## Singular Integral Operators and Convergence of Fourier Series

I’m Peter “Viking” Luthy, a journeyman graduate student at Cornell. I’m an analyst, and my current research goals are in harmonic analysis with applications to and from ergodic theory. To avoid being called a hypocrite, Greg asked me to post on occasion and spread my analytic gospel — this isn’t the Everything-but-Analysis Seminar, after all.

My goal in this post is to go through the initial setup of a deep theorem of Carleson dealing with the convergence of Fourier series on $L^p$. This theorem is almost universally interesting in and of itself. Additionally, it will give ample reason as to why people — myself included — care about objects called singular integral operators. This will also provide some impetus for some future posts as well, particularly one which will outline a famous construction of Fefferman and give some reasons why harmonic analysis in higher dimensions is distinctly harder than in dimension 1.

The Fourier series of a function has produced a myriad of applications, both within and without mathematics. While it is based upon relatively simple intuition, speaking rigorously about Fourier series has proven quite a challenge. For a function $f$ on $T$, the circle (where $T$ stands for 1-torus), the partial sums $S_N(f)(x)$ of the Fourier series are given by

$\displaystyle{S_N(f)(x)=\sum_{n=-N}^N \hat{f}(n)e^{2\pi inx}}$

where the $\hat{f}(n)$ are the Fourier coefficients, namely

$\displaystyle{\hat{f}(n) = \int f(x)e^{-2\pi i nx} dx.}$

Most especially, mathematicians have struggled to understand what

$\displaystyle{\lim_{N\rightarrow\infty} S_N(f)(x) \textrm{ or } \lim_{N\rightarrow\infty} \sum_{n=-N}^N \hat{f}(n)e^{2\pi inx}}$

mean when $f$ is a function satisfying some additional conditions. One can consider the convergence of the above expression in a variety of different contexts; in the spirit of the Lebesgue integral, we’re interested in knowing when it converges almost everywhere to $f(x)$ when $f\in L^p(T)$. Obviously this is the strongest kind of statement one could hope for in the modern framework of analysis.

In the early 1920s, Kolmogorov constructed an integrable function for which the Fourier series converges to $f(x)$ only on a set of measure zero. This was, in fact, his first mathematical publication — as if the billion or so theorems bearing his name weren’t enough reason to think highly of him. However, the question remained open for $p>1$ until 1966 when Carleson proved that the Fourier series of an $L^2$-function converges almost everywhere. The general $p>1$ case was proved later by Hunt by generalizing Carleson’s methods.

Carleson’s proof is pure genius but not at all elementary. Much effort has gone into simplifying the proof — for example, papers by Fefferman and Lacey–Thiele — but all the proofs I’ve ever heard of prove uniform boundedness of a class of singular integral operators. I’m going to explain how the problem translates from pointwise convergence to operator bounds, leaving the much more complicated remainder as a topic for another day — or at least as an exercise left to the reader (ha!).

One first thing to notice is that Carleson’s theorem is completely trivial for smooth functions. Applying integration by parts twice shows that the $n^{\textrm{th}}$ Fourier coefficient decays like $1/n^2$ so that the Fourier series converges absolutely. That the uniform limit of continuous functions is continuous together with the fact that continuous functions having the same Fourier coefficients are equal gives immediately that

$\displaystyle{\|\limsup_{N\rightarrow\infty}|S_N(f) - f|\|_p=0}$

for smooth functions $f$.

Using the density of smooth functions in $L^p$ and an $\epsilon/3$-argument, then, we would be done if we could show that $\|\sup_N(S_N(f-g))\|_p$ is small whenever $\|f-g\|_p$ is small. Specifically, we want to show the more quantitative statement that $\sup_N(S_N(\cdot))$ is a bounded linear operator from $L^p(T)$ to itself.

Now, $S_N(\cdot)$ can be written in a closed form using the Hilbert transform. I’ll define the Hilbert* transform of a function by

$\displaystyle{\widetilde{H}f = -i\sum_{n=-\infty}^\infty \textrm{signum}(n)\hat{f}(n)e^{2\pi inx}.}$

The function $\textrm{signum}(n)$ is $n/|n|$ for nonzero $n$ and zero when $n=0$. The careful reader may notice that this is actually not the Hilbert transform proper — this definition is convolution by $\cot(\pi x)$ or something like that — but it differs from the usual Hilbert transform by a bounded error since the order of the singularity of $\cot(x)$ at $x=0$ is the same as $1/x$.

So how is this helpful? Well, consider the following heuristic. Adding $i$ times the identity from $\widetilde{H}$ produces an operator which kills all the negative frequency Fourier coefficients and scales the positive frequency coefficients by $-2i$, leaving $\hat{f}(0)$ alone, i.e.

$\displaystyle{\widetilde{H}+iI -i\hat{f}(0) = -2i \sum_{n=1}^\infty \hat{f}(n)e^{2\pi inx}}$

Now, observe that
$\displaystyle{e^{-2\pi iNx}(\widetilde{H}+iI -i\hat{f}(0))(e^{2\pi iNx}f(x))}$

$\displaystyle{= -2i e^{-2\pi iNx}\sum_{n=1}^\infty \hat{f}(n-N)e^{2\pi inx}}$

$\displaystyle{= -2i \sum_{n=1}^\infty \hat{f}(n-N)e^{2\pi i(n-N)x}}$

$\displaystyle{= -2i \sum_{n=-N+1}^\infty \hat{f}(n)e^{2\pi i(n)x}.}$

By performing a very similar procedure, one can get the same sum going from $N$ to $\infty$. Taking the difference of those two quantities then gives you $S_{N-1}$. One should note that we didn’t really need to include the $-i\hat{f}(0)$ term since it will drop out anyway when we take the difference.

Let’s view functions on the torus as periodic functions on the line. As we stated before, the operator $\widetilde{H}$ has the same fundamental singularity as the regular Hilbert transform $H$ (convolution with $1/2\pi x$), and so it is fairly trivial to pass from $\widetilde{H}$ to $H$. Also, multiplication by $e^{-2\pi iNx}$ does not increase the norm at all. Putting all of this together, it is enough to understand

$\displaystyle{\sup_{N\in \mathbb{Z}}|H(f(x)e^{2\pi Nx})| \le \sup_{N\in \mathbb{R}}|H(f(x)e^{2\pi Nx})|.}$

The Fourier transform of a convolution is the product of the Fourier transforms. So, by applying the Fourier transform and it’s inverse, the above expression is a constant multiple of

$\displaystyle{\sup_{N\in \mathbb{R}}|\int_{\mathbb{R}}\textrm{signum}(\xi - N)\hat{f}(\xi)e^{2\pi x\xi}|,}$

since the Fourier transform of the Hilbert transform is just the $\textrm{signum}$ function. The Carleson–Hunt theorem is that this operator is bounded from $L^p$ to $L^p$ for $p>1$.

The thing sitting inside the supremum is a singular integral operator. One definition of such a thing is: an operator $T$ (operating on $L^p(\mathbb{R}^n$)) is a singular integral operator if we can write

$\displaystyle{Tf(x)=\int_{\mathbb{R}^n}K(x,y)f(y)dy}$

where $K$ is a measurable function with plenty of nice qualities but has some kind of singularity — in the sense that $K$ has a discontinuity or blow up. In practice, many singular integral operators are convolutions with some kind of singular function, e.g. the Hilbert transform. In this case, a singular integral operator can be written as

$\displaystyle{Tf(x)=\int_{\mathbb{R}^n}m(\xi)\hat{f}e^{2\pi ix\xi}d\xi.}$

for a measurable function $m$ which will also have lots of nice properties but which has some kind of singularity.

Of course, convergence of Fourier series is not the only place these operators come up. If a harmonic function $u$ on the unit disk has boundary values given by $f$ in $L^p$ then the operator $\widetilde{H}$ mentioned above gives the boundary values of the harmonic conjugate $v$ of $u$, where $v(0)=0$. The boundedness of the Riesz transforms, given by

$\displaystyle{\widehat{R_jf}(\xi) = -i\frac{-\xi_j}{|\xi|}\hat{f}(\xi),}$

is how one proves that the Laplacian controls the size of all the mixed second order derivatives. The Beurling transform on $\mathbb{R}^2$ has $m(\xi,\eta) = (\xi-i\eta)/(\xi+i\eta)$. This operator arises when one tries to solve the Beltrami equation

$\displaystyle{\partial f =\mu \bar{\partial} f}$

which comes up in the study of quasiconformal maps and Teichmuller Theory.

I will leave you with the following concluding remark. Carleson’s proof presents a common theme in harmonic analysis: some difficult problem translates to a quantitative statement about the boundedness of operators. Another famous instance of this is the Kakeya conjecture which translates a statement about the Hausdorff dimension of a class of sets (hard!) to a statement about linear operators. This new setting is presumably less difficult as there are many more theorems about operators than Hausdorff dimension. None-the-less it remains an open problem.

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### 5 Responses to “Singular Integral Operators and Convergence of Fourier Series”

1. Terence Tao Says:

Nice post! I just wanted to comment though that there is also a (partial) reverse implication: if a.e. pointwise convergence for Fourier summation was true in $L^p$ for some $1 \leq p \leq 2$, then there must be a weak $(p,p)$ inequality for the corresponding maximal operator. This is known as Stein’s maximum principle and is proven in this 1960 paper of Stein:

http://www.jstor.org/view/0003486x/di961782/96p0007b/0

It relies primarily on the translation invariance of the situation and on Khintchine’s inequality for random signs. Unfortunately the principle is known to fail for some operators when $p>2$ but it is still good intuition to think of pointwise convergence and maximal operator bounds as being morally equivalent.

For the easier question of norm convergence, we have the uniform boundedness principle, which shows that if we have $T_n f$ converging in $L^p$ to f for all f in $L^p$, then the $T_n$ must be uniformly bounded in the $L^p$ operator norm. Of course, it is easier to bound each of the $T_n$ individually than it is to bound the maximal operator $\sup_n T_n$, which is why pointwise convergence questions tend to be significantly harder than norm convergence questions. (The dominated convergence theorem also gives another indication why this should be the case.)

2. Peter Luthy Says:

I’m glad that you enjoyed it, Terry. Thanks for the link to Stein’s paper – I had never actually read this myself, though I am familiar with the result. As is usually the case with what he produces, it’s remarkably well-written.

Khinchin’s (I always use the shortest romanization since there are so many) inequality is a great trick to have around. We used it several times in a course I took from my adviser (Camil Muscalu) last term. As far as the intuition of pointwise convergence and boundedness of maximal operators being equivalent problems is concerned, I’ve been reading your paper on the interplay between maximal operators and ergodic theory which certainly provides additional motivation for believing such a claim:

http://arxiv.org/abs/math/0510581

3. Ars Mathematica » Blog Archive Says:

[…] intrigued by the beginning of a new series of posts at the Everything Seminar about harmonic analysis. This particular post talks about the […]

4. wlfdiver Says:

Oh my God you have eroded part of my brain!

5. rakshith Says:

assume we have a fourier series f(x) = x……..we get an in finite series in terms of sines…….now suppose we want to prove the converse how do we do it??????please reply to my mail or post the reply here itself……….