These are the lecture notes for a harmonic analysis reading group talk I gave October 14th, 2019 at UT Austin. They are based off of the book *Classical and Multilinear Harmonic Analysis Vol. 1 by Muscalu and Schlag*

Throughout the following notes, we will be working on . Our Fourier transform will be defined to be

One of the fundamental objects in classical harmonic analysis is the partial Fourier series, defined for by

In view of the Fourier inversion formula, it is natural to ask when , where this limit is taken in or pointwise a.e. We have the following basic result for convergence:

Theorem 1The following statements are equivalent for

- For all ,

*Proof:* One direction follows from density of trigonometric polynomials in , and the other follows from the uniform boundedness principle.

By using weak to bounds, duality, and Marcinkiewicz interpolation on the Hilbert transform, one in fact has the following result for .

Theorem 2For any , .

This result can be further extended to pointwise a.e. convergence, although it is much harder.

Theorem 3 (Carleson ’66, Hunt ’68)For any and , for a.e. .

The proof of this theorem is exceptionally complicated, and requires an in depth analysis of both the frequency and spacial variables simultaneously.

Amazingly the above convergence results are false on ! To show this, let us assume for simplicity that . For type convergence, this follows from the representation of , where is the Dirichlet kernel

It is an exercise to show that , and this implies that as . By Theorem 1, this implies failure of convergence of the partial Fourier sums.

For the pointwise a.e. convergence, we have the following theorem by Kolmogorov:

Theorem 4 (Kolmogorov ’23)There exists an such that does not converge a.e. as .

The rest of this talk will be devoted to proving this theorem.

Recall for a moment the proof of the Lebesgue Differentiation theorem via bounding the Hardy-Littlewood maximal function. These types of maximal functions are fundamental in proving convergence a.e. results. Calderon, Zygmund, and Stein noticed that a bound on relevant maximal functions is in fact *necessary* for showing convergence a.e. The exact statement is the following:

Theorem 5Let be a sequence of translation invariant bounded linear operators on . If the maximal function

satisfies for each trigonometric polynomial , then the following implication holds: If for any , we have , then there exists such that

for all and .

*Proof:* We proceed by contradiction. Assume there exists and such that and

We can make a few simplifications. First, note that for each , there exists such that

Next, we may assume that each is a trigonometric polynomial. The Cesaro means in , and boundedness of implies that as in . Since is a trigonometric polynomial for each , we have not lost any generality assuming is a trigonometric polynomial.

Now, for each , pick such that . Then we have by construction that

By a Borel-Cantelli type argument, one can find a set of translations for and such that lies in infinitely many for a.e. . In other words, the set

is infinite for a.e. . From these and , we can construct a function which will furnish a contradiction.

Let be a probability space supporting an iid sequence of coin flips . Define the random variables

and define the random variable

Note that by the triangle inequality, for all :

We wish to pick random values such that for a.e. . To this end, let us fix an such that is infinite. Note first that by absolute convergence of the sums and the boundedness of the operators , we have that up to a measure zero set in ,

Now, since the are translation invariant, so is the maximal function . Hence by definition, for each we have

and hence there exists an such that

Plugging this into the above expression, we get

From here, we use the randomness of the coefficients to obtain some nice bounds. Note that in particular for one of the terms in this sum, . We have the following picture;

Hence, we have by independence that

Since the set is infinite, and , we have that

It can be shown that the event above is a tail event. That is,

Hence by Kolmogorov’s 0-1 Law

Hence, since the set of which this argument was applied to had full measure, we obtain by Fubini that

a.s. in (and hence for at least one ). This contradicts the main hypothesis, so we are done.

With this in hand we have the following corollary for the maximal operator

Corollary 6Assume that there exists a set of positive measure such that converges for every . Then for any complex Borel measure on there exists a constant such that

for any .

*Proof:* This follows from considering the Vallée de Poussin’s kernel convoluted with the measure and applying the previous theorem.

With this corollary in hand, one now constructs a which grows with in such a way that does not satisfy the weak bound above. To do this, we simply consider measures of the form

where is close to . We calculate

By picking points such that the numerator of the above is the same sign, we may obtain that

To pick these points, we prove a variant of Kronecker’s Theorem

Lemma 7Assume that is incommensurate. That is, assume for all we have . Then the orbit is dense.

*Proof:* It suffices to show for all we have

To see this, assume that the orbit is not dense. Picking a “bump function” on an open subset of the complement of the orbit would imply the above limit is false.

To show the above limit holds, simply calculate

and use the decay of to show that the sum converges to as .

With this lemma, we can now construct our .

Lemma 8There exists a sequence of probability measures on such that for each ,

for a.e. for some universal constant .

*Proof:* For each and , pick such that

and the vector is incommensurate. This can be done since the set of incommensurate vectors forms a set of measure zero in . Note that the set of such that is a commensurate vector is at most countable. Therefore for a.e. the previous lemma shows that

is dense, and so

is also dense. Hence for a.e. , we can pick infinitely many such that for each ,

Therefore for these infinitely many ,

So we have by a bit of thinking that

where is universal. The right hand side is bounded below by

This finishes the proof.

We now finish with the proof of Kolmogorov’s Theorem.

*Proof:* By the Corollary 6, if we did have partial Fourier convergence on , then we would obtain a weak type bound

This implies , a clear contradiction.