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], 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?

  • 3
    $\begingroup$ Firstly, I highly recommend studying the proof of the analogous result in the case of Walsh series before studying the Fourier case. In the Walsh case, all of the technical details involving cut-off functions, Schwartz tails, etc. disappear and one can navigate the (truly beautiful) argument without these technicalities obscuring the way. [I will offer the counter-intuitive claim that the pedagogical path to Carleson through first studying the Walsh case will be shorter than the path directly to Carleson despite the deceptively short modern write-ups, like Lacey and Thiele's MRL paper.] $\endgroup$
    – Mark Lewko
    Apr 16, 2020 at 9:52

1 Answer 1


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 $\mathcal{D}$.

  • 1
    $\begingroup$ that is perfectly what I was looking for, you provided a lot of intuition behind this discretization! Thank you. $\endgroup$
    – J.Mayol
    Apr 21, 2020 at 6:11
  • $\begingroup$ is it possible to clarify two points? First, how is it "natural" to use the $P_^2$ trick? Instinctively I would only use $P_{-}f= \sum_k \varphi _k * f$ and not get the same as you. This is a very clever trick. Secondly: where does the $\int_2^4 \hat\varphi_k ^2 (t\xi)dt/dt$ trick comes from? By reverse engineering I would say that the $dt/t$ is invariant under dilations, which is really helpful, but how does it comes out naturally? I mean, it is effective for average on dilations, but at the point of your thought process you just said "average on dilations would help" $\endgroup$
    – J.Mayol
    May 16, 2020 at 18:32
  • $\begingroup$ and I do not really understand why (again, in the end, it works, but I am really focusing on how one comes up with that). I remarked that in the recent blog post of T. Tao (last set of notes) he uses the very sames tricks as you, so this must not be a coincidence. Thank you! $\endgroup$
    – J.Mayol
    May 16, 2020 at 18:33
  • 1
    $\begingroup$ I do like Tao's post, very clarifying in many respects. How natural $P_{-}^2$ is, I don't know, I just read it in other instances, and it burned into my head; a less sophisticated though perhaps more "natural" trick here is $\phi_1\phi_2 = \phi_2$, where $\phi_1=1$ in the support of $\phi_2$. The measure $dt/t$ is the Haar measure of the multiplicative goup in $\mathbb{R}_+$, so as the Lebesgue measure to $\tau+\xi$, so $dt/t$ to $\lambda\xi$. Thiele and Christ proof appeared only "recently", so if the steps were all natural, it'd be done many years ago. $\endgroup$
    – user90189
    May 17, 2020 at 8:33

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