Harmonic Analysis: A Gentle Introduction

In applications, a multiresolution analysis is a mathematical framework for decomposing a signal into components of different scales. The spaces V correspond to these scales. If any V,, is given then by (iv), any other V,,, can be found because Hence, a multiresolution analysis is completely {f (2"`-"t) I f E determined by the scaling function V (sometimes called the "father wavelet"). Because the translates of cp give us an orthonormal basis for Ve, the translates of p (2"t) gives us an orthonormal basis for each V". Unfortunately, however, the union of these bases is not, in general, a basis for L2 (R). To obtain an

6. Wavelets

109

orthonormal basis of wavelets, we construct a new sequence of subspaces as follows. For each fixed n E Z let Wn = {w E Vn+1 I `iv-, v) = 0 for all v E Vn}; Wn is called the orthogonal complement of V,, in Vn+1 Every vector in Vn+1 can be written, in just one way, as a sum of one vector from Vn and one from Wn (proved below). To indicate this, we write Vn+1 = Vn ® Wn.

We can repeat this process. Vn = Vn-1 ® Wn_1 and so Vn+1 = Vn-1 03 Wn-1® Wn. Continuing in this way we find that every vector in L2 (R) can be written as an infinite sum of vectors one from each of the spaces Wn. In symbols

L2 (R) = ® Wn. These spaces are mutually orthogonal and f (t) E W,, if, n=-oo

and only if, f (2-"t) E W. We see from this last fact that once we find a function whose translates give us an orthonormal basis for WO, we can obtain

from that function an orthonormal basis for every Wn, and because of the mutual orthogonality of the Wn the union of these bases is an orthonormal basis for L2 (R).

The problem of finding a wavelet can be now outlined as follows. Given a multiresolution analysis {V,,}, and recall this means we are given a scaling function V whose translates constitute an orthonormal basis for VO, construct the sequence {Wn } and find a function 0 whose translates constitute an orthonormal basis for WO. The function , is a wavelet. It turns out that such a 1/i exists for every multiresolution analysis, and the proof of this fact actually gives a method of constructing 0. We have:

Theorem 1. Let {V,,} be a multiresolution analysis with the scaling ao

(-1)k hl_k'P (2t - k), where hk =

function gyp. Then the function z
f

-00

oo

p (t) V (2t - k)dt is a mother wavelet; i.e., {291{' (2nt - m) I m, n E Z} 00

is an orthonormal basis for L2 (R).

Perhaps the simplest example of the process discussed above is found by

using the Haar scaling function H (t), where H (t) = 1 for 0 < t < 1 and H (t) = 0 for all other t. It is easy to see that the integer translates of H (t) give us an orthonormal set in L2 (R). These translates generate the space VO. The function H (t) has its support in [0, 1], whereas the function (2tt- n) H has its support in f 2 , n 2 111. These overlap when n = 0 in the set 10,11 and

when n = 1 in the set

r 12,111

. Thus for the Haar function ho = hl = 1 and

(x) = w (2x) - p (2x - 1). The function 0 (x) is called the Haar wavelet. As we have already said, there are many others and not all of them have compact support. For more about this fascinating subject and for a proof of Theorem 1, we refer the reader to the references listed in Notes.

Chapter Three

110

Fourier Series in Hilbert Space

We end this section with a proof of a property of Hilbert spaces that played a key role in the construction above.

Lemma 1. Let (H. <, >) be a Hilbert space, let M be a closed linear subspace of H, and let y E H\M. Then there is a unique vector into E M such that 11V - ;no ll =inf { 11 y - mll f in E MI. Furthermore, ( y - trco, m) = 0 for all

mEM. Proof. Let d= inf { I y- m I I in E M} and choose a sequence { nt }

such that d =

CM

Then, by Exercises 3, problem 1b, we may write

11 y - mn 11

T$

1

-

-;nr)II2

II(y -mk) +

-

I2

211 y

- mk I + 211 y - mt 112. We may write this as -mkl12+211

JIM1-mkll=211

-hill'-II2y-(m1+ink)112.

The very last term in this expression can be written as 4

M1 +Ink y

in, +'nt.

2

is in M. It

2

mr

follows that 4 II y

mk II

2

Imr-'nkll2<2Iy-..

> 4d2 and so V2+211

-.nr112-4d2.

Recalling the way {mn} was defined we see that our last inequality implies that {m++} is a Cauchy sequence (Section 2, Definition 3). Because H

is a Hilbert space this sequence converges, because {mn} C M and M is closed, its limit, say m0, is in M (Section 2, Lemma 1). Thus 11 y

lim ly -

- mOII

d.

Let us now show that m0 is unique. Suppose we had two vectors m1i m2, in M such that II y X11. Then, as above 11 y

=d= 1 I(y -m1) + (y -m2)112+II(y -m,)

-

2

11V

-

m'111

2 11-y

- (V -»m2)112

- m2112. This may be written as

6. Wavelets

II"a2-ml112 =4d2-4

of the proof 4

ml + m2 Y

, but as we discussed in the first part

2

ml + m2 Y

111

2

2

> 4d2 and so IIm2 - ml II

2

< 0, giving us

m2=ml. Finally, we must show that (y - mo, m) = 0 for all in E M. Let us set

y o = y - mo. Note that for any in E M and any scalar 6, ino + dm is in M, and so

d2IIp-(rno+bm)II2-IIp0-bm1I2=(po-bmo-bm>

=

p0II2-6(m,po) -3(po,m) +161211'nl12. I2

Because I I p ° I I2

= p - mo II

I

= d2 this becomes

(*) 0 < -b (m, po) - b (po.'n) + Now suppose

(p0,

m)

1612

IImII2.

0 for some m E M. Then m could not be zero

(Exercises ,3roblem p la) and so we may set 6 = (p0'm ) IImII2

Putting this into (*) we obtain 0<

-2l

which is a contradiction. IImII2

Iiml2

Imll2

Theorem 2. Let (H, (,)) be a Hilbert space and let M be a closed linear subspace of H. Then M1 = 1; E H I < ; , m >= 0 for all m } is a closed linear subspace of H. Furthermore, (a) M n M1 = J0 }, and (b) each v E H can be written in one, and only one, way as a sum of a vector in M and a vector

in M1. Proof. The proof that M1 is a closed linear subspace we leave to the reader (Exercises 6, problem 1). If v E MnM1, then (v, v> = 0, which gives v = 0 (Section 3, Definition 1c).

For any v E H\M, there is a unique moM such that II

° - mo = II

mf { II v - m II I m E M } and, as we showed in Lemma 1,

(v - mo, m) = 0 for all m E M. Thus v - mo is in M1, i.e., v - ma = p for

112

Chapter Three

Fourier Series in Hilbert Space

some vector p E M1. It follows then that mo + p, where mo E M, p E M1. Suppose now that v = mo+ where nao E M. p' E M. Then + p,

- m')

(p - p). Call this vector h. Clearly h E M, = because h = rrao and h E M1, because h = p' - p. Thus, by (a) h = 0 and we haveino=rao, p = p mo + p' so (me

Exercises 6 1. Let (H, (,)) be a Hilbert space and let M be a closed linear subspace of H.

(a) Show that M1 = (w E H I (w, v) = 0 for all V E m} is a closed linear subspace of H.

(b) Define M11 to be (M1)1 and show that M11 = M.

Notes 1. A discussion of the foundations of quantum mechanics can be found in Introduction to Hilbert Spaces with Applications, by Lokenath Dehnath and Piotr Mikusinski (Academic Press, 2nd edition, New York, 1999). See Chapter 7. 2. A proof of Theorem 1 (Section 5) can be found in Functional Analysis, by Carl L. DeVito (Academic Press, New York, 1978), pp. 92-93.

3. Wavelets are discussed in the book listed in note 1 above. See Chapter 8, and for a proof of Theorem 1 (Section 6) see page 441 (their notation is slightly different from ours.) Wavelets are also treated in Wavelets and Operators, by Yves Meyer (Cambridge University Press, Cambridge, 1992), in A Mathematical Introduction to Wavelets, by P. Wojtaszczyk (Cambridge University Press, Cambridge, 1997), and in Wavelets-Algorithms and Applications, by Yves Meyer, translated and revised by Robert D. Ryan (Society for Industrial and Applied Mathematics, Philadelphia. 1993). This last reference contains an interesting discussion of the

functions of Riemann and Weierstrass (see the discussion at the end of Section 4, Chapter 1) and their connections to wavelets. See pages 115118.

Chapter Four

The Fourier Transform

In the last three chapters we have been concerned with the Fourier series and the problem of recapturing a function from its Fourier series. Here we shall change our point of view and investigate the properties of a certain mapping between pairs of "function spaces." By the term "function space" we mean a collection of functions that is a vector space over C when addition and scalar multiplication are defined in the "usual" way. The mapping, called the Fourier transform, is the object of interest here.

There are a number of natural questions that arise at this point. Why change the emphasis in this way? What, if any, is the advantage in doing so? At this early stage these questions are hard to answer. Suffice it to say that the Fourier transform plays a central role in many areas of engineering, of physics, of chemistry, and even, in recent times, of astronomy. We discuss some of these applications later. So the effort involved in understanding this mapping is well justified. Moreover, the associated mathematics is fascinating and very beautiful.

1.

The Fourier Transform on Z

We recall the space e'(Z) = {a : Z --, C I EIa(n)I < oo} defined in Chapter Three, Section 1, example iv, and the space C(T) of continuous functions on the circle. There is a natural mapping between these spaces: For each a E t' (Z)

we set ¶ (a) = E' a(n)eine and note that because E I a(n)I < oo, the series converges uniformly over T to an element of C(T). The map 3 is called the Fourier transform on Z. Its properties are easily proved, and so our results here provide us with a model for the discussion of this mapping in more complicated settings; i.e., we shall seek analogues of the theorems proved here for the Fourier transform on R and on Z (defined later). 113

Chapter Four

114

The Fourier Transform

Lemma 1. The map 3 : fl(Z) -' C(T) is linear and has an inverse. Proof. For any a, b in fl (Z) we have (a + b) (n) = a(n) + b(n) for all n E Z. Thus 3'(a + b) = E[(a + b)(n)Jeine = E[a(n) + b(n)Jeine = E[a(n)eine + b(n)einoJ

= Ea(n)eine + Eb(n)eine =%a) +%b)

It is easy to see that given A E C, 3(Aa) = l(a) for every a E t'(Z) and so 3' is a linear mapping. We have already noted that for any a E CI (Z) the series Ea(n)eine converges uniformly over T. If we denote this sum by a(8), then (Chapter One, Section 5) it follows that

a(n) =

r a(8)e-inede,

for every n E Z

Thus 3` is one-to-one and, in fact, a'' exists and is given by -' (a{B}J = {

f_a(e)e-t9do}_oo 27r

It is convenient to denote 3(a) by a(8), or just a, and we shall often use this notation below.

Recall that fl(Z) and C(T) both have norms (Chapter Three, Section 2, Definition 1). For each a E £'(T), IIaIII = E Ia(n)I, and for each f E C(T), Illll - = maXOET If(a)1

Lemma 2. The map 9' : fl (Z) - C(T) is continuous and, in fact, is norm decreasing; i.e., Ilalla 5 IIaIII for all a. Proof

I E a(n)einel <_ > la(n)I Ilein°II. 5 > Ia(n)I = IIaIII The continuity of

now follows from Theorem 1 of Chapter Three, Section 2.

There is a very useful operation on tI (Z) called "convolution." It is a kind of multiplication.

Definition 1. For any a, b in £'(Z) and any n in Z, we define (a * b)(n) to be Zm=_x a(n-m)b(rn). This sequence {(a*b)(n)}n _00 is called the convolution of a and b, or a convolved with b, and is denoted by a * b.

1. The Fourier Transform on Z

115

We must show that a * b is in L1(Z) whenever a and b are both in this space. The next lemma proves that and gives us a bound for the norm of the convolution.

Lemma 3. For any a, bin tl (Z) the convolution, a*b, is in @1 (Z) and IIa*bII 1 IIaII1

Ilbill.

Proof. Because Ilalll = E O Ia(n)I we have, for any fixed m, Ilalll = r_ Ia(n m)I. Now

IIa*bII1=EI(a*b)(n)I =EIEa(n-m)b(m)I -00

n

m

Ia(n - m) I Ib(m)I = E Ib(m)I(E Ia(n - m)I) m

m

n

n

Ib(m)I = IIaIIIIlbIIi

= Ilall1 M

The function e(n) = 1 when n = 0, e(n) = 0 when n 54 0 is clearly in t' (Z). Note that a * e = e * a = a for all a and, further, 'e(O) = 1 for all 9 E T.

Two functions f, g in C(T) can be multiplied. We simply set (f g)(9) _ f (9)g(9) for all 9 E T and observe that f g E C(T). The Fourier transform, besides preserving sums and scalar multiples (Lemma 1), also preserves "multiplication." More precisely:

Lemma 4. For any a, b in P1(Z), 3'(a * b) = Q (a) 8'(b). In another notation: Proof. In the expression

* b(9) = 1

{an

- m)b(m) } eine

we set k = n - m to obtain

E E a(k)b(m) I ei(k+m)o = >2 eike {>a(k)b(m)etme} =

{a(k)e'} {>b(m)eIme} = a(B) ,

(9)

m

To see how all of this might be useful in the study of Fourier series, consider

the following result of N. Wiener: Suppose that f E C(T) is never zero, and

116

Chapter Four

The Fourier Transform

the Fourier series of f is absolutely convergent. Then the Fourier series of 1 is absolutely convergent. As one might expect, a proof of this result using classical methods is quite difficult. The reader may want to try to find one. Our discussion of the Fourier transform, however, enables us to reformulate the theorem as follows:

It is clear that a function in C(T) is in the range of ) if, and only if, its Fourier series is absolutely convergent. Thus the hypothesis of Wiener's theorem

tells us that there is an a E e' (Z) such that a = f. Suppose now that we can find bE e' (Z) for which a * b = e. Then £(a * b) = a b = :3`(e) = 1; i.e., a(9) . b(O) = f (9) . b(9) = 1 for all 0. Because f (O) is never zero, this says that b = 1. Thus 1 is in the range of !a, it is equal to b, and hence its Fourier series is absolutely convergent. In other words Wiener's theorem, a theorem about Fourier series, is equivalent to the following: If a E e'(Z) and a is never zero, then there is a b E t'(Z) such that a * b = e; i.e., if a E e'(Z) and a never vanishes, then a has a "multiplicative inverse" in e' . We shall investigate this result more fully in the next section. To begin, let us define our terms.

Definition 2. An element a of e' (Z) is said to be invertible if there is an element b in e' (Z) such that a * b = b * a = e. We call b the (see Exercises 1, problem 2(e)) inverse of a and denote it by a'.

It is clear that if a E e' has an inverse, then a is never zero, because a* a-' = a(a-') = F = 1. What is difficult is showing that when a is never zero, a has an inverse. We end this section with a proof that (e' (Z), II II 1) is a Banach space; i.e., every Cauchy sequence of points in this space converges to a point of this space (Chapter Three, Section 2, Definition 3).

Lemma 5. The space (e' (Z), II 11 1) is a complete normed space. -

Proof. Let {a'}'1 be a Cauchy sequence in e' (Z) and, given e > 0, choose N so that IIa"-am1I1 <e whenever both m and n exceed N. Then for any fixed jo and any fixed k with jo < k, we have k

00

Ia"(jo) -a' (jo) I <_ E Ia"(.?) - am(J)I 5 E Ia"U) - am(j)l< E

j=-k

00

whenever both m and n exceed N. It follows that {a"(j)}n is a Cauchy 1

sequence in K for each fixed j. Because K is complete, this sequence converges.

1. The Fourier Transform on Z

117

Let b(j) = liiun_x a" (j) for each j E Z. Now things get a little bit Inessy. We let

bk =I ..,O.b(-k),b(-k+l), .b(O),b(1),. ,b(k),0,. -} and we let, for each it.

bk = {... ,O,a"(-k),... a"(0) ... a"(k) 0 ...} Then bk and bk are in f'(Z) and we have k

IIbkIII <- Ilbk-bkII1+IIbkIII <-E+:Ia"(j)l J= -k

whenever it > N. If we fix n > N and let k tend to infinity we find that 00

_x

Ib(J)I < e + IIa"III

showing that b = {b(j)}' 1o is in f'(Z). Finally, we shall show that line".-.00 IIa" - b111 = 0. As we saw above, for any fixed k, k

j=-k

la"(j) - am(j)l < e

whenever both in and n exceed N. Letting in go to infinity we get k

E la"(j) - b(j)I < e

j=-k

whenever it > N. Now choose, and fix, it > N and let k go to infinity. We find

IIa" - bIlI se whenever n > N.

Exercises 1

1. Let a E f'(Z) and, for a fixed in E Z. let a...(n) = a(n - in.) for all n. Show that a,,,(0) = e+1mea(g)

118

Chapter Four

The Fourier Transform

* 2. For any a, b, c in e' (Z) and any A E C show that:

(a) a*b=b*a: (h) A(a * b) = (Aa) * b = a * (Ab);

(c) a*(b*c)=(a*b)*c;

(d) a*(b+c)=a*b+a*c. (e) Show that the inverse of a, when it exists, is unique.

* 3. Let {b,, } C e' and suppose that lim b = bo for the e'-norm. Show that lima * b = a * bo for any a E 0. If a E 1' is fixed, show that the map Ta : f' --+ V defined by TQ(b) = a * b is linear and continuous.

4. Let C E f' be defined as follows: c(n) = 1 if n = 0 or n = 1, and c(n) = 0 for all other n.

(a) Show that (c * a)(n) = a(n) + a(n - 1), here a E t' is arbitrary. (b) Let c' = /c, c2 = c * c, .... Show that for any fixed m > 0 Cm(n) _ (n) = n.(+n, n .' 5. Define g E 1'(Z) by letting g(1) = 1, g(n) = 0 for n # 1.

* (a) Let g' = g, gz = g * g, ... and show that fork fixed, gk(n) = 1 when n = k and gk(n) = 0 when n 34 k.

* (b) Let g`' (n) = I when n = -1, g- (n) = 0 for n # 1. Show that g-' is the inverse of g.

* (c) Compute (g-')k for each positive integer k.

* (d) For any fixed a E 1' show that a = Eo c a(n) gn where convergence is for the e'-norm; here go = e. (e) Given a, b in 1' use the series representation of these elements given in (d) to find a series representation for a * b.

(f) Use the representation given in (d) to show that to every continuous linear map cp : t' (Z) - C there corresponds a unique sequence {cn } such that

i.
6. If f E C(T) is never zero, show that 1 is in C(T).

2. Invertible Elements in el (Z)

2.

Invertible Elements

119

in £'(Z)

We recall that e(n) = 0 when n 54 0, e(0) = 1, and an element a of 1l is invertible if there is an element a-1 E P1 such that a * a-1 = a-1 * a = e. For a E Q1, a 0 0, we set a° = e, a1 = a and, for any integer k > 1, we set ak=a*ak-1

Theorem 1. Let a E 1'(Z) and suppose that Ilalll < 1. Then the element (e - a) is invertible. Furthermore, if a 36 0,

(e -a)-' = E a j=0

where convergence is for the 11-norm.

Proof. For each n > 1 let us set bn =

j=0 aj and consider the sequence

{bn} C 6. When r > n we have Ilbm-bn111=11

a'1115 j=n+1

Ila!111

j=n+1

M

E Ilalli (Lemma 3 of Section 1 of this chapter) n+1

E Ilalli < lllallllalll n+1

because the series of real numbers Eo Ilalli is geometric, and because Ilalll < 1,

it converges to 1'u

But, again because Ilalll < 1, limn_.a, Ilalli+1 = 0

showing that {bn} is a Cauchy sequence in (11, II Ill). By Lemma 5 of Section 1, this sequence converges to an element of 0 that we shall denote by cc

F,

aj

j=0

We shall now show that this sum is the inverse of (e - a). We have

(e-a)*

(ai) =(e-a)* (Iima3) r

=slim {(e - a) * -00

(Exercises 1, problem 3)

0

n

n

= noo lim

E aj]

aj j=0

aJ+1

j=0

lim (e. - an+1) = e

120

Chapter Four

The Fourier Transform

By Exercises 1, problem 2(a), reversing the factors give the same result.

Corollary 1. Let a E e1(Z) be invertible and suppose that b E e1 satisfies IIa - bll I < T' . Then b is invertible. IF,

Proof. We have Ile-a-1*bII, = IIa-1*(a-b)II, 5 IIa-1II,IIa-bllI < 1 by Lemma 3 of Section 1 and our hypothesis. Thus, by Theorem 1, [e- (e-a'1 *b)] = a-1 *b is invertible. Call this c. So c = a-1 * b and c is invertible. But then b = a * c, and this last convolution is invertible (its inverse is c-1 * a-1). It follows from Corollary 1 that the set of all invertible elements of e1 is open

and hence that its complement, what is sometimes called the set of singular elements of e1, is a closed set.

Definition 1. Recall that a linear map from e1(Z) into C is called a linear functional on e1 (Z). A linear functional gyp, that is not identically zero, is called a multiplicative functional if cp(a * b) = V(a) Wp(b) for all a, b in t1 (Z). If yp is a multiplicative functional, then W(a) 34 0 for some a. Consequently, cp(a) = cp(a * e) = v (a) ,p(e), and we see that W(e) must be one. Furthermore, if b is invertible, then Wp(b) 54 0 and, in fact, cp(b-1) = w . Theorem 1 tells us

more about these functionals.

Corollary 2. If V is a multiplicative functional on e1 (Z), then for every a in this space we have IV(a)I S I1all1. In particular, every multiplicative functional on e1 (Z) is continuous on this space (Chapter Three, Section 2, Theorem 1).

Proof. We argue by contradiction. Suppose that for some a we have Icp(a)I > llall1. Set A = ,p(a) and note that because I X 1 < 1, Theorem 1 tells us that 0 showing that because O is a (e - j4 0 is invertible. But then cp (e linear map, V (z) 34 1. This contradicts the definition of A. I

I I

Let us now choose, and fix, a multiplicative functional cp on e1 and consider M(cp) = {a E e1 I v(a) = 0}. The properties of M(W) are easily proved but worth listing.

(a) M(V) is closed. C M(c,) and lima,, = ao. Then Suppose clearly this gives us cp(ao) = 0; i.e., ao E M(ap)-

and

(b) M(V) is a linear subspace of e1, M(ap) 0 e1, and if N is a linear subspace for which M(V) 9 N, then either M((p) = N or N = V.

2. Invertible Elements in 1'(Z)

121

Because c is linear, it is clear that M(W) is a linear subspace of e', and

because p is not identically zero, M(W) 54 L'; in fact, we may choose ao E el such that p(ao) 0. Now given b E V set a = o . Clearly v(b - aao) = 0 so b - aao = c E M(AO), and we see that b = c + aao. Now suppose M(ap) C N. If M(W) = N, we are done. If there is a b E N\M(cp), then b = c + aao for some scalar a and some c E M. Because a 34 0 and

N is a linear subspace, we see that ao E N. But then N = el as claimed. (c) If a E M(V) and b E t', then b* a E M(ap). Because v(b * a) =,p(b) v(a) this is obvious.

Definition 2. A nonempty subset I of e' (Z) is said to be an ideal in el (Z) if (a) 1 is a linear subspace; (b) whenever a E I and b E t' (Z), b * a is in I. An ideal I is called a proper ideal if 1 # L'(Z). Finally, a proper ideal is called a maximal ideal if the only ideals that contain it are itself and V (Z).

We observe that a proper ideal cannot contain e nor can it contain any invertible element. Thus if I is a proper ideal, then I is contained in the set of singular elements of el (defined just before Definition 1). Because this last set is closed, we see that the closure (Chapter 3, Exercises 2, problem 2) of I is contained in it. We leave it to the reader to show that the closure of an ideal is an ideal and we have shown:

Lemma 1. The closure of a proper ideal in t' is a proper ideal in el. In particular, every maximal ideal in t' is closed. Let A be a nonempty family of nonempty sets. A subfamily N of A is said to be a nest if for any sets A, B in N either A C B or B C A. A nest M is called a maximal nest if the conditions (i) N is a nest and (ii) M C N together imply that M = N. So a maximal nest M is as large as possible; more precisely, if

A¢ M, then there must be a set B E M such that neither A C B nor B C A is true. There is a useful equivalent form of the axiom of choice (Chapter Zero, Section 2) that was first proved by Hausdorff. Maximum Principle: Given any

nest N in A there is a maximal nest M in A such that NC M.

Lemma 2. Every proper ideal in £1(Z) is contained in a maximal ideal. Proof. Let I be a proper ideal in el(Z) and let Z be the family of all proper ideals in this set. Then {I} is a nest in Z. By the maximum principle there is a maximal nest M in Z that contains {I}. It is convenient to index the ideals in M (Chapter Zero, Section 2, just after Definition 3). So we may write

M = {I,, Ia E A}, and we know that because {I} C M, I = IQ for some a E A. Now let J = IJ0EA IQ (Chapter Zero, Section 2, just after Definition 3). It is clear that I C J. We shall show that J is a maximal ideal.

122

Chapter Four

The Fourier Transform

Our first task is to show that J is a linear subspace of t' (Z). So given a, b in J we must show that any linear combination of these two vectors is in J. But

we have a E Ia and b E Is for a,,3 in A. Because M is a nest, either Ia C 1q or Is C I,,. Let us suppose the former. Then a and b are in II, and because Is is a linear subspace of e' (Z), any linear combination of a and b are in Iq. But Ip C J so we see that J is a linear subspace of e' (Z). Next we must show that for any a E J and any c E el(Z), a * c is in J. Now a E Ia for some a and because I, is an ideal a * c is in this set. Thus a * c E IQ C J; and we have shown that J is an ideal in t' (Z) that contains I. Now .1 = VQEA to and each Ia is a proper ideal. Thus e ¢ Ia all a E A, and so e tf J and we see that J is a proper ideal. Finally, we must show that J is a maximal ideal. Suppose J' is a proper

ideal in tl(Z) and suppose that J C X. Then M U {J'} is a nest in I and M C M U {J'}. Because M is a maximal nest, we must have M = M U {J'};

thus J' E M and so J' = I.y for some ry E A. Clearly, we have reached a contradiction because Iy C J. Our discussion above, see (b) and (c), shows that the null space of any multiplicative functional is a maximal ideal in el (Z). Using some abstract algebra and some complex function theory (Liouville's theorem), one can show that the converse is true. Because we are not supposing that our readers are familiar with these topics we shall only state: (*) Any proper closed ideal of i'(Z) is contained in the null space of a multiplicative functional on el (Z).

Theorem 2. Given a maximal ideal M in el(Z), there is a 0 E T such that

M= {aEel(Z)Ia(O)=0}. Proof. Let cp be a multiplicative functional on el such that M = {a E e' cp(a) = 0}. Such a cp exists by (*). Now recall the element g defined in Exercises 1

1, problem 5: g(1) = 1, g(n) = 0 for n 34 1. Because Iw(a)I S IIafl, for every a Eel and because light = 1, we see that A = (p(g) must satisfy IAI < 1. But IIg'II i = 1 so 111 < 1 also and we conclude that A = W(g) = e+i6 for some fixed 0 E T.

Now for any a E t' we have

p(a) =

Ckim

k-00

a(n)gnJ k

= lim

k-00

00

a(n)ene

a(n)W(g)" =

-k

(-k a(n)ger/

-00

Thus V(a) is just a evaluated at 8, and so M = {a I cp(a) = 0} = {a E Il a(6) = 0} as claimed.

3. The Fourier Transform on IR

123

Theorem (N. Wiener). If f E C(T) is never zero and if the Fourier series of f converges absolutely, then the Fourier series of 1 converges absolutely. Proof. We cannot claim to have here a complete proof because we simply stated (*) and it is crucial. We are only giving a sketch. As we have seen we can choose

b E t' such that b = f and our hypothesis tells us that b is never zero on T. Suppose that b does not have an inverse in tl, and let I = {a * b I a E tl}. Clearly, I is an ideal and because e ¢ I, it is a proper ideal in 1'. Thus there is a multiplicative functional V such that I C M = {a E t' I W(a) = 0}. Then for some 0 E T we must have M = {a E t1 I a(9) = 0}. But this says b(9) = 0,

which is a contradiction. It follows that b is invertible in t1; i.e., there is a b-1 E 11 such that b-1 * b = e. Clearly, (b-1 * b) = F or (b'1) f = 1. Exercises 2

*1. For any fixed aEtl(Z),letI(a)={a*bIbEi'(Z)}. (a) Show that I(a) is an ideal in t'(Z) that contains a. (b) Show that I(a) is a proper ideal if, and only if, a is not invertible. (c) Show that the closure of a proper ideal is a proper ideal. 2. For any a E e1(Z), let o(a) = {A E CI (a - \e) is not invertible). We call a(a) the spectrum of a, and we call C\o(a) _- p(a) the resolvent set of a. (a) Show that o(a) is a closed set by showing that p(a) is an open set.

(b) Show that any \ E C such that Al I> Ilalli is in p(a) and hence a(a) c {A E Cl Al I<- IIalli}.

3.

The Fourier Transform on IR

The theory here is very much like that for the case of Z except that the definitions, and the proofs of the theorems, involve subtleties not found in the discrete case. Let us begin informally, pursuing, as much as possible, the analogy with V (Z). Because this space consists of all a : Z - C such that la(n)l < oo; we "should" define V(R) to be

{f

1700 If(t)Idt
Assuming that this is a vector space we set, for all f E L' (R),

IIfIli =

lf(t)ldt

Chapter Four

124

The Fourier Transform

With a "proper" interpretation of the integral, we can show that (L' (R), 11

11 1)

is a Banach space. (See Chapter Three, Exercises 5, problem 3(c) for a construction of this space.) The integral needed is that of Lebesgue, an extension of the classical integral of Riemann. We shall not assume a knowledge of this integral and so our proofs of the theorems stated below are only "sketches"; i.e., they are incomplete. For each f E L1(IR) we define

f (x) = f 00f (t)e{xtdt Observe that f is also defined for all real numbers. This is a problem, and it is wise to distinguish the domain of f, which is R, from that of f . We shall do so by agreeing that f is defined on a "second" real line that we shall denote by IR.

For each a E f1(Z), a is a continuous function on the unit circle; i.e., a is an element of C(T). Let C(R) be the vector space of all continuous functions on A. Then we have:

Lemma 1. For each f E L1(R), ?E C(A). Moreover, the map :3` that takes each f to f is linear and one-to-one. Proof. Let f E L1(R) be given. Then 00

rx

f (x + h) = f f(t)e&(:+h)tdt =

J

00

Hence

etnt f(t)es=tdt

x

x

f (x + h) - f(x) = f (eiht - 1)e'xtf(t)dt and

x If(x+h) - f(x)I <_ f lf(t)I

Iesht

- lldt

As h --+ 0 so does Ic'ht - lI and, one can show, so does the integral. Thus f E C(R).

It is easy to see that a : L' (R) - C(R) is a linear map. To show that 3 is one-to-one, we need only show that one can recapture f E L1(R) from a knowledge of its Fourier transform f . To do this one uses a kind of Cesaro convergence for integrals. We have R

f (t) = lim

,

`l7f

fR

(1

I

lf J

(x)e-iztdr

3. The Fourier Transform on R

125

where convergence is for the Li-norm (compare this to Chapter One, Exercises 6, problem 3).

Note. When f happens to be integrable this formula gives

f

O0

M) =

27r

f (x)e-'=tdx

which is valid in most practical applications.

Lemma 2. For any f E L' (R),

f (x) = 0. Furthermore,

maxIf(x)I 5 Ill Iii

zER

XER

Proof. The first statement is called the Riemann-Lebesgue theorem. It generalizes the theorem of Riemann given in Section 5 of Chapter 1 (Theorem 1). We have

f (X) = J f (t)e''tdt and

- f (X) =

r x etz(t+;) f(t)dt = r

J 0

00

e4--f(t - - )dt X

J

Hence

21f(x)l 5 f

If(t) - f(t - -)Idt

For any f E L'(R) this last integral tends to zero a IxI - 00. We leave the second statement to the reader. Let us illustrate our results by calculating the Fourier transform of some specific functions. In these examples a is a fixed, positive, real number.

(a) Let U(t) = 1 when t > 0, U(t) = 0 when t < 0. Set f (t) = e-atU(t).

f

00

If (t)Idt =

f

00

e-atdt =

1

a

o

so f is in L'(R). Now 00

f (X) = f e-atU(t)etztdt = f 00 e(-a+'=)tdt 00

= ar-. lim

o

e(-a+iz)M

l -a+ix I

-

1

1

-a+ixj

=

1

a-ix

Note that f is continuous and that liml=I-00 If(x)I = line

a

1

+z

= 0.

126

Chapter Four

The Fourier Transform

(b) Let la(t) = 1 for ItI

a, 1a(t.) = 0 for Iti > a. eiax - e- sax

ja

la(x)

e'xtdt =

= 2sinax x

ix

a

Again we note that the Fourier transform is continuous and tends to zero as 1X I - oc.

(c) Referring to example (b), let k(t) = (1 - ItI)lI(t). Because k(-t) = k(t) we see that

k(x) =

J

I

k(t)cosxtdt = 2 J I (1 - t)cosxtdt

I

o

(1 - cosx1 4 y =2I x2 J- x2 sin

stn(a)l2 xl k2) - L x/2 J

(d) Consider the Abel kernel f (t) =

f(x) =

J

x x fu e(a+:Z)tdt + e-ahuIetxedt _ e(-a+u)tdt J x x fo 1

= lim M-.-oc La+ix

-

e(a+ix)M

a+ ix

I + liIn

e(-a+ix)N

1

N-oc -a + ix - I\-a+ix

1

J

2a

a2+x2 (e) Here we simply note that when f is differentiable we have

d f (x)

x = dx J

f

jut f °`

which is the Fourier transform of i t f (t). More generally, Fourier transform of (it) n f (t).

(t)]etdt

I (x) is the

In the Space PI (Z) we found a kind of multiplication. There is an analogous operation here.

Definition 1. For any f, g E L' (R) we define (f * 9)(x) = foc f (x - y)9(y)dy Lemma 3. For any f ,gin LI (R) the function f *g is in L I (R) and Ill * g (I I < Ilf III 119111.

3. The Fourier Transform on R

127

Proof

Of * 9II = /1(1 * g)(x)ldx = <

f f dx

ff I

f(x - y)9(y)dyl ds

If(x-y)II9(y)Idy= f I9(y)Idy I If(x - y)ldx

= IIgIIl fCQ If (x - y)ldx = IIgII1 Ill Iii

assuming the interchange in the order of integration is valid.

Lemma 4. For any f, g in L1(R) we have `£(f *g) = t3(f)a(g) or, in another notation, (f * g) = f . 9 Proof (f *9)(x) = f (f * 9)(t)e,=tdt = f e42C (f f (t - u)g(u)du) dt setting w = t - u this becomes

f

f

etx(w+u)du f f (w)g(u)dw =

e`xug(u)du / etxw f(w)dw

=9(x)-f(x)

J

A typical application of the Fourier transform is to solve differential equations. As an illustration of this consider

- t t2 + n2u(t) = f (t) where f (t) is assumed known. Applying !J' we get

3 [- dt2 +

n2u]

(:x)2u + a2u =

(x2

+ a2)u = f

Thus u(x)

= (x2 + a2) f (X) = g(x)f(X)

where, from (d),

g(t) =

1

e-aItl

2a

It follows, then, from Lemma 4 that u (t)

e-a2nltl

1

* f(t)

2n

foo

e-alt alf(s)ds

128

Chapter Four

The Fourier Transform

Exercises 3 1. Let f (t) = e2*t I I (t). Compute f (x).

2. Let f E L' (R) and let a, h be real numbers. Show that (a) For a j6 0, (b)

Q[f(at)] = Tjf(p);

f(t - It)] = e+imhf(x);

(c) Qr[eiat f(t)] = f (x + a); (d)

at) f (t)) =' f (x - a) +'f (x + a).

3. For any f, g, h in L' (R) and any A E C show that

(a) f *g=g*f; (b) A(f * g) _ (Af) * g = f * (Ag);

(c) f * (g * h) _ (f * g) * h;

(d) f *(g+h)= f *g+f *h. *4. Show that there is no function e(t) E L'(R) such that e * f = f * e = f for all f E Ll (R). Hint: If e exists, then e * e = e. Now apply the Fourier transform.

4.

Naive Group Theory

We can get a deeper understanding of the Fourier transform by using some ideas from modern algebra. These are usually discussed abstractly, but, we believe, it may be better to begin with some concrete examples.

Definition 1. A nonempty subset A of C is said to be an additive group if (a - b) eA whenever both a and b are in A.

It is clear that C, R, Q, Z and even {0} are additive groups. The last of these is called the trivial group. We might also recall that, for any function f : R -+ K, the set P (f) is an additive group (Chapter One, Section 2, Lemma 1). It is a nontrivial group when, and only when, the function f is periodic. Many of the properties of P (f) are shared by any additive group.

Lemma 1. Let A be an additive group. Then (i) (kA; (ii) for any aeA the number (-a) is in A; (iii) when a and b are in A so is (a + b); (iv) for any aeA the set {na I neZ} is contained in A.

4. Naive Group Theory

129

we must have (a - a) Proof. We may assume that A is nontrivial. Then for in A, which proves (i). Now that we know OeA, (0 - a) is in A for every aeA, proving (ii). To prove (iii) we note that if a, b are in A, then a, -b are in A, and

so a - (-b) is in A. Finally, let aeA and suppose that na is not in A for some positive integer n. Let no be the smallest positive integer such that noa /A.Then by (iii) we see that a+a = 2a,-A; hence no > 2. It follows that (no - 1) aeA because of the way no was defined. But, then, again using (iii), (no - 1) a + a = noa is in A, which is a contradiction. We have shown that naEA for every positive integer n. By (i), however, 0 a is in A, and by (ii) (-na) = (-n) a is in A for every n. Thus {na I nEZ} C A as claimed. When A, B are additive groups and B C A we shall say that B is a subgroup of A. In particular, when aeA the set {na I nEZ}, which is clearly an additive group, is a subgroup of A.

Clearly, every additive group is a subgroup of C, but it is often useful to specify that we are working with a subgroup of R or a subgroup of Z.

Corollary 1. Let A be a subgroup of the additive group R. If A contains a smallest positive member, say ao, then A = {naa I neZ}. Proof. We have seen that {nao I neZ} C A. Given b&A suppose, first, that b is positive. Then ao < b, and we may write b = kao+r where k is a positive integer and 0 < r < aa. Because both b and kao are in A, it follows that rcA and hence that r = 0. Thus be {nao I neZ}.When b is negative, (-b) is a positive member

of A and so (-b) = kao for some integer k. But then b = (-k) ao and we are done.

Corollary 2. If A is a subgroup of the additive group Z, then either A = {0} or A = {ne I neZ} where a is the smallest positive member of A.

Definition 2. A nonempty subset M of C is said to be a multiplicative group if xy-1eM whenever both x and y are in M. It is clear that C\ {0} , R\ {0}, and Q\ {0} are multiplicative groups. So are the sets { 11 1, -1 and T (the unit circle). We have an analogue of Lemma 1.

Lemma 2. Let M be a multiplicative group. Then (i) If M; (ii) for any xEM the number x-1 is in M; (iii) when x and y are in M so is xy; (iv) for any xeM the set {x" I neZ} is contained in M.

When M, N are multiplicative groups and N C M, we shall say that N is a subgroup of M. In particular, when xeM the set {x" I neZ}, which is clearly a multiplicative group, is a subgroup of M.

130

Chapter Four

The Fourier Transform

It is now necessary to specify the arithmetic operation under which a given set is a group. Unless it is clear, from the context, what operation we mean, we shall denote an additive group by (A, +) and a multiplicative group by (M, .).

The exponential function, f (x) = e2, maps R onto the positive real numbers. The set of all positive real numbers, let us denote this by R+, is clearly a multiplicative group. Thus f maps (IR, +) onto (IR+, ) and, furthermore, f (x + y) = f (x) f (y) for all x, y in R. This particular function is, of course, both one-to-one and onto. Let us look at this kind of mapping in a slightly more general context.

Let (A, +) be an additive group, let (M, ) be a multiplicative group, and let f : A M be a function such that f (a + b) = f (a) f (b) for all a, b in A.

Because 0 ' M and f (a) = f (a + 0) = f (a) f (0), we see that f (0) = 1.

But then 1 = f (0) = f (a + (-a)) = f (a) f (-a), giving us the fact that f (-a) = f (a)-1. Using these observations we can show that the range of f is a subgroup of (M, .). To do this suppose x, y elements in the range of f are given. We must show that xy-' is in the range of f. But we have a, b in A such that f (a) = x and f (b) = y. Thus (a - b) fA, because A is an additive group,

and f (a - b) = f (a + (-b)J = f (a) f (-b) = xy-1, proving our claim. Let us now show that {arA I f (a) = 1} is a subgroup of (A, +). Given a, b in this set (i.e., given a, b in A such that f (a) = 1 = f (b)) we must show that a - b is in this set (i.e., we must show that f (a - b) = 1). But f (a - b) = f (a + (-b)) = f (a) f (-b) = f (a) f (b)-1 = 1 and we are finished. Exercises 4 * 1. Prove Lemma 2.

+ 2. The function f (x) = In x maps IR+ onto It Moreover, for any x, y in IIt+ we have f (xy) = f (x) + f (y). This particular function is one-to-one and onto. Now let (M, ) be a multiplicative group, let (A, +) be an additive

group, and let f : M - A satisfy f (xy) = f (x) + f (y) for all x, y in M. (a) Show that the range of f is a subgroup of (A, +). (b) Show that {xfM I f (x) = 0} is a subgroup of (M, ). (c) Show that U4 = {±1,±i} is a multiplicative group. (d) Let f : Z -. U4 be defined as follows: f (n) = in for every n. Show that f (m + n) = f (m) f (n) for all m, n in Z. Identify the range of

f and If (n)=1}. (e) Let g : Z -' U4 be defined as follows: g (n) = (-1)" for every n. Show that g (m + n) = g (m) g (n) for all m, n in Z. Identify the range of g and {nfZ I g (n) = +1} .

5. Not So Naive Group Theory

131

* 3. Let n be a fixed positive integer. Let U,, = {zeC I z" = 1} .

(a) Show that U. is a multiplicative group; in fact, it is a subgroup of T.

(b) Let p (x) be a given polynomial. The complex number r is a root of the equation p (x) = 0 if, and only if, p (r) = 0. We recall that

r is a root of the equation p (x) = 0 if, and only if, x - r is a factor of the polynomial p (x). The largest positive integer k such that (x - r)' is a factor of p (x) is called the multiplicity of the root r. A root of multiplicity one is called a simple root. Show that any root of p (x) = 0 that has multiplicity greater than one is also a root of p' (x) = 0 (the derivative of p (x)). Conclude that every root of x" - 1 = 0 is simple and hence that U,, contains n distinct numbers.

4. Let Q (f) = (a + b V2- I a. b are in Q 1. Clearly, Q (v C R.

(a) Show that Q (f) is a subgroup of (R, +). (b) Show that Q (v'-2) \ {0} is a subgroup of (R\ {0} , .) .

# 5. When (M, ) is a multiplicative group and

the set {x" I neZ} C M,

as we have seen. If there is an xeM such that {x" I ncZ} = Al, then we say that (M, ) is a cyclic group and we say that x is a generator of M.

(a) Show that (U4, ) is cyclic and find all of its generators. (b) Show that (U,,, ) is a cyclic group and identify a generator. Hint: Every element of U" is an element of T and hence can be written as e'0.

(c) An additive group (A, +) is said to be cyclic if there is an such that A = {na I neZ}. We have seen that any subgroup of Z is cyclic. Is Z cyclic?

5.

Not So Naive Group Theory

We have seen that for any f : R - K, the set P (f) = {aeR I f (x + ne) = f (x) for all x) is an additive group. Observe that for real numbers x, y we have f (x) = f (y) when x - y is in P (f ); for then x - y = aeP (f) and f (x) = f (y + a) = f (y). Thus f is constant on certain subsets of R. We can get a better understanding of the structure of these sets by making use of the group theory we discussed in the last section.

Definition 1. Let A be an additive group and let C be a subgroup of A. We shall say that the elements a, b of A are congruent modulo C, and we shall write a m b mod C, if (a - b) EC.

132

Chapter Four

The Fourier Transform

We shall give many examples of this very soon. For now we observe that congruence modulo C gives us a relation on the set A; it is in fact, as we shall now show, an equivalence relation (Chapter Zero. Section 2).

Lemma 1. For any additive group A and any subgroup C of A, congruence modulo C is an equivalence relation on A.

Proof. We must show that this relation is reflexive, symmetric, and transitive. To prove that it is reflexive we must show that for any aEA, a - a mod C; i.e., we must show that for any aeA, (a - a) is in C. But this is immediate because a - a = 0 and 0 is in any subgroup of A (Section 4, Lemma 1, (i)).

Next we must show that for any a, b E A such that a - 6 mod C, we also have b - a mod C. Again this is immediate because (a - b) E C and so - (a - b) = b - a is in C by Lemma 1. (ii), of Section 4. Finally, suppose that a. b, c are in A and a - b mod C, b - c niod C. We must show that a - c mod C. The first of these gives us (a - b) E C, and the second gives (b - c) E C. By Lemma 1, (iii), of Section 4, we may add two elements of C. Thus (a - b) + (b - c) = (a - c) E C. This, of course, says

a - cmodC. For any aeA, [a] = {bfA I a - b mod C} is called the equivalence class containing a (Chapter Zero, Exercises 2, problem 1). The set of equivalence classes is denoted by A/C; read "A mod C." We began our discussion by noting that for any f : R - K, the set P (f ) is a subgroup of the additive group R and that f (x) = f (y) whenever x =y mod P (f ). Thus f is constant on the equivalence classes of R; i.e., f is defined on the elements of R/P (f ). As an illustration of this, suppose f has smallest positive period 27r. Because P (f) is then {(2-,r) n I neZ }, we see that x - y mod P (f) simply means x - y = (2ir) k for some integer k. But any real number x can be written as x = (2a) k+r where k is an integer and 0 < r < 27r. Thus each equivalence class contains a number (and clearly only one such number) from the set (0, 2ir). So we may regard f as being defined on this last set. This is, of course, just another way of saying that f is a function on the unit circle. For subgroups of the additive group Z the construction of the equivalence

classes is simple and illuminating. To be specific, we take A = Z and C = {6n [ neZ}. Then a - b modC means a - b is a multiple of six. Now for any aEZ we may write a = 6n + r, where n is an integer, r is an integer, and 0 < r < 6. Thus any aEZ is equivalent to one, and clearly only one, of the numbers 0, 1,2,3,4,5. Thus A/C = {[0] , [1], [21, [31, [41, [5]}, where [0] _

{nEZI n-OmodC}.... The choice of six is simply one of convenience. We should note here that C = {6n I rzEZ} is usually denoted by (6) and the set A/C = Z/ (6) is often denoted by Z6. Furthermore, a - b mod C is often written a 6 mod 6.

5. Not So Naive Group Theory

133

Given any subgroup C of the additive group Z, we have seen that C = {ne I ncZ) where a is the smallest positive member of C (Corollary 2 to Lemma

1 of Section 4). We shall denote this subgroup by (e). Note that Z/C = Z/ (t) consists of the a equivalence classes [0], [1] , ..., [t - 1]. This set is denoted by Zt. Returning now to the case of an additive group A and a subgroup C of A, let us note that there is a natural "addition" on the elements of A/C. Given [a], [b] in this set we choose an element of [a], let us take a, and an element of [b], take b, form their sum in A and set [a] + [b] equal to the equivalence class containing a + b. To show that this is well defined, we must show that if a'f (a) and b'f (b), then a' + b' and a + b are in the same equivalence class; i.e., we

must show that (a + b) - (a' + b') mod C. This is very easy. Because a'f [a] and b'f [b] we have a - a' in C and b - b' in C. Thus (a - a') + (b - b') is in C (Section 4, Lemma 1, (iii)), which says (a + b) - (a' + b') mod C. Let us illustrate this operation for the case of Z6 = Z/ (6). To add [3] and [4] we take any element of [3], let's take 3 (you can take 45 if you prefer). and any element of [4], we'll take 4 but you can take 52 if you wish, and add them.

So, because 3 + 4 = 7, [3] + [4] = [7] = [1] because 7 - 1 mod 6 (note that 45 + 52 = 97 so [3] + [4] = [97], but 97

1 mod 6).

Following this pattern we can construct an "addition table" for Z6. It is customary, and convenient, to denote the elements of this set by 0, 1, 2, 3, 4, 5. We have: 0

1

2

3

4

0

0

1

2

3

4

5

1

1

2

3

4

5

0

2

2

3

4

5

0

1

3

3

4

5

0

1

2

4

4

5

0

1

2

3

5

5

0

1

2

3

4

5

We should stress that the elements of Z6 are not complex numbers and the

operation + is not ordinary addition. The table shows that 0 is the identity element of Z6. We mean a + 0 = 0 + a = a for every afZ6. Furthermore. given any afZ6 there is a bfZ6 such that a + b = 0. We call b the additive inverse of a. So we can define a kind of subtraction: a - b is taken to mean a plus the additive inverse of b. For example, 3 - 2 is

to mean 3 + 4 because 4 is the additive inverse of 2. Thus 3 - 2 is 1 while

3-4=3+2=5.

There is a natural map
remainder one gets when n is divided by 6. So cp6 (12) = 0, cps (11) = 5, 1, etc. Observe that, for any in, n in Z, cps (m + n) = cp6 (m) + cps (n).

134

Chapter Four

Thb Fourier Transform

We warn the reader that in writing m + n we are using ordinary addition, whereas in writing cpr, (m) + cps (n) we are using the operation defined in Zs. Interpreting this sum as an ordinary addition leads to erroneous results.

Exercises 5 * 1. Let M be a multiplicative group and let N be a subgroup of M.

(a) For any s, t in M define s =_ t mod N to mean st-'cN. Show that congruence modulo N is an equivalence relation on M. (b) The set of equivalence classes of M under the relation defined in (a) is denoted by M/N. For any Is), [t] in M/N define [s) * It] as follows: choose se [s] , to [t] and set [s] * [t] = [st]. Show that this operation is well defined; i.e., if s'e [s] and t'e [t] show that s't' - st mod N. (c) Show that the element [1] EM/N is a multiplicative identity for this set; i.e., show that [s] * (1) = [11 * (s] = Is] for all Is) in MIN. (d) For each Is] eM/N show that there is a It] eM/N such that [s] * [t] _ I11.

(e) Let 119 be the multiplicative group of all nonzero complex numbers and let N = T (the unit circle). Identify the elements of M/N.

2. Compute the addition table for Z9. Define V9 (n) to be the remainder one gets when n is divided by 9. Show that when n > 0, cpq (n) can be computed as follows: Add the digits of n. If the result is greater than 9, add its digits. Continue until you get a number between 0 and 9. 3.

(a) We refer to problem 2. Show that for all m, n in Z, cp9 (m + n) _ 'ps (m) +,p9 (n) (b) Compute W9 (2315) + V9 (7489). Show that this is Sp9 (9804).

(c) Find

I cp9 (n) = 01.

* 4. Let A be an additive group and let C be a subgroup of A. Define V : A -y A/C as follows: cp (a) = [a] for every aeA. (a) Show that cp (a + b) = cp (a) +,p (b) for all a, beA.

(b) What is {aeA I p (a) _ [0])?

6. Finite Fourier Transform

6.

135

Finite Fourier Transform

The Fourier transform plays a fundamental role in modern digital signal processing (see Notes). This application requires the handling of enormous amounts of numerical data. I have been told that a color TV picture, for example, requires the processing of about 8 million bits of data per second. What is clearly needed here is an efficient numerical technique for computing the Fourier transforms involved because, in most cases, they cannot be found analytically. The fast Fourier transform algorithm is such a technique. Its efficacy is remarkable. In some cases 1 billion computations can be replaced by 1 million, one one-thousandth of the original number. What one does is replace the Fourier transform of a function defined on R by the Fourier transform of a function defined on Ze for large e. The algorithms are then applied to this replacement. We shall concern ourselves with the problem of how the Fourier transform can be defined on Zt. Once we have done that, we investigate its properties and give some applications. In the case of the additive group Z we considered

e'(Z)={a:Z-ClIa(n)1
_

J

x a (n - m) b (m) and note we need

For a, b in e' (Z) we defined (a * b) (n) _

here the group operation in Z; i.e., n - m = n + (-m) where (-m) is the additive inverse of m. We have seen that subtraction is possible in Zt and so we can define convolution for functions on this set. Let us do that and work out a simple case in detail. To begin with let t-1

e'(Zr)= a:Zt-CjEja(n)1
Clearly, every function on Ze is in e' (Z,), so our notation is a little silly. We use it to emphasize the analogy between what we are doing here and what we did on Z.

Definition 1. For any a, bee' (Zt) we define the convolution of a and b, a * b, 1-1

a (k - m) b (m).

as follows: (a * b) (k) = m=o

Let us stress that the subtraction is to take place in Z1. To get a "feeling" for this operation, let us work out the convolution of two elements a, bin e' (Z3).

136

Chapter Four

The Fourier Transform

We have: 2

(a * b) (0) = E a (0 - j) b((j) = a (0 - 0) b (0) + a (0 - 1) b (1) + a (0 - 2) b (2) j=o

=a(0)b(0)+a(2)b(1)+a(1)b(2) because 0 - 1 is the additive inverse of 1 in Z3 and this is 2, while the additive inverse of 2 is 1. Similarly, 2

(a *b)(1) _ Ea(1 - j)b(j) =a(1 -0)b(0) +a(1 - 1)b(1)+a(1 -2)b(2) 3-o

=a(1)b(0)+a(0)b(1)+a(2)b(2) because, in Z3, 1 - 2 = 1 + 1 = 2. 2

(a*b)(2) =Ea(2- j)b(j) = a (2 - 0) b (0) + a (2 - 1) b (1) + a (2 - 2) b (2) j-o =a(2)b(0)+a(1)b(1)+a(0)b(2) Observe that if we write a and b as (ao, a,, a2) and (bo, bi, b2), respectively, then a * b = (aobo + alb, + alb2, albo + aob, + a2b2, a2bo + albs + aob2).

This last expression clearly shows that 11 (Z3) contains an identity. The function e = (1, 0, 0) satisfies a * e = e * a = a for every a. This operation is sometimes referred to as "cyclic convolution." Its basic properties are easily established.

Lemma 1. For any a, b, c in e1 (Ze) we have

(a) a*b=b*a; (b) a*(b*c)=(a*b)*c; (c) o(a*b)_(aa)*b=a*(ab)for anycEC; (d) a*(b+c)=a*b+a*c; (e) There is an element eet' (Ze) such that a * e = e * a = a for every a. The Fourier transform on Z takes each function in el (Z) to a continuous function on T; the unit circle in the complex plane. Recall that for aee1 (Z), d (9) _ a (n) e"`9. Let us examine, a little more closely, eiie for 9 fixed. We can define ce : Z T by setting ce (n) = e'ne for each neZ; remember 9 is fixed. Observe that ce (m + n) = et(m+")e = eime . ein9 = ce (m) , ce (n) for all m. n in Z. Maps like this have a name.

6. Finite Fourier Transform

137

Definition 2. A map c : Z -. T such that c (tit + n) = C (tit) c (n) for all in, nFZ is Called a character of Z.

Lemma 2. To every character c of Z there corresponds a unique real number Bc[0, 21r) such that c (n) = (cie)" for every nfZ. Proof. Because c (n) is never zero and c (n + 0) = e (n) c (0), we see that c (0) = 1. Furthermore, c (n) = c (1 + 1 + ... + 1) = c (1)" for every natural number it.

Combining these observations we have I = c(0) = c. (n + (-n)) = c(n) c(-n) and so r (n) = c (1)" for every integer it. Now c (l) E T and so c (1) = coy for some real number y. If y V [0,27r) we may write y = (2a) k + 0 where k is an integer and 0 E (0, 2a). Clearly. then c(n) = (c ,lo)" = ri(2kv+9)n = (..i2knff WO)" = (c'B)" for every n E Z. This proves the lemma because it is clear that no two 0's in (0, 2a) can give the same character of Z.

Corollary 1. There is a one-to-one correspondence between the characters of Z and the points of T. Proof. Because there is a one-to-one correspondence between the points of T and the set [0. 2r), this follows immediately from our lemma.

For aef t (Z) , a (0) = F°`,., a (n) ei"B. If we think of 0 as fixed, then we

are taking the character eie and applying it to each n. co (n) = e'r'e. then multiplying by a (n) and finally summing over the domain of a: i.e.. summing over Z. We will do something similar for functions in P (Z,). To begin, however, we must first identify the "characters of Z,."

Definition 3. Let f be a natural number and consider the set Z,. A reap (n) for all m. n y : Z, T is called a character of Z, if y (m + n) = y (m) in Z,.

Lemma 3. There is a one-to-one correspondence between the characters of Z, and the solutions to the equation zt = 1. Proof. Let ur be a complex number such that wf = 1. Define a map y : Zr - T by setting y (k) = wk for each keZ,. Clearly, y (tit + it) = ittr"k" = urn w" = y (tit) -y (it) for any in, it in Z,. Furthermore, if in - it mod F, then y (tit) = y (it). so -y is well defined and is clearly a character of Z.

Suppose now that y is a given character of Z,. Then because y (k + 0) = y (k) y (0) we see that y (0) = 1. Also. for any kfZ, we have .y (k) = y (1 + ... + 1) = )(1)k . Thus y is completely determined`t.by y (1). But y (F) = y (0) = y (1)' = I and soy (1) is a root of the equation

= 1.

Chapter Four

138

The Fourier Transform

Definition 4. Let f be a natural number. The set of all solutions to the equation zt = 1 is denoted by Ut. The elements of this set are called fth roots of unity.

We have seen (Exercises 4, problem 3) that Ut is a multiplicative group containing f elements. Let us look more closely at U4 and its multiplication table: 1

-1

i

-i

1

-1

i

-i

-1

-1

1

-i

i

-i

-i

i

1

-1

1

It is, of course, true that 14 = 1, but 4 is not the smallest positive integer for which this is true. Clearly, 11 = 1; we say 1 has order 1. Similarly, (-1)4 = 1,

but the smallest positive integer for which this is true is 2; we say (-1) has order 2. The remaining elements i and -i each has order 4; 4 is the smallest positive integer n such that i" = 1 and it is the smallest positive integer n each

that (-i)" = 1. Note that {i" I neZ} is just {1, -1, i, -i}; i.e., it coincides with U4. Thus U4 is a cyclic group and i is a generator of this group (Exercises 4, problem 5). Clearly, -i is also a generator of this group.

Definition 5. Let t be a natural number and let

The order of a, O (a), is the smallest natural number k such that ak = 1. When 0 (a) = f we say that a is a primitive fth root of unity.

Clearly, e2*'lt = cos + isin T is a primitive fth root. When f = 3, this gives us cos z3 + i sin s3 = -+i. When f = 4, we get cos 24 + i sin 21 = i.

Lemma 4. The order of each element of Ut is a divisor of f. An element w of U, is a primitive fth root if, and only if, it generates the group Ut.

and 0 (a) = k, then clearly k < f. Now f = qk + r where q and r are integers and 0 < r < k. Thus 1 = at = qk+r = (ak)° ar = (1)° ar = a'', which says. because k is the smallest natural number such that ak = 1, that Proof. If

r=0.

Now suppose that wfUt has order f. Then Ut 2 {w' I 1 < j < f} so to show that these sets are equal we need only show that the last one contains f distinct members. But if wp = w9 for 1 < p < q:5 f, then w4_y = 1 and 0 < q - p < f. This contradicts the fact that w has order f. Thus w generates Ut.

6. Finite Fourier Transform

139

Finally, let w be a generator of Ur. If 0(w) = k < f, then{w' I jeZ} = { tn' . w2, w3, ..., wk } .

and this contains k < I elements. So the order of w must

be e.

Definition 6. Let f be a fixed natural number and let ur be the primitive tth For any aff' (Zr) we define a (wk) = F'a(j) (wk)3 and call this root o?=a.0

function, its domain is Ur, the Fourier transform For the case of a function adfl (Z3) this becomes 2

a (w°) _

a (j) (w°)' = a (0) + a (1) + a (2) J_° 2

a(w) = Ea.(j)vi = a(0)+a(1)w+a(2)w2 j=0 2

a(w2) =Ea(.I)(w2)' =a(0)+a(1)w2+a(2)w4 J=O

A more convenient way to write this is

f a (w°) & (w)

1=

8 (w2)

1

1

1

1

w

w2

1

w2

w'

a (0) a (1) a (2)

Of course. because w = e2i'13(Us, 111 4 = w.

Recall that the map.F takes f' (Z) into C (T). So if act' (Z), F (a) = a is defined on T. Furthermore- F is linear, invertible, and "preserves" convolution.

The map Fr takes each function ad' (Z1) and transforms it into a function a on (It. However, it is customary, or at least very common, to regard a as also having domain Zr. This is done as follows: Recall y (k) = wk is a map from Zr to Ur that satisfies y (k + ru) = y (k) y (m) all k, m. in Zr. When w is primitive, as it is in this discussion, y is both one-to-one and onto. Thus we may identify (Zr. +) with (Ur, ). With this understanding a (w°) = a (0).

a(w) = a(1).....a W- 1) = a(f - 1). Our Fourier transform may now be written Fr (a) (k) = a (k) _ r-1

Ea(j)(III k)' _ E a (j) (c'P

in matrix notation:

a=°

j=0

a (0)

1

1

l

...

1

a (0)

a(1)

1

u}

w2

...

wr-'

a(1)

a((- 1)

1

(ur-')

(t+'1-1)2

(wr-t)r '

a(f - 1)

140

The Fourier Transform

Chapter Four

Theorem 1. For each fixed l the map .fit is linear and preserves convolution; i.e., Ft (a * b) = .fit (a) Ft (b) Proof a-1

a-1

t-1

Ft(a*b)(n) =>(a*b)(k)(wn)k = k=0

t-

=i(j)

[a(i)b(k_i)j

k=0 j=0

j)

j=0

(wn)k

(wn)k

k=0

E-1

a (9) (wn)j E

1=O b (k -

j)

(wn)k-j

= Fit (a) (n) .Ft (b) (n)

a (j) (Wk)j. Then

Let us set -'F;-1 (a) = 7

t-1

Theorem 2. For any act' (Ze) we have a (k) = .Pt (a) (k) _

EaU) (alkyl j=0

t-1 and a (k) = Ft 1 (a) (k) = 71E,& U) (YU

k)-1.

j=0

Proof. In this last sum we replace a (j) by its definition obtaining 1

F

[-1

Ea(j) j=0

(wk)J

t-1

= , j ( 'n=o- ' a (n) (wf)n

(wk)J

j=o

t-1

E a (n)

a-1

(w,i-k)i

n= n=0

When n = k this inner sum is e, and when n 96 k it is zero (Exercises 6, problem 1). Thus

l t-1 a (j) (+o j=0

k)J

= a(k)

6. Finite Fourier Transform

141

We have seen that the Fourier transform on Z,, the finite Fourier transform, maps functions defined on Z, to functions defined on the set of characters of Z,, the set Z,. Now Zl can be identified with U,, and this last set can be identified with Z,. So the transform 3r maps functions defined on Z, to functions defined on Zc. The transform is linear and so in this, finite, case it can be represented by a matrix. Moreover, the transform preserves convolution and is invertible. We shall see (Exercises 6, problem 1) that when w E U1 (i.e., when w is

an fth root of unity) and w y6 1, then E?_ow1 = 0. Now suppose that w is a primitive fth root of unity. Then for any m, 0 < m < e - 1, wm is in U, and, because w is primitive. wm 1. Hence Et_o(wm)J = 0. Let us work out the case e = 3. Recall:

(f) = f =

1

1

1

1

w

w2

f

w2 wd

1

where, of course, w4 = w. Consider now 1

1

1

1

1

1

1

w

w2

1

w

w2

1

w2

w4

1

=

w2 w°

3

1+MT+w2

1+w2+W4

1 + w+ w2

3

1+ w+,D2

1 + w2 + w4

1 + w + w2

3

we have used the fact that sv = 1/w (Exercises 6, problem 2(b)). This last matrix is just

3

1

0 0

0

1

0

0

0

1

and, it is easy to see, reversing the factors gives us the same result. Thus to invert `), we take its matrix and replace each entry by its complex conjugate and multiply the result by 3. In another notation: 2

2

%(f)(k) = f(k) _ Ef(J)(Wk)4 _ f(J)(e+ZP )k' J=0

J=0

Chapter Four

142

The Fourier Transform

and 2

2

9(j)('Dk)j = 3

(9)(k) = 3 J=o

9(3)(e-

211

)ki

i=o

Remark on Notation. Some writers define the Fourier transform in a slightly different way to obtain a symmetry between the transform and its inverse. They set

and obtain 1

t-1

72: t g(j) (CVk)j i=0

This is also done in connection with the Fourier transform on R. One writes

`3(f) = f(x) =

1 tar

r00

J

, f(t)e+ixtdt

so that l-1(9)

=

1

00 9(x)e-'xtdx

tar J_,o

We see little advantage in doing this, but the reader should be aware that it is not uncommon.

Exercises 6 + 1. Let w E Ut, w 76 1. One of the properties of tth roots of unity that helps make the fast Fourier transform algorithm so efficient is the following: E-a = 0. Follow the steps below to give two proofs of this. wJ-

(a) Set r = rJ_ow). Show that wr = r and hence, if r

0, we have a

contradiction.

(b) Factor n - 1. 2. Let W E Ut. Show that (a) w-1 = wt-1; (b) Gr = 1/w; (c) if t is a prime (an integer larger than one that has no factors other than itself and one), then every element of U, except +1 is a primitive tth root of unity.

7. An Application

143

3. Show that e'(Zt) contains an element e such that e * f = f for every f E e1(ZZ). Compute cci(e). An element f of t'(Z1) is said to be invertible if there is an element g E e' (Z,) such that f * g = e.

(a) Show that f is invertible if, and only if, Z (f) is never zero.

(b) If f is not invertible find all g such that f * g = 0. (c) f is called a divisor of zero if there is g y6 0 such that f * g = 0. Show that f is a divisor of zero if, and only if, f is not invertible. (d) If f is invertible we have an element f -' E e' (Z1) such that f * f -' = e. Show that f is unique and that -1)(n) = 1(f)(in) for

every nEZ,. 4. Write the Fourier transform on t' (Z2) in matrix form and then invert the matrix. 5.

(a) Let f E e1(Z2) be given by f = (0, 1). Compute !ZMf). Similarly, compute 3'3 (f) where f = (0, 0, 1). (b) Characterize all g E e' (ZZ) such that g * g * terms of 3't(g).

* g (e factors) is e. in

6. Given f = (ao, a,, a2), 9 = (bo, b1, b2) in e'(Z3) compute:

(a) f and

(b) f 9; (c) f * 9;

(d) (f * 9) 7. Continuous characters of R. Let c : R - T be a continuous mapping such that c (x + y) = c (x) c (y) for all x, y in R. Show that there is an element a E R such that c (x) =

(et°)= for all x E R. Hint: Recall that Q is dense in R (Chapter Zero, Section 3, Corollary 2 to Theorem 1).

7.

An Application

There are some interesting, and surprising, connections between the finite Fourier transform and the theory of polynomial equations. These were found by Lagrange in his study of the cubic and quartic. The calculations involved are a bit messy, and we shall try to present the underlying ideas first so as not to lose sight of them in the technical details. Let us start with the simplest case of all, the well-known quadratic.

144

Chapter Four

The Fourier Transform

Suppose we are given the equation

ax2+bx.+c=0

(1)

where a 96 0 and the roots are xl, 22. The French mathematician Vieta first noted the relation between the roots and coefficients. If this case they are .r1 + x2 =

(2)

a

a

(3) a As functions of x1i x2 these are clearly symmetric, meaning they are invariant when we interchange the two variables. It turns out that any symmetric function of xl, 22 can be represented in terms of the two functions given in (2) and (3); i.e., any symmetric function of the roots can be expressed in terms of the coefficients. We shall prove this only for the special cases we need for our discussion. As an example, consider 0 = (xl - 22)2. This is clearly invariant x1=1'2 =

when we reverse our variables. We can write 0 in terms of a, b, and c as follows:

A = (x1 - x2)2 = xj + x2 - 251x2 = xj +X 2 + 221x2 - 42122 = (Si + x2)2 4x122 c -b 2 b2 - 4ac

(4)

where we have used (2) and (3). Combining (2) and (4) we have the system b (5)

XI +X2 XI

a

- x2 =

a

This is easily solved, but note that it is equivalent to: b 1

1

21

a

x2

b2 - 4ac

(6) 1

-1

a

1

The coefficient matrix in (6) is the finite Fourier transform obtained using the primitive square root of unity. We may solve (1) by solving (5) or (6). In the latter case

(7)

7. An Application

145

Let us turn now to the cubic

x3+Px2+qx+r=0

(8)

with roots xl, x2, x3. Lagrange chose, and fixed, a primitive cube root of unity w and considered the form Al = (x1 + wx2 + w2x3)3

(9)

He noted that this is not symmetric. Three of the six permutations of x1, x2, x3 do leave it invariant, but three transform it into A2 = (xl +w2x2 +wx3)3

(10)

Similarly, the three permutations that leave (9) invariant also leave (10) invariant, whereas the other three transform it into (9). This means that Al +L2 and A I A2 are invariant under all six permutations and hence can be expressed

in terms of p, q, and r; we shall do this explicitly below. It follows then that Al and A2 can be found by solving the equation

X2-(01+A2)X+(01A2)=0 and once we have those numbers we may write

xl+x2+x3=-P XI + wx2 + w2x3 =

01

(12)

xl + W2x2 + WS3 = 3 A2 or 1

1

XI

w

w2

X2

w2

IC

1(

X3

=

(13)

OrN2

Again we see that the coefficient matrix is the finite Fourier transform. As we have seen in Section 6, the system (13) can be solved by inverting the matrix. We find that the solutions to (8) are xi

X2

X3

(14)

146

Chapter Four

The Fourier Transform

We now show how one can express Al and 02 in terms of p, q, and r. Recall xl, x2, x3 are roots of

x3+px2+qx+r=0

(8)

The Vieta formulas in this case are

X1+x2+x3=-p

(15)

xlx2 + xlx3 + X2X3 = q

(16)

X1X2z3 = -r

(17)

Observe that (-p)2 = (XI +x2+x3)2 = x1 +z2+x3+2(x1x2+xlx3+x2x3) and hence, using (16), we have

xi+x2+x3=p2-2q

(18)

Now if we put xI, x2i x3 into (8) and add the results we get

x1+xz+x3+p(x1+x2+x3)+q(xl+x2+x3)+3r=0 and putting (15) and (18) into this equation we find

xi+xZ+x3=-p(p2-2q)+pq-3r=-p3+3pq-3r

(19)

We are trying to express 2102 and Al + A2 in terms of p, q, and r. The results obtained above enable us to do this for 01A2 rather easily. Let 6 = (xl + wx2 + W2x3)(xl + W2 X2 + WX3)

= xi + x2 + x3 + w(xlx3 + xlx2 + x2x3) + w2(xlx3 + x1x2 + x2x3) (20) _ (p2 - 2q) + (w + w2)q

where we have used (18) and (16). Now w is a primitive cube root of unity, hence not equal to 1. Thus w2 + w + 1 = 0 (Exercises 6, problem 1). Using this in (20), we get

6 =p2 -2q-q = p2 - 3q and so 01A2 = (x1 + wx2 + w2x3)3(x1 + w2x2 + wx3)3 = (p2 - 3q)3

(21)

The expression for Al + A2 in terms of p, q, and r is more difficult to obtain. We need a preliminary result first. Let

A=xix2+xlx2+xix3+xlxa+x2x3+x2x3

(22)

7. An Application

147

By equations (15) and (16) we may write -pq = (x1 + X2 + x3)(xlx2 + XI x3 + x2x3) = X2

2 + xjx3 + X1X2X3

+ XIX2 + xlx2x3 + x2x3 + x1x2x3 + x1x3 + x2x3

=.1-3r Hence

A= -pq+3r

(23)

A direct calculation shows that

(a+b+c)3=a3+b3+c3+3(a2b+ab2+a2c+ac2+b2c+bc2)+6abc (24)

Using this we find that

Al = (x1 + wx2 + w2x3)3 = xi + x3 + x3 + 3[xjwx2 + x1(wx2)2 + xiw2x3 + x1(w2x3)2 + (wx2)2w2x3 + (wx2)(w2x3)21 + 6x1x2x3

= x1 + x2 + x3 + 3w(xjx2 + x1x3 + x2x3) + 3w2[xix2 + xjx3 + x2x3 + 6x1x2x3

and, similarly, we find that

A2 = (x1 + w2x2 + wx3)3 = x1 + x2 + x3 + 3w2[xlx2 + xjx3 +x2x3 + 3w[xlx2 + xjx3 + x2x31 + 6x1x2x3

Adding and again using the fact that w2 + w + 1 = 0 we find that

Al + A2 = 2(x1 + x2 + x3) - 3(4x2 + x1x2 + xjx3 + x1x3 + x2x3 + x2x3) + 12x1x2x3

Finally, using (19), (23), and (17) we get Al + A2 = (x1 + wx2 + w2x3)3 + (x1 + w2x2 + wx3)3

(25)

= 2(-p3 + 3pq - 3r) - 3(-pq + 3r) - 12r = -2p3 + 9pq - 27r Equation (11) is then

72-[A1+A21X+AlA2=X2-[-2p3+9pq-27r]X+(p2-3q)3=0

148

Chapter Four

The Fourier Transform

Rom this we can solve the cubic as we showed earlier.

Note. In applying this method we must choose cube roots of Al and A2. We must be careful to do that in such a way that 'VA I A2 = p2 - 3q. To illustrate how this method may be used, we shall solve an equation with known roots. We take

x3-6x2+11x-6=0 whose roots are easily seen to be 1, 2, and 3. Clearly, p = -6, q = 11, and r = -6. Thus 01A2 = (p2 - 3q)3 = (36 - 3(11)]3 = 27 and

Al+L2=-2p3+9pq-27r = -2(-6)3 + 9(-6)(11) - 27(-6) = 0 Hence our quadratic is simply

X2+27=0 and, solving this, we find that

A2 = -i 33/2

Al = i 33/2,

We must take the cube root of each of these quantities. Of the three cub roots of i, we select -i. With this choice we find that

'01=-if and '02=if = -i23 = 3 = p2 - 3q. The roots of our equation,

Note that

x1i x2, x3, are given by x1

1

1

x2

1

w

2

X3

1

w2

w

1

-p

V Y2

7. An Application

149

which leads to (taking w = 11 + i$)

X1 = 3(-P+ Y O1 +

X2 = 3(-p+w2' = X3=

9

A2) = 3[-(-6) - if +if] = 2

01 +w

V A-2)

= 3[6-w2i\ 3+wiV3]

3[6+if(w-w2)] = 3[6+(if)(if)]

=

3[6-3] = 1

3(-P+w' O1 +w2 3 D2) = 3[6+w(-iV3)+w2(f)]

= 3[6 + (w2 - w)if] = 3[6 - (if)(if)] = 3(9) = 3 Exercises 7 1. Use Lagrange's method to solve:

(a) x3-1=0 (b) x3-5x2+8x-4=0 2. A permutation of a finite set is a one-to-one map from this set onto itself. There are six permutations of the set {1, 2, 3}. These may be written as

a=

d=

1

2

3

2

3

1

1

2

3

2

1

3

b=(1 2

3

2

1

'

II\\

3

e=

_ C=

1

2

3

1

3

2

f-C 1

2

3

3

1

2

where we understand that a(1) = 2, a(2) = 3, a(3) = 1, etc. Let us call x1 + wx2 + w2x3i y, and let a(y) = xa(1) + wxa(2) + W2xa(3) _ x2+wx3+W2T.1. (a) Show that a(y) = w2y, and f (y) = wy. (b) Show that b(y) = w2c(y), and d(y) = wc(y). (c) Conclude that y3 = (x1 + wx2 + w2x3)3 is unaffected by a, e, and f whereas b, c, and d transform it into (x1 + w2x2 + wx3)3.

3. Let A = (x1 - x2)(x1 - x3)(x2 - x3). Referring to problem 2. take a(0) = [xa(1) - xa(2)][xa(1) - xa(3)][xa(2) - xa(3)] and similarly for the other permutations. Which permutations map A to A (these are called even) and which map A to -A (these are called odd)?

150

8.

Chapter Four

The Fourier Transform

Some Algebraic Matters

We have seen that, for a fixed natural number n, the multiplicative group (U,,, ) is generated by any primitive nth root of unity. In particular, if w is a primitive

6th root of unity, then w'1 = 1, six is the smallest positive integer for which this is true, and U6 = {wo, w, w2, w3, w4, w5 }. It is easy to see that w2, which is

a 6th root of unity, is also a cube root of unity. The same is true of

w4. The

element w3 is a square root of unity, whereas ws is primitive.

There is a problem here that we shall solve in this section. A complex number z is an nth root of unity if z" = 1. But, for such a number z, we may have zk = 1 for k < n. The smallest positive integer for which this is true was called the order of z (Section 6, Definition 5), 9(z), and we have seen that 9(z) must be a divisor of n. What we shall do is figure out how to compute the order of every nth root. We have already seen how to begin. We choose a primitive nth root w and note that because U. is equal to {art 10 < j < n - 1), every nth root of unity is equal to a" f o r some j = 0,1, 2, .. , n - 1. So all we need do is figure out how to compute the order of '. The problem can be solved by recalling a little number theory. When a, b are integers we say a is a factor, or a divisor, of b if there is an integer in such that am = b. When this is the case we say a divides b (or divides evenly into b). and write alb. A useful observation is this:

Lemma 1. If a, b, c, and d are integers, if a + b = c, and if d divides any two of the numbers a, b, c, then it divides the third one. Definition 1. Let a, b be two positive integers. A positive integer d is said to be the greatest common divisor of a and b if (i) dia and dub; (ii) any integer c that divides both a and b also divides d.

So. for example. the numbers 18 and 24 have 1, 2, 3, and 6 as common divisors. The greatest of these is, of course, 6.

Lemma 2. Any two positive integers a, b have a unique greatest common divisor d. Furthermore, there are integers xo and yo such that d = axe + byo. Proof. Let .11(a, b) = {ax + by x. y are in Z}. Clearly, a = a 1 + b 0 and b = a - 0 + b 1 are in M(a, b), and because (ax, + by,) - (ax2 + by2) = a(x, - X2) + b(y, - y2), this set is an additive group (Section 4, Definition 1). Hence. by Corollary 2 to Lemma 1 (Section 4), there is a smallest positive I

integer d E M(a, b) and M(a, b) = {kd I k E Z}. It follows that dia and lib. However, because d E M(a, b), there are integers xo and yo such that d = axo + Mjo. From this, and Lemma 1, we see that any common divisor of a and b must divide d.

To see that d is unique we simply note that any two greatest common divisors of a and b must divide each other. Because they are positive integers they must then be equal.

8. Some Algebraic Matters

151

Definition 2. Given two positive integers a and b, we denote their greatest common divisor (their g.c.d.) by (a, b). We shall say that a and b are relatively prime when (a, b) = 1.

From Lemma 2 we see that when (a, b) = 1 there are integers xo and yo such that 1 = axo + byo. The converse is also true. If, for two natural numbers a and b, there are integers x0 and yo such that 1 = axo + by0, then, by Lemma 1, a and b must be relatively prime. This has a useful consequence. Suppose

t = (a, b), so e = ax + by for some integers x and y. Then 1 = (f )x + (t )y. showing that the natural numbers 7 and b are relatively prime. If a, b, c are natural numbers and albc we cannot conclude that alb or that aic. For example, 6124 so 618 3, but 6 does not divide either 8 or 3. This observation gives meaning to our next result.

Lemma 3. Let a, b, c be natural numbers and suppose that a and b are relatively prime. If albs, then a1e.

Proof. We have I = (a, b) and so 1 = ax + by for some integers x and y. Thus c = (ac)x + (bc)y and hence, by Lemma 1, ale. Recall that a prime number is an integer greater than 1 whose only factors

are itself and 1. If p is a prime and a is a natural number, then either pla or (p, a) = 1. Thus:

Corollary 1. Let a, b be natural numbers, let p be a prime, and suppose that plab. Then either pea or plb. We are now ready to compute the order of wk, where w is a primitive nth

root of unity and 0 < k < n - 1. Recall that the order of wk is the smallest positive integer d such that (wk)d = 1. We have seen that if for some positive integer m, (wk)"' = 1, then dim. It is easy enough to repeat the argument here. Because d < m. m = qd + r for integers q and r with 0 < r < d. But (Wk) m = 1 and (wk)d = 1 so (wk)'' = 1, showing r must be zero.

Theorem 1. Let w be a primitive nth root of unity. Then the order of wk is k.i

; i.e., the integer obtained by dividing n by the g.c.d. of n and k. In

particular, wk is primitive if, and only if, n and k are relatively prime.

Proof. Suppose that the order of wk is d and that e = (k, n). We want to show that d = . Because these are positive integers we need only show that they divide each other. Now (wk)n/t = (wn)k/l = (1)k/f = 1 and so, by the observation we made immediately above, Because (wk)d = 1, nikd. It follows that (7)l(7).

Chapter Four

152

The Fourier Transform

which shows () I () d. But ' and k are relatively prime. Thus ()Id and we are done.

The number theory we have discussed here has some further interesting applications that we shall now explore. We first note that any additive subgroup of Z has an unusual property. Given a subgroup (t) of Z we have (a b) E (e)

whenever b E (e), and this is true for all a E Z. It is this property that enables us to introduce a new operation, a kind of multiplication, onto the set Z1. First we need:

Lemma 4. Let (e) be any additive subgroup of Z and suppose that a - a' mod (e), b =_ b' mod (e). Then, a b = a' b' mod (e). Proof. Because a - a' mod (t), we have a = a' + kt for some integer k. Also, b = b'+ me for some integer m. Hence a b = a'. Y+ ((a'm)e+ (b'k)e+ (mke)e], which proves our result.

Definition 3. Let a be a fixed natural number greater than 1. For any U, b in Zt we define a b to be (ab); note here that a denotes the equivalence class, modulo 1, that contains a.

It is clear, from our lemma, that the binary operation is well defined. Moreover, because (ab) = (ba) we have a b = b a, because (a(bc)J = (ab)c] we

a=a

have

Let us compute the multiplication tables in two cases: 0 0

0

1

0

2 0

3 0

0

1

2

3

4

2

0

2

4

1

3

4

0

0

3 4

1

3

(Z5, )

4 2

1

2

3

4

5

0

0

0

0

0

0

1

0

1

2

3

4

5

2

0

2

4

0

2

4

3

0

3

0

3

0

3

4

0

4

2

0

4

2

5

0

5

4

3

2

1

0

1

3

0 0

4

2 1

(Za, )

These tables display some interesting features of this new operation, but before we can discuss them we need some additional terminology. To begin with, the element 1 is called the multiplicative identity of Zt. An element b of Zt is said

to be invertible if it has a multiplicative inverse; i.e., an element a E Zt such 1. that The table for (Z5, ) shows that every nonzero element of this set has a multiplicative inverse. This is not true in Ze. There we see that only I and 5 have inverses, whereas 2,!i, i do not. Even stranger things happen in (Z6, ).

8. Some Algebraic Matters

153

The table shows that 2.3 = 0 and that 3.4 = 6, while none of the factors is zero. We have developed the number theory we need to explain these "weird" results. Again we must start with some terminology.

Definition 4. Two elements a, 6 of Z, are said to be proper divisors of zero if a54 034

We observe that a proper divisor of zero could not have a multiplicative inverse (see Exercises 8, problem 2).

Theorem 2. The set (Zt, ) contains proper divisors of zero if, and only if, f is a composite integer (i.e., an integer greater than one that is not a prime). k f 1 < n < f.. Clearly then, 10 i 0 a whereas (kn) = 0, showing that Zt contains proper

Proof. Suppose first that a is composite, sot =

divisions of zero. Now suppose that there are proper divisions of zero in Zt. Let us say k, Iii

are nonzero, yet k n is zero. Then kn = mi for some integer m. If a were a prime it would follow that elk, giving us k = 0, or fIn and implying that n = 0. Thus f must be a composite integer.

Corollary 1. When a is a prime the set Zt does not contain proper divisors of zero.

Our next result tells us which elements of Zt have, and which do not have, multiplication inverses. Its proof also shows the connection between the latter elements and the proper divisors of zero.

Theorem 3. An element k of Zt has a multiplicative inverse if, and only if, the integers k and t are relatively prime.

Proof. Suppose that (k, t) = 1. Then there are integers x and y such that 1 = kx + £y. Thus kx =_ 1 mod (P) showing that a is the multiplicative inverse of k.

Next we suppose that k and t are not relatively prime; i.e., (k, 1) = n > 1.

Then k = k1n and i = Pin where 1 < kl < k and 1 < Pi < t. It follows that k f i = (kin) ei = k1(n1i) = kiI E (1). Thus k and ti are proper divisors of zero.

Remark 1. The proof of Theorem 3 shows that when k E Z1, k 0 6 and k is not invertible, then k is a proper divisor of zero. More precisely, there is an ti E Zt such that k and ti are proper divisors of zero.

154

Chapter Four

The Fourier Transform

Corollary 1. When t is a prime, every nonzero member of Zt has a multiplicative inverse. Moreover, for a, 6 in Zt we have a. b = b if, and only if, one of these factors is zero.

Remark 2. Given a natural number rn let E(m) = {k E NJ k < m and (k, m) = 1}. Euler defined a function