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I have redone this question:

On $\mathbb R^n$ the Carleson Operator if defined by $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions).

  • For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' $$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$
  • For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$ for $p\neq 2$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work).
  • The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture

I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons

  1. The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$.
  2. The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult.

Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups?

I'll make it a community wiki so people can fix it if it has errors.

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I've just deleted the comments, which all referred to a previous incarnation of the question, dumping a copy at tea.mathoverflow.net/discussion/1083/comment-cleanup –  Scott Morrison Jul 13 '11 at 21:22
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2 Answers

up vote 3 down vote accepted

There is a p-adic analogue of Carleson's theorem. This was worked out by Hunt and Taibleson and you can find an exposition in Taibleson's book "Fourier analysis on local fields". (NB Taibleson did a lot of work extending Euclidean harmonic analysis results such as Carleson's theorem and boundedness of singular integrals to local fields.) I don't know if an adelic Carleson's theorem has been worked out.

The notion of smoothness in the p-adics is different than in real spaces since the p-adics are totally disconnected. Instead of say differentiability (which can be defined at least formally), one looks at locally constant functions. For more general groups, you need to have an appropriate notion(s) of smoothness to ask how changing the smoothness affects your problem.

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Assuming your question is actually about Carleson's theorem:

Carleson's theorem about the pointwise convergence of Fourier series is really a one-dimensional statement. It makes use of the fact that one can write the (symmetric) partial sum operators in terms of the Hilbert transform. After some manipulations, one arrives at showing that the maximal operator

$\displaystyle\sup_{N\in\mathbb{R}}\left|\int\textrm{signum}(\xi-N)\hat{f}e^{2\pi ix\xi}d\xi\right|$

is bounded from $L^2(\mathbb{R})$ to itself (or weak $L^2$), where the signum function simply outputs the sign of the input. Indeed, proving this operator is bounded is quite difficult, and Carleson's proof has not been drastically simplified in the last 55 years, although I admit this is hearsay as I have never read Carleson's original proof.

If one moves from the line to the plane, how does one change the signum function? Does it become the characteristic function of some half-space? Can the set of discontinuities be the intersection of two lines? Can it be curved? These variations of the problem can drastically change the answer as well as the difficulty of the problem. For instance the question of whether the partial sums of the Fourier series of a function on the 2-torus converges pointwise is a bit more delicate: if you take the sum over bigger and bigger squares in $\mathbb{Z}^2$, the answer is yes, but if one sums over bigger and bigger discs, the question is open.

The one counterexample to the general trend that things outside of $\mathbb{R}$ are harder (which I know of, anyway) is the finite field variant of the Kakeya conjecture: it eluded harmonic analysts for many years but was recently solved by a graduate student in computer science in about two pages. This was a pretty big deal. So anyway, for your particular problem (if it is actually about Carleson's Theorem), I would be surprised if the answer is known but not the top search on google for 'Carleson's Theorem for Adeles'. The result will almost certainly be group-specific if it is known.

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Sorry for changing the question on you. Thanks for the response. –  Taylor Dupuy Jul 9 '11 at 12:43
    
I wanted to accept both answers. Is it possible to do that? –  Taylor Dupuy Nov 21 '11 at 19:26
    
I don't think you can accept more than one answer, but that's ok. I just hope what I wrote helped in some way! –  Peter Luthy Nov 23 '11 at 1:01
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