Topics in Fourier analysis

Akos Magyar

1 Finite Fourier series.

We discuss two applications of finite exponential sums. One is Roth’ theo-rem which can be considered part of geometric Ramsey theory and whosesubsequent generalizations has been very influential in combinatorics, er-godic theory and also in harmonic analysis. The other is the so-called FastFourier Transform which, despite its simplicity, is arguably one of the mostapplied result in analysis.

1.1 Ramsey type results.

May be the simplest problem of this type, suitable even for elementary schoolstudents, is as follows;

Imagine there is a party of six people. Show that there are either threeof them who are mutual acquaintances or there are three who are completestrangers. Also, this statement is not true if there are only five people atthe party.

Indeed, to every person, there are either at least three acquaintances orat least three who are stranger to him. Assuming the first case, if two ofthose three are acquaintances then we’re done. Otherwise the three peopleare complete strangers (to each other). The second case is analogous.

Let us give a more mathematical formulation of the generalization of theabove; Let G be a graph on n vertices. Let E be the set pairs of the vertices,called the edges of the graph. Color the each edge by one of r possible colors.In 1930 Ramsey proved the following result

Theorem 1.1.1 For given r, k natural numbers, there exists a numbern(k, r), such that if n ≥ n(k, r) and if G is a graph of n vertices whoseedges are colored by r colors, then there are k vertices such that all edgesbetween them are colored by the same color.

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The smallest possible such number n(k, r) is called the Ramsey number.Notice that the first statement about the dinner party means that n(3, 2) =6. However to find the Ramsey numbers for larger values of k and r becomesextremely difficult. Indeed n(4, 2) = 18, but already n(5, 2) is unknown. Thebest result is the estimate: 43 ≤ n(5, 2) ≤ 49.

Here we describe a simple argument which shows that n(k, 2) ≤ 22k. LetG be a graph with n = 22k vertices whose edges are colored by two colors,say red and blue. It is helpful to assume that the vertices are ordered, thatis there is first (or smallest) vertex, a second and so on. Now let x1 be thefirst vertex, then by the pigeonhole principle there is a set of vertices G1 ofsize at least 22k−1 such that every edge from x1 to G1 has the same color.Let x2 be the smallest vertex in G1 and there is a subset G2 ⊂ G1 of size atleast 22k−2 such that every edge from x2 to G2 has the same color again.

Continuing this process we get a sequence of sets G1 ⊇ G2 ⊇ . . . ⊇ G2k

and a sequence of vertices x1, . . . , x2k such that xi ∈ Gi−1 and every edgefrom xi to Gi has the same color. The point is that if i < j then the colorof the edge between xi and xj depends only on xi which we can define to bethe color of the vertex xi. By the pigeonhole principle again, there are atleast k of the points x1, . . . , x2k of the same color, and thus all of the edgesjoining these points have the same color too. 2.

Note that although the argument is simple nevertheless it uses two basicideas, the pigeonhole principle and the recursive structure of the problem.Next we show the lower bound: 2k/2 ≤ n(k, 2), which is due to Paul Erdosand is a nice example of his ”probabilistic method”. First we present itwithout the language of probability, as a counting argument. These twobounds together tells us the order of magnitude of the Ramsey numbersn(k, 2).

Consider all colorings of the edges of a graph on n vertices. Since there

are: (n2

) = n(n−1)2 edges, the number of all possible colored graphs is: N =

2(

n2

)

. Let us denote these graphs by G1, . . . , GN . For each colored graphcount the number of monochromatic subgraphs of k vertices and denote thisnumber by ni. The idea is evaluate the average:

A =1N

N∑

i=1

ni (1)

and show that A < 1 if n ≤ 2k/2. Indeed then there must be a graph Gi

2

which contains 0 monochromatic subgraphs of k vertices. If k vertices arefixed then then there are two monochromatic graphs on them. Each one is

contained in 2(

n2

)−(k2

)

colored graphs, since the

rest of the edges can be colored arbitrarily. Since there are (nk

) sets of

k vertices, we have:

N∑

i=1

ni = 2 · 2(

n2

)−(k2

)

· ( nk

) (2)

and thus the average number of monochromatic subgraphs is

A = 21−(

k2

)

· ( nk

) (3)

Now it is easy to check that this number is less then 1 if n ≤ 2k/2 2

This argument simplifies considerably if one uses the - usual - language ofprobability. Lets consider the collections of all colored graphs as a ”randomgraph” constructed in way such that each edge is colored independently, redwith 1/2 probability (and then also blue with 1/2 probability). For a set ofk vertices, the probability that all edges between them are of the same color

is 21−(

k2

)

and hence the expected number of monochromatic subgraphs is

A = 21−(

k2

)

· ( nk

). If this number is less then 1, then there must be a

coloring with no monochromatic subgraph on k vertices. 2

The probabilistic method has become extremely useful in combinatoricsoften rendering hard problems to almost trivial. We discussed it since it willgive us the right heuristics in Roth’ theorem.

1.2 Van der Waerden’s theorem.

A result of similar nature, which is essentially more difficult but was provedearlier (in 1927), roughly says that if the natural numbers are colored byfinitely many colors then there are arbitrarily long monochromatic arith-metic progressions. The precise statement due to Van der Waerden is asfollows;

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Theorem 1.2.1 For given r, k natural numbers, there exists a numberW (k, r), such that if N ≥ W (k, r) and if the first N natural numbers arecolored by r colors, then there is an arithmetic progression of k numbers ofthe same color.

In order not to go into combinatorics too deeply we only present theestimate W (3, 2) ≤ 330, that is if the first 330 natural numbers are coloredby 2 colors, then there are 3 numbers of the same color forming an arithmeticprogression. We don’t claim that this is even close to the truth, however theorder of magnitude of the numbers W (k, 2) is far from known. In fact thebest bounds obtained by combinatorial arguments, due to Shelah (in 1987!) can be described as follows. Define a tower function T inductively byletting T (1) = 2 and T (k) = 2T (k−1) for k > 1. Then define a function Rby R(1) = 2 and R(k) = T (R(k − 1)). Shelah obtained the bound of theform: W (k, 2) ≤ R(Ck) with C > 1 being some fixed number. Note thateven with C = 1, R(4) = T (216) which is a 216 high tower of powers of 2,and this was already a huge improvement on previous bounds!

Going back to our estimate for W (3, 2), partition the first 330 numbersinto 66 blocks, where each block consists of 5 consecutive numbers. Eachblock can be colored in 25 = 32 ways, thus by the pigeonhole principle,there must be two of the first 33 blocks which are colored exactly the sameway. Let us denote them by B1 and B2, and at the same time let B3 denotethe block for which the first numbers of the 3 blocks form an arithmeticprogression.

Arguing indirectly, in any block the colors cannot alternate, since thenthe first, third and fifth entries in the block would be of the same color.Thus there must be two consecutive numbers in the block of the same colorand then the adjacent number must have the opposite color. Let us assumethat these 3 consecutive numbers in blocks B1 and B2 have colors Red, Blueand Blue, the other cases are completely analogues. But then the sameentries in block B3 must have colors Blue, Red and Red, to avoid the samecoloring of the same entries in the 3 blocks, which would form an arithmeticprogression. Finally notice that the third entry from B1, the second from B2

and the first from B3 are all colored Blue, forming an arithmetic progression,which is a contradiction. 2

To notice the connection between the two results, note that Ramsey’stheorem can be translated in the following way. If the edges of a full graphare partitioned into finitely many classes, then one class also contains a fullsubgraph, while that of Van der Waerden’s says that if a long arithmetic

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progression is partitioned into finitely many classes then one class also con-tains an arithmetic progression, that is a similar (but a much shorter) copyof itself. In general Ramsey type results are roughly about to show, thatregularity of a certain structure is preserved after partitioning it into finitelymany classes.

In 1937 Erdos and Turan studying the various proofs of Van der Waer-den’s theorem, arrived to the conjecture that in every sufficiently densesubset of integers must contain an arithmetic progression of length k. Thiswould be a strengthening of the above theorem, as if the numbers from 1 toN are colored by r colors then the set of numbers of some color must havedensity of at least 1/r, the density δ of a set A ⊂ {1, . . . , N} is defined tobe: δ = |A|

N .Surprisingly even for k = 3, the first proof due to Roth, appeared in

1952, and used Fourier analysis instead of combinatorics. Later in 1974,Szemeredi solved the general conjecture, and in 1976 Furstenberg reprovedit by using ergodic theory. Finally in 1998 Growers proved it again by usingharmonic analysis, obtaining for the first time exponential type bounds forboth Van der Waerden’s and Szemeredi’s theorem (for which he won theFields Medal!). After some preparations we discuss his version of the proofof Roth’ theorem.

1.3 Finite Fourier series.

Let ZN denote the set of residue classes (mod N), which can be identifiedwith the set {1, . . . , N}. Addition (mod N) makes ZN a commutativegroup, which is also cyclic, as 1 generates it, with 0 being the identityelement.

The set FN of complex valued functions: f : ZN → C can be identifiedwith the n- dimensional complex Euclidean space by assigning the vector:vf = (f(1), . . . , f(n)) to the function f . We recall some basic facts aboutthis space which remains true for the infinite dimensional function spaces aswell, to be discussed later.

Definition 1.3.1 The inner product of the functions f, g ∈ FN is definedto be

(f, g) =N∑

n=1

f(n)g(n) (4)

(where z sands for the complex conjugate of z). This has the usual properties

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Proposition 1.3.1 If f, g, h ∈ FN and λ, µ are complex numbers then

i) (f, g) = (g, f)

ii) (f, f) ≥ 0 and (f, f) = 0 if and only if f = 0.

i3) (λf + µg, h) = λ(f, h) + µ(g, h).

The proof is immediate from the definition. Note that (f, f) =∑N

n=1 |f(n)|2is the square of the length of the complex vector vf = (f(1), . . . , f(N)). Thuswe define the l2- norm of the function f by

Definition 1.3.2 For f ∈ FN let ‖f‖2 = (f, f)1/2

Next we give a proof of the cauchy-Schwartz inequality which generalizeto other function spaces too.

Proposition 1.3.2 If f, g ∈ FN then one has

|(f, g)|2 ≤ (f, f)(g, g) (5)

Proof. One can assume g 6= 0 and hence (g, g) > 0. We use that(f + tg, f + tg) ≥ 0 for all complex numbers t. Writing out this product onegets

(f, f) + 2Re t(g, f) + |t|2(g, g) ≥ 0 (6)

Taking the derivative formally with respect to t (forgetting about the realpart and the absolute value), and setting it to zero, suggests to evaluate thisexpression at the spacial value: t = − (f,g)

(g,g) . This gives

(f, f)− |(f, g)|2(g, g)

≥ 0 (7)

which we wanted to prove. 2.

The l2- norm has the usual properties

Proposition 1.3.3 For f, g,∈ FN and λ ∈ C one has

i) ‖λf‖2 = |λ|‖f‖2

ii) (triangle inequality) ‖f + g‖2 ≤ |‖f‖2 + |‖g‖2 where equality holds ifand only if f = λg or g = λf for some complex number λ.

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Here i) is obvious and ii) is obtained by evaluating ‖f+g‖22 = (f+g, f+g)

and applying the Cauchy-Schwartz inequality.

Next, we introduce a special basis for the space FN consisting of theso-called characters of the group ZN . For 1 ≤ m ≤ N , let em : ZN → C bedefined by em(n) = e2πi mn

N . Also let ω = e2πiN denote the n-th root of unity,

and note that em(n) = ωmn.From the definition it is immediate

Proposition 1.3.4

i) em(0) = 1 and |em(n)| = 1 for all m and n

ii) em(k + l) = em(k) · em(l)

i3) The functions em (1 ≤ m ≤ N) form an orthogonal basis of the spaceFN , more precisely one has

(em, en) =

{N if m = n0 otherwise

(8)

Proof. Parts i) and ii) are obvious from the definition and so is part i3)when m = n. Note, that if m 6= n then ωm−n 6= 1, and the inner product isthe geometric series:

(em, en) =N∑

k=1

ω(m−n)k =ω(m−n)N − 1ωm−n − 1

= 0 (9)

It follows that the functions em are linearly independent, indeed if∑Nm=1 λmem = 0 then taking the inner product of this sum with en one gets

Nλn = 0. They form a basis since the space FN has dimension N . 2

Note that if the δ0 denotes the Delta function, that is δ0(0) = 1 andδ0(m) = 0 if m 6= 0, then (8) can be written in the more compact form:(en, em) = δ0(n−m).

Definition 1.3.3 The Fourier transform f , of the function f : ZN → C isdefined by

f(n) = (f, en) =N∑

k=1

f(k)e−2πi nkN =

N∑

k=1

f(k)ω−nk (10)

Of fundamental importance are the following:

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Theorem 1.3.1 Let f, g ∈ FN . Then one hasi) Fourier inversion formula

f(n) =1N

∑m

f(m)ωmn (11)

ii) Parseval’s formula

∑n

f(n)g(n) =1N

∑m

f(m)g(m) (12)

i3) Plancherel’s formula

∑n

|f(n)|2 =1N

∑m

|f(n)|2 (13)

Proof. All three formulae follows from (9).Indeed, substituting the definition of f(m) in (11), one gets:

1N

∑m

∑

k

f(k)ω−mkωmn =∑

k

f(k)∑m

1N

ωm(n−k)

=∑

k

f(k) δ0(n− k) = f(n)

The same way the right side of (12) becomes

1N

∑m

∑

n,k

f(n)g(k)ωm(n−k) =∑

n,k

f(n)g(k) δ0(n− k) =∑n

f(n)g(n)

Finally, (13) is the special case of (12) obtained by taking g = f . 2

The expansion (11) of a function f is called its Fourier series, usingω = e

2πiN this takes the usual form:

f(n) =∑m

f(m)e2πi mn

N (14)

This is a finite trigonometric sum, indeed e2πi mn

N = cos(2πi mnN )+i sin(2πi mn

N )moreover if f(m) = am + ibm then both the real and the imaginary part ofthe sum in (14) takes the form:

f(n) = c0 +N−1∑

m=1

cm cos(2πi mn

N) + dm sin(

2πi mn

N) (15)

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One of the reasons that the Fourier transform: F : f → f has wide rangeof applications is that it has many algebraic properties. Among them is thefact that it takes convolutions into pointwise multiplication.

Definition 1.3.4 Let f, g ∈ FN . The convolution f ∗ g is defined by

f ∗ g(n) =∑

k

f(k)g(n− k) (16)

The summation, as always in this section is taken over elements of ZN ,unless specified otherwise. Note that f ∗ g = g ∗ f as can be seen by makingthe substitution: k := n− k in the sum.

Proposition 1.3.5 One has

f ∗ g(m) = f(m)g(m) (17)

Proof. The left side of (17) is of the form

∑n

∑

k

f(k)g(n− k)ω−kmω−(n−k)m =∑

k

f(k)ω−km ·∑n

g(n)ω−nm =

= f(m)g(m)

2

In fact we’d also need a ”twisted” version of this fact, let us define the”twisted” convolution of the functions f and g by (which has nothing to dowith the twisted convolution arising in the Heisenberg group):

f ∗ g(n) =∑

k

f(k)g(k − n) (18)

The similarly to (17) one has

Proposition 1.3.6 One has

f ∗ g(m) = f(m)g(m) (19)

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1.4 Roth’ theorem.

Hopefully by now we have both the appreciation and the preparation forthis result. It says

Theorem 1.4.1 Let δ > 0 be given, and let A ⊂ 1, 2, . . . , N be set of size|A| = δN . If N > exp exp (C

δ ) for some absolute constant C then A con-tains an arithmetic progression of length 3.

Let us tell first the idea of the proof. If x, y, z are natural numbers and ifx+y = 2z then they form an arithmetic progression of length 3, which we’llcall briefly a 3-progression. Instead we’ll count first the numbers of triplesx, y, z chosen from the set A, which satisfy the congruence: x + y ≡ 2z(mod N) , that is the equation in ZN , with the aid of the Fourier transform.This can be done as follows; let S0 denote the number of triples solving theabove congruence, then by formula (9) one has

S0 =∑

x∈A

∑

y∈A

∑

z∈A

1N

N−1∑

m=0

ωm(2z−x−y) = (20)

=1N

N−1∑

m=0

(∑

x∈A

ω−mx

) ∑

y∈A

ω−my

(∑

z∈A

ω2mz

)=

=1N

N−1∑

m=0

A2(m)A(−2m) =

= δ3N2 +1N

N−1∑

m=1

A2(m)A(−2m)

where A(m) denotes the Fourier transform of the characteristic functionof the set A. Note, that A(0) = |A| = δN gives the first term on the lastline of formula (20).

This term is quite instructive as it is the number of solutions of thecongruence if the set A is random, obtained by selecting each natural numberfrom 1 to N independently with probability δ. Indeed then after choosingsay x and z arbitrarily from the set A, which can be done δ2N2 ways, theprobability that y ≡ 2z − x (mod N) is in A is equal to δ. Thus if A isvery uniformly distributed among the numbers 1, 2, . . . , N then we expect alot 3-progressions (mod N) in it.

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Definition 1.4.1 Let δ > 0 be given, and let A ⊂ 1, 2, . . . , N be set of size|A| = δN . Moreover if α > 0 and if

|A(m)| ≤ αN for all 1 ≤ m ≤ N − 1 (21)

the set A is called α- uniform.

Lemma 1.4.1 Assume that α < δ2/2 and that A is α- uniform. ThenS0 ≥ δ3N2/2, that is A contains at least δ3N2/2 3-progressions ( mod N ).

Proof. Note that in Roth’ theorem one can assume N is odd, otherwisesimply consider A ⊂ {1, 2, . . . , N + 1}, δ changes only slightly. Then fromPlancherel’s formula (13) and from (21) one has:

N−1∑

m=1

|A2(m)A(−2m)| ≤ αNN |A| = αδN3 ≤ δ3/2N3

The Lemma follows from the last line of formula (20). 2

The second part of the proof is to show that if the set A is not α-uniform, then there is a (long) arithmetic progression in which its density isincreasing. More precisely we’ll show

Lemma 1.4.2 Let N > 0, α > 0 be given and assume that N > 16/α.Moreover let A ⊂ ZN and suppose that |A(r)| ≥ αN for some r 6= 0. Thenthere exists an arithmetic progression P ⊂ {0, 1, . . . , N−1} of length at least(αN)1/2/30 such that: |A ⋂

P | ≥ (δ + α/4)|P |.

Let us introduce the following notation, if n is a natural number thenlet:

|n|N = mins≡n (mod N)

|s|

that is the magnitude of the smallest element of the residue class of n(mod N). It is immediate to check that

i) |t + s|N ≤ |t|N + |s|Nii)|ts|N ≤ |t|N |s|NThe proof of the above Lemma is based on the following

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Proposition 1.4.1 Let N > 0, α > 0 be given and assume that N > 16/αand let 0 < r < N .

Then the set {0, 1, . . . , N −1} can be partitioned into arithmetic progres-sions P1, . . . , Pm such that for every j:

(i) 18(αN)1/2 ≤ |Pj | ≤ 1

4(αN)1/2

(ii) If n1, n2 ∈ Pj then |n1r − n2r|N ≤ αN8π

Proof. Let 0 < t < N be a parameter whose value will be chosenlater. Consider the set: {0, r, . . . , (t − 1)r} ⊂ ZN ' {0, 1, . . . N − 1}. Bythe pigeonhole principle, there are 0 ≤ k1 < k2 < t such that the distancebetween k1r and k2r is at most N/t. Thus for k = k2 − k1 one has

|kr|N ≤ N/t

Now, divide the set {0, 1, . . . N − 1} into residue classes (mod k).If P is an arithmetic progression consisting of M consecutive elements

of a residue class, then for n1, n2 ∈ P one has n1 − n2 = sk with |s| ≤ Mand thus

|n1r − n2r|N ≤ |s| · |kr|N ≤ (MN)/t (22)

To satisfy (ii) we need that:

M ≤ αt

8π(23)

The number of elements in each residue class mod k is [N/k] (or [N/k] + 1) thus to be able to divide each into sub-progressions of approximately Mconsecutive elements we need only that

2M ≤ N/t (24)

since we can concatenate the last two sub-progressions. If we choose:

t = 4(

Nα

)1/2and M = 1

8(αN)1/2 then both (23) and (24) are satisfied andthe Proposition is proved. 2

Proof of Lemma 1.4.2 If χA stands for the characteristic function of Aand 1 is the constant 1 function, then let us call fA = χA− δ1 the balancedfunction of the set A. Since 1 = δ0 and |A| = δN we have: fA(0) = 0 andfA(r) = A(r) for r 6= 0.

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Now, let P1, . . . , Pm be the partition of {0, 1, . . . , N − 1} into arithmeticprogressions given in Proposition 1.4.1. Then

αN ≤ |fA(r)| = |∑n

fA(n)ω−nr| ≤m∑

j=1

|∑

n∈Pj

fA(n)ω−nr| (25)

The points is that for n1, n2 ∈ Pj one has

|ω−n1r − ω−n2r| = |1− e2πi(n1−n2)r

N | ≤ 2π

N|n1r − n2r|N ≤ α

2(26)

using ω = e2πN and (ii) of Proposition 1.4.1. Thus for every j

|∑

n∈Pj

fA(n)| ≥ |∑

n∈Pj

fA(n)ω−nr| − α

2|Pj | (27)

It followsm∑

j=1

|∑

n∈Pj

fA(n)| ≥ α

2N (28)

On the other handm∑

j=1

∑

n∈Pj

fA(n) =∑n

fA(n) = 0 (29)

thus the sum in j of the non-negative sums:∑

n∈PjfA(n) is at most

αN/4. The there must be an index j for which:∑

n∈Pj

fA(n) = |A ∩ Pj | − δ|Pj | ≥ α

4|Pj | (30)

and this proves the Lemma. 2

We have to pass from (mod N) to true arithmetic progressions, thisis based on the following observation and a slight modification of Lemma1.4.1.

Proposition 1.4.2 Let |A| = δN be a subset of the first N natural numbers.Let A0 = A ∩ [N/3, 2N/3), A1 = A ∩ [0, N/3), A2 = A ∩ [2N/3, N). Theneither

(I) |A0| ≥ δ4N

or(II) |Ai| ≥ 3δ

8 N for i = 1 or i = 2.

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Proof. Since |A0| + |A1| + |A2| = N then if (I) is not the case then:|A1|+ |A2| ≥ 3N/4 and (II) follows. 2

Note that if x + y ≡ 2z (mod N) and x, z ∈ A0 and y ∈ A then also:x + y = 2z thus x, y, z is a true 3-progression. Indeed, then y ≡ 2z − x(mod N) and 0 < 2z − x < N . Let S denote the number of such triplesx, y, z.

Lemma 1.4.3 If A is α- uniform with α ≤ δ2

32 and if |A0| ≥ δ4N then:

S ≥ δ3N2

8 .

Proof. One writes - as in Lemma 1.4.1

S =1N

N−1∑

r=0

A0(r)A(r)A0(−2r) =

=δ3N2

16+

1N

N−1∑

r=1

A0(r)A(r)A0(−2r) =δ3N2

16+ E

one uses the Cauchy-Schwartz inequality and Plancherel’s formula toestimate the error term E:

|E| ≤ α

(∑r

|A0(r)|2)1/2 (∑

r

|A0(−2r)|2)1/2

≤ αδN2 ≤ δ3N2

32

and the Lemma follows. 2

Note that so far we haven’t excluded the trivial 3-progressions x = y = z,fortunately there are only |A| = δN of them. If N > 32

δ2 then the aboveLemma guarantees the existence of a non-trivial 3-progression. Let us sum-marize what we have shown so far.

Lemma 1.4.4 Let δ > 0 and N be a natural number such that N > 128δ2 .

Let A ⊂ {1, 2, . . . , N} with |A| = δN .Then either A contains a non-trivial 3-progression, or there exists an

arithmetic progression P ⊂ {1, 2, . . . , N} such that

(i) |P | ≥ δ50N1/2 and

(ii) |A ∩ P | ≥ (δ + δ2

128)|P |

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Proof. We can assume |A0| ≥ δN4 otherwise take P = [0, N/3) or

P = [2N/3, N) by Proposition 1.4.2. If A is α = δ2

32 - uniform then Acontains a nontrivial 3-progression. Otherwise - by Lemma 1.4.2 - thereexists an arithmetic progression P satisfying both (i) and (ii). 2

Now an easy iteration argument finishes the proof of Roth’ theorem -with constant C = 310 -

Proof of Theorem 1.4.1 Let us argue indirect. Assume that N >exp exp (300

δ ) and that A does not contain any 3-progression (which isequivalent of saying that it does not contain any arithmetic progression atall).

Let P be the arithmetic progression given by Lemma 1.4.4. Let usidentify P ' {1, 2, . . . N1} and A1 ' A ∩ P - by simply enumerating theelements of P in increasing order. It is clear that A1 does not contain a(non-trivial) 3-progression as well. If |A1| = δ1N1 then by Lemma 1.4.4 onehas

(i) N1 ≥ δ50N1/2 thus: log log (N1) ≥ log log (N)− 1

since N is extremely large w.r.t. 1/δ (Check it!).

(ii) δ1 ≥ δ + δ2

128

Now, using Lemma 1.4.4. for the set A1, and then repeatedly k = 128δ

times we arrive to a set Ak ⊂ Pk with density: δk ≥ δ + δ = 2δ since thedensity at each step is increasing by at least δ2

128 .Then after another 128

2δ steps the density is increasing from 2δ to 4δ. Itis clear that in less then 256

δ = 128δ + 128

2δ + . . . steps the density is increasingbeyond 1 (in fact to infinity - similarly as in Zenon’s paradox !) which isabsurd, unless the process stops because the length’s Nk of the arithmeticprogressions Pk become too small - we need Nk > 128

δ2 in each step. Howeverfrom (i) it follows that:

log log (Nk) ≥ log log (N)− k ≥ 64δ

if k ≤ 256δ since log log (N) ≥ 310

δ by our assumption. This proves theTheorem. 2

15

1.5 The Fast Fourier Transform.

If N is a natural number, then let FN denote the Fourier transform offunctions defined on ZN , that is:

FNf(m) =N−1∑

n=0

f(n)ω−nmN (31)

where ωN = e2πiN is the N -th root of unity.

If N is large an important practical problem arises which is to computethe Fourier transform using as few elementary operations (such as multipli-cations and additions) as possible.

If M(N) denotes the minimum number of multiplications needed to com-pute FNf of any function f : ZN → C, then a naive count tells us thatM(N) ≤ N2. Our aim is to discuss the following

Theorem 1.5.1 (Cooley-Tukey 1965) Let N = 2k be a power of two. Thenthe Fourier transform FNf of any function f : ZN → C can be computedby performing at most: M(N) ≤ N(log2 n − 1) multiplications.

The algorithm behind the above theorem is the so-called Fast FourierTransform (FFT), and has turned out to be extremely useful in applica-tions, such as in signal processing. It has been implicitly used by manymathematicians, arguably even by Gauss in 1805!

It is based on the following

Lemma 1.5.1 One has M(2N) ≤ 2M(N) + 2N .

Proof. For f : {0, 1, . . . , 2N − 1} → C lets denote its restrictions toeven and odd numbers by: fe, fo : {0, 1, . . . , N − 1} → C where

fe(n) = f(2n) , fo(n) = f(2n + 1)

Also since ω22N = ωN one has by definition if 0 ≤ m < N :

F2Nf(m) =N−1∑

n=0

f(2n)ω−2nm2N +

N−1∑

n=0

f(2n + 1)ω−(2n+1)m2N = (32)

N−1∑

n=0

fe(n)ω−nmN + ω−m

2N

N−1∑

n=0

fo(n)ω−nmN = FNfe(m) + ω−m

2N FNfo(m)

16

while for N ≤ m′ < 2N by writing m′ = N + m one gets exactly thesame way (using ωN

2N = −1)

F2Nf(m′) = FNfe(m)− ω−m2N FNfo(m) (33)

Now to compute FNfe and FNfo, one needs 2M(N) multiplications andthen 2N additional multiplications are needed to compute ω−m

2N and theproducts ω−m

2N FNfo(m). This proves the lemma. 2

Proof of Theorem 1.5 Let N = 2k and proceed by induction on k.For k = 1 one has: F2f(0) = f(0) + f(1) and F2f(1) = f(0) − f(1) so

M(2) = 0For the induction step k → k + 1 one has by the above lemma:

M(2k+1) ≤ 2M(2k) + 2k+1

M(2k+1)2k+1

≤ M(2k)2K

+ 1 ≤ k − 1 + 1 = k

and this is what we wanted to show. 2

Next we discuss a ”cheap” way of multiplying polynomials of large degreeand also large numbers using the FFT.

Let p(x) =∑P

n=0 anxn and q(x) =∑Q

m=0 bmxm be polynomials of degreesP and Q. Then their product r(x) = p(x)q(x) is of the form: r(x) =∑R

k=0 ckxk with R = P + Q and for 0 ≤ k ≤ R one has

ck =∑n

anbk−n where 0 ≤ n ≤ P and 0 ≤ k − n ≤ Q (34)

Again a by naive count, to compute the coefficients ck one would firstcompute all the products anbm using (P + 1)(Q + 1) multiplications. Theidea is that formula (34) looks like the convolutions of functions on ZN

defined in (16) which is transformed into point-wise multiplication by theFourier transform.

Let N be a power of 2 such that: P +Q < N ≤ 2(P +Q) and define thefunctions: f, g : ZN → C by:

f(n) =

{an if 0 ≤ n ≤ P0 if P < n ≤ N − 1

and similarly

g(m) =

{bm if 0 ≤ n ≤ Q0 if Q < n ≤ N − 1

17

Then it is easy to check that

ck = f ∗ g(k) =N−1∑

n=0

f(n)g(k − n)

Note that the difference k−n is computed in ZN that is (mod N), so if k < nthen k − n := N − (n− k). Using the fact that: f ∗ g(m) = f(m)g(m) onecomputes the convolution f ∗ g by applying the FFT twice to get f and g,then using N multiplications one gets f ∗ g and by one more application ofthe FFT gives f ∗ g. Thus we have

Corollary 1.5.1 The product polynomial r(x) = p(x)q(x) can be computedby using no more than

M ≤ 3N log2 N − 2N (35)

multiplications, where N ≤ 2(P + Q).

18

EXERCISES.

Problem 1. Let a1, a2, . . . , aN be a sequence of positive numbers, andlet:

S =1N

N∑

j=1

aj and S2 =1N

N∑

j=1

a2j

a) Show that

S2 − S2 =1N

N∑

j=1

(aj − S)2

and thus the Cauchy-Schwarz inequality.

b) Assume that the inequality is almost sharp, that is there is a (small)β > 0 such that: S2 − S2 ≤ β. Show that:

|{n : |an − S| ≥ β1/4}| ≤ Nβ1/2

that is for most n’s, an ≈ S.

Problem 2. Consider the Fourier transform on functions: f : ZN → C.Define translation and modulation as follows: Tkf(n) = f(n−k), Mkf(n) =ωnkf(n).

a) Show that: ˆTkf = M−kf , Mkf = Tkf

b) (Uncertainty Principle) Let I ⊂ {1, . . . , N} be an interval. We say fis supported on I if f(n) = 0 for all n /∈ I.

Assume that f is supported on I and f is supported on J . Show that|I| · |J | ≥ N/4.

Hint: Prove indirect. Use translations and modulations to obtain morethen N disjoint translates of the rectangle I×J . Show that the correspond-ing functions are linearly independent.

Problem 3. Prove Roth’ theorem for δ > 2/3 by an elementary argu-ment. In general try to prove the theorem for as small δ as you can (withoutusing the Fourier transform).

Problem 4. Instead of Proposition 1.4.1, one can use the followingargument to finish the proof of Roth’ theorem.

19

Let r 6= 0 and assume that |f(r)| ≥ αN . Choose M = [αN1/2]. Bythe pigeonhole principle there is a k ≤ N1/2 such that |kr|N ≤ N−1/2. LetP = {k, 2k, . . . ,Mk}. Show that

a) |P (r)| ≥ M/2

b) If f denotes the balanced function of the set A, then

f ∗ χP (m) = |A ∩ (m + P )| − δ|P |

, where ∗ denotes the twisted convolution and m + P = {m + k : k ∈ A}.

20

2 Infinite Fourier series.

In this section we develop the basic theory of Fourier series of periodicfunctions of one variable, but only to the extent that is needed for themain applications discussed below. These will include linear and non-linearpartial differential equations of interest, such as the non-linear Schrodingerequation (GNLSE), and Ergodic theorems.

2.1 Basics of Forier series.

Let f : R → C be a continuous function, we say it is periodic of period Tif f(x + T ) = f(x) for all x. If I = [a, a + T ] is an interval of length Tand f is continuous on I such that f(a) = f(a + T ) then f has a uniquelyextends to a periodic continuous function on R. Also by a simple scaling:f(x) = f(xT

π ), f becomes a function of period π.Now let C(2π) denote the space of continuous functions of period π,

identified with those of f : [−π, π] :→ C such that f(−π) = f(π).Of special importance are the complex exponentials:

en ∈ C(2π), en(x) = einx = cos(nx) + i sin(nx) (36)

Similarly to the discrete case one has

(en, em) = δ(n−m) =

{1 if m = n0 otherwise

(37)

where (f, g) denotes the inner product of the functions f and g

(f, g) =12π

∫ π

−πf(x)g(x) dx

Indeed if n 6= m then

(en, em) =12π

∫ π

−πen(x)em(x) dx =

12π

∫ π

−πei(n−m)x dx =

ei(n−m)x

n−m|π−π = 0

Note that i) − i3) of Proposition 1.3.1 is valid for the inner product(f, g) of functions in C(2π), and so does the Cauchy-Schwarz inequality(Proposition 1.3.2), since it follows only from properties i)− i3) of the innerproduct. Again analogously to the discrete case we define the L2- norm by

21

Definition 2.1.1 Let f ∈ C(2π), then the L2- norm of f is defined by

‖f‖2 = (f, f)1/2 = (12π

∫ π

−π|f(x)|2 dx)1/2 (38)

This norm satisfy i) − ii) of Proposition 1.3.3, that is the triangle in-equality and is homogeneous. The analogy with the discrete case is takenfurther to define

Definition 2.1.2 Let f ∈ C(2π), then the for k ∈ Z the k-th Fourier coef-ficient of f is given by

f(k) = (f, ek) =12π

∫ π

−πf(x)e−ikx dx (39)

and the function f : Z → C is called the Fourier transform of f .

The usefulness of the Fourier transform again is because it has many nicealgebraic properties (p.e. it diagonalizes differential operators, see below),and thus certain problems become ”simpler in Fourier space”, involvingf , g, ... instead of f, g, .... But to use this one has to have a way of expressinga function in terms of its Fourier transform. To see how such a formula lookslike we introduce the spaces of trigonometric polynomials

Definition 2.1.3 A function s(x) ∈ C(2π) of the form

s(x) =∑

|k|≤N

ckeikx =

a0

2+

N∑

k=1

ak cos (kx) + bk sin (kx) (40)

is called a trigonometric polynomial of degree (at most) N . The space oftrigonometric polynomials of degree N is denoted by MN .

Note that by the above definition MN is the linear subspace of C(π)spanned by the exponentials ek(x) −N ≤ k ≤ N . These are orthogonaland hence are linearly independent, thus they form an ortho-normal basis(o.n.b) of MN .

Proposition 2.1.1 Let f(x) = MN , then one has

f(x) =∞∑

k=−∞f(k)eikx (41)

22

Proof. Let f(x) ==∑|k|≤N cke

ikx then by orthoganility:

f(l) =∑

|k|≤N

ck(ek, el) = cl

for |l| ≤ N and equal to 0 otherwise, and this is the content of the proposi-tion. 2

The left side of (41) is the so-called Fourier series of the function f(x),and the question: to what extent formula (41) remains true for all continuousfunctions is of basic importance in harmonic analysis. To be more precisewe define

Definition 2.1.4 Let Letf ∈ C(2π), its N -th partial Fourier series aredefined by:

SNf(x) =∑

|k|≤N

f(k)eikx (42)

It is known for a long time that there are functions f ∈ C(2π) suchthat SNf(x) diverges at an infinite number of points x, however it is onlyrelatively recent that the set of these exceptional points - where SNf(x) →f(x) as N →∞ is NOT true - is of measure zero, i.e. it can be covered bya countably many intervals whose total length is arbitrary small. This wasproved by L. Carleson in 1967, and its subsequent proofs (by C. Fefferman,Lacey-Thiele) had a major influence on the development of Fourier analysis.

We will discuss some (much easier) results in this direction, namely that‖SNf − f‖2 → 0 as N →∞ and that f(x) can be recovered by an easy wayfrom the sequence SNf(x). Moreover assuming that f(x) has a continuousderivative, the pointwise convergence: SNf(x) → f(x), is true at everypoint.

We start by some simple consequences of orthogonality.

Proposition 2.1.2 Let f ∈ C(2π) and s ∈MN . Then

‖f − s‖2 ≤ ‖f − SNf‖2 (43)

Proof. Write f − s = f − SNf + SNf − s. Note that by definition

(SNf, ek) = f(k) = (f, ek)

for all |k| ≤ N and thus

(f − SNf, ek) = 0

23

for all |k| ≤ N , which then remains true for any trigonometric polynomialin place of ek. In particular

(f − SNf, SNf − s) = 0

and thus by the ”Pythagorean theorem”

‖f − s‖22 = ‖f − SNf‖2

2 + ‖SNf − s‖22

which proves the proposition. 2.

The geometric meaning of the above is that SNf is the orthogonal pro-jection of f to the linear subspace MN . As a corollary we get

Theorem 2.1.1 (Bessel’s inequality) One has for every N

‖f‖22 ≥

∑

|k|≤N

|f(k)|2 (44)

Proof. Taking s = 0 in the above proposition, one gets

‖f‖22 ≥ ‖SNf‖2

2 = (SNf, SNf) =

=∑

|k|≤N, |l|≤N

f(k)f(l) (ek, el) =∑

|k|≤N

|f(k)|2

by orthogonality. 2

Corollary 2.1.1 (Riemann-Lebesgue lemma) Let f ∈ C(2π), thenf(k) → 0 as |k| → ∞.

Indeed this is immediate from the fact the the series:∑

k∈Z |f(k)|2 isconvergent. Note that the series

∑k∈Z |f(k)| may very well be divergent,

however this is not the case if make some ”smoothness assumptions” on f .

Definition 2.1.5 Let f ∈ C(2π), we say f is n-times continuously dif-ferentiable if the n-th derivative f (n) ∈ C(2π) as well. The space of suchfunctions will be denoted by C(n)(2π).

The Fourier coefficients of functions in C(n)(2π) have characteristic de-cay, based on the following important property, which is also the startingpoint of the applicability of Fourier analysis to differential equations;

24

Lemma 2.1.1 Let f ∈ C(n)(2π). Then one has

f (n)(k) = (ik)nf(k) (45)

Proof. We proceed by induction. For k = 1 using integration by parts,one has

f ′(k) =12π

∫ π

−πf ′(x)e−ikx dx = − 1

2π

∫ π

−πf(x)(e−ikx)′ dx =

= ik12π

∫ π

−πf(x)e−ikx dx = ikf(k)

then one applies the same argument to f (k)(x). 2

Now, we discuss our first convergence result, due to Chernoff

Theorem 2.1.2 Let f ∈ C(1)(2π), then there exists a function S(x) ∈ C(π)such that for every x The Fourier series

S(x) =∑

k∈Z

f(k)eikx (46)

is absolutely convergent.

25