I'm having troubles to understand the philosophy behind the modern proof of Carleson's theorem. For convenience, let me state precisely what I am asking for. 


For any $f \in L^2(\mathbb{R})$, let $\mathcal{Cf}:=\sup\_{N \in \mathbb{Z}} \left\vert P\_{-}(e^{iN\cdot}f)\right\vert$ be the maximal Carleson operator; where $P\_{-}$ is the projection on negative Fourier spectrum $\{\xi <0\}$. The Carleson theorem essentially states that:



> 
> **Theorem (Carleson).** $|\{\mathcal{C}f > \lambda\}|\_{L^2} \lesssim \lambda^{-2}\|f\|^2\_{L^2}$. 
> 
> 
> 


In the modern proof of this theorem [I'm reading [this](https://arxiv.org/pdf/math/0307008.pdf)], one instead studies the operators defined by 
$Q\_{\xi}f:= \sum\_{s \in T} \mathbf{1}\_{\omega\_s^+}(\xi) \langle f, \varphi \_s\rangle \varphi \_s$, where the notation is as follows:


* $T$ denotes the set of all tiles $I\_s \times \omega\_s$ such that $I\_s, \omega\_s$ are dyadic intervals, such that the area of the tile $I\_s \times \omega\_s$ is one.
* $\omega\_s^+$ stands for the upper half of the interval, and the $\varphi\_s$ are functions such that $\hat \phi\_s$ have Fourier support inside $\omega\_s^{-}$ (the lower half of the interval)


It is not difficult to pass from $Q\_{\xi}$ to the Carleson operator, one can take averages and get that the operator:
$$Q:=\lim\_{Y \to \infty} \frac{1}{Y^2}\int\_{[1,2] \times [0,Y]^2} Dil^2\_{2^{-\lambda}}Tr\_{-y}Mod\_{-\xi}Q\_{\xi} Mod\_{\xi}Tr\_yDil\_{2^{\lambda}}^2d\lambda dy d\xi\,,$$
commutes with translations and dilations and its Kernel is made of functions with Fourier support lying on $\{\xi >0\}$, thus this operator is $P\_{-}$. 


My question is then:



> 
> How is it somewhat 'natural' to come up with the operators $Q\_{\xi}$? How one can guess that such an operator has a similar behavior to that of $\mathcal C$? 
> 
> 
> 


I think there is some 'discretization' idea behind but I do not see how is it natural in any sense. Put in another way my question is: starting from $\mathcal C$ and $P\_{-}$ how does one introduces the operators $Q\_{\xi}$?


Does anybody has some good insights?



Idea behind Carleson's theorem modern proof "intitial reductions"
I have my own confusions here, but let me share my thoughts.


As you mention, there is a discretization here. If you want to decompose the operator $P\_-$, you use the standard decomposition $\sum\_k\hat{\varphi}\_k = 1\_{(-\infty,0]}$, where $\hat{\varphi}\_k(\xi) := \hat{\varphi}(\xi/2^k)$ is supported at frequencies $\vert\xi\vert\sim 2^k$. You can think of $\varphi\_k$ as attached to the tile $I\_s\times\omega\_s = [-2^{-k-1},2^{-k-1}]\times[-2^k,0]$ ---this tile doesn't belong to the mesh $\mathcal{D}$, but let's ignore these "unfortunate technicalities", as Fefferman put it.


Since $P\_-^2f = P\_-f$, then we get the decomposition 
$$
P\_- f = \sum\_{k,k'} \varphi\_k\*\varphi\_{k'}\*f.
$$
We may assume that $\varphi\_k\*\varphi\_{k'} = 0$ unless $k=k'$; you can take Fourier transform to see that this is morally true. For each term in the series we get
$$
\begin{align}
\varphi\_k\*\varphi\_k\*f(x) &= \int f(z)\varphi(y-z)\varphi(x-y)\,dydz \\
&= \int \varphi\_k(y)\int f(z)\varphi\_k(x-z-y)\,dzdy \\
&= \int \textrm{Tr}\_y\varphi\_k(x)\langle f,\textrm{Tr}\_y\varphi\_k\rangle\,dy \\
&= \sum\_{\vert I\vert = 2^{-k}}\frac{1}{2^{-k}}\int\_{-2^{-k-1}}^{2^{-k-1}}2^{-\frac{k}{2}}\textrm{Tr}\_{y+c(I)}\varphi\_k(x)\langle f,2^{-\frac{k}{2}}\textrm{Tr}\_{y+c(I)}\varphi\_k\rangle\,dy
\end{align}
$$
In the third identity we used $\overline{\tilde{\varphi}} = \varphi$, where $\tilde{\varphi}(x) = \varphi(-x)$, because $\hat{\varphi}$ is real. In the last term, let's define $\textrm{Tr}\_{c(I)}\textrm{Dil}\_{2^{-k}}^2\varphi = \varphi\_s$, where $s$ denotes the tile $(c(I)+[-2^{-k-1},2^{-k-1}])\times [-2^k,0]$. We rewrite then the last integral as the average
$$
\varphi\_k\*\varphi\_k\*f(x) = \frac{1}{2Y}\int\_{-Y}^{Y}\sum\_{\vert I\vert= 2^{-k}}\textrm{Tr}\_y\varphi\_s(x)\langle f,\textrm{Tr}\_y\varphi\_s\rangle\,dy,
$$
where $Y = 2^{-k-1}$; however, you can modify the argument above to see that actually you can take the limit $Y\to\infty$. Summing up we have
$$
P\_-f(x) = \lim\_{Y\to\infty}\frac{1}{2Y}\int\_{-Y}^Y\sum\_s \langle f,\textrm{Tr}\_y\varphi\_s\rangle\textrm{Tr}\_y\varphi\_s(x)\,dy,
$$
where the tiles $s$ are those here constructed.


We left open the assumption $\varphi\_k\*\varphi\_{k'}=0$ unless $k=k'$. In the paper they took $\hat{\varphi}$ supported in an interval of length $\frac{1}{4}$, but it is then impossible, I think, to get $\sum\_k\hat{\varphi}\_k = 1\_{(-\infty,0]}$. I suspect the the average in dilation helps to solve the problem here; in fact, I would try to find a function $\varphi$ such that $\sum\_k \int\_2^4\hat{\varphi}\_k^2(t\xi)\frac{dt}{t} = 1\_{(-\infty,0]}$, and such that the supports of $\hat{\varphi}\_k$ and $\hat{\varphi}\_{k'}$ are disjoint, but not sure. The square $\hat{\varphi}\_k^2$ is to use the same trick $P^2\_-$.


In any case, we are reduced to the operator
$$
Tf := \sum\_s\langle f,\varphi\_s\rangle\varphi\_s.
$$
Now if we try to use it in the Carleson operator, we have to deal with
$$
\vert T(e^{iN\cdot}f)\vert = \vert\sum\_s\langle f,\textrm{Mod}\_{-N}\varphi\_s\rangle\textrm{Mod}\_{-N}\varphi\_s\vert.
$$
Now there is another technicality, for the frequency support of $\textrm{Mod}\_{-N}\varphi\_s$ doesn't belong to the mesh $(j2^k,(j+1)2^k)$. The way out from this nuisance is to average over translations of the mesh by using $\textrm{Mod}\_\xi$, much like we did above for the intervals $I$. 


Excuse me if I do not complete all the computations, but I think the idea it is more or less clear. The point is that the operator $Q\_\xi$ allows to get rid from many technicalities, but mainly with problems with the relative position of the mesh $\mathcal{D}$.


