Partial Fourier Sums and Some Probability

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 {\mathbb{T}^d : = [0,1]^d}. Our Fourier transform will be defined to be

\displaystyle \widehat{f} (n) : = \int_{\mathbb{T}^d} f(x) e^{2\pi i n \cdot x} dx.

One of the fundamental objects in classical harmonic analysis is the partial Fourier series, defined for {N \in \mathbb{N}} by

\displaystyle S_N f(x) : = \sum_{|n| \leq N} \widehat{f}(n) e^{2\pi i n \cdot x}.

In view of the Fourier inversion formula, it is natural to ask when {S_N f \rightarrow f}, where this limit is taken in {L^p(\mathbb{T}^d)} or pointwise a.e. We have the following basic result for {L^p} convergence:

Theorem 1 The following statements are equivalent for {p \in [1,\infty)}

  • {\sup_{N\in {\mathbb N}} \Vert S_N \Vert_{p \rightarrow p} < \infty}
  • For all {f \in L^p( \mathbb{T}^d)}, {\Vert S_N f - f \Vert_p \rightarrow 0}

Proof: One direction follows from density of trigonometric polynomials in {L^p(\mathbb{T}^d)}, and the other follows from the uniform boundedness principle. \Box

By using weak {L^1} to {L^2} bounds, duality, and Marcinkiewicz interpolation on the Hilbert transform, one in fact has the following result for {p \in (1, \infty)}.

Theorem 2 For any {p \in(1,\infty)}, {\sup_{N\in {\mathbb N}} \Vert S_N\Vert_{p \rightarrow p} < \infty}.

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

Theorem 3 (Carleson ’66, Hunt ’68) For any {p \in (1, \infty)} and {f \in L^p(\mathbb{T}^d)}, {S_Nf(x) \rightarrow f(x)} for a.e. {x}.

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 {L^1(\mathbb{T}^d)}! To show this, let us assume for simplicity that {d =1}. For {L^1} type convergence, this follows from the representation of {S_Nf = D_N * f}, where {D_N} is the Dirichlet kernel

\displaystyle D_N(x) : = \frac{\sin ((2N+1) \pi x) }{\sin(\pi x) }.

It is an exercise to show that {\Vert D_N \Vert_{1 \rightarrow 1} \sim \log N}, and this implies that {\Vert S_N \Vert_{1 \rightarrow 1 } \rightarrow \infty} as {N \rightarrow \infty}. By Theorem 1, this implies failure of {L^1} 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 {f \in L^1( \mathbb{T})} such that {S_Nf} does not converge a.e. as {N \rightarrow \infty}.

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 5 Let {T_n} be a sequence of translation invariant bounded linear operators on {L^1(\mathbb{T})}. If the maximal function

\displaystyle Mf(x) : = \sup_{n \in {\mathbb N}} |T_n f(x)|

satisfies {\Vert M f \Vert_\infty < \infty} for each trigonometric polynomial {f}, then the following implication holds: If for any {f \in L^1(\mathbb{T})}, we have {|\{ x \in \mathbb{T} : Mf (x) < \infty \} | > 0}, then there exists {A > 0} such that

\displaystyle |\{ x \in \mathbb{T} : Mf(x) > \lambda \} | \leq \frac{A}{\lambda} \Vert f \Vert_1

for all {f\in L^1(\mathbb{T})} and {\lambda > 0}.

Proof: We proceed by contradiction. Assume there exists {\{f_j\} \subset L^1 ( \mathbb{T})} and {\{ \lambda_j \} \subset (0,\infty)} such that {\Vert f_j \Vert_1 = 1} and

\displaystyle |E_j| = |\{ x\in \mathbb{T} : Mf_j(x) > \lambda_j \} | > \frac{2^j}{\lambda_j}.

We can make a few simplifications. First, note that for each {j}, there exists {M_j > 0} such that

\displaystyle |\{ x\in \mathbb{T} : \sup_{1 \leq k \leq M_j} |T_kf_j(x) > \lambda_j \} | > \frac{2^j}{\lambda_j}.

Next, we may assume that each {f_j} is a trigonometric polynomial. The Cesaro means {\sigma_Nf \rightarrow f} in {L^1}, and boundedness of {T_k} implies that {T_k \sigma_N f\rightarrow T_k f} as {N\rightarrow \infty} in {L^1}. Since {\sigma_N f} is a trigonometric polynomial for each {N \geq 1}, we have not lost any generality assuming {f_j} is a trigonometric polynomial.

Now, for each {j \in {\mathbb N}}, pick {m_j\in {\mathbb N}} such that {m_j \leq \lambda_j/2^j < m_j +1}. Then we have by construction that

\displaystyle \sum_{j=1}^\infty m_j |E_j| = \infty

By a Borel-Cantelli type argument, one can find a set of translations {x_{j,l} \in \mathbb{T}} for {j\in {\mathbb N}} and {1 \leq l \leq m_j} such that {x - x_{j,l}} lies in infinitely many {E_j} for a.e. {x \in \mathbb{T}}. In other words, the set

\displaystyle \mathcal{J}_x : = \{ j \in {\mathbb N} : x - x_{j,l} \in E_j , \exists l \in \{1, \dots, m_j\} \}

is infinite for a.e. {x \in \mathbb{T}}. From these {x_{j,l}} and {f_j}, we can construct a function which will furnish a contradiction.

Let {(\Omega, \mathcal{F}, \mathbb{P})} be a probability space supporting an iid sequence of coin flips {X_n \in \{\pm 1\}}. Define the random variables

\displaystyle \epsilon_n : = \frac{1}{n^2 m_n} X_n,

and define the random variable

\displaystyle f(x) : = \sum_{j=1}^\infty \sum_{l =1}^{m_j} \epsilon_j f_j(x - x_{j,l}).

Note that by the triangle inequality, {f \in L^1( \mathbb{T})} for all {\omega \in \Omega}:

\displaystyle \Vert f \Vert_1 \leq \sum_{j=1}^\infty \sum_{l=1}^{m_j} \frac{1}{j^2 m_j}= \sum_{j=1}^\infty \frac{1}{j^2} <\infty

We wish to pick random values such that {Mf(x) = \infty} for a.e. {x}. To this end, let us fix an {x \in \mathbb{T}} such that {\mathcal{J}_x} is infinite. Note first that by absolute {L^1} convergence of the sums and the boundedness of the operators {T_k}, we have that up to a measure zero set in {\mathbb{T}},

\displaystyle T_n f(x) = \sum_{k=1}^\infty \sum_{l=1}^{m_k} \epsilon_k T_n f_k(x - x_{k,l})

Now, since the {T_k} are translation invariant, so is the maximal function {M}. Hence by definition, for each {j \in \mathcal{J}_x} we have

\displaystyle Mf_j(x - x_{j,l}) > \lambda_j,

and hence there exists an {n(j,x) \in {\mathbb N}} such that

\displaystyle |T_{n(j,x)} f_j(x - x_{j,l})| > \lambda_j.

Plugging this {n(j,x)} into the above expression, we get

\displaystyle T_{n(j,x)} f(x) = \sum_{k=1}^\infty \sum_{l=1}^{m_k} \epsilon_k T_{n(j,x)} f_k(x - x_{k,l}).

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, {|\epsilon_{j} T_{n(j,x)} f_{j} (x- x_{j,l})| >\frac{\lambda_j}{j^2 m_j}}. We have the following picture;

Hence, we have by independence that

\displaystyle \mathop{\mathbb P}\Big( \{ |T_{n(j,x)} f(x) | > \frac{\lambda_j}{j^2 m_j} \} \Big) \geq \frac{1}{2}

Since the set {\mathcal{J}_x} is infinite, and {\frac{\lambda_j}{j^2 m_j} \rightarrow \infty}, we have that

\displaystyle \mathop{\mathbb P} \big( \{ Mf(x) = \infty \} \big) \geq 1/2.

It can be shown that the event {F = \{ Mf(x) = \infty \} \in \mathcal{F}} above is a tail event. That is,

\displaystyle F \in \bigcap_{k=1}^\infty \sigma \Big( \bigcup_{n = k}^\infty \sigma(X_n) \Big)

Hence by Kolmogorov’s 0-1 Law

\displaystyle \mathop{\mathbb P} \big( \{ Mf(x) = \infty \} \big) =1.

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

\displaystyle |\{ x \in \mathbb{T} : Mf(x) = \infty \} | = 1

a.s. in {\Omega} (and hence for at least one {\omega \in \Omega}). This contradicts the main hypothesis, so we are done. \Box

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

\displaystyle Cf(x) : = \sup_{N \in {\mathbb N}} |S_Nf(x)|.

Corollary 6 Assume that there exists a set {G \subset \mathbb{T}} of positive measure such that {S_Nf} converges for every {f \in L^1(\mathbb{T})}. Then for any complex Borel measure {\mu} on {\mathbb{T}} there exists a constant {A} such that

\displaystyle |\{x \in \mathbb{T} : C \mu(x) > \lambda \} | \leq \frac{A}{\lambda} \Vert \mu \Vert

for any {\lambda > 0}.

Proof: This follows from considering the Vallée de Poussin’s kernel convoluted with the measure {\mu} and applying the previous theorem. \Box

With this corollary in hand, one now constructs a {\mu} which grows with {S_N} in such a way that {C\mu} does not satisfy the weak {L^1} bound above. To do this, we simply consider measures of the form

\displaystyle \mu_N : = \frac{1}{N} \sum_{k=1}^N \delta_{x_{k,N} },

where {x_{j,N}} is close to {j/N}. We calculate

\displaystyle S_n \mu_N (x) = \frac{1}{N} \sum_{k=1}^N \frac{ \sin( (2n +1)\pi ( x- x_{k,N} )) }{\sin(\pi( x - x_{k,N}))}

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

\displaystyle |S_n \mu_N (x)| \approx \frac{1}{N} \sum_{k=1}^N \frac{ 1}{|\sin(\pi( x - x_{k,N}))|} \approx \frac{1}{N} \sum_{k=1}^N \frac{ 1}{j/N} \approx \log N

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

Lemma 7 Assume that {\theta \in \mathbb{T}^d} is incommensurate. That is, assume for all {n \in {\mathbb N}^d} we have {n \cdot \theta \notin {\mathbb Z}}. Then the orbit {\{ k \theta \mod {\mathbb Z}^d : k \in {\mathbb Z}\} \subset \mathbb{T}^d} is dense.

Proof: It suffices to show for all {f \in \mathcal{C}^\infty( \mathbb{T}^d)} we have

\displaystyle \frac{1}{N} \sum_{k=1}^N f(k \theta) \rightarrow \int_{\mathbb{T}^d} f(x) dx.

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

\displaystyle \frac{1}{N} \sum_{k=1}^N f(k \theta) = \frac{1}{N } \sum_{k=1}^N\sum_{\nu \in {\mathbb Z}^d } \widehat{f}(\nu) e^{2\pi i k \theta \cdot \nu} = \widehat{f}(0) + \sum_{\nu \in {\mathbb Z}^d\setminus \{0\}} \frac{1}{N} \Bigg( \frac{1- e^{2\pi i (N+1) \theta \cdot \nu}}{1- e^{2\pi i \theta \cdot \nu } }\Bigg)

and use the decay of {\widehat{ f} (\nu)} to show that the sum converges to {0} as {N\rightarrow \infty}. \Box

With this lemma, we can now construct our {\mu_N}.

Lemma 8 There exists a sequence {\{\mu_N\}_{N\in {\mathbb N}}} of probability measures on {\mathbb{T}} such that for each {N},

\displaystyle (\log N)^{-1} \limsup_{n \rightarrow \infty} | S_n \mu_N(x)| \geq \alpha > 0

for a.e. {x \in \mathbb{T}} for some universal constant {\alpha}.

Proof: For each {N} and {1 \leq j \leq N}, pick {x_{j,N} \in \mathbb{T}} such that

\displaystyle \Bigg |x_{j,N} - \frac{j}{N} \Bigg| \leq \frac{1}{N^2}

and the vector {(x_{j,N})_{j=1}^N \in \mathbb{T}^d} is incommensurate. This can be done since the set of incommensurate vectors forms a set of measure zero in {\mathbb{T}^d}. Note that the set of {x \in \mathbb{T}} such that {(2( x- x_{j,N}))_{j=1}^N} is a commensurate vector is at most countable. Therefore for a.e. {x \in \mathbb{T}} the previous lemma shows that

\displaystyle \{ 2k (x - x_{j,N})_{j=1}^N \mod {\mathbb Z}^N : k \in \mathbb{Z} \} \subset \mathbb{T}^N

is dense, and so

\displaystyle \{ (2k +1) (x - x_{j,N})_{j=1}^N \mod {\mathbb Z}^N : k \in \mathbb{Z} \} \subset \mathbb{T}^N

is also dense. Hence for a.e. {x \in \mathbb{T}}, we can pick infinitely many {k= k(x) \in {\mathbb N}} such that for each {1\leq j \leq N},

\displaystyle \sin( (2k+1)\pi(x- x_{j,N})) \geq 1/2.

Therefore for these infinitely many {k},

\displaystyle | S_{k}\mu_N (x) | \geq C \frac{1}{N} \sum_{j=1}^N\frac{1}{ |\sin(\pi ( x- x_{j,N} ) ) | }

So we have by a bit of thinking that

\displaystyle \frac{1}{N} \sum_{j=1}^N\frac{1}{ |\sin(\pi ( x- x_{j,N} ) ) | }\geq C \frac{1}{N} \sum_{j=1}^N \frac{1}{| [ x - j/N] | + N^{-2} } \geq C \frac{1}{N} \sum_{j=1}^N \frac{1}{j/N + N^{-1} + N^{-2} }

where {C} is universal. The right hand side is bounded below by

\displaystyle \geq C\sum_{k=1}^N \frac{1}{j+2} \geq C \log(N-1).

This finishes the proof. \Box

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

Proof: By the Corollary 6, if we did have partial Fourier convergence on {L^1}, then we would obtain a weak type bound

\displaystyle 1= |\{x \in \mathbb{T} : C\mu_N (x) > \alpha \log N \}| \leq \frac{A}{\log N } \Vert \mu_N\Vert = \frac{A}{\log N }

This implies {\log N \leq A\rightarrow \infty}, a clear contradiction. \Box

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: