\documentclass[reqno,11pt,letterpaper]{amsart}
% math packages
\usepackage{amsmath,amssymb,amsthm,mathtools,amsfonts,cases}

% color and hyperlink
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage[colorlinks=true,linkcolor=Red,citecolor=Green]{hyperref}

% Draw pictures
\usepackage{tikz}
\usepackage{siunitx}

% \usepackage{wasysym} % for "\varhexagon" macro
% \usepackage{graphicx}% for "\scalebox" macro
% \usepackage{subcaption} % for subpicture

% mathbb{1}
\usepackage{bbold}


% new def
\def\smallsection#1{\smallskip\noindent\textbf{#1}.}
\def\indic{\operatorname{1\hskip-2.75pt\relax l}}
\let\epsilon=\varepsilon % sorry Knuth

% newcommand notations
\newcommand{\RR}{{\mathbb R}}
\newcommand{\NN}{{\mathbb N}}
\newcommand{\CC}{{\mathbb C}}
\newcommand{\QQ}{{\mathbb Q}}
\newcommand{\TT}{{\mathbb T}}
\newcommand{\ZZ}{{\mathbb Z}}
\newcommand{\EE}{{\mathbb E}}
\newcommand{\ML}{{\mathcal{L}}}
\newcommand{\MT}{{\mathcal{T}}}
\newcommand{\msH}{{\mathscr H}}
\newcommand{\msV}{{\mathscr V}}
\newcommand{\msU}{{\mathscr U}}
\newcommand{\msG}{{\mathscr G}}
\newcommand{\msE}{{\mathscr E}}
\newcommand{\bfn}{{\textbf{n}}}
\newcommand{\Spec}{{\operatorname{Spec}}}

\DeclareMathOperator{\const}{Const}
\DeclareMathOperator{\diag}{Diag}
\DeclareMathOperator{\Imag}{Im}
\let\Im = \Imag
\DeclareMathOperator{\Real}{Re}
\let\Re = \Real
\DeclareMathOperator{\sgn}{Sgn}
\DeclareMathOperator{\supp}{Supp}
\DeclareMathOperator{\Tr}{Tr}

% newcommand operators
\newcommand{\floor}[1]{\lfloor #1 \rfloor}
\newcommand{\ceil}[1]{\lceil #1 \rceil}

% newcommand for authors 
\newcommand{\xz}[1]{\color{JungleGreen}{#1}}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Specific for this project

% Operators

\newcommand{\Lax}{{\Lambda_{\alpha,x}}}
\newcommand{\Hax}{{H_{\alpha,x}}}

\newcommand{\tpsi}{\widetilde\psi}
\newcommand{\tPsi}{\widetilde\Psi}
\newcommand{\ntwo}{\floor{\frac{n_1 - x + 1}{\alpha}}}
\newcommand{\ax}{a^{(x)}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Define environment
\newtheorem{theorem}{Theorem}
\newtheorem{proposition}{Proposition}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{example}{Example}
\newtheorem{ex}{Exercise}
\newtheorem{lemma}[proposition]{Lemma}
\newtheorem{corollary}[proposition]{Corollary}
\newtheorem{remark}{Remark}[section]
% \newtheorem{conjection}{Conjecture}
% \newtheorem{assumption}{Assumption}
% \newtheorem{notation}{Notation}
\newtheorem*{claim}{Claim}

\numberwithin{equation}{section}
\mathtoolsset{showonlyrefs}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title[Harmonic analysis]{Harmonic analysis}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

% \begin{abstract}
% \end{abstract}
\maketitle
\section{Basic facts about Fourier transform}
We denote the torus by $\mathbb T^d:=\mathbb R^d/(2\pi \ZZ)^d$ with Lebesgue measure.
For $f \in L^1(\mathbb T^d)$ we define the Fourier transform for $n \in \ZZ^d$
\[ \hat{f}(n):=(2\pi)^{-d} \int_{\mathbb T^d} f(x) e^{-ixn} \ dx.\]
The inverse Fourier transform is defined for $f_n \in \ell^1(\mathbb Z^d)$ by 
\[ \check{f}(x):=\sum_{n \in \mathbb Z^d} f_n e^{ixn}.\]

\begin{theorem}[Parseval theorem]
Let $f \in L^2(\mathbb T^d)$, then $\hat{f} \in \ell^2(\ZZ^d)$ and 
\[ \Vert \hat f \Vert_{\ell^2}^2 = (2\pi)^{-d} \Vert f \Vert_{L^2}^2.\]
Moreover, the Fourier transform is bijective.
\end{theorem}
 \begin{ex}
 Give a proof of this result using the Stone-Weierstrass theorem.
  \end{ex}

Basic question in Harmonic Analysis: Can we get rates on the convergence?

We claim that for any Lipschitz function $f : \mathbb T \to \mathbb C$ we have 
\[ \vert \hat{f} (n)\vert \le \vert n \vert^{-1} \Vert f \Vert_{\text{Lip}},\]
where $\Vert f \Vert_{\text{Lip}} := \sup_{x \neq y} \frac{\vert f(x)-f(y)\vert}{\vert x-y \vert}.$
This follows from writing 
\[ \begin{split} 2\pi \hat{f}(n) 
&= \frac{1}{2} \int_{\mathbb T} e^{-inx} f(x) + e^{-in(x+\pi/n)} f(x+\pi/n) \ dx\\
&= \frac{1}{2} \int_{\mathbb T} (f(x)-f(x+\pi/n)) e^{-i nx } \ dx.
 \end{split}
 \]
 \begin{ex}
 Generalize the above argument to arbitrary dimensions!
 \end{ex}
 More generally, let $f$ be a H\"older function, i.e. 
 \[ \Vert f \Vert_{\Lambda_{\alpha}}:= \sup_{x \neq y} \frac{\vert f(x)-f(y)\vert}{\vert x-y \vert^{\alpha}},\]
 then $\hat{f}(n)=\mathcal O(1/\vert n \vert^{\alpha}).$

We define the continuous Fourier transform as 
\[ \hat{f}(\xi):=\int_{\mathbb R^d} f(x)e^{-i x \xi} \ d\xi\]
and also 
\[ \check{f}(\xi):=\int_{\mathbb R^d} f(x)e^{i x \xi} \ d\xi.\]
\begin{theorem}[Plancherel theorem]
For any $f \in L^2(\mathbb R^d)$, $\Vert \hat f \Vert_2 = (2\pi)^{d/2} \Vert f \Vert_2.$ Moreover, $f = (2\pi)^{-d} \check{\hat{f}}. $Finally, $f \mapsto (2\pi)^{-d/2} \hat{f}$ is an isometry of $L^2(\mathbb R^d)$ onto $L^2(\mathbb R^d).$
\end{theorem}





\subsection{Schwartz functions and the space of tempered distributions}
We start by giving the definition of the Schwartz space
\begin{definition}
$\mathcal S$ is the space of all $f \in C^{\infty}(\mathbb R^d)$ for which each of the following quantities 
\[ \Vert f \Vert_{m,n} :=\sup_{x \in \mathbb R^d} \langle x \rangle^n \sum_{\vert \alpha \vert \le m} \vert \partial^{\alpha}f(x)\vert\]
is finite. 
\end{definition}
It is complete metrizable space $d(f,g) = \sum_{m,n} 2^{-\vert (m,n)\vert} \frac{\Vert f-g\Vert_{m,n}}{1+ \Vert f-g\Vert_{m,n}}$ whose locally convex topology is defined by finite intersections of open sets 
\[ V_{f,m,n,\varepsilon} :=\{ g\in \mathcal S; \Vert f-g\Vert_{m,n}\le \varepsilon\}. \]
Thus, if $f_n \to g$ in $\mathcal S$ then this is equivalent to having $\Vert f_n-g\Vert_{m,n}\to 0$ for all $m,n.$
\begin{definition}
The set  $\mathcal S'$ is the space of all continuous linear functionals from $\mathcal S$ to $\mathbb C.$
\end{definition}
We can find a topology such that $\varphi_j \to \varphi$ in $\mathcal S'$ is equivalent to $\varphi_j(f) \to\varphi(f)$ for every $f \in \mathcal S.$
\begin{definition}
$\mathcal S'$ is a topological vector space for which a neighbourhood base of $\varphi \in \mathcal S'$ is the collection of all finite intersections of sets 
\[ V_{\varphi, f,\varepsilon} :=\{ \psi \in \mathcal S': \vert (\varphi-\psi)(f)\vert \le \varepsilon \}.\]
\end{definition}
\begin{remark}
A locally finite Borel measure $\mu$ is called \emph{tempered} if there exist $C,N>0$ such that 
\[ \mu\{ x\in \mathbb R^d; \vert x \vert \le R\} \le C R^{N} \text{ as } R \to \infty.\]
The set of tempered $C^{\infty}$ functions $f$ with measure $d\mu = \vert f(x)\vert dx$ turns out to be dense in $\mathcal S'$. Hence, the space $\mathcal S'$ got its name: The space of tempered distributions.
\end{remark}

\begin{lemma} To any continuous linear transformation $T: \mathcal S \to \mathcal S$, there is an associated continuous linear transformation $T': \mathcal S' \to \mathcal S'$ defined by 
\[ (T'\varphi)(f) = \varphi(Tf).\]
\end{lemma}
\begin{proof}
By linearity, it suffices to show continuity at $0$. Let $V$ be a neighbourhood of $0$ in $\mathcal S'$. We must show there is a neighbourhood $U$ of $0 \in S'$ such that $T'(U) \subset V$.
There is $\varepsilon>0$ and finitely many $f_j \in \mathcal S$ such that $V \supset \cap_{j=1}^n V_j$ where 
\[V_j = \{ \varphi \in \mathcal S'; \vert \varphi(f_j)\vert < \varepsilon\}.\]
Define $U_j :=\{ \psi \in \mathcal S'; \vert \psi(T f_j) \vert <\varepsilon\}$ and $U:=\cap_j U_j.$
If $\psi \in U$ then $\vert T'(\psi(f_j))\vert = \vert \psi(T f_j)\vert <\varepsilon $ for each $j$ so $T'(\psi) \in V.$
\end{proof}
\begin{theorem}
The Fourier transform is a continuous homeomorphism from $\mathcal S$ onto $\mathcal S$ and therefore also between $\mathcal S'$ and $\mathcal S'.$
\end{theorem}
\begin{proof}
We assume basic familiarity with the Fourier transform on $\mathcal S$ which imply that it is bijective. 
To show continuity, we notice that 
\[ \xi^{\alpha} \partial_{\xi}^{\beta} \hat{f} = (-i)^{\vert \alpha \vert} \mathcal F(\partial_x^{\alpha}((-ix)^{\beta} f).\]
Moreover, 
\[  \vert \xi^{\alpha} \partial_{\xi}^{\beta} \hat{f} \vert \le \int_{\mathbb R^d} \langle x \rangle^{-d-1} \ dx \sup_{x} \langle x \rangle^{d+1} \vert \partial_x^{\alpha}(x^{\beta} f)(x)\vert\]
This shows that $\Vert \hat{f} \Vert_{k,k} \le C \Vert f \Vert_{k,k+d+1}.$ 
\end{proof}

Every $f \in L^p$ can be naturally associated with a tempered distribution by defining 
\[\varphi_f(g) = \int fg \]

We thus have that for $\delta_{\xi}(f)=f(\xi)$ that 
\[ \hat{\delta_{\xi}}(f)=\delta_{\xi}(\hat{f})= \hat{f}(\xi)=\int_{\RR^d} f(x)e^{-ix \xi} \ dx\]
which justifies $\hat{\delta_{\xi}} = e^{-ix \xi}.$

\subsection{Convolution}

Let $f \in L^1(\mathbb T)$ and we may ask if $f$ is equal to $\sum_{n \in \ZZ} \hat{f}(n) e^{inx}$, that is whether 
\[ S_N f(x) = \sum_{\vert n\vert\le N} \hat{f}(n) e^{inx}\] 
converges to $f$.
We compute 
\[\begin{split}
 S_N f(x) &=  \sum_{\vert n\vert\le N} e^{inx} (2\pi)^{-1} \int_0^{2\pi} f(y) e^{-iny} dy \\
 &=\sum_{\vert n\vert\le N} (2\pi)^{-1} \int_0^{2\pi} f(y) e^{-in(x-y)} dy \\
 &=(2\pi)^{-1} \int_0^{2\pi} f(y) D_N(x-y) \ dy,
 \end{split}
 \]
 where $D_N(x) = \sum_{\vert n\vert\le N} e^{inx} = \frac{\sin((N+\frac{1}{2})x)}{\sin(x/2)}$ is the \emph{Dirichlet kernel}.
 
 The convergence of the Fourier series is therefore equivalent to the asymptotic properties of a certain integral operator!
 
Using a suitable partitioning that we leave as an exercise, one readily verifies that even though $\int D_N = 2\pi$, the $L^1$ norm diverges as $N \to \infty.$

We might get our hopes up that the Dirichlet kernel is just an approximate identity which would imply converges of the Fourier series in every $L^p$ space with $p<\infty.$ However, 
\begin{ex}
There is $c>0$ such that 
\[ \Vert D_N\Vert_{L^1} \ge c \log(N) \text{ for all }N.\]
\end{ex}


 

On $\mathbb R^d$ we define the convolution by 
\[ (f* g)(x) = \int_{\mathbb R^d} f(x-y) g(y) \ dy.\]
In addition one has Young's inequality which states that for $1+1/r = 1/p+1/q$
\[ \Vert f*g \Vert_r \le \Vert f \Vert_p \Vert g \Vert_q.\]

Basic question in Harmonic Analysis: What is the sharpest constant in this inequality? What are the optimizers?\footnote{For Young's inequality this has been studied by Beckner and Brascamp-Lieb.}


Analogously, one defines a convolution on the torus by setting
\[ (f* g)(x) =(2\pi)^{-d} \int_{\mathbb T^d} f(x-y) g(y) \ dy.\]

If we want to convolve with a (finite) measure $\mu$ we replace $g(y) \ dy$ by $d\mu(y).$ The convolution of two measures is the measure 
\[ (\mu*\nu)(E):=(\mu \times \nu)\{ (x,y); x+y \in E\}\]
such that 
\[ \int f d(\mu*\nu)= \int \int f(x+y) \ d\mu(x) \ d\nu(y).\]
One can even convolve two tempered distributions, under one constraint: indeed let first $\varphi \in \mathcal S'$ and $f \in \mathcal S$ then 
\[ \varphi*f(x)=\varphi(f(x-\bullet))\]
is a $C^{\infty}$ function, e.g. $\delta^{(n)}_0*f(x)=f^{(n)}(x).$ However, in general this is not a Schwartz function, we therefore have to assume that $\psi$ is compactly supported\footnote{This means that $\psi(f)=0$ for all $f$ that are supported away from a certain compact set} and define 
\[(\varphi*\psi)(f):=\psi(\varphi*f(-x)).  \]
One application of convolutions are approximate identities 
\begin{definition}
A sequence $(\varphi_j)_j$ is called an approximate identity if 
\begin{itemize}
\item $\int \varphi_j=1$ 
\item $\Vert \varphi_j\Vert \le C<\infty$
\item $\int_{\vert x \vert>\delta} \vert \varphi_j(x)\vert \to 0 $ as $j \to \infty$ for all $\delta>0.$
\end{itemize}
\end{definition}
An examples are centered Gaussians whose variance tends to zero. In fact, any normalized positive $\phi \in L^1$ naturally induces an approximate identity by setting $\varphi_j(x)=j^d \phi(jx).$ Approximate identities have their name because of the following property
\begin{theorem}
\item For any $f \in C_0(\mathbb R^d)$ we have $f *\varphi_j \to f$ uniformly.
\item For any $f \in L^p(\mathbb R^d)$ we have $f*\varphi_j  \to f$ in $L^p$ for $p<\infty.$
\end{theorem}


\subsection{Uniform convergence of Fourier series} 
While we know that for $f \in C^0(\mathbb T^d)$ we get $L^2$ convergence of the Fourier series, we may ask whether for $f \in C^0(\mathbb T^d)$ we get uniform convergence. This is false by a very elegant observation argument using the uniform boundedness principle. 
\begin{theorem}
There exists $f \in C^0(\mathbb T)$ such that $(S_Nf(0))$ diverges.
\end{theorem}
\begin{proof}
If false, then define $\ell_n f = S_n f(0) \in \mathbb C$ for every $f \in C^0(\mathbb T)$. Also $\ell_n: \mathbb C^0 \to \mathbb C$ is a bounded linear functional, since 
\[ \ell_n f = (2\pi)^{-1} \int f(y) D_n(-y) \ dy\]
and $D_n \in L^1$. 
Banach-Steinhaus implies that if $\sup_{n} \vert \ell_n(f)\vert <C_f$ for every $f$, then $\sup_n \Vert \ell_n \Vert<\infty$. This is however false since $\Vert \ell_n \Vert \to \infty$ since $\Vert D_n \Vert_{L^1} \to \infty.$
\end{proof}
The situation improves by assuming slightly more regularity.
\begin{theorem}
For any $\alpha \in (0,1)$ and every $f \in \Lambda_{\alpha}$, $S_N f \to f$ uniformly as $N \to \infty.$

Moreover, there exists a constant $C_{\alpha}<\infty$ such that 
\[\Vert S_Nf -f\Vert_{\infty} \le CN^{-\alpha} \log(N) \Vert f \Vert_{\Lambda_{\alpha}}. \]
\end{theorem}
\begin{proof}
Writing 
\[ S_N(f)(x) -f(x) = (2\pi)^{-1} \int (f(x-y) - f(x) )D_N(y) \ dy,\]
we can decompose this integral into 
\[ \vert S_N(f)(x) -f(x)\vert \le C \int_{\vert y \vert<\delta} |y|^{-1} |f(x-y)-f(x)| dy +\vert \int_{\vert y \vert > \delta} \sin((N+1/2)y) g(y) \ dy\vert,\]
where $g(y)=(f(x-y)-f(x))/\sin(y/2).$
Setting $x=0$ for simplicity, the first term is majorized by choosing $\delta= \mathcal O(1/N)$ by 
\[ \int_{\vert y \vert \le \delta}  \Vert f \Vert_{\Lambda^{\alpha}} \vert y \vert^{\alpha-1} \ dy =\mathcal O( \Vert f \Vert_{\Lambda^{\alpha}}\delta^{\alpha} )= \mathcal O(\Vert f \Vert_{\Lambda^{\alpha}} N^{-\alpha}).\]

To estimate the second term, we observe that up to errors of order $\mathcal O(N^{-\alpha})$ it can be written using a substitution as 
\[  \int_{\vert y \vert> \delta} \sin((N+1/2)y) (g(y)-g(y-\pi/(N+1/2))) \ dy.\]
It remains to show that for $\vert y \vert \ge \delta$
\[ \vert g(y)-g(y-\pi/(N+1/2))\vert \le C N^{-\alpha}\vert y \vert^{-1}.\]
Let $y' = y-\pi/(N+1/2). $ Then, $(C_0-\pi)N^{-1} \le \vert y'\vert \le \pi+ \pi/N.$
This implies that using
\[ g(y)-g(y')= \frac{f(y)-f(0)}{\sin(y/2)} + \frac{f(0)-f(y')}{\sin(y'/2)}=\frac{f(y)-f(y')}{\sin(y/2)} + \Big( \frac{f(y')-f(0)}{\sin( y/2)-\sin(y'/2)}\Big)\]
and H\"older continuity in the second line
\[ \begin{split} \vert g(y)-g(y')\vert &\le \vert f(y)-f(y')\vert \vert \sin(y/2)\vert^{-1} + \vert f(y')-f(0)\vert \frac{\vert \sin(y/2)-\sin(y'/2)\vert}{\vert \sin(y/2)\sin(y'/2)\vert} \\ 
&\le C \Big(  \vert y-y'\vert^{\alpha} \vert y \vert^{-1} + \vert y' \vert^{\alpha} \frac{\vert y-y'\vert}{\vert y y'\vert}\Big).
\end{split}\]
Finally, since $\vert y-y'\vert \lesssim \vert y-y'\vert^{\alpha}N^{\alpha-1}$ the second term on the right is majorized by the first term. 





\end{proof}

\subsection{$L^p$ convergence of Fourier series}
We are now asking: Let $f \in L^p$ and $p \in [1,\infty]$, do we have $\Vert f - S_N(f)\Vert_{p} \to 0$ for all $f$?
The convergence fails for general $f \in C(\mathbb T).$
It also fails for $p=1$,since convergence in $L^1$ at least requires that 
$\sup_n \Vert S_n f\Vert<\infty$, but then by Banach-Steinhaus also $\sup_n \Vert S_n \Vert<\infty.$
However, the operator norm of $\Vert S_n \Vert = \Vert D_n \Vert \to \infty$ as $n \to \infty.$
Indeed, recall that by using a positive approximate identity $\Vert S_n \varphi_j  -D_n \Vert_{L^p} \to 0.$ Thus, for $p=1$ we have $\Vert S_n \Vert \ge \Vert D_n \Vert_{L^1}$, since $\Vert \varphi_j \Vert_{L^1} =1,$ and the converse inequality follows by Young. 




To see this recall that 
\[ S_nf(x) = \int D_n (x-y) f(y) \ dy.\]
where the first inequality follows from Young's inequality and the limit since $F_n$, the so-called Fej\'er kernel
\[F_n=(n+1)^{-1}\sum_{i=0}^n D_i = \frac{\sin((N+1)/2x)^2}{(n+1) \sin(x/2)^2} , \]
is an approximate identity, since it is positive and integrates up to $1$. In fact, the Fej\'er kernels form an approximate identity (show this!).
In particular, this implies that 
\[ \sigma_N = (N+1)^{-1} \sum_{n=0}^N S_n f \]
satisfy 
\begin{proposition}
For any $f \in C^0(\mathbb T)$ we have $\sigma_N f \to f$ uniformly as $N \to \infty.$ For any $p \in [1,\infty)$ and any $f \in L^p$ we have $\Vert \sigma_N f - f \Vert \to 0$ as $N \to\infty.$  
\end{proposition}

Since the convergence is true for $p=2$ (Parseval), we are left with studying $p\in (1,\infty)\setminus \{2\}.$

\subsection{$L^p$ convergence of the Fourier series}
\begin{definition}
A Banach lattice of measurable functions is a Banach space $X$ such that whenever $g \in X$, $f$ is measurable and $\vert f \vert \le \vert g \vert$ a.e., then $f \in X$ and $\Vert f \Vert_X \le \Vert g\Vert.$
\end{definition}
Examples of Banach lattices are the $L^p$ spaces, but Sobolev spaces are e.g. no Banach lattices in general. 



Let $\mathcal P$ denote the set of trigonometric polynomials and define 
\begin{definition}
We define the operator $P$ on $\mathcal P$ whose Fourier transform is the projection
\[\widehat{Pf}(n) = \begin{cases} \hat{f}(n) &n \ge 0 \\ 
0 & n<0. \end{cases}\]
\end{definition} 
An operator $T$ on $L^2(\mathbb T)$ satisfying 
\[ \widehat{Tf}(n) = a_n \hat{f}(n)\]
with $a_n$ a bounded sequence is called a Fourier multiplier operator and defines itself a bounded linear operator. 

It is closely related to the so-called discrete Hilbert transform $\hat{Hf}(n)=-i\sgn(n) \hat{f}(n)$ such that 
\[ \frac{1}{2}(I+iH)f = Pf - \frac{1}{2}\hat{f}(0).\]
Thus, $P$ extends to an $L^p$ bounded operator if and only if $H$ does. 

We then have 
\begin{proposition}
Let $X \subset L^1(\mathbb T)$ be a Banach lattice and suppose that $\mathcal P$ is dense in $X$. If $P: \mathcal P \to \mathcal P$ extends to a bounded linear operator $P:X \to X$ then $\Vert S_n f - f \Vert_X \to 0$ as $n \to \infty$ for every $f \in X.$

\end{proposition}

One then shows that 
\begin{theorem} 
$P$ and $H$ extend to bounded linear operators on $L^p(\mathbb T)$ for every $p \in (1,\infty).$
\end{theorem}

\begin{corollary} 
Let $p \in (1,\infty)$ then the Fourier series converges for $f \in L^p(\mathbb T)$ also in $L^p(\mathbb T).$
\end{corollary}

\section{Hardy-Littlewood Maximal function}

We start by recalling basic properties.

The distribution function $\lambda_f$ is defined as 
\[ \lambda_f(\alpha):=\mu\{x \in X;\vert f(x)\vert>\alpha\}.\]
One then has for any measurable $f:X \to \mathbb C$ and any $p\in (0,\infty)$
\[ \int_X \vert f \vert^p d \mu = p \int_0^{\infty} \alpha^{p-1} \lambda_f(\alpha) d \alpha.\]
In addition one has Chebyshev's inequality for any $p \in (0,\infty)$ and $f \in L^p$
\[ \lambda_f(\alpha) \le \alpha^{-p} \Vert f \Vert_p^p.\]
\begin{definition}
For each $p \in [1,\infty)$ we define $L^{p,\infty}$ the weak $L^p$ space of all measurable $f$ for which there is a constant $C>0$ such that 
\[ \lambda_f(\alpha) \le \alpha^{-p} C^p.\]
\end{definition}
The infimum of all such constants is defined to be $\Vert f\Vert_{p,\infty}.$ An example of a function that is in the weak space but not the full space is $\vert x \vert^{-d/p}.$

The above definition defines a quasi-norm in the sense that the triangle inequality holds for some $C_p<\infty$ such that 
\[ \Vert f+g\Vert_{p,\infty} \le C_p(\Vert f \Vert_{p,\infty}  + \Vert g \Vert_{p,\infty}).\]
For $p \in (1,\infty)$ there actually exists a genuine norm on this space. 
\begin{definition}
An operator $T$ is said to be of weak type $(p,q)$ if it maps $L^p$ to $L^{q,\infty}$ and satisfies 
\[\Vert Tf \Vert_{q,\infty}\Vert \le C \Vert f \Vert_{p}.\]
\end{definition}

Finally, we define a key object for our following studies 
\begin{definition}
Let $f \in L^1_{\text{loc}}(\mathbb R^d)$, we define the \emph{Hardy-Littlewood Maximal Function}
\[ Mf(x):=\sup_{r >0} \vert B_r(x)\vert^{-1} \int _{B_r(x)} \vert f(y)\vert dy.\]
\end{definition}
We then have the following Theorem
\begin{theorem}
\label{theo:int}
For each $p \in (1,\infty]$ there is $C(p,d)<\infty$ such that $\Vert Mf \Vert_p \le C \Vert f \Vert_p$. Moreover, for any $f \in L^1$ and $\alpha>0$ we have 
\[ \vert \{x ; Mf(x)>\alpha \} \vert \le C \alpha^{-1}\Vert f \Vert_1.\]
\end{theorem}
However, $M$ fails to map $L^1$ to $L^1.$ In fact, if $Mf \in L^1$ then $f \equiv 0.$
To see this, we observe that if $ \int_{B_r(0)} \vert f \vert >0$, then for any $x$, we find $\int_{B_{\vert x \vert+2R}(x)} \vert f \vert \le  \int_{B_r(0)} \vert f \vert. $
We deduce that 
\[ Mf(x) \gtrsim \langle x \rangle^{-d} \]
which is not integrable.

In addition, we note that the bound $\Vert Mf \Vert_{\infty} \le \Vert f \Vert_{\infty}$ is obvious. In fact, we even have 
\[ \lim_{r \to 0} \vert B_r(x) \vert^{-1} \int_{B_r(x)} f(y) \ dy = f(x)\]
by Lebesgue's differentiation theorem, which implies $Mf(x) \ge \vert f(x)\vert.$
Our approach to show Theorem \ref{theo:int} will be to show the $L^{1} \to L^{1,\infty}$ bound and then to use interpolation. 

We have the Vitali covering Lemma
\begin{lemma}
For each $d \ge 1$, there is $C_d<\infty$ such that for any measurable $E \subset \RR^d$ of finite measure and any collection of balls $\mathcal B$ such that 
\[ E \subset \bigcup_{B \in \mathcal B} B,\]
there is a collection $\mathcal B'$ of disjoint elements of $\mathcal B$ such that 
\[\vert E \vert \le C_d \sum_{B'\in \mathcal B'} \vert B' \vert.\]
\end{lemma}

The $L^{1} \to L^{1,\infty}$ bound can then be deduced as follows.

Let $f \in L^1$ and $\alpha>0$ be given. Define $E_{\alpha}:=\{x ; Mf(x) >\alpha\}.$ Define $\mathcal B$ to be the balls $B$ satisfying 
\[ \vert B \vert^{-1} \int_B \vert f \vert >\alpha.\]
The union of all those contains $E_{\alpha}$. 
Then using the Vitali covering Lemma, we conclude 
\[\vert E_{\alpha}\vert \le C_d \sum_{B' \in \mathcal B'} \vert B'\vert \le C_d \sum_{B' \in \mathcal B'} \alpha^{-1} \int_{B' }\vert f \vert \le C_d \alpha^{-1} \Vert f \Vert_1.\]

We shall now turn to the proof of the covering Lemma 
\begin{proof}
Choose $K \subset E$ compact with $\vert K \vert \ge \vert E \vert /2$. Choose a finite subcovert $\mathcal B'' \subset \mathcal B$ that covers $K$. Write $\mathcal B''=\{B_1,B_2,...\}$ ordering the balls so that $\vert B_j \vert \ge \vert B_{j+1}\vert.$
We then define $\mathcal B'$ as follows:
Select $B_1$. If $B_N$ is disjoint from all previously selected one, we select it, otherwise we discard it. 
This way $\mathcal B'$ has only pairwise disjoint elements. 
We find that for any $B_m \in \mathcal B'' \setminus \mathcal B'$ there is $B' \in \mathcal B'$ such that $B_m \subset ( B')^*$ where $(B')^*$ denotes the ball concentric with $B'$ having three times as large a radius. Finally,
\[ \vert K \vert \le \vert \bigcup_{B \in \mathcal B''} B \vert \le \vert \bigcup_{B' \in \mathcal B'} (B')^* \vert  = 3^d \sum_{B' \in \mathcal B'} \vert B'\vert.\]

\end{proof}

The interpolation result we need is the Marcinkiewicz Interpolation Theorem
\begin{definition}
An operator $T$ is said to be sublinear if it satisfies $\vert T(f+g)\vert \le \vert Tf\vert+ \vert Tg\vert$
\end{definition}
Let $p_{\theta}^{-1}=(1-\theta) p_0^{-1} + \theta p_1^{-1}$ and $q_{\theta}^{-1}=(1-\theta) q_0^{-1} + \theta q_1^{-1}.$
The Marcinkiewiz interpolation theorem then states 
\begin{theorem}
Having, $\Vert Tf\Vert_{q_j,\infty} \le C \Vert f \Vert_{p_j}$, we conclude that 
\[\Vert Tf \Vert_{q_{\theta}} \le \tilde C \Vert f \Vert_{p_{\theta}}.\]
\end{theorem}
The main advantage of the Marcinkiewicz theorem is that, unlike the Riesz-Thorin theorem, it only requires weak estimates at the end-points. 
\begin{proof}
To keep it simple, we will just the proof the case that we need: 
Let $p_0=q_0=1$ and $p_1=q_1 = \infty.$ Suppose that 
\[ \Vert Tf \Vert_{\infty} \le C_1 \Vert f \Vert_{\infty}, \Vert Tf \Vert_{1,\infty} \le C_0 \Vert f \Vert_{1}\] 
as well as $\vert T(f+g)\vert \le C_2 (\vert Tf \vert  + \vert Tg \vert).$

Given $\alpha>0$ we split $f=g+h$ where $h(x)=0$ if $\vert f(x)\vert \le \alpha/(2C_1C_2)$ and $h(x)=f(x)$ otherwise. 
Then $\Vert g \Vert_{\infty} \le \alpha/(2C_1C_2)$ so $\Vert Tg \Vert_{\infty} \le \alpha/(2C_2)$. This implies that 
\[ C_2 \vert Th \vert + \alpha/2 \ge C_2 \vert Th \vert  +  C_2 \Vert Tg \Vert_{\infty} \ge C_2(  \vert Th \vert + \vert Tg \vert) \ge \vert Tf \vert \]
Hence, if $\vert Tf \vert \ge \alpha$, then $\vert Th \vert \ge \alpha/(2C_2).$

This implies by the monotonicity of measures, since $\lambda_f(\alpha):=\mu\{x \in X; \vert f(x)\vert >\alpha \}$ we find that
\[ \lambda_{Tf }(\alpha) \le \lambda_{Th}(\alpha/(2C_2)).\]
This implies, using that $\Vert h \Vert_p = p \int_0^{\infty} \alpha^{p-1} \lambda_h(\alpha) \ d\alpha$ and the definition of $h$
\[ \begin{split} 
\lambda_{Tf}(\alpha) &\le  \lambda_{Th}(\alpha/(2C_2)) \le 2C_2 \alpha^{-1} \Vert Th \Vert_{1,\infty}  \\ 
&\le  2C_0 C_2 \alpha^{-1} \Vert h \Vert_{1}  \\ 
&=  C \alpha^{-1}\int_0^{\infty} \lambda_h(\beta) d \beta   \\ 
&=  C \alpha^{-1}\int_0^{\infty} \min(\lambda_f(\beta) , \lambda_f(\alpha/(2C_1C_2))) \ d \beta  \\ 
& \le. C \alpha^{-1} \int_{\alpha/(2C_1C_2)} \lambda_f(\beta) \ d\beta + C \lambda_f (\alpha/(2C_1C_2)).
\end{split}\]
Thus, for any $p \in (1,\infty)$
\[ \begin{split} 
\Vert Tf \Vert_p^p
&= p \int_0^{\infty} \alpha^{p-1} \lambda_{Tf}(\alpha) \ d\alpha \\
& \le C \int_0^{\infty} \alpha^{p-1} \left(\alpha^{-1} \int_{\alpha/(2C_1C_2)}^{\infty} \lambda_f(\beta) \ d \beta + C\lambda_f(\alpha/(2C_1C_2)) \right) \ d\alpha \\
& = C \int_0^{\infty} \lambda_f(\beta) \int_0^{2 C_1 C_2 \beta} \alpha^{p-2} \ d\alpha \ d\beta + C \int_0^{\infty} \alpha^{p-1} \lambda_f(\alpha/(2C_1C_2)) \ d\alpha \\
& \le C \int_0^{\infty} \gamma^{p-1} \lambda_f(\gamma) \ d\gamma. \end{split} \]
\end{proof}

\section{Singular Integral Operators}

\begin{definition}
A Calderon Zygmund kernel is a continuous function on $\mathbb R^d \times \mathbb R^d \setminus \Delta$, where $\Delta = \{(x,y); x=y\}$ is the diagonal such that 
\[ \vert K(x,y) \vert \le C \vert x-y\vert^{-d}\]
and there is $\delta \in (0,1]$ such that whenever $\vert y-y'\vert \le 1/2\vert x-y\vert$ then 
\[ \vert K(x,y) - K(x,y') \vert + \vert K(y,x)-K(y',x)\vert \le C \vert y-y'\vert^{\delta} \vert x-y\vert^{-d-\delta}. \]
\end{definition}

\begin{definition}
A continuous linear operator $T: \mathcal D \to \mathcal D'$ is associated to a kernel $K \in L^1_{\text{loc}}(\mathbb R^d \times \mathbb R^d \setminus \Delta)$ if for every pair $f,g \in \mathcal D$ of disjoint support, we have  
\[\langle Tf,g\rangle = \int \int K(x,y) f(x) g(y) \ dy \ dx.\]
\end{definition}
An operator has at most one kernel but a kernel does not uniquely define an operator, e.g. $K=0$ corresponds to both the identity and the first derivative operator. 

One then has 
\begin{theorem}[Calderon Zygmund]
\label{theo:CZ}
Suppose that for some $q \in (1,\infty)$ $T$ is a bounded linear operator on $L^q(\mathbb R^d)$ and $T$ is assocaited with a CZ kernel. Then $T$ extends to a bounded linear operator for all $q \in (1,\infty)$ and is of weak $(1,1)$ type, i.e. 
\[\Vert Tf \Vert_{1,\infty} \le C \Vert f \Vert_1.\]
\end{theorem}

The essential step in the proof is the following result
\begin{proposition}
\label{prop:weak11}
Under the assumptions of Theorem \ref{theo:CZ}, the operator $T$ is of weak $(1,1)$ type.
\end{proposition}

We can now give the proof of Theorem \ref{theo:CZ}
\begin{proof}
Using Marcinkiewicz interpolation we can conclude that $T$ is bounded for every $p \in (1,q).$ If $q=\infty$ then we are good. If not, then we study the transpose operator $T' \in L(L^{q'})$ defined by $\int f T'g = \int Tf g$ with kernel $K'(x,y) = K(y,x).$ Applying the Proposition,  $T'$ is bounded for every $r \in (1,q')$. This however implies that $T$ is bounded on all $L^p$ with $p \in (1,\infty).$
\end{proof}

To prove our Proposition, we need another tool that is commonly referred to as the \emph{Calderon-Zygmund decomposition}
\begin{proposition}
Let $f \in L^1(\RR^d)$ and $\alpha>0$. Then $f$ can be written as $g+b$ with $\Vert g \Vert_{\infty} \le \alpha$ and $b = \sum_j b_j$ with each $b_j$ supported on a dyadic cube $Q_j$ \footnote{ A cube of sidelength $2^k$ for some $k \in \ZZ$ with vertices in $\ZZ 2^k$ }and 
\begin{itemize}
\item $Q_i \cap Q_j = \emptyset$ for $i \neq j.$
\item $\int b_j =0$
\item $\Vert b_j \Vert_1 \le 2^d \alpha \vert Q_j \vert$
\item $\sum_j \vert Q_j \vert \le \alpha^{-1} \Vert f \Vert_1.$
\item $\Vert b \Vert_1 + \Vert g \Vert_1 \le C \Vert f \Vert_1.$
\end{itemize}
\end{proposition}

We can now state the proof of Prop. \ref{prop:weak11}. 
\begin{proof}
Let $\vert Tf(x)\vert >\alpha$ then using the same notation as in the CZ decomposition 
\[ \vert \{x ; \vert Tf(x)\vert >\alpha \}\vert \le \vert \{x ; \vert Tg(x)\vert >\alpha/2 \}\vert  + \vert \{x ; \vert Tb(x)\vert >\alpha/2 \}\vert .\]
We also have that by the CZ decomposition again.
\[ \Vert g \Vert_{q}^{q} \le \Vert g \Vert_{\infty}^{q-1} \Vert g \Vert_1 \le C \alpha^{q-1}\Vert f \Vert_1.\]
Thus, by Chebyshev
\[ \vert \{x; \vert Tg(x)\vert >\alpha/2\}\vert \le 2^q \alpha^{-q} \Vert Tf \Vert_q^q \le C \alpha^{-q} \Vert g \Vert_q^q \le C \alpha^{-1}\Vert f \Vert_1.\]
This is the weak (1,1) boundedness for $g$, now we also need this for $q.$
Let $Q_j^*$ denote the ball concentric with $Q_j$ whose radius is twice the diameter of $Q_j$.

We define the exceptional set 
\[ E = \bigcup_j Q_j^*,\]
then using the CZ decomposition
\[ \vert E \vert \le  C \sum_j \vert Q_j \vert \le C \alpha^{-1} \Vert f \Vert_1.\]
This implies that 
\[\vert \{x ; \vert Tb(x)\vert > \alpha/2 \} \vert \le \vert E \vert + \vert \{x \notin E ; \vert Tb(x)\vert > \alpha/2 \} \vert \le \vert E \vert + 2\alpha^{-1} \Vert Tb \Vert_{L^1(\RR^d \setminus E)}.\]
The term $\vert E \vert $ has already been estimated two lines above. We now focus on $ \Vert Tb \Vert_{L^1(\RR^d \setminus E)}$ and use that 
\[ \Vert Tb \Vert_{L^1(\RR^d \setminus E)} \le \sum_j \Vert Tb_j \Vert_{L^1(\RR^d \setminus E)} \le \sum_j \Vert Tb_j \Vert_{L^1(\RR^d \setminus Q_j^*)}.\]
We now need an additional Lemma that shows that 
\begin{equation}
\label{eq:aux_bound}
\Vert Tb_j \Vert \le C \Vert b_j \Vert.
\end{equation}
This then allows us to show that 
\[ \sum_j \Vert Tb_j \Vert_{L^1(\RR^d \setminus Q_j^*)} \le C \sum_j \Vert b_j \Vert_{L^1} = C \Vert b \Vert_1 \le C \Vert f \Vert_1.\]
\end{proof}
We now show \eqref{eq:aux_bound}.
Let $y_0$ denote the center of $Q_j$, then since $\int b_j=0$ we have for $x \notin Q_j^*$
\[ Tb_j(x) = \int (K(x,y)-K(x,y_0))b_j(y)  \ dy.\]
We conclude 
\[ \begin{split}
\int_{x \notin Q_j^*} \vert Tb_j(x)\vert \ dx &= \int_{x \notin Q_j^*} \vert \int_{y \in Q_j} (K(x,y)-K(x,y_0)) b_j(y) \ dy \vert dx \\
&= \int_{y \in Q_j}  \int_{x \notin Q_j^*}  \vert K(x,y)-K(x,y_0)\vert \ dx \vert b_j(y) \vert \ dy  \\
&\le \Vert b_j \Vert_1 \sup_{y \in Q} \Vert K(\bullet,y) - K(\bullet,y_0)\Vert_{L^1(\RR^d \setminus Q_j^*)}.
\end{split}\]
On the other hand, for $\ell$ the side-length of the cube $Q_j$
\[ \vert Tb_j(x)\vert \le  C \vert x-y_0\vert^{-d-\delta} \int_{Q_j} \vert y-y_0 \vert^{\delta} \vert b_j(y)\vert \ dy \le C \vert x-y_0\vert^{-d-\delta} \ell^d \Vert b_j \Vert_1.\]
Using that for $y \in Q_j$ and $x \in Q_j^*$ we have $\vert x-y_0 \vert \ge 2 \vert y-y_0 \vert$ we have by the properties of the CZ kernel
\[ \vert K(x,y)-K(x,y_0)\vert \le C \vert y-y_0\vert^{\delta} \vert x-y_0 \vert^{-d-\delta}.\]
Integrating then, we find 
\[ \int_{\RR^d \setminus Q_j^*} \vert x-y_0\vert^{-d-\delta} \ dx \le \int_{\vert x-y_0\vert \ge 2\ell} \vert x-y_0\vert^{-d-\delta} \ dx = c\ell^{-\delta}\]
Thus, one finds 
\[ \Vert Tb_j \Vert_1 \le C \Vert b_j \Vert_1.\]
\bibliographystyle{unsrt}
\bibliography{ref}


% \begin{thebibliography}{0}
% \bibitem[DS84]{DS84} F. Delyon and B. Souillard. \emph{Remark on the continuity of the density of states of ergodic finite difference operators}, Communications in Mathematical Physics volume 94, pages289–291 (1984)
% \end{thebibliography}

\end{document}