial sums of the Fourier series converge almost everywhere to $f\in L_p({\mathbb T})$ $(1<p\le \infty )$. Since the proof can be found in many papers and books (see, e.g., Carleson [51], Hunt [174], Arias de Reyna [9], Grafakos [143], Muscalu and Schlag [242], Lacey [192], or Demeter [80]), we present the result without proof.

Definition 1.2.11

We denote by

$$ s_*f:=\sup _{n \in {\mathbb N}} \left| s_{n}f \right| $$

the maximal operator of the partial sums.

Theorem 1.2.12

If $f\in L_p({\mathbb T})$ for some $1<p< \infty$, then

$$ \left\| s_{*}f \right\| _p \le C_p \left\| f\right\| _p $$

and if $1<p\le \infty$, then

$$ \lim _{n\rightarrow \infty } s_{n}f=f \qquad \text{ a.e. } $$

The inequality of Theorem 1.2.12 does not hold if $p=1$ or $p=\infty$, and the almost everywhere convergence does not hold if $p=1$. Du Bois Reymond [84] and Fejér [108] proved the existence of a continuous function $f\in C({\mathbb T})$ and a point $x_0\in {\mathbb T}$ such that the partial sums $s_nf(x_0)$ diverge as $n\rightarrow \infty$. Kolmogorov gave an integrable function $f\in L_1({\mathbb T})$, whose Fourier series diverges almost everywhere or even everywhere (see Kolmogorov [186, 187], Zygmund [367], or Grafakos [143]).

Since there are many function spaces contained in $L_1({\mathbb T})$ but containing $L_p({\mathbb T})$ $(1<p\le \infty )$, it is natural to ask whether there is a “largest” subspace of $L_1({\mathbb T})$ for which almost everywhere convergence holds. The next result, due to Antonov [7], generalizes Theorem 1.2.12.

Theorem 1.2.13

If

$$\begin{aligned} \int _{{\mathbb T}} |f(x)| \log ^+|f(x)| \log ^+ \log ^+ \log ^+ |f(x)| \, dx <\infty , \end{aligned}$$

(1.2.6)

then

$$ \lim _{n\rightarrow \infty } s_{n}f=f \qquad \text{ a.e. } $$

Note that $\log ^+u=\max (0,\log u)$.  It is easy to see that if $f\in L_p({\mathbb T})$ $(1<p\le \infty )$, then f satisfies (1.2.6). If f satisfies (1.2.6), then of course $f\in L_1({\mathbb T})$. For the converse direction, Konyagin [188] obtained the next result.

Theorem 1.2.14

If the non-decreasing function $\phi :{\mathbb R}_+\rightarrow {\mathbb R}_+$ satisfies the condition

$$ \phi (u)= o\Big (u \sqrt{\log u}/\sqrt{\log \log u}\Big ) \qquad \text{ as } u\rightarrow \infty , $$

then there exists an integrable function f such that

$$ \int _{{\mathbb T}} \phi (|f(x)|) \, dx<\infty $$

and

$$ \limsup _{n\rightarrow \infty } s_{n}f(x)=\infty \qquad \text{ for } \text{ all } x\in {\mathbb T}, $$

i.e., the Fourier series of f diverges everywhere.

For example, if $\phi (u)=u\, {\log ^+\log ^+ u}$, then there exists a function f such that its Fourier series diverges everywhere and

$$ \int _{{\mathbb T}} |f(x)| {\log ^+\log ^+ |f(x)|} \, dx<\infty . $$

1.3 Hardy-Littlewood Maximal Function and Lebesgue Points

Before continuing our investigations about the convergence of the Fourier series, we have to introduce the Hardy-Littlewood maximal function. We will prove that it is bounded on $L_p({\mathbb T})$ for $1<p\le \infty$ and it is of weak type (1, 1). Using this result, we obtain Lebesgue’s differentiation theorem and the theorem about the Lebesgue points.

Definition 1.3.1

For $f\in L_1({\mathbb T})$, the Hardy-Littlewood maximal function is defined by

$$ Mf(x):= \sup _{x\in I} \frac{1}{|I|}\int _{I} |f| \,d\lambda \qquad (x\in {\mathbb T}), $$

where the supremum is taken over all open intervals I containing x.

We can also define the centered maximal function,

$$ M_cf(x):= \sup _{h>0} \frac{1}{|I(x,h)|}\int _{I(x,h)} |f| \, d\lambda \qquad (x\in {\mathbb T}), $$

where I(x, h) $(x\in {\mathbb T},h>0)$ denotes the interval with center x and radius h:

$$ I(x,h):=\{y\in {\mathbb T}: |x-y|<h\}. $$

Obviously, $M_cf \le Mf$. If $x\in I(y,h)$, then $I(y,h)\subset I(x,2h)$ and so $Mf \le 2M_cf$. Let $rI(x,h):=I(x,rh)$ for $r>0$.

Lemma 1.3.2

(Vitali covering lemma) Let be given finitely many open intervals $I_j$ and let $E=\bigcup _j I_j.$ Then there exists a finite subcollection $I_1,\ldots ,I_m$ of disjoint intervals, such that

$$ \sum _{k=1}^m |I_k| \ge \frac{|E|}{3}. $$

Proof

Let $I_1$ be an interval of the collection $\{I_j\}$ with maximal radius. Next choose $I_2$ to have maximal radius among the subcollection of intervals disjoint with $I_1$. We continue this process until we can go no further. Then the intervals $I_1,\ldots ,I_m$ are disjoint. Observe that $3I_k$ contains all intervals of the original collection that intersect $I_k$ $(k=1,\ldots ,m)$. From this, it follows that $\cup _{k=1}^m 3I_k$ contains all intervals from the original collection. Thus

$$ |E| \le \left| \bigcup _{k=1}^m 3I_k \right| \le \sum _{k=1}^m \left| 3I_k\right| \le 3 \sum _{k=1}^m \left| I_k\right| , $$

which shows the lemma.    $\blacksquare$

Theorem 1.3.3

The maximal operator M is of weak type (1, 1), i.e.,

$$\begin{aligned} \sup _{\rho>0} \rho \lambda (Mf > \rho ) \le 3 \Vert f\Vert _1 \qquad (f \in L_1({\mathbb T})). \end{aligned}$$

(1.3.1)

Moreover, if $1<p \le \infty$, then

$$\begin{aligned} \Vert Mf\Vert _p\le C_p \Vert f\Vert _p \qquad (f \in L_p({\mathbb T})). \end{aligned}$$

(1.3.2)

Proof

Let $E\subset \{Mf>\rho \}$ be a compact subset. For each $x\in \{Mf>\rho \}$, there exists an open interval $I_x$ such that $x\in I_x$ and

$$\begin{aligned} \rho <\frac{1}{|I_x|} \int _{I_x} |f| \,d\lambda . \end{aligned}$$

(1.3.3)

Since $x\in I_x$, we can select a finite collection of these intervals covering E. By Lemma 1.3.2, we can choose a finite disjoint subcollection $I_1,\ldots ,I_m$ of this covering with

$$ |E|\le 3 \sum _{k=1}^m |I_k|. $$

Since each $I_k$ satisfies (1.3.3), adding these inequalities, we obtain

$$ |E| < \frac{3}{\rho } \sum _{k=1}^m \int _{I_k} |f| \,d\lambda \le \frac{3}{\rho } \int _{\{Mf>\rho \}} |f| \, d\lambda . $$

Taking the supremum over all compact sets $E\subset \{Mf>\rho \}$, we conclude

$$ \lambda (Mf>\rho ) \le \frac{3}{\rho } \int _{\{Mf>\rho \}} |f| \, d\lambda \le \frac{3}{\rho } \int _{{\mathbb T}} |f| \, d\lambda , $$

which gives exactly (1.3.1).

For $p=\infty$, obviously

$$ \frac{1}{|I|} \int _{I} |f| \,d\lambda \le \left\| f\right\| _\infty , $$

and so

$$ \Vert Mf\Vert _\infty \le \Vert f\Vert _\infty \qquad (f \in L_\infty ({\mathbb R})). $$

Now the theorem follows easily for $1<p<\infty$ by interpolation (see, e.g., Bergh and Löfström [33]).    $\blacksquare$

The boundedness on $L_\infty ({\mathbb T})$ and the weak type (1, 1) boundedness of M imply a finer version of (1.3.1).

Theorem 1.3.4

We have

$$ \rho \lambda (Mf> 2\rho ) \le 3 \int _{\rho }^ \infty \lambda (|f|> t) \, dt \qquad (\rho >0). $$

Proof

Let us decompose f into the sum of $f_0\in L_1({\mathbb T})$ and $f_1\in L_\infty ({\mathbb T})$ as follows. For an arbitrary $\rho > 0$, set

$$ f_{1,\rho }(t):= \left\{ \begin{array}{ll} f(t), &{} \hbox {if } |f(t)|\le \rho ; \\ \rho / \mathrm{sign \, }f(t), &{} \hbox {otherwise.} \end{array} \right. $$

and

$$ f_{0,\rho }(t)=f(t)-f_{1,\rho }(t). $$

Then $\Vert f_{1,\rho } \Vert _{\infty } \le \rho$. Since

$$ Mf\le Mf_{0,\rho }+Mf_{1,\rho } \qquad \text{ and } \qquad \left\| Mf_{1,\rho }\right\| _ \infty \le C_\infty \left\| f_{1,\rho }\right\| _ \infty \le \rho , $$

we have

$$ \{Mf>2\rho \} \subset \{Mf_{0,\rho }>\rho \} \cup \{Mf_{1,\rho }>\rho \} = \{Mf_{0,\rho }>\rho \}. $$

Hence

$$\begin{aligned} \rho \lambda (Mf>2\rho )\le & {} \rho \lambda (Mf_{0,\rho }>\rho ) \\\le & {} 3 \left\| f_{0,\rho }\right\| _1\\= & {} 3 \int _{\{|f|> \rho \}} (|f|-\rho ) \, d\lambda \\= & {} 3 \int _{{\mathbb T}} \int _0^\infty 1_{\{|f|>t>\rho \}} \, dt\, d\lambda \\= & {} 3 \int _{\rho }^\infty \lambda (|f|>t) \, dt \end{aligned}$$

as desired.    $\blacksquare$

It is known that inequality (1.3.2) does not hold for $p=1$. However, we can prove that

$$ \left\| Mf\right\| _{1} \le C+ C \left\| |f| \left( \log ^+|f|\right) \right\| _{1}, $$

where $\log ^+u:=\max (0,\log u)$. We generalize this inequality as follows.

Theorem 1.3.5

For every $k \in {\mathbb P}$ and $f \in L_1(\log L)^{k}({\mathbb T})$,

$$ \left\| Mf \left( \log ^+Mf\right) ^{k-1}\right\| _{1} \le C+ C \left\| |f| \left( \log ^+|f|\right) ^{k}\right\| _{1}. $$

Proof

First, we handle the case $k>1$. Observe that

$$\begin{aligned} \left\| |f| \left( \log ^+|f|\right) ^{k-1}\right\| _1&= \int _0^ \infty \lambda (|f|>\rho ) \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho \nonumber \\&= \int _1^\infty \lambda (|f|>\rho ) \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho . \end{aligned}$$

Theorem 1.3.4 implies

$$\begin{aligned}&\int _1^ \infty \lambda (|M f|>\rho ) \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho \\&\qquad \le \int _1^ \infty \frac{6}{\rho } \int _{\rho /2}^ \infty \lambda (|f|> t) \, dt \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho \\&\qquad = 6 \int _{1/2}^ \infty \lambda (|f| > t) \int _1^{2 t} \frac{1}{\rho } \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho \, dt. \end{aligned}$$

Since

$$ \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho }= (\log ^+ \rho )^{k-1}+(k-1) (\log ^+ \rho )^{k-2}, $$

we conclude

$$\begin{aligned} \int _1^{2 t} \frac{1}{\rho } \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho&= \frac{1}{k} (\log ^+ (2t))^{k}+ (\log ^+ (2t))^{k-1} \\&= \frac{1}{2k} \frac{d(2t (\log ^+ 2t)^{k})}{dt}. \end{aligned}$$

Therefore

$$\begin{aligned}&\int _1^ \infty \lambda (|M f|>\rho ) \frac{d(\rho (\log ^+ \rho )^{k-1})}{d\rho } \, d\rho \nonumber \\&\qquad \le \frac{3}{k} \int _{1/2}^ \infty \lambda (|f| > t) \frac{d(2t (\log ^+ 2t)^{k})}{dt} \, dt \nonumber \\&\qquad = C \left\| 2 |f| \left( \log ^+|2f|\right) ^{k}\right\| _1 \nonumber \\&\qquad \le C + C \left\| |f| \left( \log ^+|f|\right) ^{k}\right\| _1, \end{aligned}$$

(1.3.4)

which completes the proof for $k>1$.

Let $k=1$ and notice that $\lambda (Mf\le 1)\le 1$. Then

$$\begin{aligned} \int _{\{Mf>1\}} Mf(t)\, dt= & {} \int _0^ \infty \lambda (Mf> \max (\rho ,1)) \, d\rho \\= & {} \int _1^ \infty \lambda (Mf>\rho ) \, d\rho + \lambda (Mf>1). \end{aligned}$$

Moreover,

$$\begin{aligned} \lambda (Mf>1)\le & {} 3 \Vert f\Vert _1 = 3 \int _{\{|f|\le e\}} |f| \, d\lambda + 3 \int _{\{|f|>e\}} |f| \, d\lambda \\\le & {} C + C \int _{{\mathbb T}} |f| \log ^+|f| \, d\lambda . \end{aligned}$$

Since (1.3.4) holds for $k=1$, too, we obtain

$$\begin{aligned} \Vert Mf\Vert _1&= \int _{\{Mf \le 1\}} Mf(t)\, dt + \int _{\{Mf>1\}} Mf(t)\, dt \\&\le C + C \int _{{\mathbb T}} |f| \log ^+|f| \, d\lambda , \end{aligned}$$

as we stated in the theorem.    $\blacksquare$

Now we present a density theorem due to Marcinkiewicz and Zygmund [234]. Let $L_0({\mathbb T})$ denote the set of measurable functions and $X\subset L_0({\mathbb T})$. Let the operators $T, T_n :X\rightarrow L_0({\mathbb T})$ $(n\in {\mathbb N})$ be given. Moreover, we introduce the maximal operator by

$$ T_*f(x):=\sup _{n\in {\mathbb N}} \left| T_nf(x)\right| \qquad (f\in X, x\in {\mathbb T}). $$

Theorem 1.3.6

Let X be a normed space of measurable functions and $S\subset X$ be dense in X. Suppose that T and $T_n$ $(n \in {\mathbb N})$ are linear operators and

$$ \lim _{n\rightarrow \infty } T_nf = Tf \qquad \text{ a.e. } $$

for all $f \in S$. If

$$\begin{aligned} \sup _{\rho>0} \rho \,\lambda (|Tf|>\rho ) \le C\Vert f\Vert _X \qquad (f \in X) \end{aligned}$$

(1.3.5)

and

$$\begin{aligned} \sup _{\rho>0} \rho \,\lambda (T_*f>\rho ) \le C\Vert f\Vert _X \qquad (f \in X), \end{aligned}$$

(1.3.6)

then for every $f \in X$,

$$ \lim _{n \rightarrow \infty } T_nf = Tf \qquad \text{ a.e. } $$

Proof

Fix $f \in X$ and set

$$ \xi := \limsup _{n \rightarrow \infty } |T_nf-Tf|. $$

It is sufficient to show that $\xi =0$ a.e. Choose a sequence $f_m \in S$ $(m \in {\mathbb N})$ such that

$$ \lim _{m\rightarrow \infty } \left\| f-f_m\right\| _X=0. $$

By the triangle inequality,

$$ \xi \le \limsup _{n \rightarrow \infty } |T_n(f-f_m)| + \limsup _{n \rightarrow \infty } |T_nf_m-Tf_m| + |T(f_m-f)| $$

for all $m \in {\mathbb N}$. Since $f_m \in S$, we have

$$ \limsup _{n \rightarrow \infty } |T_nf_m-Tf_m| = \lim _{n \rightarrow \infty } |T_nf_m-Tf_m| = 0 \qquad \text{ a.e. }, $$

so

$$ \xi \le T_*(f_m-f) + |T(f_m-f)| \qquad \text{ a.e. } $$

Applying inequalities (1.3.5) and (1.3.6), we obtain

$$\begin{aligned} \lambda (\xi>2\rho )\le & {} \lambda (T_*(f_m-f)>\rho ) + \lambda (|T(f_m-f)|>\rho ) \\\le & {} C\rho ^{-1} \Vert f_m-f\Vert _X + C \rho ^{-1} \Vert f_m-f\Vert _X \end{aligned}$$

for all $\rho >0$ and $m \in {\mathbb N}$. Since $f_m \rightarrow f$ in the X-norm as $m \rightarrow \infty$, we get that

$$ \lambda (\xi >2\rho ) =0 $$

for all $\rho >0$. This implies immediately that $\xi =0$ almost everywhere.    $\blacksquare$

The next theorem can be proved in the same way.

Theorem 1.3.7

Let X be a normed space of measurable functions and $S\subset X$ be dense in X. Suppose that $T_n$ is a sublinear operator for every $n \in {\mathbb N}$ and

$$ \lim _{n\rightarrow \infty } T_nf = 0 \qquad \text{ a.e. } $$

for all $f \in S$. If

$$ \sup _{\rho>0} \rho \,\lambda (T_*f>\rho ) \le C\Vert f\Vert _X \qquad (f \in X), $$

then for every $f \in X$,

$$ \lim _{n \rightarrow \infty } T_nf = 0 \qquad \text{ a.e. } $$

Now we can state Lebesgue’s differentiation theorem mentioned before.

Corollary 1.3.8

For all $f\in L_1({\mathbb T})$,

$$ \lim _{r\rightarrow 0}\frac{1}{2r}\int _{x-r}^{x+r}f(t) \,dt=f(x)\qquad \text{ a.e. } x\in {\mathbb T}. $$

Proof

Let $r_n>0$ $(n\in {\mathbb N})$ and $\lim _{n\rightarrow \infty }r_n=0$. Define

$$ Tf(x):=f(x) \qquad \text{ and } \qquad T_nf(x):=\frac{1}{2r_n}\int _{x-r_n}^{x+r_n}f(t) \,dt \qquad (x \in {\mathbb T}). $$

These operators are linear and

$$ \sup _{\rho>0} \rho \,\lambda (|Tf|>\rho ) = \sup _{\rho>0} \rho \,\lambda (|f|>\rho ) \le \sup _{\rho>0} \int _{\left\{ |f|>\rho \right\} } \left| f\right| \, d\lambda \le \Vert f\Vert _1 $$

implies (1.3.5). Inequality (1.3.6) follows from Theorem 1.3.3. The result obviously holds for continuous functions. If S denotes the set of continuous functions, then S is dense in $L_1({\mathbb T})$. Now Theorem 1.3.6 implies Corollary 1.3.8.    $\blacksquare$

Similarly, we get

Corollary 1.3.9

For all $f\in L_1({\mathbb T})$,

$$ \lim _{x\in I, |I|\rightarrow 0} \frac{1}{|I|} \int _I f \,d\lambda = f(x)\qquad \text{ a.e. } x\in {\mathbb T}. $$

Corollary 1.3.8 implies that $\left| f(x)\right| \le Mf(x)$ for almost every $x\in {\mathbb T}$, and so the converse of (1.3.2) is also true:

$$ \Vert f\Vert _p \le \Vert Mf\Vert _p \qquad (1\le p \le \infty ). $$

Now we introduce the concept of Lebesgue points. Corollary 1.3.8 can be written in the form

$$ \lim _{h\rightarrow 0} \frac{1}{2h} \int _{-h}^{h} f(x-t) \,dt = f(x) $$

for almost every $x\in {\mathbb T}$ and $f\in L_1({\mathbb T})$. Thus

$$ \lim _{h\rightarrow 0} \frac{1}{2h} \int _{-h}^{h} \left( f(x-t)-f(x) \right) \,dt = 0 $$

for almost every $x\in {\mathbb T}$, which is equivalent to

$$ \lim _{h\rightarrow 0}\frac{1}{2h} \left| \int _{-h}^{h}\left( f(x-t)-f(x)\right) \, dt \right| =0 $$

for almost every $x\in {\mathbb T}$. Though the definition of the Lebesgue point is a stronger condition, we prove in the next theorem that almost every point is a Lebesgue point.

Definition 1.3.10

A point $x\in {\mathbb T}$ is called a Lebesgue point of $f \in L_1({\mathbb T})$ if

$$ \lim _{h\rightarrow 0} \frac{1}{2h} \int _{-h}^{h} \left| f(x-t)-f(x)\right| \,dt =0. $$

Theorem 1.3.11

Almost every point $x\in {\mathbb R}$ is a Lebesgue point of $f\in L_{1}({\mathbb T})$.

Proof

For all rational numbers q let

$$ G_q:= \left\{ x\in {\mathbb R}: \lim _{h\rightarrow 0}\frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-q\right| \,dt= \left| f(x)-q\right| \right\} . $$

Applying Corollary 1.3.8 to the function $\left| f(\cdot )-q\right|$, we can see that $B_q:={\mathbb R}\setminus G_q$ is of Lebesgue measure 0. Observe that f is almost everywhere finite. Set $N:=\{x\in {\mathbb R}: |f(x)|=\infty \}$. Then the set

$$ B:= N \bigcup \left( \bigcup _{q\in \mathbb {Q}} B_q \right) $$

has Lebesgue measure 0. We show that the points of $G:={\mathbb T}\setminus B$ are Lebesgue points. Let $\epsilon >0$ and $x\in G$ be arbitrary. Choose $q\in \mathbb {Q}$ such that

$$ \left| f(x)-q\right| < \frac{\epsilon }{2}. $$

Then

$$\begin{aligned}&\frac{1}{2h}\int _{-h}^{h} \left| f(x-t)-f(x)\right| \,dt \qquad \nonumber \\&\qquad \quad \le \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-q\right| \, dt + \frac{1}{2h}\int _{-h}^{h}\left| q-f(x)\right| \, dt \\&\qquad \quad = \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-q\right| \, dt + \left| q-f(x)\right| . \end{aligned}$$

Since $x\notin B_q$, we have

$$ \limsup _{h\rightarrow \infty }\frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-f(x)\right| \,dt \le 2 \left| f(x)-q\right| <\epsilon . $$

Thus, every $x\in G$ is a Lebesgue point of f.    $\blacksquare$

Lemma 1.3.12

If x is a Lebesgue point of $f \in L_1({\mathbb T})$, then f(x) and Mf(x) are finite.

Proof

f(x) is clearly finite. For $\epsilon =1$ there exists $\delta >0$ such that for all $|h|<\delta$,

$$ \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-f(x)\right| \,dt <1. $$

Thus

$$ \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)\right| \,dt \le \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)-f(x)\right| \,dt + |f(x)| <1 + |f(x)|. $$

On the other hand,

$$ \frac{1}{2h}\int _{-h}^{h}\left| f(x-t)\right| \,dt \le \frac{1}{2 \delta } \Vert f\Vert _1 $$

for all $|h| \ge \delta$.    $\blacksquare$

1.4 Summability of One-Dimensional Fourier Series

Though Theorems 1.2.10 and 1.2.12 are not true for $p=1$ and $p=\infty$, with the help of some summability methods they can be generalized. Obviously, summability means have better convergence properties than the original Fourier series. Summability is intensively studied in the literature (see, e.g., the books Stein and Weiss [293], Butzer and Nessel [47], Trigub and Belinsky [319], Grafakos [143] and Weisz [332, 346], and the references therein).

One of the first investigated summability methods is the Fejér method. In 1904, Fejér [107] investigated the arithmetic means of the partial sums, the so-called Fejér means $\sigma _{n}f$. He proved for an integrable function $f \in L_1({\mathbb T})$ that if the left and right limits $f(x-0)$ and $f(x+0)$ exist at a point x, then the Fejér means converge to $(f(x-0)+f(x+0))/2$, that is,

$$\begin{aligned} \lim _{n\rightarrow \infty }\sigma _{n}f(x)= \frac{f(x-0)+f(x+0)}{2}. \end{aligned}$$

(1.4.1)

One year later Lebesgue [197] extended this theorem and obtained that the convergence holds for every $f \in L_1({\mathbb T})$ and every Lebesgue points, i.e.,

$$\begin{aligned} \lim _{n\rightarrow \infty }\sigma _{n}f(x)=f(x) \end{aligned}$$

(1.4.2)

at each Lebesgue point of f, thus almost everywhere. In this section, we generalize these results.

Definition 1.4.1

For $f\in L_1({\mathbb T})$ and $n \in {\mathbb N}$, the nth Fejér means $\sigma _nf$ of the Fourier series of f and the nth Fejér kernel $K_n$ are introduced by

$$ \sigma _n f(x) := \sum _{k=-n}^n \left( 1-\frac{|k|}{n} \right) \widehat{f}(k)e^{\imath kx} $$

and

$$ K_n(t) := \sum _{k=-n}^n \left( 1-\frac{|k|}{n} \right) e^{\imath k t}, $$

respectively.

One can see that

$$ \sigma _n f(x) = \frac{1}{2\pi }\int _{{\mathbb T}} f(x-t) K_n(t) \, dt $$

(see Fig. 1.2). We will prove the next result in Lemma 1.4.12.

Fig. 1.2

Fejér kernel $K_n$ for $n=5$

Lemma 1.4.2

For $f\in L_1({\mathbb T})$ and $n \in {\mathbb N}$, we have

$$ \sigma _n f(x) = \frac{1}{n} \sum _{j=0}^{n-1} s_{j}f(x) $$

and

$$ K_n(t) = \frac{1}{n} \sum _{j=0}^{n-1} D_j(t). $$

Lemma 1.4.3

For $n \ge 1$ and $t \in {\mathbb T}$, $t\ne 0$,

$$\begin{aligned} \sum _{k=0}^{n-1} \sin (k+1/2)t= \frac{1-\cos (nt)}{2\sin (t/2)} \end{aligned}$$

(1.4.3)

and

$$\begin{aligned} \sum _{k=0}^{n-1} \cos (k+1/2)t= \frac{\sin (nt)}{2\sin (t/2)}. \end{aligned}$$

(1.4.4)

Proof

Adding the equalities

$$ 2\sin (t/2) \sin (k+1/2)t = \cos (kt)- \cos (k+1)t $$

and

$$ 2\sin (t/2) \cos (k+1/2)t = \sin (k+1)t-\sin (kt), $$

we obtain the lemma.    $\blacksquare$

Lemma 1.4.4

For $n \ge 1$ and $t \in {\mathbb T}$, $t\ne 0$, we have

$$ K_{n}(t)= \frac{1}{n} \left( \frac{\sin (nt/2)}{\sin (t/2)} \right) ^2. $$

Proof

By Lemmas 1.2.3 and 1.4.2,

$$\begin{aligned} K_n(t)&= \frac{1}{n} \sum _{k=0}^{n-1} D_k(t) = \frac{1}{n} \sum _{j=0}^{n-1} \frac{\sin ((k+1/2)t)}{\sin (t/2)}. \end{aligned}$$

The lemma follows from (1.4.3).    $\blacksquare$

Corollary 1.4.5

For $n \ge 1$ and $- \pi \le t \le \pi$, $t\ne 0$,

$$ |K_n(t)|\le 2n-1 \qquad \text{ and } \qquad |K_n(t)|\le \frac{C}{n|t|^{2}}. $$

Proof

The inequalities follow from Lemmas 1.2.4 and 1.4.4.    $\blacksquare$

Now we generalize the Fejér summability.

Definition 1.4.6

For  $\alpha \ne -1,-2,\ldots$, let $A_{-1}^{\alpha }:=0$ and

$$ A_{n}^{\alpha } := \left( {\begin{array}{c}n+\alpha \\ n\end{array}}\right) = \frac{(\alpha +1)(\alpha +2)\cdots (\alpha +n)}{n!} \qquad (n \in {\mathbb N}). $$

Obviously, $A_0^{\alpha }=1$ and if $\alpha =0$, then $A_n^{0}=1$, if $\alpha =1$, then $A_n^{1}=n+1$ $(n \in {\mathbb N})$.

Lemma 1.4.7

For any $n\in {\mathbb N}$, $\alpha ,\beta \ne -1,-2,\ldots$, we have

$$\begin{aligned} A_n^{\alpha +\beta +1} = \sum _{k=0}^n A_{k}^{\alpha } A_{n-k}^{\beta }. \end{aligned}$$

Proof

It is known that, for any $x \in {\mathbb C}$, $|x|<1$,

$$ (1-x)^{- \alpha -1} = \sum _{n=0}^{\infty } \left( {\begin{array}{c}- \alpha -1\\ n\end{array}}\right) (-x)^{n}. $$

From this, it follows that

$$\begin{aligned} (1-x)^{- \alpha -1} = \sum _{n=0}^{\infty } \left( {\begin{array}{c}n+(-n- \alpha -1)\\ n\end{array}}\right) (-x)^{n} = \sum _{n=0}^{\infty } A_n^{\alpha } x^{n}. \end{aligned}$$

(1.4.5)

Similarly,

$$ (1-x)^{- \beta -1} = \sum _{n=0}^{\infty } A_n^{\beta } x^{n} $$

and

$$ (1-x)^{- \alpha - \beta -2} = \sum _{n=0}^{\infty } A_n^{\alpha +\beta +1} x^{n}. $$

However, the last series can be obtained also by multiplying the first two:

$$ (1-x)^{- \alpha - \beta -2}=(1-x)^{- \alpha -1}(1-x)^{-\beta -1} = \sum _{n=0}^{\infty } \left( \sum _{k=0}^{n} A_k^{\alpha } A_{n-k}^{\beta }\right) x^{n}, $$

which implies the desired result.    $\blacksquare$

Lemma 1.4.8

For any $n\in {\mathbb N}$, $\alpha \ne -1,-2,\ldots$, we have

$$\begin{aligned} A_n^\alpha = \sum _{k=0}^n A_{k}^{\alpha -1}, \qquad A_n^\alpha -A_{n-1}^\alpha = A_n^{\alpha -1}. \end{aligned}$$

Proof

We obtain the first equality by replacing $\alpha$ by $\alpha -1$ and $\beta$ by 0 in Lemma 1.4.7. The second one follows easily from the first one.    $\blacksquare$

Lemma 1.4.9

For any $n,N\in {\mathbb N}$, $-N<\alpha \le N$ and $\alpha \ne -1,-2,\ldots$, there exist $c_N, C_N>0$ such that

$$\begin{aligned} c_N n^\alpha< |A_{n}^{\alpha }| <C_N n^\alpha . \end{aligned}$$

(1.4.6)

Proof

By Taylor’s formula, for $x \in (-1,1)$ there exists $\xi \in (0,x)$ such that

$$ \ln (1+x)=x- \frac{1}{2(1+\xi )^{2}} x^{2}. $$

This implies that $\ln (1+x)=x+O(x^{2})$ if $-N/(N+1)<x<1$. Then

$$\begin{aligned} \ln |A_n^{\alpha }|&= \sum _{k=1}^{n} \ln \left| 1+\frac{\alpha }{k}\right| \\&= \sum _{k=1}^{N} \ln \left| 1+\frac{\alpha }{k}\right| + \alpha \sum _{k=N+1}^{n} \frac{1}{k} + \alpha ^{2} \sum _{k=N+1}^{n} O \left( \frac{1}{k^{2}}\right) \\&= \sum _{k=1}^{N} \ln \left| 1+\frac{\alpha }{k}\right| +\alpha \Big ( \ln n +O(1) \Big ) + \alpha ^{2} O(1), \end{aligned}$$

that is,

$$ A_n^{\alpha } = n^{\alpha } O(1). $$

This proves the lemma.    $\blacksquare$

Definition 1.4.10

For $f\in L_1({\mathbb T})$, $n \in {\mathbb N}$ and $\alpha \ge 0$, the nth Cesàro means $\sigma _n^{\alpha }f$ of the Fourier series of f and the nth Cesàro kernel $K_n^{\alpha }$ are introduced by

$$ \sigma _n^{\alpha } f(x) := \frac{1}{A_{n-1}^{\alpha }} \sum _{k=-n}^n A_{n-1-|k|}^{\alpha } \widehat{f}(k) e^{\imath k x} $$

and

$$ K_n^{\alpha }(t) := \frac{1}{A_{n-1}^{\alpha }} \sum _{k=-n}^n A_{n-1-|k|}^{\alpha } e^{\imath k t}, $$

respectively.

Note that the Cesàro means are also called $(C,\alpha )$-means.  Obviously, for $\alpha =1$, we get back the Fejér means and for $\alpha =0$, the partial sums. The definition of the Cesàro kernels implies

Lemma 1.4.11

For $\alpha \ge 0$ and $n \in {\mathbb N}$, we have

$$ \frac{1}{2\pi }\int _{{\mathbb T}} K_n^{\alpha }(t) \, dt =1. $$

One can see that

$$\begin{aligned} \sigma _n^{\alpha } f(x)&= \frac{1}{2\pi } \frac{1}{A_{n-1}^{\alpha }} \sum _{k=-n}^n A_{n-1-|k|}^{\alpha } \int _{{\mathbb T}} f(t) e^{\imath k(x-t)} \, dt \nonumber \\&= \frac{1}{2\pi }\int _{{\mathbb T}} f(x-t) K_n^{\alpha }(t) \, dt \qquad (n \in {\mathbb N}). \end{aligned}$$

(1.4.7)

Lemma 1.4.12

For $f\in L_1({\mathbb T})$, $\alpha >0$ and $n \in {\mathbb N}$, we have

$$ \sigma _n^{\alpha } f(x) = \frac{1}{A_{n-1}^{\alpha }} \sum _{j=0}^{n-1} A_{n-1-j}^{\alpha -1} s_{j} f(x) $$

and

$$ K_n^{\alpha }(t) = \frac{1}{A_{n-1}^{\alpha }} \sum _{j=0}^{n-1} A_{n-1-j}^{\alpha -1} D_j(t). $$

Proof

By Lemma 1.4.8,

$$\begin{aligned} K_n^{\alpha }(t)= & {} \frac{1}{A_{n-1}^{\alpha }} \sum _{k=-n}^n A_{n-1-|k|}^{\alpha } e^{\imath k t}\nonumber \\= & {} \frac{1}{A_{n-1}^{\alpha }} \sum _{k=-n}^n \sum _{j=|k|}^{n-1} A_{n-1-j}^{\alpha -1} e^{\imath kt} \\= & {} \frac{1}{A_{n-1}^{\alpha }} \sum _{j=0}^{n-1} A_{n-1-j}^{\alpha -1} D_j(t), \end{aligned}$$

which shows the lemma.    $\blacksquare$

The following lemma shows that if the $(C,\alpha )$ means $(\alpha >-1)$ are convergent then the $(C,\alpha +h)$ means $(h>0)$ are convergent, too.

Lemma 1.4.13

For $\alpha >-1$ and $h>0$, we have

$$ \sigma _n^{\alpha +h}f = \frac{1}{A_{n-1}^{\alpha +h}} \sum _{k=1}^{n} A_{n-k}^{h-1} A_{k-1}^{\alpha } \sigma _{k}^{\alpha } f. $$

Proof

Indeed, by Lemmas 1.4.7 and 1.4.12,

$$\begin{aligned} \frac{1}{A_{n-1}^{\alpha +h}} \sum _{k=1}^{n} A_{n-k}^{h-1} A_{k-1}^{\alpha } \sigma _{k}^{\alpha } f&= \frac{1}{A_{n-1}^{\alpha +h}} \sum _{k=1}^{n} A_{n-k}^{h-1} \sum _{j=0}^{k-1} A_{k-1-j}^{\alpha -1} s_{j} f \\&= \frac{1}{A_{n-1}^{\alpha +h}} \sum _{j=0}^{n-1} s_{j} f \sum _{k=j+1}^{n} A_{n-k}^{h-1} A_{k-1-j}^{\alpha -1} \\&= \frac{1}{A_{n-1}^{\alpha +h}} \sum _{j=0}^{n-1} A_{n-1-j}^{\alpha +h-1} s_{j} f, \end{aligned}$$

which gives the result.    $\blacksquare$

For Cesàro means, instead of inequalities (1.4.3) and (1.4.4), we will use the following lemma.

Lemma 1.4.14

For $0<\alpha \le 1$, $n \ge 1$, and $t \in {\mathbb T}$, $t\ne 0$,

$$\begin{aligned} \left| \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} \sin ((k+1/2)t) \right| \le \frac{C}{|\sin (t/2)|^\alpha } + \frac{Cn^{\alpha -1}}{|\sin (t/2)|} \end{aligned}$$

(1.4.8)

and

$$\begin{aligned} \left| \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} \cos ((k+1/2)t) \right| \le \frac{C}{|\sin (t/2)|^\alpha } + \frac{Cn^{\alpha -1}}{|\sin (t/2)|}. \end{aligned}$$

(1.4.9)

Proof

The inequalities follow from Lemma 1.4.3 for $\alpha =1$. Let $0<\alpha <1$. Suppose that $- \pi \le t \le \pi$. Then

$$ \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} \sin ((k+1/2)t)= \mathfrak {I}\left( \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} e^{\imath (k+1/2)t}\right) $$

and

$$\begin{aligned} \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} e^{\imath (k+1/2)t}&= e^{\imath (n-1/2)t}\sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} e^{\imath (k+1-n)t} \nonumber \\&= e^{\imath (n-1/2)t} \sum _{j=0}^{n-1} A_{j}^{\alpha -1} e^{-\imath jt}, \end{aligned}$$

(1.4.10)

where $\mathfrak {I}$ denotes the imaginary part of the function. We know that

$$\begin{aligned} \sum _{j=0}^{\infty } A_{j}^{\alpha -1} x^j = (1-x)^{-\alpha } \end{aligned}$$

(1.4.11)

for $x \in {\mathbb C}$, $|x|<1$. However, this holds also for $|x|=1$, $x \ne 1$. Indeed, $A_n^{\alpha -2} \le 0$ and so by Lemma 1.4.8, $(A_n^{\alpha -1})_{n \in {\mathbb N}}$ is non-increasing. The left-hand side of (1.4.11) is convergent because Abel rearrangement implies that

$$\begin{aligned} \left| \sum _{j=n}^{m} A_{j}^{\alpha -1} x^j \right|&= \left| \sum _{j=n}^{m-1} \left( A_{j}^{\alpha -1}-A_{j+1}^{\alpha -1}\right) \left( \sum _{i=n}^{j}x^i\right) + A_{m}^{\alpha -1} \left( \sum _{i=n}^{m}x^i\right) \right| \nonumber \\&\le A_n^{\alpha -1} \sup _{n \le j \le m} \left| \frac{x^{n}-x^{j+1}}{1-x} \right| \nonumber \\&\le \frac{2A_n^{\alpha -1}}{|1-x|} \rightarrow 0 \end{aligned}$$

(1.4.12)

as $n\rightarrow \infty$. The last convergence follows from Lemma 1.4.9. Similarly, if $0<r<1$ is near to 1, say $r_0<r<1$, then

$$ \sum _{j=0}^{\infty } A_{j}^{\alpha -1} r^{j}x^j = (1-rx)^{-\alpha } $$

and

$$\begin{aligned} \left| \sum _{j=n}^{m} A_{j}^{\alpha -1} r^{j}x^j \right|&\le \frac{2A_n^{\alpha -1}}{|1-rx|} \le \frac{4A_n^{\alpha -1}}{|1-x|} \rightarrow 0 \end{aligned}$$

for $|x|=1$, $x \ne 1$. This implies that

$$\begin{aligned}&\left| (1-x)^{-\alpha } - \sum _{j=0}^{n-1} A_{j}^{\alpha -1} x^j \right| \le \left| (1-x)^{-\alpha } - (1-rx)^{-\alpha } \right| \\&\qquad +\left| (1-rx)^{-\alpha } - \sum _{j=0}^{n-1} A_{j}^{\alpha -1} r^{j}x^j\right| + \left| \sum _{j=0}^{n-1} A_{j}^{\alpha -1} r^{j}x^j-\sum _{j=0}^{n-1} A_{j}^{\alpha -1} x^j\right| \\&\qquad< 2 \epsilon +\left| (1-rx)^{-\alpha } - \sum _{j=0}^{n-1} A_{j}^{\alpha -1} r^{j}x^j\right| <3 \epsilon \end{aligned}$$

if r is near enough to 1 and n is large enough. Thus (1.4.11) is true for $|x| \le 1$, $x \ne 1$.

Using (1.4.10), (1.4.11), (1.4.12), and (1.4.6), we get that

$$\begin{aligned} \left| \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} e^{\imath (k+1/2)t} \right|&= \left| (1-e^{-\imath t})^{-\alpha } - \sum _{j=n}^{\infty } A_{j}^{\alpha -1} e^{-\imath jt} \right| \\&\le |(1-e^{-\imath t})^{-\alpha }| + 2 A_{n}^{\alpha -1}|(1-e^{-\imath t})^{-1}|\\&\le \frac{C}{|t|^\alpha } + \frac{Cn^{\alpha -1}}{|t|}, \end{aligned}$$

which proves (1.4.8). Inequality (1.4.9) can be handled similarly.    $\blacksquare$

The derivatives of the left-hand sides of (1.4.8) and (1.4.9) can be estimated as follows.

Lemma 1.4.15

For $0<\alpha \le 1$, $n \ge 1$ and $t \in {\mathbb T}$, $t\ne 0$,

$$ \left| \sum _{k=0}^{n-1} k A_{n-1-k}^{\alpha -1} \sin ((k+1/2)t) \right| \le \frac{Cn}{|\sin (t/2)|^\alpha } + \frac{Cn^{\alpha }}{|\sin (t/2)|} $$

and

$$ \left| \sum _{k=0}^{n-1} kA_{n-1-k}^{\alpha -1} \cos ((k+1/2)t) \right| \le \frac{Cn}{|\sin (t/2)|^\alpha } + \frac{Cn^{\alpha }}{|\sin (t/2)|}. $$

Proof

We apply Lemma 1.4.14 to obtain

$$\begin{aligned} \left| \sum _{k=0}^{n-1} k A_{n-1-k}^{\alpha -1} \sin ((k+1/2)t) \right|&\le \sum _{j=1}^{n-1} \left| \sum _{k=j}^{n-1} A_{n-1-k}^{\alpha -1} \sin ((k+1/2)t) \right| \\&\le \sum _{j=1}^{n-1}\left( \frac{C}{|\sin (t/2)|^\alpha } + \frac{Cj^{\alpha -1}}{|\sin (t/2)|}\right) \\&\le \frac{Cn}{|\sin (t/2)|^\alpha } + \frac{Cn^{\alpha }}{|\sin (t/2)|}. \end{aligned}$$

The second inequality can be shown in the same way.    $\blacksquare$

The following theorem will be used several times in this book. It can also be found in Zygmund [367].

Theorem 1.4.16

For $0<\alpha \le 1$, $n \ge 1$ and $- \pi \le t \le \pi$, $t\ne 0$,

$$\begin{aligned} \left| K_n^\alpha (t) \right| \le 2n-1 \qquad \text{ and } \qquad \left| K_n^\alpha (t) \right| \le \frac{C}{n^{\alpha }|t|^{\alpha +1}}. \end{aligned}$$

(1.4.13)

Proof

For $\alpha =1$, the theorem is exactly Corollary 1.4.5. Let $0<\alpha <1$. The first inequality follows from Lemma 1.2.4 and 1.4.12. By Lemmas 1.2.3 and 1.4.12,

$$\begin{aligned} K_n^{\alpha }(t)&= \frac{1}{A_{n-1}^{\alpha }} \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} D_k(t) \\&= \frac{1}{A_{n-1}^{\alpha }} \sum _{k=0}^{n-1} A_{n-1-k}^{\alpha -1} \frac{\sin ((k+1/2)t)}{\sin (t/2)}. \end{aligned}$$

Now, by Lemma 1.4.14,

$$ |K_n^{\alpha }(t)| \le \frac{1}{A_{n-1}^{\alpha }|\sin (t/2)|} \left( \frac{C}{|t|^\alpha } + \frac{Cn^{\alpha -1}}{|t|}\right) \le \frac{C}{n^{\alpha }|t|^{\alpha +1}} + \frac{C}{n |t|^{2}}. $$

If $|t| \ge 1/n$, then

$$ \frac{1}{n |t|^{2}} \le \frac{1}{n^{\alpha }|t|^{\alpha +1}}. $$

If $|t| < 1/n$, then the first inequality of (1.4.13) implies the second one.    $\blacksquare$

1.5 Convergence at Lebesgue Points of the Cesàro Means

Now we are ready to generalize Lebesgue’s theorem given in (1.4.2) for Cesàro summability. But first we introduce the Herz spaces which, as we will see later, are very closely connected to the concept of Lebesgue points.

Definition 1.5.1

The Herz space $E_\infty ({\mathbb T})$ contains all functions f for which

$$ \left\| f\right\| _{E_\infty }:= \sum _{k=-\infty }^0 2^{k} \left\| f 1_{P_k}\right\| _\infty <\infty , $$

where $P_k:=I(0,2^k \pi )\setminus I(0,2^{k-1} \pi )$, $(k\in {\mathbb Z})$.

Recall that $I(x,h):=\{y\in {\mathbb T}: |x-y|<h\}$. The Cesàro kernels are all in $E_\infty ({\mathbb T})$ for $0<\alpha \le 1$.

Theorem 1.5.2

If $0<\alpha \le 1$, then $K_n^{\alpha } \in E_\infty ({\mathbb T})$ and

$$ \sup _{n \in {\mathbb N}} \left\| K_n^{\alpha }\right\| _{E_\infty } \le C_\alpha . $$

Proof

By Theorem 1.4.16, $|K_n^{\alpha }(t)| \le g_n^{\alpha }(t)$, where

$$ g_n^{\alpha }(t):= C \min \left( n,\frac{1}{n^{\alpha }|t|^{\alpha +1}}\right) . $$

Since $g_n^{\alpha }$ is non-increasing and integrable, we obtain

$$\begin{aligned} \left\| K_n^{\alpha }\right\| _{E_\infty }&= \sum _{k=-\infty }^0 2^{k} \left\| K_n^{\alpha } 1_{P_k}\right\| _\infty \le C_\alpha \int _{{\mathbb T}} g_n^{\alpha }(t)\, dt \\&\le C_\alpha \int _{0}^{1/n} n \, d \lambda + C_\alpha n^{-\alpha }\int _{1/n}^{\pi } |t|^{-\alpha -1} \, dt \le C_\alpha , \end{aligned}$$

which shows the desired result.    $\blacksquare$

In the same way, we obtain

Corollary 1.5.3

If $0<\alpha \le 1$, then $K_n^{\alpha } \in L_1({\mathbb T})$ and

$$ \sup _{n \in {\mathbb N}} \left\| K_n^{\alpha }\right\| _{1} \le C_\alpha . $$

Theorem 1.5.4

If $0<\alpha <\infty$, then

$$ \lim _{n\rightarrow \infty }\sigma _n^\alpha f(x) = f(x) $$

for all Lebesgue points of $f\in L_1({\mathbb T})$.

Proof

First suppose that $0<\alpha \le 1$. Set

$$ G(u):= \int _{-u}^{u} |f(x-t)-f(x)| \, dt \qquad (u>0). $$

Since x is a Lebesgue point of f, for all $\epsilon >0$, there exists $m\in {\mathbb Z}$ such that

$$\begin{aligned} \frac{G(u)}{2u} \le \epsilon \qquad \text{ if } \qquad 0<u\le 2^{m}. \end{aligned}$$

(1.5.1)

It follows from Lemma 1.4.11 and (1.4.7) that

$$ \sigma _n^{\alpha } f(x)-f(x) = \frac{1}{2\pi }\int _{{\mathbb T}} (f(x-t)-f(x)) K_n^{\alpha }(t) \, dt. $$

Thus

$$\begin{aligned} |\sigma _n^{\alpha } f(x)-f(x)|\le & {} C \int _{{\mathbb T}^d} |f(x-t)-f(x)| \left| K_n^{\alpha }(t) \right| \, dt \\= & {} C \int _{-2^{m}\pi }^{2^{m}\pi } |f(x-t)-f(x)| \left| K_n^{\alpha }(t) \right| \, dt \\&{}+ C \int _{{\mathbb T}\setminus (-2^{m}\pi ,2^{m}\pi )} |f(x-t)-f(x)| \left| K_n^{\alpha }(t) \right| \, dt \\=: & {} A_1(x)+A_2(x). \end{aligned}$$

We estimate $A_1(x)$ by

$$\begin{aligned} A_1(x)= & {} C\sum _{k=-\infty }^{m} \int _{P_{k}} |f(x-t)-f(x)| \left| K_n^{\alpha }(t) \right| \, \, dt\\\le & {} C\sum _{k=-\infty }^{m} \sup _{P_{k}} \left| K_n^{\alpha } \right| \int _{P_{k}} |f(x-t)-f(x)| \, dt \\\le & {} C\sum _{k=-\infty }^{m} \sup _{P_{k}} \left| K_n^{\alpha } \right| \, G(2^{k}\pi ). \end{aligned}$$

Then, by (1.5.1),

$$ A_1(x) \le C \epsilon \sum _{k=-\infty }^{m} 2^{k} \sup _{P_{k}} \left| K_n^{\alpha } \right| \le C\epsilon \left\| K_n^{\alpha }\right\| _{{E}_\infty ({\mathbb T})} \le C\epsilon . $$

On the other hand, Theorem 1.4.16 implies

$$\begin{aligned} A_2(x)\le & {} C \sup _{{\mathbb T}\setminus (-2^{m}\pi ,2^{m}\pi )} \left| K_n^\alpha \right| \int _{{\mathbb T}\setminus (-2^{m}\pi ,2^{m}\pi )} |f(x-t)-f(x)| \, dt \\\le & {} \frac{C}{n^{\alpha } 2^{m(\alpha +1)}} \left( \Vert f\Vert _1+|f(x)|\right) , \end{aligned}$$

which tends to 0 as $n\rightarrow \infty$. Finally, for $1< \alpha <\infty$, the result follows from Lemma 1.4.13.    $\blacksquare$

We can weaken the definition of Lebesgue points and we can suppose that

$$\begin{aligned} \lim _{h\rightarrow 0+0} \frac{1}{h} \int _{0}^{h} \left| f(x-t)+f(x+t)-2f(x)\right| \,dt =0. \end{aligned}$$

(1.5.2)

By a triangle inequality, it is clear that if x is a Lebesgue point then (1.5.2) holds. The following result can be proved similar to Theorem 1.5.4.

Theorem 1.5.5

If $0<\alpha <\infty$, $f\in L_1({\mathbb T})$ and (1.5.2) holds for a point $x \in {\mathbb T}$, then

$$ \lim _{n\rightarrow \infty }\sigma _n^\alpha f(x) = f(x). $$

Proof

Using (1.4.7), Lemma 1.4.11 and that the function $K_n^{\alpha }$ is even, we obtain

$$\begin{aligned} \sigma _n^{\alpha } f(x)-f(x)&= \frac{1}{2\pi }\int _{-\pi }^{\pi } (f(x-t)-f(x)) K_n^{\alpha }(t) \, dt \\&= \frac{1}{2\pi }\int _{0}^{\pi } (f(x-t)+f(x+t)-2f(x)) K_n^{\alpha }(t) \, dt. \end{aligned}$$

Thus

$$\begin{aligned} |\sigma _n^{\alpha } f(x)-f(x)|&= \frac{1}{2\pi }\int _{0}^{\pi } \left| f(x-t)+f(x+t)-2f(x) \right| \left| K_n^{\alpha }(t) \right| \, dt \end{aligned}$$

and the proof can be finished as in Theorem 1.5.4.    $\blacksquare$

Now we can generalize Fejér’s theorem given in (1.4.1).

Corollary 1.5.6

Suppose that $0<\alpha <\infty$, $f\in L_1({\mathbb T})$ and that the left and right limits $f(x-0)$ and $f(x+0)$ exist at a point x. Then

$$ \lim _{n\rightarrow \infty }\sigma _n^\alpha f(x) = \frac{f(x-0)+f(x+0)}{2}. $$

Proof

Choosing

$$ f(x):=\frac{f(x-0)+f(x+0)}{2}, $$

we can easily see that (1.5.2) holds. The corollary follows from Theorem 1.5.5.    $\blacksquare$

If f is continuous at a point x, then we get

Corollary 1.5.7

Suppose that $0<\alpha <\infty$, $f\in L_1({\mathbb T})$ and f is continuous at a point x. Then

$$ \lim _{n\rightarrow \infty }\sigma _n^\alpha f(x) = f(x). $$

In the next theorem, we verify the norm convergence of the Cesàro means.

Theorem 1.5.8

Suppose that $0<\alpha <\infty$ and $1 \le p< \infty$. If $f\in L_p({\mathbb T})$, then

$$ \sup _{n \in {\mathbb N}} \left\| \sigma _{n}^{\alpha }f\right\| _p \le C_\alpha \left\| f\right\| _p $$

and

$$ \lim _{n\rightarrow \infty } \sigma _{n}^{\alpha }f=f \qquad \text{ in } \text{ the } L_p({\mathbb T})\text {-norm}. $$

Proof

Again, it is enough to show the result for $0< \alpha \le 1$. By (1.4.7), Minkowski inequality and Corollary 1.5.3,

$$ \left\| \sigma _n^\alpha f(\cdot )\right\| _p \le \frac{1}{2\pi } \int _{{\mathbb T}} \left\| f(\cdot -t) \right\| _p \left| K_n^{\alpha }(t)\right| \, dt \le C_\alpha \Vert f\Vert _p. $$

The convergence obviously holds for all trigonometric polynomials and so it holds also for all $f\in L_p({\mathbb T})$ $(1 \le p<\infty )$ by density.    $\blacksquare$

We get the next corollary with the same proof.

Corollary 1.5.9

If $0<\alpha <\infty$ and $f\in C({\mathbb T})$, then

$$ \sup _{n \in {\mathbb N}} \left\| \sigma _{n}^{\alpha }f\right\| _\infty \le C_\alpha \left\| f\right\| _\infty $$

and

$$ \lim _{n\rightarrow \infty } \sigma _{n}^{\alpha }f=f \qquad \text{ uniformly. } $$