Fourier and Wavelet Analysis

n , and consider the vectors

=

i

with

i2

f.

=

Wi = ( 0, 0, . . , i2 , .

. , O, - (n + 1) 2 , 0, 0, . . . .

appearing in the ith position and

( + 1 )st position. Since n

.

ai =

ai E

) , 1 � i � n,

- (n + 1 ) 2

2 ! (Wi ) = ( Wi ' Z) = �Zi2 - (n(n ++ 1)1) 2 = 0,

appearing i n the

98

3. Bases

it is clear that each

Wi E M. Since

Since this implies that each

y

.1.. M, it follows that

ai = 0 it follows that

y

= o.

0

If M is a nontrivial closed subspace of a Hilbert space X , then X = (top) , so any vector in X has a unique representation of the form = + n where E and n E M by the projection theorem (3.2.4) . Since X M EEl (top) , we know that the projection on M along M 1. is a continuous linear map . This special type of projection is called the O RTHOGONAL PROJECTION ON M. It is easy to see that the range of is M and that if is applied twice, nothing happens the second time: for every E X . The property is called IDEMP O TEN CE . Let I denote the identity map 1-+ of X onto X . Since (I = n, I is seen to be the orthogonal projection on An orthogonal projection on a closed subspace of a Hilb ert space behaves like an orthogonal proj ection in in several imp ortant respects. A con­ sequence of 3.2. 7( c) is that the orthogonal projection on a one-dimensional space is just z z

M EEl M 1.

x m

x m M M1.

=

1.

P P P(Px) = p 2 x = Px -P -P)x = x-Px

x

PM

x x

p2 = P

M1. .

R3 ( x, / II z ll ) / II z ll .

[z]

3.2.7 ORTHO GONAL PROJECTION BEST ApPROXIMATION Let M be a closed lin ear su bspace of the HilbfOrt space X , let be the orthogonal pro­ iection on M , and let E X. Then: (a) .1.. M, and is the only vector with this property. (b) is the best approximation to x from M (see Definition 3. 1. 1): M ) and for all =f:. < in M. } (c) ORTHO G ONAL PRO JECTION AND INNER PRO DUCT Let ... be an orthonormal set of vect ors in X. Let = . . , n ] . (Since any finit e- dimensional su bspace of a normed space is closed, M is closed.) Then the orthogonal proiection P on M is given by = [For an infinite-dimensional version, see 3.3 .4( d).]

x - Px Px Il x - Px ll = d (x,

Proof.

x

P

Px m Px II x - Px ll Il x - mil {X l , X 2 , , xn M [X l , X 2 , . x Px :L�== l (x, X i ) X i .

x

x Px M M 1.; m M mo m Px M. Px x m II x - ll II il :L�== [X x - l (x, Xi ) x i l , X 2 , . . . , x n], :L�== l { x, X i } X i . 0

x - Px x-m Px = mo Px

For any E X , = E M 1. . +nE EEl therefore, .1.. M By (3 . 1 .3) the only vector E with the property that is the best approximation to x from M. It follows that and that A direct computation for all =f:. in < = shows that .1.. so, by 3 . 1 .3 again, The context " closed subspace of a Hilb ert space" is imp ortant in Example 3.2.7. In Example 3.2.6, the space

x m

m

M

X

M1.

3. Bases

99

Example 3.2.8 BEST MEAN ApPROXIMATIONS Approxima tion with re­ spect to in the Hilb ert space is called MEAN APPROXIMATIO N o r APPROXIMATION IN THE MEAN. (a) Sh ow that there is a uniqu e best mean approximation to sin t in by a polynomial of degree ::; (b) What is this approximation ?

L2 [a, b]

1 ·1 2

L2 [0 , 1]

x=

5.

M

5

Sol ution . Clearly, the space of polynomials of degree ::; is of dimension 6. By 1 .5.4 and Example 2.6.2( a) , is a 6-dimensional Hilbert subspace of 1] . By the projection theorem 3.2.4 and 3 .2.7, it follows that the orthogonal projection on M is the best approximation to sin by a polynomial of degree ::; 5 : For all v

L2[0 ,

M

Px t E M, I x - Px II 2 = l 1 Isin t - px(t) 1 2 dt � l 1 I sin t - (t ) 1 2 dt. To see what Px is, use the Gram- Schmidt process 1 .4.3 to convert the polynomials {1, t, . . . , t5} into an orthonormal basis {X l , X 2 . . . , X6} for M. Px is then given by 2:f= l [fo1 Xi (t) sin t dt] Xi . 0 v

Exercises 3 . 2

1 . PRO PERTIES O F ORTHOCO MPLEMENTS For any subsets S and an inner product space X, show that

T of

S C S1.1. . [S]) 1. = S-L . ([S] denotes the linear span of S. ) S C T ===} T1. C S1. . (d) ( S1.1. ) 1. = S1.. 2 . If M and N are linear subspaces of the inner product space X, show that M1.1. + N1. C (M + N) l..L . 3 . WAY TO COMP UTE ORTHOGONAL PRO JECTION Let M = [X l , X 2 , . . . , (a) (b) (cl (c)

l.

xn ] be the Xlinear span of the subset {X l " ' " Xn } of a Hilbert space Px of x on The point of this exercise is to show that one way to do it is by solving a certain set of linear equations. Since X - Px M, it follows that (PX, Xj) = (x , Xj) , j = 1 , 2 , . . . , n. Since Px E M, there must b e scalars that satisfy equation (3 .2) below. It turns out that these are the only scalars that satisfy equation M?

X, and let E X. How do we compute the orthogonal projection

.1

Ci

(3.2) . The imp ortance of this metho d of computing the orthogonal proj ection is illustrated in Exercise 4 below .

100

3. Bases

(a) If the scalars

C 1 , C2 , , Cn satisfy the equations

(t;=1 Ci Xi , Xj ) = ti= l Ci {Xi , Xj} = (X, Xj ) , •

.

•

j

= 1 , 2 , . . . , n,

(3.2) is the orthogonal projection of on

2:7= 1 Ci X i {C 1 , C2 , . . . , cn {X l , X 2 , . . X n {X l , X 2 , Xn Ci = (X, Xi } / (Xi , Xi } , i =

X

then show that M. (b) Show that a solution } of equation (3.2) must exist . Show that it is unique if the vectors . , } are linearly independent ; if } is orthogonal, then show that ..., 1 , 2, . . . , n .

4.

SOLVING THE UNSO LVABLE We apply the results of Exercise 3 to the problem of "solving" incompatible linear equations m

L Cj X ij = bi ,

X ij

j=l

i = 1 , 2, . . . , n, j

= 1 , 2, . . . , m,

(3.3)

Ci

and bi are known scalars, the are unknowns and the where equations are incompatible . For such a system of equations, we use the best mean approximation , since no exact solution is available. Let A be the matrix i 1 , 2, . , n. with column vectors = Equation (3.3) Let x be the column vector (b;) E (n) thus becomes m

(X ij )

Xj (X ij), = = (Kn , 11·11 2 ) ' L CjXj = X. j=l '-2

.

.

(3 .4)

We take as the best solution of equation ( 3. 4 ) the best approximation , the orthogonal projection E M , ], is minimal. We know by Exercise 3a that it suffices because to find that satisfy equation (3 .2) to find the orthogonal projection We illustrate in some special cases.

Px.

l i z - x II 2 Ci

z = Px = 2:7=1 Cj Xj

= [X l , . . . x n

(a) Find the best solution z to the system of equations

a + 2b 2a + 4b

=

and compute the percentage error

X= ( ! ).

1, 3,

(3.5)

li z - x I1 2 / l i x il

x

100, where

(b) Find the best solution z to the system of equations

2a + 3b C a + b + 3c 3a + 4b + 2c

5,

9,

15.

(3 .6)

Compute the percentage error l i z

UJ xl, X , . . . , Xn

-

x II 2 I II x l!

x

3. Bases

101

1 00,

x=

where

2 = [X , X , . . . , Xn]. a , . . , an l l 2 X . . . , n? ( x, Xj) = aj

belong to the Hilbert space X and consider the 5. Let subspace M be scalars. When is . Let for j = 1 , 2 , there E X such that As we discuss below, when there is such an x, there is one whose norm is minimal .

( a)

Show that if the set is orthogonal, then "'n .·_ l .... { a ....=-r is a solution to the system. � - \X" Z.l

{X l , . . . , xn}

Xi

x=

{X I , . . . , X n } aI , (c) Suppose that for scalars a I , . . . , a n there exists x E X such that (x, X i ) = ai , for i = 1 , 2, . . . , n . Show that the orthogonal pro­ jection of onto M is the vector z of minimal norm such that ( z, Xj) = aj for j = 1 , 2, . . . , n. 6 . BEST MEAN ApPROXIMATIO N TO t 4 Use the technique of Example 3.2.8 to compute the best mean (i.e . , 11 · 11 2 ) approximation to x = t 4 by a polynomial y of degree :::; 3 . Sketch your solution and see whether

(b) Show that if the vectors are linearly dependent , then there exist scalars . . . , a n such that there is no solution x.

x

it looks like a good approximation .

N

7. SUMS O F COMP LETE SUBSPACES If M and are complete orthog­ onal subspaces of a Hilbert space X , then (a) M + is closed; (b) the orthogonal projection PM + N PM + PN .

=

N

8 . SUMS O F CLOSED SUBSPACES Unlike what happens in a Hilbert space, as in the preceding exercise, the sum of closed orthogonal subspaces of an inner product space need not be closed. Let (0, 0, . . . , 0, 1 , 0 , . . . ) , E N , be the standard basis vectors for let . . ] . Let M be the linear sub­ ( l/i), and let X space of X whose odd entries are 0, and N the subspace whose even entries are 0. Show that M and N are closed orthogonal subspaces of X but that M + N is not closed.

i = [x, e l , e 2 , . . . , e n ,

x=

ei = £2 ,

.

9 . Let cp be the space of finite sequences of Example 3.2.6 endowed with the inner product of Let

£2 .

=

Show that M1. cpo By Exercise 2.4-8(b) , M is closed in therefore a closed subspace for which M =P M

1.1. .

cp o

M is

102

3. Bases

Hints 3(a) . Show that x - z

{X l , X 2 , , n

1. M and use the result of 3.2.7.

3 ( b) If x } is linearly independent, then there is only one way to write the orthogonal projection x on M. .

4(a) .

.

•

•

P = E?=l Ci X i The system has no solution because 2a + 4b = 2 (a + 2b) but 3 =J 2 · 1 .

In vector form the system may be written

By equation (3.2) , we want to find scalars a, b such that

and

Since the second equation is merely double the first, it suffices to solve the first one. To get a solution, set b 0 to get a = 7/5. Thus, the orthogonal projection of

(!)

=

on M

=

[ ( ; ) , ( � )]

=

[( � )] ( ; ) + O · ( � ) = ( ( ! ) , � ( ; ) ) (� ( � ) ) , (i) is given by

( 7/5)

where the term on the right is more obviously the projection of on M. The percentage error is 14.1%.

{X I , . . , X m } is a maximal linearly independent subset of {X l , " " xn }. Then,. for each k > m, X k can be written as a linear combination of {X l , . , x m } . Therefore , for k > m, each a k can be written as a linear combination of { a I , . . . , a m } . Choose a k so that it

5 ( b) . Suppose

.

.

.

violates this latter condition.

5(c) . Px and ( 1 - P) x denote the orthogonal projections on M and M l. , respectively. Show that V [xl + M l. contains all vectors such that j = 1 , 2, . . , n . By 3 . 1 . 3 and Example 3.2.7, x ­ (1 = Px is the vector of minimal norm in V.

= {z, Xj} = aj , - P) X By 3 . 1 .3, PM + N X is the only vector such that X - PM + N X Show that X - PM X - PNx M + N.

z

.

7.

1.

1.

M+

N.

3. Bases

3.3

103

Orthonormal Sequences

In R3 one routinely expresses a vector as the vector sum of its projections onto the standard unit basis vectors i, j, k, along the coordinate axes . The standard unit basis vectors may be replaced by an orthonormal sequence in many imp ortant spaces �uch as [ - 11", 11"] . We consider the basic properties of orthonormal sequences in this section. We considered Bessel 's equality and inequality in finite-dimensional spaces in Section 1 .4 and Exercise 1 .4-6 .

L2

(xn ) bein anX theorthonor­ series

3.3.1 BES SEL'S EQU ALITY A N D INEQU ALITY L e t mal sequence in an inner produ ct space X. For a n y converges and satisfies

x

LneN I{ x, Xn ) 1 2 (3.7) L: I{ x, X n ) 1 2 :::; I x II 2 (Bessel's inequality) , neN and for convergent L neN (x , X n ) X n , 2 x - L: (x, Xn ) Xn + L: I{x, Xn )1 2 = I x I 2 (Bessel'S equality) , neN neN (3.8)

) x = L: (X, Xn ) Xn L: l{x,xn ) 1 2 = I x l 2 ( Parseval's entity n eN neN

Moreover,

<==>

.d

1

.

.

(3.9)

The last equality is one version of Parse val's identity or PARSEVA L'S EQUAL­ ITY ; another version is given in 3.3.4(f) .

by Example , . ' " X(xn . XForXany. TheX, Pythagorean X l , X 2L?: x- L?=l (x, Xi ) Xi i) i , l II x I 2 Il x - L:;= l (x, Xi ) Xi 1 22 + 1 L:: 1 (X, Xi ) X i l 1 2 Il x - L:;= l (x, Xi )2Xi 1 + L:;= l I { x, xi) 1 2 > I L;= l {X,Xi) Xi I1 = L:;=1 1{x,Xi) 1 2 , which yields the convergence of L nEN l{ x, x n ) 1 and Bessel 's inequality. If Ln e N (x, Xn ) Xn = x, then we may utilize the continuity of the norm and take limits on the second line to obtain the Bessel equality ; equation (3.9) follows immediately from equation (3.8) . If, conversely, L I{ x, Xn ) 1 2 Xn = I x I 2 , n EN

Proof.

Consider the first n vectors 3.2.7, is orthogonal to relation therefore implies that for any n,

104

3 . Bases

then taking limits on the second line shows that

x = L nEN (x, Xn) Xn. 0

Example 3.3.2 CLA SSICAL FO URIER COEFFICIENTS G O TO 0

The sequen ces

(Xn) = CJ;t ) and ( Yn) = ( s�t ) are orthonormal sequences in L [-11", 11"] . The FOURIER COEFFICIENTS of x E L 2 [-1I", 11"] are ao = ;: f�1< x(t)2 dt, an = -11"1 1 1< X(t ) cos nt dt = 1 (x, x n) , n E N , _ 1< and 2 bn 11"1 1 1< x (t) sm. nt dt = 1 (x, Yn) , n E N. r;;; y 1l"

=

By Bessel's inequality,

li�

r;;; y 1l"

-

0

1: x(t) nt dt sin

=

0

and li�

11< x(t) cos nt dt = 0 , 1<

so an , bn 0, a fact known as the RIEMANN-LEBESGUE LEMMA; for a much more general version see 4.4. 1 . 0 �

Our next result illustrates the strong similarity between orthonormal sequences and the usual basis vectors in R3 . A big differ­ ence is that not every vector in X can necessarily be represented as x LnEN (x, xn) Xn -only those in c l [X l, X 2 . . . , Xn, . . .] can. 3.3.4 ORTHO N O RMAL SEQUENCES IN HILBERT SPACES For any or­ thonormal sequence (Xn) in a Hilbert space X : ( a) ( RIESZ-FIS CHER) Ln EN anxn converges if and only if L n E N l an l 2 < 00 , i. e., (an) E 1.2 • This is equivalent to: For any orthogonal sequence ( Yn ), Ln E N Yn converges if and only if Ln EN ll Yn 11 2 < 00 . (b) If L n EN anxn = x, then an = (x, xn) for all n. (This is true in any inner product space.) It follows that the coefficients an are uniqu e. ( c ) For all x E X, the series Ln N (x, xn) Xn = Y converges-not nec­ essarily to x, but to something in cl [(xn ) ] = cl [X l, X 2 , " " X n , . . .]. (d) For all x E X, x - Ln ( x, Xn) Xn J.. cl [( xn )) so Ln EN (x, Xn) Xn = 3. 2. 7. Px, the orthogonal projectionENof x on M = cl[(x n)), by Example (e ) If x E cl [(xn )) , then x = LnEN (x, xn) Xn. (f) PARSEVAL'S IDENTITY ( equation (3.9) is also known as Parseval 's identity) . For convergent series x = L n EN anxn = L nEN (x, xn) Xn and Y = LnEN bnxn = Ln EN ( y, xn) Xn, Example 3.3.3

=

(x, y) = L anbn = L (x, xn) ( y, xn) nEN nEN

(Parseval's identity) .

(3.10)

3. Bases

105

( g) PYTHAG OREAN THEOREM If(Yn) is an orthogonal sequence of nonzero L n e N Yn, then (3. 1 1 ) II x II 2 L ll Yn 11 2 . neN Proof. ( a) Let Sn = L 7= 1 ai X i· The sequence ( L7= 1 l ail 2 ) is Cauchy if and only if (sn) is Cauchy since n 2 = sn s II m ll =L l ai l 2 . i m+ l ( b) Let Sn L7= 1 ai Xi . For any n > j , {Xj , sn } = aj so limn {Xj , sn} = aj . By the continuity of the inner product , it follows that {Xj , x} aj for each j. ( c ) By 3 .3 . 1 , LneN I {x, Xn} 1 2 converges in any inner product space. Therefore , L neN {x, Xn} Xn converges by ( a ) . As a limit of vectors in [(Xn)], this sum belongs to cl[ ( xn )) . ( d) It is trivial to verify that if x S, then x cl S. Therefore , it suffices to show that x - L n e N ( x, xn) Xn [(X n)). This follows from the fact that for any j, using the Kronecker delta 6n j , vectors and x =

=

=

=

..L

.1

.1

( e ) If x E cl [( xn )] , then by ( d ) , Px = x Ln eN { x, xn} xn. (f) Let x = Ln e N anXn and Y = L ke N b k x k . It follows from the mutual orthogonality of the (X n ) and the continuity of the inner product that =

To get the previous version, equation (3.9) , of the Parse val identity, let

x. (g ) Rewrite L n e N Yn as Ln eN ( II Yn ID ( Yn/ ll Yn I! ) Since ( Yn/ ll Yn II ) is an orthonormal sequence, it follows from (b) that {x, Yn/ II Yn ll} = II Yn l! ' n E N . It follows from (f ) that II x I1 2 = L ll Yn 11 2 . 0 n eN y =

.

106

3. Bases

Exercises 3 . 3

1.

(xn) is an orthonormal sequence from L 2 [0, 2 ], 1r

CHANGE O F SCALE If show that the functions

are orthonormal on

[0, b).

(xn) and conx Ln EN anXn Y Ln EN bn xn in a Hilb ert (x, y) = L anbn n EN by Parseval's identity 3.3.4 ( f) . Show that Ln EN anbn converges ab­

2. ABS OLUTE PARSEVAL For an orthonormal sequence vergent series and = space

solutely.

x = Ln EN Xn and ( y, Xn) = 0 for E N, ( y, x ) 4. Let (xn) be a sequence of vectors in a Hilbert space X whose linear span [(xn)) is all of X . Show that X is finite-dimensional. 5 . Let (xn ) be an orthonormal sequence in an inner product space X . ( a) For any > 0, show that the set { Xn : II x II 2 < f I (x, X n) 1 2 } 3. In any inner product space, if every n show that = O.

finite.

{

IS

( b) Improve the result of (a) . Show that for any k E N the set

has at most

k elements .

Ld -1r, x E L 2 [-1r, ] x E Ld ] 7. If {X l , X 2 , . . . , Xn} is an orthonormal set in L 2 [a, b), show that Yij ( t) = Xi ( 8 ) Yj (t) , i, j E N , is an orthonormal set in L 2 ([a, b) [a, b]) .

1r) We know that the Fourier 6 . RIEMANN-LEBESGUE LEMMA FOR 1r go to 0 by Example 3.3.2. Show that the coefficients of same result holds for - 1r , 1r . x

8,

3. Bases

107

Hints 2. Replace

an and bn by l an l and I bn l .

4. Use the Gram-Schmidt process ( 1 .4.3) to orthornormalize an infinite linearly independent subset of to get an orthonormal sequence ( Yn ) , and consider

(xn) LnEN { l /n)Yn.

5 . Use 3 .3 . 1 for both parts. 6 . The step functions are dense in

Ld - 1r, 1r]. Or peek ahead to 4 .4 . 1 .

7. Write the double integral as an iterated integral.

3.4

Orthonormal Bases

Let X be an inner product space. We now know that orthonormal sequences in X behave much like the usual basis vectors in in that we may write for many vectors x E X. The defect is that there x = (x , may not be sufficiently many for every vector to be expressible in terms of them. The notion we consider next characterizes orthonormal sets that are large enough to make this representation universally possible . Orthonormal sets that are not properly contained in any other or­ thonormal set are special. Such a maximal orthonormal set is called an O RTHONORMAL BASE ( BASIS ) or a COMPLETE ORTHONORMAL SET . We prove in 3 .4.8 that orthonormal bases in separable Hilbert spaces are the infinite-dimensional analogue of the standard basis vectors in If A is an orthonormal sequence but not an orthonormal base , then it is properly contained in another orthonormal set, so there must be some unit {x d maximal? If not , there must be Is vector such that some unit vector such that { x d . An inductive style argument like this, using Zorn's lemma ( Bachman and Narici 1966, pp. 149-150) , shows that

L nEN Xn) Xn

R3

Xn

S

Rn .

Xl

• •

•

X2

X l J.. A . A u X 2 J.. A u

Orthonormal bases exist in any inner product space. Any orthonormal subset is a subset of an orthonormal base . We also say that an orthonormal subset can be extended to an orthonormal base . All orthonormal bases are of the same cardinality ( Bachman and N ar­ ici 1966, p . 166) . This common cardinality is called the ORTH O G O N A L DIMENSION of the space; it is the same as the lin ear dimension, the cardinality of any Hamel base , only in finite-dimensional spaces .

108

3. Bases

3.4.1 ORTHO N O RMAL BASES IN INNER PRODUCT SPACES A n orthonor­ mal subset S of an inner product space X is an orthonormal base (a) if and only if SJ. = {O}. (b) if the lin ear span [5'] is dense in X.

Proof.

S

{O} ,

{O}.

(a) If J. =p there exists x E SJ. Since S is properly contained in the orthonormal set S , S is not an orthonormal base. Conversely, if SJ. = then S is not a proper subset of another orthonormal set. (b) Suppose that is dense in X. If x 1. clearly then .1 . Since is dense in X, there is a sequence from that converges to x. Since .1 n = 0 for every n. By the continuity of the inner product, = Thus = 0 , and the completeness it follows that 0 = limn follows from (a) . 0 -

u {xl I Ix ll }

{O},

[S] x [5'] , ( x, x )

[S]

( x,

(x n ) xn) II x 1l 2 .

S, [5'] x

x [5']

In a Hilb ert space X, cl = X constitutes a criterion for completeness of orthonormal sets S. We prove it for denumerable S below .

[5']

3.4.2 COMP LETE ORTHONORMAL SEQUENCES IN HILBERT SPACES A n orthonormal sequence (xn) i n a Hilbert space X is a n orthonormal base if and only if any one of the equivalent conditions below is satisfied. (a) cl [(xn)] = cl [X l X 2 . . . , Xn, .] = X .

(b) = (c) The Parseval identity = for every (d) = ,

.

.

[(xn )]J. {O}. II x I1 2 � n E N I (x, Xn) 1 2 holds for all x E X. x E X. x �n E N (x, xn) Xn Proof. (a) If cl [( xn )] = X , then (Xn) is an orthonormal basis by 3 .4. 1(b). Conversely, if (x n ) is an orthonormal basis, let M = [X l ,X 2 . . . , X n , . . .]. Then M J. = {O}. Therefore , by 3.2.5, cl M = MJ.J. = X . (b) A linear subspace M of a Hilbert space is dense if and only if M J. = {O} by 3.2.5(b) . Therefore cl[(xn )] = X if and only if [(xn)]J. = {O}. ( c ) Assume that Parseval's identity holds for all x E X . I f x [(xn)], then II x II 2 = L I (x, Xn )1 2 = 0 n EN s o X = 0 and (Xn) i s an orthonormal base b y (b) . Conversely, i f (x n ) i s an orthonormal base , then cl [ ( xn )] = X by (a) . Therefore Parseval's identity holds by 3 .3 .4(e, f) . (d) The validity of Parseval's identity for all x E X is equivalent to the ability to write every x = � n EN (x, xn) Xn by 3.3. 1 ; the desired result follows from (c) . 0 1.

We utilize the criteria of the previous theorem in the next two examples.

3. Bases

109

Example 3 .4.3 RADEMACHER FUN CTIO NS NOT A BASE

(1'n), 0, L 2 [0, ] 1 [0, 1] ' 1'1 1 [0, 1'2 [0, [�, 1]. 1'3 , [0, 1]

The functions n � from I are known as the RADEMACHER = on FUNCTION S : = on t) and - I on Next, divide into four equal parts and define = I on t ) and [t , �), and To get = - I on [ t , t ) and divide into eighths and let be on the first eighth, - I on the second, and so on. A compact description is afforded by means of the SIGNUM FUN CTI O N sgn t , which is for � ° and for < O . The Rademacher functions are then given by = sgn ( sin 2 7r , ° � �

1'2

[0, 1]

1'0

1

1

t 1'n (t)

-1

t n t)

[t , l].

1'3

t 1.

05 ·0.

-0 5 . -1 .

The Rademacher function

r2

The Rademacher functions are an orthonormal set but not an orthonor­ = To see that the mal base: Clearly, are mutually orthogo­ and n, p E N . On any subinterval nal, consider k 2 is constant , on which must change sign an even number of times. The integral of the product on all such subintervals is therefore = To see that they are incomplete, and it follows that we use the criterion of 3 .4.2( d) and exhibit a function x E for which x i= (x , Consider the function x that rises linearly from o to on then falls linearly from to ° on Now consider and Since the First, compare what happens on integrands represent triangles, we can compute the areas by geometry :

1'n f01 1'; (t) dt 1. 1'n 1'n +p , [(k - 1) /2n , / n ) 1'n +p 1'n 1'n 1'n p 1'( n , 1'n +p ) + 0. 0, L 2 [0, 1] 1' 1' Ln>o n) n· 1 [0, f] 1 [t, l]. 1 [0, i] [� , 1] . f0 X(t)1'2 (t) dt. 1/4 [Jo X(t)1'2 (t) dt 2"1 · 41 · 41 ' while 1 1 1 1 13/ 4 X(t)1'2 (t) dt = - -2 . -4 . -4 , =

110

3 . Bases

clude that

I; 1 . Thus ,

o for all n �

I1 /; x(t)r2 (t) dt

- I:/24 x(t)r2 (t) dt . We con­ x(t)r2 (t) dt = O. The same idea shows that x(t)rn (t) dt

so their sum is O. Similarly,

L ( x , rn) rn =

n ?O

=

[1 1 x(t)ro (t) dt] ro � ro 0

=

I01

=

=

�.

Since Ln>o ( x , rn ) rn is constant, it is certainly not equal to x; the Rademacher functions are therefore not an orthonormal base by 3.4.2( d) .

0

The following result looks like a strong form of continuity. Its corollary 3 .4.5 is that even if Parseval's identity holds only for all x in a dense subset of a Hilbert space, then the orthonormal sequence (xn ) is an orthonormal basis. 3 .4.4 PA SSAGE TO THE LIMIT Let (xn ) be an orthonormal sequence in the inner product space X . If Yn -- Y and

Yn

L (Yn , Xk) xk kEN

=

for every n, then Y=

L (y, Xk) Xk. kEN

Proof. Given € > 0 , choose N such that l lYn - y ll < € / 3 for n � N. By hypothesis YN = L k E N (YN , Xk ) Xk , so there exists M such that n

L (YN , Xi) x ; - YN i= 1

< €/3 for n � M.

By the triangle inequality,

I L�= l (y , Xi) Xi - L�= l (YN , Xi) Xi I + II L�= l (YN , Xi) Xi - YN I I + l l YN - Y I I · (3. 12) By the orthonormality of the Xi and the Bessel inequality of 3 .3 . 1 we have I L�= l (y, Xi) Xi - y l n

<

L (y - YN , Xi) Xi

i= l

2

n

=

I { y - YN , x ; } 1 2 ::; l I y - YN I I 2 , L ;=

1

so the first term in inequality (3. 12) is also less than €/3.

0

111

3. Bases

3.4.5 PARSEVAL O N A DENSE SUBSET Let (x n ) b e a n orthonormal se­ quence in a Hilbert space X. If Parseval 's identity, equation (3. 9) of 3. 3. 1,

lI y ll 2 L I {y, x k } 1 2 , keN =

or, equivalently,

Y = L (y, X k ) x k , keN

holds for each y in a dense subset D of X , then basis for X.

Proof.

(xn) an IS

orthon ormal

We use the criteria of 3.4.2(b, c) to show that Parseval's identity holds for all in X . Let E D be such that -+ By hypothesis

Yn y. Yn Yn L (Yn, X k ) X k ke N for every n, so the desired result follows from 3.4.4. 0 y

=

y'2;i) , The density criterion for completeness of 3 .4.2( a) implies that n E Z, is an orthonormal base for L 2 [ - 1I" 1I" , as outlined in Example 3 .4.6{b) .

( e i nt /

, )

Example 3 .4.6 ORTHO NORMAL BASES FOR

(a) STANDARD BASIS IN

ei

=

£2 ( n )

£2 AND L 2 [ - 1I", 1I"]

The standard basis vectors

(0, 0 , . . . , 0, 1 , 0, . . . , 0) , 1 ::; i ::;

n,

Kn , N, { a e ) ai . x

with a 1 i n the ith position , are an orthonormal base for as are the sequences = (0 , 0, . . . , 0, 1 , 0 , . . . ), i E for £ 2 . For any = E £2 (n) , Thus, if is ::; n ::; 00 , ( j ) , i = orthogonal to the linear span [( then = 0 . /V2i) , n E Z, is an orthonormal basis for L 2 [ - 1I", 11"] . (b) Orthogonality follows from the fact that for n i m ,

ei x (aj ) (e i nt

1 ei )],

111' ei(n - m)t dt - 11'

=

z

.

n - m )t l 1l' i( e _ (n - m) 1

1I'

=

0.

[(e i nt /V2i»), n E N , of

As noted in Example 2.4.5(c) , the linear subspace trigonometric polynomials n E is dense in the desired result follows from 3 .4.2(a) . (c) Essentially, the facts used in (b) prove that

L:�= _n a k e i k t ,

N,

x

L 2 [-1I", 1I"] , so

, ( / J1i) sin t, ( 1 / J1i) sin 2t, . . . , (1/ J1i) cos t, (1/ J1i) cos 2t , . . . is also an orthonormal base for real L 2 [ - 1I" , ] 0 /

1 .,f2; 1

11" .

Example 3 .4.7 demonstrates the fragile nature of completeness for or­ thonormal sets.

112

3 . Bases

Example 3.4.7 REMOVE ONE-NoT A BASE

Consider

S = { sin nt j.fi : n 2: 2 } U { cos nt j.fi : n 2: 1 } u {lj.J2;} L 2 [-'II" , 'II"] . Since sin t S, it follows that S1. # {O} and S is a not an orthonormal base. 0 C

.1

In a separable inner product space, orthonormal bases are countable. 3.4.8 SEPARABLE <=> COMPLETE ORTHONORMAL SEQUENCE a An inner product space X is separable if and only if it has a complete

()

(xn), i. e., a sequence (xn) that is an orthonormal (b ) ( FO URIER SERIES ) In a separable Hilbert space any x E X can be

orthonormal sequence base.

written uniqu ely in the form

x = L (x, Xn) Xn n EN for any orthonormal basis (xn). The values (x, xn) are called the FO URIER C O EFFICIENTS of x. The series E nEN (x, x n) Xn is called the FO URIER for x. Thus, in separable Hilbert spaces we recover the ability to write a vector as the "sum of its projections "on the basis vectors.

SERIES

11

11

Proof. ( a) If (Xn) is an orthonormal base , then linear combinations of the Xn with rational or Gaussian rational (a + bi, where a and b are rational) coefficients constitute a countable dense subset of X . Conversely, if X is sep­ arable , let {x n } be a countable dense subset . Let { Yn } be the sequence ob­ tained from the {x n } by the Gram-Schmidt process ( 1 .4.3 ) . Since the linear spans [{X n }] and [{ Yn }] are equal, it follows that cl [{yn }] = cl [{x n }] = X j therefore , {yn} is an orthonormal basis b y 3.4 . 1 ( b ) . (b ) Let X be a separable Hilbert space. By ( a) there exists an or­ thonormal basis (xn) in X . By 3.4.2 ( a) , cl [(xn)] X . By 3 .3.4 ( e ) , x EnEN (x, xn) xn · The uniqueness of the coefficients (x, Xn) follows from 3.3.4 ( b ) . 0 We characterized finite-dimensional Hilbert spaces X over K in 1 .5 .4: Essentially, K n is the only one. We chose an orthonormal basis X l, X 2 , . . . , Xn for X , then showed that the map E � 1 a i X i (ai ) was a linear isometry of X onto K n . We use a similar technique in the corollary below . As a consequence , a function x from L 2 [ - 'II" , '11"] may be characterized by "digital information ," namely the sequence «( x, x n) ). =

=

1-+

Corollary 3 .4.9 SEPARABLE HILBERT SPA CE � '- 2 Any infinite­ dimensional separable Hilbert space X is inner product isomorphic

to '-2 .

3. Bases

Proof.

£2

113

B y 1 .5 .3 w e only have t o show that X and are norm isomorphic. By 3 .4.8(a) , it follows that X has an orthonormal basis By 3 .4.8(b) it follows that any E X can be written in the form = Consider the map

x

A:

X

--

X = LnEN (X, Xn} Xn

1---+

(x n ). x LnEN (x, xn) xn.

£2 ( (X, xn})

The linearity of the inner product in the first argument yields the linearity of A. If = 0, then Since ° for every n E so 1.. is an orthonormal base , it follows that = 0, and A is seen to be injective. To see that A is surjective, suppose that < 00. By the Riesz- Fischer theorem, 3 .3.4(a) , converges to some x; by so 3.3.4(b) , a = = By the Parseval identity, equation (3.9) of 3 .3 . 1 , it follows by 1 .5.3 that

(xn)

N , x [(x n)]. x LnEN I an 1 2 L n eN anxn n ( x, x n ), Ax (an ).

Ax

(x, xn) =

II A x lI; = 1I ((x, xn}) II � = L l (x, xn } 1 2 = II x 11 2 . 0 n EN Exercises 3 . 4

1.

ORTHO N O RMAL BASES FOR - 11", 11"] Which of the following are or­ thonormal bases for -11", 11"] (a) L �= - n ak cos kt, n E N . (b) �= - ak sin kt , n E (c) L �=_ n (a k cos kt + b k sin kt) , n E

C[ , I H I 2 )?

(C [

N.

L n

N.

2. ORTHO N O RMAL BASES FOR 11"] Show that each of the sequences (cos nt) , n E U {O} , and {sin nt} , n E suitably normalized, are orthonormal bases for real 11" . Note : 11"] , not 11" .

N

L 2 [0, L 2 [0, ]

N, L 2 [0, L 2 [-1I", ] 3 . Show that any n orthonormal vectors in Rn (n E N ) form an or­ thonormal base .

(xn)

be an orthonor­ 4. CONTINUITY O F FOURIER COEFFICIENTS Let mal sequence in a Hilbert space X . Show that each of the C O EF­ FICIENT FUNCTIONALS = x, defines a continuous linear functional on X. (Hint: Cauchy-Schwarz .)

In (x) ( x n)

5 . DIRECT SUMS O F ORTHONORMAL BASES Recall (Section 2 . 1 0) that the direct sum X Ef) Y of Hilbert spaces X and Y is a Hilbert space. If are countable orthonormal bases for X and Y, re­ and 0) , (0, spectively, show that } is an orthonormal basis for

(x n)

X Ef) Y .

( Yn )

{(xn '

Yn)

114

3. Bases

x E L 2 [-7I", 71"] , show that J�'If I x (t) 1 2 dt Ln eZ 2� I J�'If x (t)2 e - int dt l 2 2 2� I J�'If I x (t) 1 2 dt l + L2 n eN � I J�'If x (t) cos nt dt l + � 1 f"'If x (t ) sin nt dt I .

6 . For any

Hints Suppose that E L 2 [0, 7I"] is orthogonal to all terms of the cosine sequence. Extend to L 2 [-7I", 71"] by defining Show that the extended function is orthogonal to both the cosine and the sine sequences.

2.

3.5

x

x

x (-t) = x (t).

The Haar Basis

The standard Fourier coefficients

'1f

an = -71"1 1 x(t) cos nt dt, n E NU {O} , and bn

1

1 k x(t) cos n t dt , n E N, 71" involve integrals of products of the function x E L 2 [-7I", 71"] with sines and - 'If

= -

a

cosines. As the reader undoubtedly knows, such integrals can be difficult to compute . One gets simpler integrals using Haar functions (Definition 3 .5 . 1 ) . Though this simplicity is desirable, the Haar functions are discontinuous; so ultimately, we find ourselves in the strange situation of expressing a continuous function in terms of discontinuous ones . We show in this section that the Haar functions form an orthonormal basis for L 2 [0, 1], a fact that we shall use in Chapter 7 on wavelets. We remind the reader that elements of L 2 [0, 1] that differ on a set of Lebesgue measure ° are considered to be the same. The function 9 = f i� called a DILATION O F the function f by The first Haar function that we introduce is

° is to convert the cozero set [ 0, 1 ) of

(t)

(bt)

l ib).

(t) =

(t)


b.

(t),

Ib):

(bt

(bt - k) = l [k / b ,(k+l)/ b) (t) .

Thus,

(t) =

(the scaling identity)

(3 . 13)

3. Bases

The MOTHER WAVELET (see Section 7.3)

1P (t) = t,b (2t) - t,b (2t - 1 ) = f ( x ) =

looks like a truncated sine wave.

{ �1,

115

[O, i) [i,

on - I on 1) elsewhere

0,

o.� o�

-0 . 2

0.2

0.4

0.6

0.8

1 .2

-0 . 5

The Haar Mother Wavelet 'ljJ

Definition 3 . 5 . 1 HAAR FUNCTIO NS

The mother wavelet

1P generates

We refer to j as the G ENERATION of 1Pj,k . The collection together with t,b is called the set of HAAR FUNCTIONS. 0

\II

of the 1Pj ,k

j I , we get two functions 1P 1,O (t) = 2 1 /2 1P (2t) and 1P 1,1 (t) 12 /For 2 1P (2t - 1 ) ; 1P 1,O and 1P , 1 have greater amplitude (so that they have norm 1) than 1P but smaller cozero sets (of length i): =

1/; 1 , 0 (t) =

-<>.

{

0.1

1

on [0 , �), Vi -..;2, - l on [ � , i ) , elsewhere, 0,

0.2

0.3

0.4

.,pI,. = 2l/2.,p (2t)

5

0,6

and

1P 1,1 (t) =

0.2

{

on [i , �) , Vi on [ � , -Vi, elsewhere. 0,

-1

0.4

0.6

.,pl,l (t) = 21/2 (2t -

0.8

1)

1 ).

116

3. Bases

j 2,

[0, 1]

At the next generation, = we divide into fourths; on each fourth (half-open interval) we define a function that is on the first eighth and on the second; the amplitudes increase from v'2 to and the cozero sets shrink from t to ! in length : are, respectively, given by , 3 ) . The cozero set of each for any k is of length The number of functions as varies from = ° to n is 1 + + + . . . + are Evidently,
2 2, -2 '1/;2 ,0 , . . . , '1/;2 ,3 2'1/; (4t) 2 '1/; (4t - 1) , 2 '1/; (4t - 2) , 2 '1/; (4t 'l/;i ,k 1/2i . n n 1 'l/;i ,k j 2 4 2 2 + - 1. 'l/;j ,k . 'l/; ,k (j, k) 1= (j', k') j 1= j' i 1= k'. j j' 1= k', 'l/;i ,k 'l/;i ,k ' 'l/;j , k 'l/;j , k ' {'I/;j , k , 'I/;j ,k /} f�oo 'l/;j , k 'l/;i ' ,k 1 dt 0. j 1= j', 'l/;i ,k 'l/;j / ,k l 'l/;j , k

L 2 [0, 1].

L 2 [0, 1],

(n2 -i , (n 1) 2 - j ) , jE N , n 0, 1, . . , 2n - 1. L 2 [0, 1] 1974, 71, 2.4.5( [iii] .
is denoted by

'1/; , '1/; 1 , 0 , '1/;1 , 1 , '1/; 2 ,0 , '1/;2 ,1 , '1/;2 , 2 , h I , h 2 ' h3 ,

.

.

.

[0, 1].

.

[\If]

..

•

We index the discontinuity points of a finite collection of them by increasing size . For example , the discontinuities of are at t , i , and we index them not in the order they appear with their functions but in = !, = = t, ascending order as = i , and = When w e add = (which is on the first fourth , 1 on the second, then to the list , we introduce a new point of discontinuity at = �. We reindex the discontinuities as 3 = = 8' = 4' = 2 ' = 4 ' and =

O, ! , h I , h 2 , h3 , h4 , a l 0, a 2 a3 a4 as 1 . 1 0) hs 1/;2 , 0 t a l 0, a 2 1 a3 1 a 4 1 as a6 1. The point of the next two results is to show that any x E C [0, 1] can be l written as a series E nEN bnhn , with bn = fo x (t)hn (t) dt, which is uni­ n formly convergent to x on [0, 1] except on { k / 2 : ° ::; k ::; 2 n , n = 0, 1, 2, . . . } , 1;

-

3. Bases

0.

117

a set of measure Since uniform convergence almost everywhere on a set of finite measure implies L 2 -convergence, it will follow that is dense in L2 and therefore that W is an orthonormal basis for L 2 by

[w] [0, 1] 3.4.2.

[0, 1] ,

Let a 1, a 2 . . . , an, an + 1 be the discontinuity points of the first n elements h 1 ' h 2 ' . . . , hn of W arranged in ascending order. If s and t are between successive discontinuity points ale , a k + 1 , then

Lemma 3.5.2

otherwise,

Proof.

2:7= 1 hi (t) hi ( ) = 0. s

1, the sum above is just h 1 (t) h 1 ( ) {O, I} = {a l , a 2 }. It is trivial to verify

We proceed by induction. For n = and the points of discontinuity are that

s

hn + 1 '

Let us move on to bigger things. Assume the theorem for n, add in and assume that the cozero set of in the existing (up lies in to set of discontinuity points; inserts a new discontinuity point d right in the middle of If one of 8 , belongs to and the other does not , then ( s ) = Thus,

[alc, a k + 1 ] hn + 1 hn + 1 [ale, a k 1]. t [alc , a k + d 0. hn + 1 (t)+hn + 1 n n +1 L: hi (t) hi ( 8 ) = L: hi (t) hi ( ) ;= 1 i =1

hn)

s

,

and the result follows from the induction hypothesis . Now suppose that both 8 and belong to Further, suppose that s and t are on opposite sides of the midp oint d, i.e . , for example , that < < Assume that = "pm ,j , so that its cozero set is an interval of length Since d is the point at which changes sign , it follows that the product

t

s ::; ak + 1 .

[alc, a k + 1 ]. hn 1 112 m . +

By induction,

so

n+1 L: h ; (t) hi ( ) = 2m - 2m = 0, ; =1 s

hn + 1

ale ::; t d [alc , a k + 1]

,

118

3 . Bases

which is the desired result, since s and t are not between cbnsecutive points of discontinuity of Finally, when s, E [ak , are on the same side of the midp oint d, then hence (s) =

h 1 , h 2 , . . . , hn, hn + 1 .

hn + 1 (t) hn + 1

t

aH 1 ] 2n - 1 ;

[1Ji] C [0, 1] Let x E C [0, 1] and bn = J01 x(t)hn (t) dt, N. Ln EN bnhn converges uniformly to on [0, 1] \ {k/2n : 0 � k � 2n , n = 0, 1, 2, . . . } , the complement of the dyadic points in [0, 1] . Proof. Consider the nth partial sum n b.h. (t) " xn (t) L.Ji= 1 L1�_ 1 ( J(o x(s)hi (s) dS) hi (t) 1 L�= 1 1 x(s)hds ) hdt) ds.

IN 3.5.3 DENSITY O F nE Then the series

x

I

I

Since x is uniformly continuous (being continuous on a closed interval) , then given f > there must exist 6 > such that

0,

It - t' l < 6

==>

0 sup I x (t) - x (t') 1 < f . /t -t' / < 6

1 a . . . , an, an + 1 h l , h 2 , . . . , hn2 h 1 , h2 , , hn

,

Since the discontinuity points a , (arranged i n ascending order) of the elements of IJi are the dyadic points j 1 k an integer m can be chosen such that for n > m, the distance between suc­ cessive discontinuity points of is less than 6. For a non dyadic point E (ak , a H 1 ) , by the lemma,

t

•

.

2

•

1] '

Note that this is the average value of x on the interval [ak , a H so that the approximation to x is formed by step functions whose value is the average value of x on the dyadic intervals. By the choice of 6 and m, it follows that

for any nondyadic point t.

0

3. Bases

119

Exercises 3 . 5

1.

[iii] L 2 [0, 1] [iii] . L 2 [0, 1]. Show that the dyadic step functions D-functions constant on dyadic intervals [m/2 k , (m + 1) 12 k ) , m, k E Z-are dense in S. Show that the linear span Hn of the first 2 n elements of [iii ]

DENSITY O F IN Here is another way t o demonstrate the density of As noted in Example 2 .4.5( d), the class S of all step functions is dense in

(a) (b)

coincides with the linear span of the dyadic step functions Dn that are constant on the intervals of length 2 -n .

Hint

1.

{t l ,i 2 . . . t

xES

(a) Let , n } be the discontinuity points of that are not dyadic points . Surround each of these points by a small interval and define a dyadic step function Y D to agree with x outside these small intervals in such a way that f. (b) Hn C Dn and dim Hn = dim Dn .

E II x - y I1 2 <

3.6

Unconditional Convergence

A REARRANGEMENT of a series L neN Xn of vectors Xn from a normed space X is a series L neN Yn in which each Xn appears as some (unique) Yj · More formally, L n e N Yn is a rearrangement of L neN Xn if there exists a bijection f : N -+ N such that Yn = Xf( n ) for every n N . Riemann made the remarkable observation that the series L neN ( - l ) n In converges, but can be rearranged so as to converge to any real number whatsoever. Virtually any advanced calculus text has the argument ; it is outlined in the hint to Exercise a) . Series that converge on condition that the order in which they appear be left intact but that may diverge if rearranged are called CONDITIO NALLY CON VERG ENT . If the series converges to the same limit no matter how the terms are permuted, then it is called COMMUTA ­ TIVELY or UNCOND ITIO NALLY CON VERG ENT . In this section we examine some connections between absolutely convergent, convergent and uncondi­ tionally convergent series. The gist of it is that in any infinite-dimensional Banach space, there is the following hierarchy of implications:

E

3(

absolute

'\./ convergent

unconditional

120

3. Bases

In finite-dimensional Banach spaces, unconditional convergence implies ab­ solute convergence-this is essentially the result of Exercise 3 . In a Hilb ert space of arbitrary dimension, convergence of a sequence of orthogonal vec­ tors implies unconditional convergence (3.6.2) . in We observed in 2.6.3(a) that an absolutely convergent series a Banach space must converge. We strengthen this now . We show next that an absolutely convergent series in a Banach space is commutatively convergent .

(xn)

3.6.1 ABS OLUTE CON VERGENCE IMPLIES UNCONDITIONAL CON VERG ENCE

A n absolutely convergent series ally convergent.

Proof.

N N

(xn) in a Banach space X is uncondition­

( Yn) (x /(n ) ) be a rearrange­ n n Sn = L Xi and tn = L Yi . i= l i= l Since each X i must eventually appear in the sequence (Yn), for any n E N we may choose m � n sufficiently large that Let f : -> be a bijection and let = ment of the sequence Consider the partial sums

(xn).

m

n

2:7= 1 Il y; l1

i s therefore bounded above , s o it must The increasing sequence converge. Since X is a Banach space, the absolute convergence of and implies that it is convergent (2.6.3(a) ) . To see that the limits of are the same, we show that -> O. Since the partial sums are a Cauchy sequence, for t > 0 there exists such that

Sn - tn

. . . , m}

mEN

tn - Sn is a sum of terms x 0

Let k = max 1- 1 ( { 1 , ). For n � k, j> it follows from ( * ) that :S t .

m;

I i tn - sn ll

2: EN Yi Sn i t n 2:7= 1 Il x ill j

with

What about the converse? Does unconditional convergence imply abso­ lute convergence? For series of real numbers, unconditional convergence implies absolute convergence (Exercise 3). Indeed, (same exercise) the ar­ gument can be extended to show that unconditional convergence implies absolute convergence in any finite-dimensional normed space. But there it stops (Example 3.6 .3) . A profound result of Dvoretzky and Rogers ( 1950) asserts that if unconditional convergence implies absolute convergence in a Banach space X, then X is finite-dimensional! The proof can be found in Swartz 1 992 , pp. 440-4, Lindenstrauss and Tzafriri 1977, p. 1 6 , or for a very different approach, Grothendieck 1 955 , p . 149, or Diestel 1984, p .

3. Bases

121

59. Warning: These arguments are not for the faint of heart. I n Hilbert a series spaces, as we show next , for orthogonal sequences converges if and only if it converges unconditionally.

L n EN Xn

(xn),

3.6.2 UNCONDITIO NAL CONVERGENCE IN HILBERT SPACES Let (x n) be an orthonormal sequence in a Hilbert space X. Suppose (an) is a sequen ce from K an d L n EN a n Xn converges. Then L n e N anXn is uncon ditionally convergent. ( Note that any orthogonal sequence (mn) of nonzero vectors can be rewritten in the form (anxn) Il mn ll (mnl li mn I! ) We can therefore say that if (Yn) is a summable sequence of orthogonal nonzero vectors, then Ln e N Yn converges unconditionally.) .

=

Proof. Let x = L n EN anxn · Since L n EN anXn converges, L n eN l an l 2 < 2 00 by 3 .3 .4(a) . SinceL n E N l an l is an absolutely convergent series of real numbers, it converges unconditionally. By (3.3.4) (a) again, it follows that any rearrangement L n EN bn Yn of L nEN anXn converges as well. Suppose Ln EN bnYn = y . Does x y? For any i, for sufficiently large m, =

X i ) ai for all i. Thus, by Y L xn ) Xn L anXn x. 0 n EN n eN

so by the continuity of the inner product, (y, 3 .3 .4(e) , = = (y,

=

=

Example 3.6.3 UNCONDITIONAL CONVERGENCE DOES NOT IMPLY AB­ S O LUTE CON VERGENCE

(xn)

Let be a denumerable orthonormal basis for a Hilbert space X. converges by Since the coefficients are square-summable , = 3.3 .4(a) ; hence it converges unconditionally by 3 .6.2. Since does not converge absolutely. the series

Ln EN xnln Ln eN II xnln ll 0

lin L n e N xnln

Ln eN l i n,

Exercises 3 . 6

1.

(xn)

b e a denumerable orthonormal basis for a Hilbert space X . Let Imitate the argument of 3.6.3 t o show that for any square-summable sequence the series is unconditionally convergent.

(an),

LnEN anXn 2. SEPARABLE HILBERT SPACE £2 (Z) In 3.4.9 we showed that any separable infinite-dimensional Hilbert space is linearly isometric to £2 ( N ) . Show that £2 ( N ) can be replaced by £2 (Z) by showing that £2 ( N ) is linearly isometric to £2 (Z) . �

122

3.

3. Bases

UNCONDITIONAL IMPLIES ABSOLUTE in

Rn R Rn E N .

(a) Show that any unconditionally convergent series in is abso­ lutely convergent . ,n (b) Show that the result remains valid if is replaced by (c) Show that the result remains valid in any n-dimensional space over

R

K.

4.

2.6.3(

CRITERION F O R COMPLETENESS We observed i n a) that if X is a Banach space, then absolute convergence implies convergence. Show the converse, that if absolute convergence implies convergence in a normed space X, then X is a Banach space.

(Xn)

5 . CRITERION FOR UNCONDITIO NAL CON VERGENCE Let be a se­ is unconditionally quence in a Banach space X. Show that convergent if and only if for all f > 0, there exists a finite subset J of such that for any subset H of N for which J n H = 0 , s: f .

EnEN Xn

II En E H Xn I

N

Hints Consider a bijection f : -+ Z such as f ( n ) = % if n is even and for n odd. By then f (n) = if (Z). I--t < 00 . Now consider the map

N 3.6.1, (an) E £2 ( N ), En EN laf ( n ) 1 2 (an) (af ( n ) ) E £2 3. (a) Prove the contrapositive , that if EnEN l an l 00 then there is a rearrangement E n EN dn that is divergent. If E nEN an diverges, there is nothing to prove so suppose that it converges. Since E n EN an < 00 , there must be infinitely many positive terms and negative terms. Let bn and Cn be the subsequences of nonnegative and nega­ tive terms of (an) in the order in which they appear . For each n, then there are positive integers p� and qn such that E �=l a i (b 1 + . . . + bp n ) + ( C l + . . · + cq J . If EnEN bn 00 then E nEN Cn must converge well because E nEN an does and therefore so would E n l an l = (b 1 + . . . + bp J - (C l + . . . + cq n ) · Consequently, L n EN bn EN00 , and, similarly, E nEN Cn -00 . Now choose enough b's to make their sum � 1, then add in enough c's to make the sum S: O. I n this way create a rearrangement that oscillates and therefore does not converge. (b) For (a i , bi ) E R2 , note that E �= da i' bi ) = ( E �= l a i , L �= l bi ). Now use Example 2.2.6(b) and the result of (a) above . 2.

l ;n

=

<

as

=

=

=

3. Bases

123

(xn) II xn - x m ll Y Xnk - Xnk_1 LnEN Yn Xnk .

4. The idea is to show that a Cauchy sequence has a convergent sub­ sequence. Choose an increasing sequence ( nk ) such that :::; and k = 2- k for n , � nk . Let = for k > 1 . The is kth partial sum of the absolutely convergent series

m

YI Xn,

5. To show sufficiency, argue as in 3.6 . 1 . Argue necessity by contradic­ tion to get , for some > 0, disjoint finite subsets (H k ) of N such that IILn EHk Xn I � for every k. Define a bijection f N N such that €

€

:

L n E N xJ ( n ) is not convergent .

3.7

-+

Orthogonal Direct Sums

We considered external and internal direct sums of a finite number of inner product spaces in Section 2 . 1 0 . We consider infinite direct sums of Hilb ert spaces in this section. If is a countable orthonormal base for a Hilbert space X, then by 3 .4.2(a) , X = cl ( ) ] , where denotes the linear span of . . . } . Furthermore, the subspaces are mutually or­ thogonal and closed. This inspires the following definition.

(xn)

{X l , . . . , Xn,

[ xn

[(x n )l [x n l

Definition 3.7.1 INTERN AL ORTHOGONAL DIRECT SUMS

ffi n EN Mn

The INTERNAL O RTHOGONAL DIRECT SUM ( HILBERT SUM ) (or (int)) of a countable family of mutually orthogonal closed subspaces of a Hilbert space X is cl

ffi n E N Mn

(Mn ) [Un EN Mn] . 0

If is an orthonormal base for a Hilbert space X , then X = Moreover, by 3 .3 .4(e) , any E can be written as in other words, can be written as an infinite series of orthogonal projec­ Our next result shows that a vector tions on the subspaces from an orthogonal direct sum may be written as the infinite sum of its orthogonal projections on the

(x n )

( x, xn) Xn

X

X ffi nEN [xnl [xnl. ffin EN Mn Mn .

ffinEN [xnl. Ln EN (x, xn) Xn ;

3.7.2 PRO PERTIES O F DIRECT SUMS (a) SQUA RE-SUMMABILITY

For (Mn) and X as in Definition 3. 7. 1, and x E ffi nEN Mn (int) there exist mn E Mn such that x = LnEN mn and II x I1 2 = LnEN li mn 11 2 . Conversely, for mn E Mn such that L n EN II mn l1 2 < 00, then L nEN mn converges. For each E N , let Pn denote the orthogonal projection of X on Mn. Then, for any x E X, x = L nEN Pnxn. n

(b) UNC O N D ITIO NALLY CONVERG ENT SUMS

ffinEN Mn (int) = L n EN Mn , mn E Mn .

the set of all unconditionally convergent sums L nEN mn with

124

3. Bases

N [Un EN Mn] . x ffin EN Mn X n Mk ( n ) I I x - xn ll X n Mi + . . . + Mk ( n ) Xn -- x; Mi , Pn X Mn . 3 2 7( ) Pn x Mn . 3.2 7( ) P Pi + . . ' + Pk (n ) ' PiX + . . . + Pk (n ) X Mi + . , . + Mk (n ) ' X n Mi + . . . + Mk (n ) ' Proof.

For each n E there = cl (a) Let E exists k (n) E N and E such that < lin . Hence there exist vectors we can assume E such that that the sum starts in provided that we take some 0 vectors . Let denote the orthogonal projection of on the complete (closed subset of a complete space) subspace By . b , is the best approximation to x from Hence By Exercise - b M1 + .. +Mk ( n ) = is the best approximation to x from Therefore , since E .

(Pnx) x L nEN Pnxn . 2 3.3.4(g) II x II Ln EN II Pnx I1 2 . mn Mn L nEN mn L n EN li mn 11 2 M 3.3.4(a). mn ffi n] [U Ln E N n EN Mn · n EN 3.6.2, x ffi n EN Mn , mn Mn x L nEN mn · (mn ) 3.6.2. x ffin EN Mn LnEN Mn . Mn mn L nEN . Ln EN x ffin EN Mn 3.3.4(a). Ln e N I I mn l1 2 0

Since is an orthogonal sequence, it follows Thus, = from that = E is such that Conversely, if < 00 , then E cl = converges by Clearly, (b) By there is no difference between convergence and uncondi­ tional convergence of series of orthogonal vectors in a Hilbert space. By (a) , if E then there exist E such that = Since is an orthogonal sequence, the series is unconditionally con­ vergent by Hence C Conversely, let = E As a convergent series of orthogonal vectors , it by follows that < 00 by Therefore, E (a) above. To define external Hilbert sums, we use square-summability again . Definition 3.7.3 EXTERN AL ORTHOGONAL DIRECT SUMS

We define the EXTERNAL ORTHOGO NAL DIRECT SUM ffi n EN Xn (or ffi n EN Xn (ext)) of a countable family (Xn , ( - , ' } n ) of Hilbert spaces to be the collection of all sequences (Xn) from the Cartesian product Il nEN Xn such that L nEN II xn ll 2 < 00 . ffi n EN Xn (ext) is also called the EXTERNAL

HILBERT SUM. We add such sequences and multiply them by scalars in the usual pointwise way to make a vector space of them. We define the inner product of two such sequences and to be

(x n ) ( Yn ) {(Xn), (Yn) } = L {xn, Yn } n ' 0 (3.14) n EN As we prove below , not only does equation 3 . 14 define an inner product , ffin eN Xn i s a Hilbert space. 3. 7.4 EXTERN AL ORTHOGONAL DIRECT SUMS

direct sum ffi n E N Xn of a countable family Hilbert space.

The external orthogonal of Hilbert spaces is a

(Xn)

3. Bases

125

Proof.

The closure with respect to the operation of addition follows from the Minkowski inequality 1 .6.3(b) with p = 2. The only remaining problem of any substance about the proposed inner product is the convergence of We show that is (absolutely) the series convergent by using the Cauchy-Schwarz inequality 1 .3.2, followed by the Holder inequality 1 .6.2(b) with p =

L n EN (Xn, Yn} n

L n EN (Xn, Yn} n ·

2:

The remaining properties o f the inner product are trivial t o verify. To see that is a Hilbert space, consider a Cauchy sequence = = Xn . Let us now "look down the columns" . . . from from at n = 3, say, at the sequence For any k and j, It follows that , with ; :S � = is a Cauchy sequence from the Hilbert space Xj . Let n fixed, and for each n. We now show that = E = limj --4 Since is a Cauchy sequence in given > 0 there exists N such that for k, < . = Taking the limit as --4 00 ,

Xl ffin N Xn (X1 n )' X 2 (XE 2n )' ffin EN X3 . (X k 3 ) kEN 2 II X k 3 - xj 3 11 Ln E N II X k n - Xj n ll II X k - Xj 11 . (X, n)jE N Xj n Yn Y ( Yj ) ffinEN Xn Xn y. (Xn) ffi2 nEN Xn , f m 2: N, II X k - x mll Lj II X kj - X mjll� f 2 m (3. 15) L II X kj - Yj ll� :S f 2 . j Thus , X k - Y E ffi n E N Xn. Since Y = (y - X k ) + X k for every k, it follows that Y E ffi n E N Xn. That Xn Y now follows from equation ( 3 . 1 5 ) . 0 Thus, the elements of external direct sums are sequences; the elements of internal direct sums are sums. Let (Mn) be a collection of mutually --4

orthogonal closed subspaces of a Hilb ert space. It routine (but tedious) to verify that the map

A : ffinEN Mn (ext) (mn)

---*

�

ffin EN Mn (int) L n EN mn

(3. 16)

is a surjective inner product isomorphism .

Exercises 3 . 7

X

1. PASTING ORTHONORMAL BASES TOGETHER Let the Hilb ert space be decomp osed as = (int) and let be an orthonormal base for for each n E Show that is an orthonormal basis for

Mn X.

X ffin EN Mn N.

Bn UnEN Bn

126

2.

3. Bases

Show that the inner product of Definition uct axioms.

3.7.3 obeys the inner prod­

3. Verify that the map of (3.16) is an inner product isomorphism . 3.8

Continuous Linear Maps

Let X and Y be normed linear spaces. A linear map : X ---> is contin­ uous if and only if it is bounded on the closed unit ball U of X by 2.3.3; i.e . , for some � 0,

A

Y

k

II A x li ::; k for all x E U {::::> II A x li ::; k II x li for all x E X. (3.17) The set L (X, Y) of all continuous linear maps (also called OPERATORS or BOUNDED OPERATORS) of X into Y is a vector space with respect to the pointwise operations: for any A, B E L (X, Y) and any scalar a, we define A + B and aA at any x E X to be ( A + B) x = Ax + Bx and (a A) x aAx.

=

With

=

II A II = sup II A (U) II , L (X, Y) is a normed space, as we verify next . Clearly, II A II � 0 and II A II 0 if and only if A = O . For any scalar and any x E U, II (aA) x II = l a l ll A x ll so aA is bounded on the closed unit ball, and since nonnegative multiples a

come "through" the sup,

lI aA I i = l a l ll A II · For A, B E L (X, Y ) , and x E U, II ( A + B) x II = II Ax + Bx ll ::; II A x li + II Bx ll · The triangle inequality follows immediately. are given in Some alternative ways to compute

II A II

3.S.1 T H E NORM OF A LINEAR MAP

For A

3.8.1.

E L (X, Y ) ,

sUPl :9 11 Ax ll = sUP l l xx ll = l I I Ax i l = inf {k � 0 II A xl l ::; k for all x E U} . Proof. Let I<. = {k � 0 for all x E U, II A x li ::; k}. Clearly II A II ::; inf I<. . Since I I A x II ::; II A II for all x E U, it follows that I I A II E I<. The refore II A II = inf I<. . Since sup i s monotonic (the bigger the set, the bigger the sup) , c = sUP l x l l = l l1A x li ::; II A II . Suppose that 0 :5 c < I IA II . If so, then there must II A ll

:

:

3. Bases

127

x

be someSincewithx 0 << 1,II xthis l < 1 such that c < I A x l i . Thus, I A (xl I x lDI l ::; c < I Ax l i . I l l means that II Ax l ::; c I Ix l < I Ax l l x l l ::; I A x l · It follows from this contradiction that c = sUPl l x l = l l 1 Ax l l = I I AI I . 0 We compute the norm of an orthogonal projection next.

Example 3.8.2 NORM OF ORTHOGONAL PROJECTION

=

1

Let M be a closed nontrivial linear subspace of the Hilbert space X . By the projection theorem, 3.2.4, X = M$M .L (top). Let Pbe the orthogonal projection x E X, there are unique elements m E M and n E M.L such that xon M.m For + By the Pythagorean theorem, I x l l 2 = I mll 2 + I n l 2 � I ml l 2 = II Px ll 2 so I I Px ll 2 ::; I x 1 l 2 ; therefore, II P II ::; 1. Since there must be unit vectors x E M for which P x = x, it follows that I I P II = 1 . 0 =

n.

3.8.3 NORM O F AN INTEGRAL OPERATOR ON REAL C [a, b] Let C [a, b] denote the real Banach space of real-valued continuous functions on [a, b]. For every s E [c, d] , let the real-valued function D (t, s) be continu ous for all t E [a, b]. For E C [a, b], define

x

Then

Ax (s) = lb x (t ) D (t, s) dt.

A is a continuous linear map of C [a, b] into C [c, d], and I A I = s esup[c ,d] Ja[b I D (t, s) 1 dt.

D is called the K ERNEL or K ERNEL FUNCTION of the integral operator A . Proof. The map A is clearly linear and maps C [a, b] into C [c, d]; A is

bounded since l Ax (s)1 = Ja[b x (t ) D (t, s) dt (s esup[c,d] Ja[b I D (t, s) 1 dt) II x l i oo (x E C [a, b] ). Thus, [c,d] f: I D (t, s ) 1 dt . In the remainder of the argu­ II Dthat= sDUPiss e the mentConsider weI Ashow the ramp functionsnorm of A. { -1 t ::; -lIn, Un (t) nt, I t I < l I n, , n E N , 1, t � l In, ::;

::;

=

'

3. Bases

128

= Un Let X n for any fixed this composite of continuous functions of is clearly continuous in For all and the product

(D (t, s)); s, t t. t s, lin xn(t, s)D (t, s) = { nI DD(t,2 (t,s)s)l , :S I D (t, s) l , II DD (t,(t, s)s) 1l >:;;: lin (t, s)

is nonnegative. Since continuous functions are Lebesgue measurable,

la b xn(t, s) D (t, s) dt

AX n (s) =

2:

{

J{tE[a , bl:ID(t ,s)l � l/n}

I D (t, s) 1 dt (3.18)

Since

lb I D (t, s) 1 dt

{

I D (t, s) 1 dt + {1D(t,s )15.l/n I D (t, s) 1 dt J 1 1ID(t,s )l ? l/n I D (t, s) 1 dt + lID(t,s )l $ l/n -n dt,

J1D(t,s )I? l/n

<

equation

( 3 18 ) implies .

b A x n (s) 2: la I D (t, s) 1 dt - -n1 (b - a)

for every n and

s. Therefore ,

b

I Axn (s) 1 2: SE[supc ,d] la I D (t, s) 1 dt. nEN,sE[ c ,d] sup

Since

IX n (t, s) 1

I IAII

sup

=

1I "'1 I 00 5.

:S

1 for all t, n and s, it follows that

sup I D (t, s) 1 dt = D, I Axn (s) 1 2: s sup II Ax ii oo 2: nEN,sE[c,d] l E[c ,d] la

b

which completes the proof.

0

Exercises 3 . 8

1. 2.

A:

X � Y b e a linear isometry between NORM OF A N ISOMETRY Let the normed spaces X and Y . Show that =

II A II 1.

Let x b e a nonzero element of the normed space X and consider the linear map g", Show that the map ...... g", is a � X , ...... linear isometry of X onto X) .

:K

a

ax. L (K,

x

3. Bases

129

loo {U l , U 2 , " " un} A : loo N . C l , . . . , Cn K. loo loo a I , a 2 , . . . , an, A ( E� l ai u i ) = E7= 1 Ci a i U i , Ci {U l , U 2 , . . . , un} II A II = maxi l ei l · 4. FINITE- DIMENSIONAL CASE : THE NORM OF A : l2 ( n ) l2 ( n ) , n E N . This seems as though it should be easy to compute, but it is not. We discuss the results and provide some details in the hints . Let { U l , . . . , un } be an orthonormal basis for l2 ( n ) and let (aij ) be the matrix representation for A with respect to this basis. We denote the rows (ai l , a i 2 , . . . , a i n) of (a ij ) as a i and assume that none of them is 0; we denote the columns by Ci . Note that this means that for x E lz ( n ) , 3.

( n ) --+ THE NORM O F ( n ) Let be a basis ( n ) by (n) , n E E Let ( n ) -+ for Define A : taking, for any scalars so that the matrix representation for A with respect to the basis is a diagonal matrix with on the main diagonal. Show that

loo

--+

Ax =

( ���: :� )

(an , x) We begin with some bounds on II A II . (a) B O UNDS Show that

n max { miax ll aill , mF ll eill } :S II A II :S L l aij l 2 . i ,j = l

(b) PARALLEL Rows If the rows are "parallel" in the sense that for then there are constants . . . , n, Thus, for example ,

VE?'j =1 I aij I 2 .

ki ' i = 2, 3, 1 1 1 2 2 2 3 3 3

a i = ki a l ,

II A II =

= )42.

(c) ORTHOGONAL Rows If the rows are mutually orthogonal (as column vectors in ( n )) , then (d) ORTHO GONAL GROUPS OF PARALLEL Rows If the rows can be grouped in sets where within each the rows are parallel but rows from different groups are orthogonal , then is a group of rows for which where

l2

II A II = maxi lI a; ll .

Ml , M2 , ' . . , Mk Mj IIAII = VEa,EMJ Il ail1 2 , VEa,EMj Il ail1 2 is maximal.

Mi

130

3. Bases

Xl, X 2 , , Xn is an orthonormal basis of £2 ( n ) of eigenvectors of A £2 ( n ) - £2 ( n ) , then its matrix representation is a diagonal ma­ trix with its eigenvalues C l , C 2 , , Cn on the diagonal. Use (d) of the previous exercise to show that II A II = maxi l ed .

5. If

.

•

.

:

.

•

•

Hints

4.

AUi = Ci for each i and conclude that II cd l :S II A II·

(a) Show that N ext, note that

bi - l Il ai II bi + l bj

bn

Il ad l :S I (ai , x) 1 lI aill :S Il x ll 1, I l ad l ll x ll Il aill for each i. This implies that II Ax l1 :S /'L 7= 1 1I aiI1 2 J'L�j = l l aij 1 2 . (b) Consider an orthonormal basis {x 1 , X , . . . , xn} for £ ( n ) in which X l a l · Thus (aj , X k ) = 0 for j, k 2:2 2. For a unit2 vector X = 'L7= 1 bi X i , the jth row of the2 column vector Ax is bj kj (a l , a l ) with k l = 1. Thus II Ax ll 2 = I bd 11 a l l1 4 'L7= 1 I ki 1 2 . This fraction is maxi­ mal when 1 � 1 2 is maximal, which-since is a unit vector-occurs when I h I = 1. Therefore , n n 2 2 2 Ik = L a j l2 . II A II = II a ll1 L =i l il i ,j = l l i (c) {a l , a 2 , . . . , an} must be an orthogonal basis for £ 2 ( n ) . For x 'L7= 1 bi a i , Ax is the column vector whose jth row is (aj , x ) = bj Il aj l1 2 . Assume that is a unit vector and that Il a l ll 2: Il aj II for all j. Show that n 2 2 = a + Ax II l1 II l ll L I bd 2 ( 11 ai11 4 - ll a d I 2 1I ad I 2 ) . i= 2 Since Il a lll is maximal, it follows that Il a il1 4 - II a ll1 2 11 a ill 2 :S 0 for i = 2, 3, . . , n . Therefore, II A x ll 2 achieves its maximum lI a l l1 2 when bi 0 for i 2: 2. for appropriate and every i (the b's depend on i) . Therefore, for each i. < = by the Cauchy-Schwarz inequality For

II A II

=

=

=

x

x

.

=

=

3. Bases

3.9

131

Dual Spaces

L

Y)

A s i n the preceding section, ( X , denotes the normed linear space of all continuous linear maps of the normed space X into the normed space i.e . , the linear forms are A special case of great interest is that of = to X ' . The normed space the linear functionals on X. We shorten (X, is called the (CONTINUOUS) DUAL or (CONTINUOUS) CONJUGATE of X. The study of the interplay between X and X' is called DUALITY THEORY . We compute some norms of linear functionals i n the next example .

Y.

Y K, L K)

X'

Example 3 . 9 . 1 NORMS OF SOME LINEAR FUN CTIONALS

(a) PROJECTION O N A VECTOR: Let z be a nonzero = vector in an inner product space X and consider the map Iz : X --+ The map Iz is linear by the linearity of the inner product in 1--+ the first argument ; it is continuous because it is bounded on the unit ball for U of X by the Cauchy-Schwarz inequality: l iz it = any E U. Hence z = sUP II�II = l l /z Since l iz follows that z (b) EVALUATIO NS. For 0 , 1 define the (linear) evaluation map i to be the map 1--+ of 0 , into Clearly, I i I = ( ) for any x, so II i ll As there are continuous functions x of norm 1 such that = 1 , II i ll = (c) INTEG RAL OPERATOR ON REAL Let be the space of continuous real- or complex-valued functions on with sup norm For let I ( ) = = sup dt . For any the linear form defined by

" z ll

II {·, z} 1I

K,

x {x , z} .

(x) 1 � II z ll ll x ll � II z ll (z) 1 II z 11 2 , (x) 1 � II z ll .

II l " II l " = II z ll · t E [ ], x x (t) C [ 1 ] K. � 1. 1. x (t)

x

I x t 1 � II x 11 00

C [a, b] C [a, b] [a, b] { l x (t) 1 : t E [a, b]), x E C [a, b]. x E C [a, b] ' D E C [a, b]

II x lioo f: x(t)

is continuous, and

c = d.

(x)

0

x

I (x) = lb x(t)D (t) dt

11/11 = f: I D (t ) 1 dt. This follows from 3 . 8.3 by taking

The following result , known as the Hahn-Banach theorem, is very im­ portant. We omit its proof but it can be found in almost any book on functional analysis (for example , B achman and Narici 1966 , p . 176f) . 3.9.2 HAHN-BANACH THEOREM If f is a bounded linear form defined on a linear subspace M of a normed space X then there is a continu ous extension F of f to X such that II F II = 11/11 .

F: X

I

I:

I M

'\.

---+

K

and

11/11 = II F II

3. Bases

132

Two imp ortant consequences of the Hahn- Banach theorem are the fol­ lowing: • •

=P {O} . If all continuous linear forms vanish on a given vector x, then x = O. The dual of a nontrivial normed space X is not trivial : X '

Information about the dual of a space can be very enlightening. Hilbert spaces X, for example, are essentially their own duals. Before we prove that, let us make an observation about the enormity of the null space of a linear functional. 3.9.3 NULL SPACE O F A LINEAR FUNCTIO NAL If f is a nontrivial linear form on the vector space X, then any algebraic complement of its null space N 1 - 1 (0) is one-dimensional, i. e., of the form [m] = Km for some vector m =P o. =

P roof. Does N have an algebraic complement? Since f =P 0, there exists m E X such that I (m) =P O. For any x E X , consider z x - (f (x) / I (m)) m. Clearly, I (z) = 0 so z E N and x z + (f (x) /f (m)) m E N + [m]. Obviously, N n [m] = {O} so [m] is an algebraic complement of N. If L is any algebraic complement of N and y is any nonzero vector in L, then X = N [y] , exactly in the argument above . Consequently, if z E L, then we can write it in the form z + ay for some E N and scalar a. Since z - ay E N and z - ay E L, it follows that z - ay 0, i.e . , that z E [y]. Therefore , L = [y] and the proof is complete. 0 We observed in Example 3.9 . 1(a) that maps fz (x) (x, z) are contin­ uous linear functionals of norm II z ll . In Hilbert spaces, these are the only continuous linear functionals, as shown next . The ability to write any con­ tinuous linear functional on a Hilbert space in the form Iz (x) = (x, z), =

=

Ef)

as

= n

= n

n

=

=

where z is uniquely determined by f, is known as the RIESZ REP RESENTA­ TION THEOREM; it is subsumed by 3.9 .4.

3.9.4 HILBERT SPACES ARE (ESSENTIALLY ) SELF-DUAL Let X be a real

or complex Hilbert space. For any z E X, the map Iz , continuous linear functional on X. If X is real,

A:

X

--+

z ....... ( . , z ) ,

is a

, Iz ,

X'

then A is a surjective linear isometry. If X is complex, then A is a "conju­ gate lin ear" isometry of X onto X ' , where by "conjugate lin ear" we mean A (a x) = aAx for any scalar a and any x E X .

A

f II A z ll = II lz II

Proof. It is trivial to verify that the map : X � X ' , Z 1--+ z is conj ugate linear or linear , depending upon whether X is real or complex, respectively. = We observed in Example 3.9 . 1(a) that is norm-preserving:

A

3.

li z I

Bases

133

, i.e . , it is an isometry. To see that A is onto, let I be a nontrivial continuous linear functional on The null space N = (0) is closed by = N $ N .L by the projection theorem 3 .2 .4. By 3.9.3, 2.4.4. Therefore, there is some unit vector such that N .L = [ . By Example 3.2.7(a, c) , we can therefore write any E X as = n + for some unique n E N. Hence

X

1- 1

X.

m

x

m] {x, m} m

x

I (x) = I (n + (x, m } m) = (x, m} / (m) = ( x ' / (m)m ) . In other words, I = Ij ( m ) m . Finally, note that I (m)m is unique, for if y were also such that

{ x , y} = I (x) = ( x, I (m)m ) , for all x E X, then y - I (m)m would be orthogonal to every vector x and would therefore have to be O. 0 Exercises 3 . 9

1.

R3 R Iz ( ) { . , }

Let I : � b e defined by . = z . z such that

(a, b , c)

�

a

3 -b+

I (x y) x, y X] a R x Let I : R3 R be the linear map (a, b, c) continuous and compute 11/11 .

c. Find the vector

I (x) I (y) X R. (ax) a l (x)

2 . Let I be a continuous, additive [i.e. , + = for all + Show that E map of the real normed linear space into J is necessarily SCALAR HOMO GENEOUS , i.e. , I for all = E and E X. 3.

�

1---+

a + b. Show that l is

4 . By Exercise 2.9-4, all linear functionals on a finite-dimensional normed space X are continuous. Let , } be a basis (not neces­ Then, for any sarily an orthonormal basis, just a Hamel basis) for Let I be a E there exist scalars such that = in the following cases: linear functional on Find

{X l , X 2 ,

x X,

•

.

.

xn X. x L� l a i X i .

ai 11 / 11 (a) Norm X by taking II x ll maxi l a i l . 1/2 (b) Norm X by taking II x ll = IIL?=l a i x i ll [L ?=1 I a i I 2 ] . 5 . On the space £ 1 of absolutely summable sequences x = (an) (i.e . , (a n ) such that Ln eN l an l < 00 ) , consider the map I defined by taking I (x) = Ln e N an · (a) With II x ll 1 = L n eN l an l , show that I is a continuous linear X.

=

form and compute its norm.

=

134

3. Bases

(b) If i 1 is normed by taking discontinuous.

6.

II (an) lI oo

=

S UPn

l an l , show that I is

DUALS ARE COMPLETE For any normed space X, show that (a) The dual X ' is a Banach space. (b) If Y is a Banach space, then so is X.

L (X, Y) for any normed space

7. THE SECOND DUAL Let x E X where X is a normed space and define --+ I f-+ I (x) . x'

: X' K,

( a) Show that x , is a bounded linear functional on X' . (b) The dual X" of X, is called the SECOND DUAL or SECOND CONJUGATE of X. With x' as in (a) , show that the map into X". --+ f-+ x ' , is a linear isometry of J : (c) If the map J of (b) is onto, then is said to be REFLEXIVE. Show that all finite-dimensional spaces and Hilbert spaces are reflexive. (We are not asking you to prove this, but we mention that the ip for 1 < p < 00 are reflexive but i 1 is not .)

= L (X', K) X X", X

8.

X

X

Let Ibe a linear functional on the Hilbert space X with null space N. (a) Show that if l is not continuous, then N is dense in (b) Prove that I is continuous if and only if N is closed .

X.

9. L e t X be a Hilbert space. If x , y E X are such that for any contin­ uous linear functional I on X, I (x) I (y) , then show that x y. =

=

(It follows from the Hahn-Banach theorem that this is true for all normed spaces. )

Hints 2. First show that I (nx) = nl (x) for integers n; then show that I (x I n) = ( l in) I (x) . Conclude that for any rational number r, I (rx) = r I (x) . Finally, use the density of the rationals in the reals .

4.

ai X i·

Let x = E?= 1 (a) I / (x) 1 � I l x I l E?= 1 1 / (xi ) l , so 1 1/1 1 � E?= 1 1/ (xi Now let ai = I/ (xi ) l /l (xi ) for I (xi) 0, = 0 otherwise. (b) By the triangle and Holder inequalities,

f. ai

a d (X i ) I < E?- 1 l ai I I I (X i ) I < I l x l l (E?- 1 I I (X i ) 1 ) 1 / 2

IE?- 1 IIxll so 1 1 / 1 1

-

IIxII

-

� (E i=n 1 1 1 (X i ) 1 ) 1/ 2 . Now let ai I (X i ). =

--

Ilxll

'

3. Bases

135

5 . (a) Use the triangle inequality. ( b ) Consider the sequence from the unit ball = ( 1 , 1 , . . . , 1 , 0, 0, . . .

xn

)

whose first n entries are 1 and 0 thereafter. 6(a) . For a Cauchy sequence im

l n In (x) .

8.

Un ) from X'

and

x

E X , define

I (x)

=

X

(a) . Prove the contrapositive. Assume that cl N =P and choose a unit vector w E (cl N) 1. . Show that 1 0 = - . ( w ) w .

(

I

)

(b) We need only prove that N closed implies f continuous. This follows from ( a) . 9 . Use 3 . 9 . 4 to show that

3 . 10 Let Fix

x - y E X 1. .

Adj oints

X be a Hilbert space and let A : X � X be a bounded linear map . y E X and consider the map Iy (x) = (Ax, y) , x E X .

As the composite of continuous linear maps , f is clearly a continuous linear functional on As such , by the Riesz Representation theorem 3 .9 .4, there exists a unique E such that

X. y* X (Ax, y) = ( x, y* ) for all x E X

This procedure induces a mapping

A* : X ---+ X, Y 1----+ y* , called the ADJOINT A* of A , where A*y = y* is defined by (3.19) (Ax, y) = (x, A*y) , for all x E X . It i s routine t o verify that A* i s linear and that , for any bounded linear maps A, B X X and any scalar a, A** A (see 3 . 1 0 . 1 ) , (A + B) * A* + B* , CiA* , (aA) * (AB) * B*A*. If A = A* , then A is called SELF-ADJ OINT. :

�

136

3. Bases

3 . 1 0 . 1 ADJOINTS Let X be a Hilbert space and lei A X bounded lin ear map. The adjoint A* is a bounded linear map with :

X be a II A* II =

II A II · Proof. We first show that adjoint is norm-reducing, i.e. , that II A* II � II A II . To this end , note that for all x E X, (A*x, A*x) ( AA*x, x) II AA* x ll · ll x ll < II A II II A*x ll ll x ll For nonzero A*x, this implies that II A * x ll � II A ll ll x ll so II A * II � II A II · Thus, since A* is bounded, we can consider its adjoint A** as defined by (A*x, y) = {x, A**y} for all X, y E X. Since (x, Ay) = (Ay, x) = ( y, A*x) = ( A * x, y) = (x, A ** y) for all and y, it follows that A**y Ay for all y and therefore that = A* * A . Since the norm of an adjoint is no larger than the norm of the x

original map , we have

=

=

II A II II A** II � II A * II .

0

We show next that the adjoint of a linear map defined by matrix multi­ plication is determined by the transpose of the original matrix in the real case; otherwise, it is the conjugate transpose . Example 3.10.2 ADJOINT OF

A : R2 R2 _

"

= " TRANSPOSE O F

A

Let a, b, d E R. In this example we make repeated use of the fact that if axi + bX 2 = 0 for all X l , x 2 E R, then a = b = O. Consider the linear map A : R2 R2 , ( : � ) ( � � ) ( : � ) (matrix multiplication) . Since A * also maps R2 into R2 , we see that there must b e a 2 2 matrix ( ; { ) such that A* ( �� ) = ( ; { ) ( �� ) for all ( � � ) E R2 . For all ( � � ) and ( �� ) , by the definition of adj oint, c,

_

f-+

x

or

(a YI + CY2 - e YI - !Y2 ) X l + (bYI + dY2 - gYI - h Y2 ) x 2

which implies that

(a - e) YI + (c - f) Y2 (b - g) YI + (d - h) Y2

0,

O.

=

0,

3. Bases

Therefore, a

= e, C =

the transpose of any

n

x

n

I, etc. We see that

( � ! ).

A*

=

137

( � � ) ( � ! )t,

We ask the reader to prove that the adj oint of

matrix is its conj ugate transpose . 0

Exercises 3 . 1 0

1.

X, A X X

EIGEN VALUES O n a vector space let : -> b e a linear map . If x is a nonzero vector and a is a scalar such that x ax , then we say that a is an EIGENVALUE of A; x is called an EIGENVECTOR of A. Suppose that is a Hilbert space and that E IS self-adjoin t .

X

A = A L (X, X)

(a) Show that all eigenvalues of A are real. (b) If x and y are eigenvectors of A associated with the distinct eigenvalues a and b, respectively, then x is orthogonal to y.

4

Fo urier Series Fourier uses his mathematics with the delightful freedom and naIvete of [one] who trusts in a mathematical providence. -E. B . van Vleck N OTE . Except for Section 4.19, we restrict consideration to real­ valued functions in this chapter. The expression "function I" . means "I : R -+ L:; [-11", 11"] and Ll [-11", 11"] denote the subspaces of real-valued elements of L 2 [ - 11", 11"] and Ll [-11", 11"], respectively. The process of understanding is always facilitated if more complicated structures are known to be synthesized from simpler ones-matter from molecules, molecules from atoms , atoms from quarks, organisms from cells , integers from primes. In mathematical analysis we don't usually get a full decomposition into the simpler things but an approximation of the more complex by the more elementary-real numbers by rationals, for example . When you truncate the Taylor series expansion of a function, you approx­ imate the function by a polynomial. The quality of the approximation de­ pends on the number of terms you take. Of course, for a function to have a Taylor series, it must (among other things) be infinitely differentiable in some interval, and this is a very restrictive condition. Sines and cosines serve as much more versatile "prime elements" than powers of t. Sines and cosines may be used to approximate not only nonanalytic functions ; they even do a good j ob in the wilderness of the wildly discontinuous. And even though sines and cosines are periodic, this does not constitute a very serious limitation. In most situations there is usually some interval in time or space on which interest centers. Functions can be truncated to the interval of in­ terest , and then periodically extended . As we shall see, sines and cosines triumphantly approximate functions I from the wide class Ll [-11", 11"] of integrable functions on [-11", 11"] in the sense that their FOURIER SERIES

R;"

�o + L an cos nt + L bn sin nt nEN nEN

(

4 .1)

(C, I), etc.) , to I, ( 4 .2)

converge somehow (in the mean, pointwise, uniformly, where ={O} , I(t) cos nt dt , n E

an

and

1

Nu 1"2 bn = -1 1 "- I(t) cos nt dt , n E N ; 11"

_ ,,-

11"

0

( 4 .3)

140

4. Fourier Series

bn

are called the FO URIER COEFFICIENTS of f. the terms a n and The attempt to approximate functions by trigonometric series began in connection with the problem of vibrations of strings, i.e . , attempts to solve the wave equation

fJ 2 w - a 2 8 2 w 8t 2 - 8x 2 ·

(4.1))

Daniel Bernoulli ( 1 753) proposed a trigonometric series (equation as a solution. Euler and a young Lagrange also made contributions t o the problem. In connection with a trigonometric series that arose in a different problem, Euler published the integral form of the and of equation ( . 1 in 1757 . The problem of how heat diffuses in a continuous medium defied the mathematics of the day in "the best of times . . . the worst of times," the early 1 800s. Jean Baptiste Joseph Fourier invented powerful new techniques to attack the problem, which included "Fourier series." (We should point out that "convergent trigonometric series" and "Fourier series" are not the same thing [see Section 4. 1 1] . ) Fourier's efforts were based on a shaky, intuitive foundation: He did not prove that his series converged . His con­ temporaries, Lagrange , Legendre, and Laplace, objected-not a pedantic complaint when you consider some of the nonsense that can occur when infinite processes are treated too casually. Fourier asserted that any peri­ odic function whatsoever could be written as a trigonometric series. Alas (Section .9 , some functions are too tempestuous to permit such a repre­ sentation. But even so, Fourier was mostly right. It is imp ossible to conceive of the signal processing upon which the modern world depends, from CDs to satellites, without Fourier series and its modern offshoots-the Fourier transform, the fast Fourier transform ( "the most valuable numerical algo­ rithm in our lifetime" [Strang 1 993 , p. 290) ) , and the wavelet transform (see Section 7 . 1 1 ) . An unintended consequence of Fourier's insightful but rash forays essentially brought analysis from its childhood to adolescence during the nineteenth century. Much of the gestation had a direct root in an attempt to j ustify Fourier's manipulations, or to solve other problems related to Fourier series. Dirichlet was the first to prove pointwise convergence of Fourier series for a large class of functions. In memoirs in 1829 and 1 837 , he showed that the Fourier series of a piecewise smooth function 1 on [-11", 11"] converges pointwise at every to the average of the limiting values

an

4 )

bn

4 )

(4.6.2)

t l (r) + / (t + ) 2

Dirichlet suspected that the theorem was true for a wider class of functions but was never able to prove it. Jordan did in 188 1 . To do so he defined functions 1 of bounded variation, and showed that the Fourier series of such an 1 converges pointwise to at every Fejer ( 1904) discovered that for any integrable function 1 o n [-11", 11"] , 1 (t)

(I (t - ) + / (t + )) /2

t (4 . 6.6).

4. Fourier Series

141

can be recovered by summing the arithmetic means of its Fourier series; even if the Fourier series for does not converge at a point the series of arithmetic means converges to /2 (4. 15.3) , provided that + the one-sided limits exist . Riemann was inspired by Fourier series too, and made deep contributions to their theory. When Riemann first considered the problems, "integral" was used in the sense given it by Cauchy in 1823: "Integrable" meant con­ tinuous. Riemann decided that a more general integral was needed, and he created the Riemann integral for functions that were required only to be bounded-no formal continuity requirement . He did this in his Habilitation­ sehrift at Gottingen in 1854 entitled Uber die Darstellbarkeit einer Funk­ tion du reh eine trigonometrisehe Reihe; it was not published until 1867, after his death. As an application of his new integral, Riemann showed that the Fourier coefficients of equations (4.2) and (4.3) of a and Riemann-integrable function approach 0 as n -+ 00 , an early version of the Riemann-Le besgue lemma 4.4 . 1 . Riemann also discovered the "localization principle (4.5. 1 1) ," that the behavior of the Fourier series of at a point depends only on the values of in an arbitrarily small open interval about t; he also proved the first version of the Parseval identity. In terms of the intrinsic character of a function whose Fourier series con­ verges "well," the smoother (continuous, continuously differentiable , etc.) the function is, the "better" its Fourier series converges (the more rapidly the coefficients approach 0) . Stokes ( 1 847) and Seidel ( 1 848) discovered the infinitely increasing slowness of convergence of the trigonometric series at a point of jump discontinuity. They observed that the discontinuity can­ not be enclosed in any interval in which the convergence of the Fourier series is von gleieh em Grade (uniformly convergent) . Through Weierstrass, and his students , especially Heine (as in "Heine-Borel" ) , the concept of uniform convergence percolated into analysis generally. In the presence of uniform convergence, certain attributes (e.g . , continuity) of each term of a sequence or series persist to the limit , and series can be integrated term by term. For example, if a series of continuous functions converges uniformly, then its limit is continuous. Heine proved in 1 870 that the Fourier series of a piecewise smooth 21r-periodic function converges uniformly in any interval that does not contain discontinuities of the function (4. 7.4(b ) ) ; in particular, if f is continuous, then its Fourier series converges uniformly and absolutely on every Closed interval (4.7.4(a) ) . Riemann's notion of integral prevailed until the arrival of the Lebesgue integral in the early twentieth century. Lebesgue's invention was also moti­ vated by problems concerning Fourier series. (Perhaps "motivated" is too strong, but he certainly was actively involved in research on Fourier series, and the first applications he made of his new integral were to the theory of Fourier series. ) Now continuity restraints were greatly relaxed (Riemann integrability is equivalent to continuity a.e.) , and there was much more free­ dom to take limits through the integral by means of Lebesgue's dominated

f

(f (t - ) f (t + ))

bn

an

t

t,

f

f (t)

f

142

4. Fourier Series

convergence theorem (4.4.2). In 1903 Lebesgue improved Riemann 's result that the Fourier coefficients an and of a bounded Riemann-integrable function j approach 0 as n � 00; he broadened it to include Lebesgue­ integrable functions j, even if j is unbounded. He also extended Rie­ mann's localization principle to Lebesgue-integrable functions. The subject of "Fourier series" -also known as harmonic analysis--- c ontinues to develop today in a much broader context known as abstract harmonic analysis. We begin our development of Fourier series in Section 4 . 1 by considering Fourier analysis in the REAL Hilbert space L� [-1I", 11"] of square-integrable functions on [-11", 11"] . Square-integrable functIons include a wide variety of functions that occur in practical applications such as the piecewise con­ tinuous functions on the closed interval b] . In the L5 [-1I", 11"] setting, we consider convergence of the Fourier series of a function E L5 [-11", 11"] "in the mean" -in the L 2 -norm 1 1 ' l b-to the limit function. Mean convergence of the Fourier series for I E L5 [-1I", 11"] signifies that if is the nth partial sum of the Fourier series, then the mean square error

bn

[a,

I

Sn

I: II (t) - Sn (t) 1 2 dt I

goes to 0 as n � 00 . For the step function of Example 4 . 1 .3, it means that the shaded area of the figure below shrinks to 0 as n � 00 .

A

square

wave and

84

Generally, mean convergence does not imply pointwise convergence (Exam­ ple 2.2.7(b)) . In Sections 4.6 and 4.7 we consider smoothness assumptions on the function that guarantee pointwise and uniform convergence of the as­ sociated Fourier series. In regard to pointwise convergence, no smoothness assumptions are really necessary. A deep result of Carleson ( 1 966) shows

4. Fourier Series

143

fE

that the Fourier series of any function L:; [- 7I', 7I'] converges pointwise a.e . on [-71', 71'] . We move to the wider context of Fourier series for functions in Section [-71', 71'] for all p � 1 , this is more general and -71', 71'] C Since quite useful. For functions we consider pointwise , uniform, and (C, l ) convergence o f the Fourier series. To illustrate how different the L l and theories are , consider the following point. Unlike the a.e. convergence of the Fourier series of any 71'] , Kolmogorov gave an example of 71'] whose Fourier series diverges at every point. We say more an f E about the phenomenon of divergent Fourier series in Section

Lp [

4.4.

Ll

L2

f E L 2 [ - 7I',

L 1 [- 7I',

4. 1

L'i

Ll

4.9.

Warmup

A real- or complex-valued function h defined on all closed intervals except possibly for a finite number of points in each of them is p-PERIODIC if there exists a p > 0 such that h p) = h (t) for all t and t + p where h is defined. If h is p-periodic, then the function 9 defined for all by 9 = h ( pt /271') is 271'-periodic : As t advances by 271', pt/271' increases by p. Periodic phenomena are commonplace; they include the rotation of the earth on its axis , the motion of the moon about it, planetary motions generally, vibrations of strings or tuning forks, pendulums, tides , and electromagnetic w aves. Any real- or complex-valued function h defined on a closed interval [-p, p] can be made into a periodic function by PERIODIC EXTENSION or 2p-PERIODIC EXTENSION as follows: Write x R as x = 2 p t for appro­ priate N and [-p, p] ; then define

(t +

nE

t

(t)

n +

E

tE

h (t

+ 2 np) = h (t) , n E Z, tE R.

Since we are considering Lebesgue integrals, if we change the values of a function at a finite number of points, this will not affe ct the integrals that produce the Fourier coefficients (equations and This gives us some options .

(4.6)

(4.7)).

Remark PERIODIC EXTEN SION AND THE ENDP OINTS

[- 71', 71']

h

If a real- or complex-valued function h is defined on and (-71') #­ h (71') , then we have choices about its periodic extension. We could omit the values of h at -71' and 71' or redefine the function so that h (-71') = h ( 71' ) and then extend h periodically. For example , if h (t) = on we might redefine it to be at - 7I'-or declare it to be undefined there-and then periodically extend it by defining h (t) = h 271') for every x. No matter what we do, if h #- h the periodic extension of h will have discontinuities at odd multiples of 71'. We consider what the Fourier series of h converges to at jump discontinuities of the periodic extension of h in 0 Section

71' (-71')

4.6.

t

(71'),

(t +

[-71', 71']

144

4. Fourier Series

A TRIG O N OMETRIC SERIES is a series of the form

IS

�O + L an cos nt + L bn sin nt , t E R, neN

neN

(an) n>O

and (bn ) n > 1 are complex sequences . It is easy to convert where the the nth partial sum n n Sn + k cos kt + bk sin kt = k= l k=l

(t)

La

�

L

of the trigonometric series into exponential form n Sn = k= - n

(t) L Ck e ik t

by taking bo

=

0, and for all k 2: 0,

Ck = a k -2 ibk

an d L k -

_

a k + ibk

(4.4)

2

ak

To convert from exponential to trigonometric form, use Ck + L k and bk = - L k ) . Generally, one says that a two-sided series (or from a normed space CON VERGES to the of vectors = x . This is not what is customarily vector x if limn , k -+oo L� = - k done in the special case of the series L e however. We say that converges (somehow) t o s ( t ) if the symmetric partial sums = L�= converge to (t) as --> 00 .

i (Ck eries) Ln eZ Xn

Xn

Xm

Ln eZ cn e i n t Un (t) n Ck e ik t

s

bis­

n Z cn e i nt , n

Definition 4 . 1 . 1 FO URIER SERIES AND COEFFICIENTS

If

I E L1 [-11", 11"] , then

1 1 "" I (t) e - z.nt dt, n E Z,

I ( n) = 211" �

_ ,,"

I,

( 4.5) e i nt I

is called the nth FO URIER COEFFICIENT for and L n e Z fen) is called the EXPONENTIAL (or COMPLEX) FORM OF THE FO URIER SERIES FOR Another name for fen) is the FINITE FOURIER TRANSFORM of evaluated at (cf. Section 5 . 1 ) . Equivalently,

n

an = :;1 1 "" I(t) cos nt dt, n E Nu {O} _,,"

and

1 I"" I(t) sin nt dt , n E N,

bn = 11"

_ ,,"

I.

4. Fourier Series

are called the nth FOURIER COSINE AND SINE COEFFICIENTS F O R spectively, and

145

f, re­

f (t) = � + L an cos nt + L bn sin nt nEN n EN

is also called the TRIG ONOMETRIC FORM O F THE FOURIER SERIES F O R are related to /( n ) by equation ( 4.4) . 0 The and

an

bn

f.

Even though the exponential form provides the most symmetric, elegant formulation of Fourier series, we use the trigonometric form throughout this chapter because we think that it is easier for the first-time reader. Fourier series provide highly effective means to investigate periodic func­ tions. Using them, you can dissect a sound into component parts called harmonics; the process is known as harmonic analysis. The is the neutral position, cos sin sin the fundamental tone , cos which go to 0 with the first octave and so on. The amplitudes and increasing n, determine the imp ortance of the overtones in toto. In essence, Fourier analysis of periodic phenomena enables reduction of the problem to determining the response to one harmonic at a time , then taking a limit of a linear combination ( superposition ) of these . Not only that , as van Vleck 19 14] observed:

al t + bl t

a o /2 a 2 2t + b2 2t bn,

an

[

. . . we mathematicians may well query . . . whether it may not have conditioned the form of physical thought itself-whether it has not actually forced the physicist often to think of compli­ cated physical phenomena as made up of oscillatory or harmonic components , when they are not inherently so composed . To what extent did Fourier series accelerate the development of the wave theories of modern physics? Who would have thought of an electron as a wave? We compute some Fourier series for functions f E 11"] . Since

L2[ - 1I", I } U { sin nt , cos nt .. n E N } B {_ y 1l" y 1l" y 211" -

ro=

;;;

;;;

11"] ( 3 .4.6 ( c )) , any is an orthonormal basis for written as the infinite series of its projections ,

L2[-1I",

f E L2[-7I",

f = L (f, e } e .

)

eEB

The terms of the series are of the form nt (f, e ) e - I\ f ' cos .Ji -

t

cos n .ji or

I\ f' sin nt ) sin.jint . .Ji

] may be

11"

146

4. Fourier Series

For historical reasons and ease of notation, we deviate from the conven­ tion (3.4.8(b)) about what we call the Fourier coefficients ) of f in the cos Hilbert space 11-]. Instead of (f, ) ft = we change the normalizing factor from 1 / J1i to 1/11". As in Definition 4 . 1 . 1 , w e take as the Fourier coefficients of f

(f, e e = ( /, e t ) J;r D" / (t) nt dt,

L2[-1I",

1 an = ;:

and

f"" f(t) cos nt dt, L

1 bn = -

11"

1

"

_"

n E NU {O}

f (t ) sin nt dt, n E N .

(4.6)

--+

(4.7)

It follows from the Riemann-Leb esgue lemma 3.3.2 that an , bn 0; in­ deed, by Parseval's identity 3.3. 1 , the sequences (a n ) and (bn ) are square­ summable (belong to £ 2 (00)). The Fourier series for f E is then the 1 I · l b-convergent series

L2[-1I", 1I"]

�o + L (an cos nt + bn sin n t ) .

(4.8)

nEN

The a n and bn are unique by 3 . 3 .4(b) ; if (an ) and (bn ) are square-summable sequences , then the trigonometric series of equation (4.8) is the Fourier series of some E by the Riesz-Fischer theorem 3 3 4 (a)

/ L2[-1I", 11"]

.. .

Historical Note THE MEANING O F DEFINITE INTEGRAL In 1 807 there was controversy about the meaning of the definite integrals in equa­ tions (4.6) and (4.7) . At the time, determination of the antiderivative to compute area was deemed essential. Since an antiderivative cannot always be found in terms of elementary functions, as with f d ? It was for example , what meaning should be attached to f12 Fourier's idea to bypass the antiderivative and define the definite in­ tegral as the area under the curve . Dirichlet later realized that not every function could be integrated even in this sense, that not every function has a meaningful area between it and the horizontal axis ; the characteristic function l Q of the rational numbers Q, now known as the DIRICHLET FUNCTION , is an example of such a function. 0

e -t 2 t

e - t 2 dt,

We always get something extra with Fourier series. If we start with a function f defined on [- 11" 11" , the Fourier series automatically produces a periodic extension of f because of the periodicity of the sine and cosine. Our attitude to this largesse is this: As long as it converges somehow to on [- 11", 11"] , who cares what it does elsewhere? Let us get some concrete experience .

, ]

f

4. Fourier Series Example 4. 1 . 2 SQUARE WAVE I

function

f

O,

(t) = {

147

Compute the Fourier series for the step

1r t

- :5 < 0, . I , 0 :5 t < 1r.

[-1r, 1r]

Solution . As noted above, the Fourier series for this step function on actually converges to a square wave on R. By equation (4.6) , its cosine = coefficients are 1 and "" = = 0, = -1 f( ) cos cos

ao

an 1r 1

t

_ ,,"

nt dt -1r1 10 "" nt dt

n E N.

cos n1r 1 = bn = -1 1 "" sm. nt dt = 1 - n1r n1r ( I - ( - l t ) , n E N .

Its sine coefficients are 1r

0

The Fourier series of f is therefore

�+� � L...J

sin nt (1 - (- It)

�+� � L...J

n - l) t .

sin (2 2n - 1

(4. 9) 7r n e N This converges to f in the Hilbert space Lli[-7r, J i.e . , it converges to f in the mean. What does it converge to pointwise? That depends on the point t. By the pointwise convergence theorem (4.6.2) , this series converges to f (t) at points of continuity. At t = 0, it converges to 1/2, the average of 2

1r ne N

n

=

2

7r ,

the limiting values f (0 - ) and f (0 + ) . Also (and this is generally true) the series converges more slowly at points where the function is less smooth, at points of discontinuity or sharp corners . 0

t -t - (t) .

t,

A real-valued function f is EVEN if f ( ) = f ( ) for all i.e . , if it is symmetric about the y-axis, as etc. , and cos t are , for example ; a real-valued function f is ODD if J ( ) = f Linear combinations of odd powers of t are odd and so are linear combinations of sin where is an integer. It is trivial to verify that the product of two even or two odd functions is even; only even x odd odd. Equally trivial to verify is the fact that the integral of an odd function over a symmetric interval is If f is an odd periodic function with period then its integral over any interval of length p is 0 (4.2.2) . The practical use of these observations is that the sine coefficients of any even function are 0, as are the cosine coefficients of any odd function .

t 2 , t4, -t

kt

=

O.

p,

Example 4 . 1 . 3 SQUARE WAVE II Find 9

( )=

t

{ - I, I,

the Fourier series for

- 1r t t

< < 0, 0 < < 7r .

k

[-c, c]

148

4. Fourier Series

an

Solution . Since 9 is odd, the cosine coefficients are all 0, whereas '11" 2 2 = - ( 1 - cos sin n 0 odd, n even . 0,

n1r) n1r n1r ' n sin (2 n - 1 ) t · serIes lor 9 IS t here£ore 4 '" Th e FO U rIer . ;: L..m E N 2n 1 bn = -1r 1 g (t) .

l"

t dt

{�

•

_

-4

0

4

A square wave and

84

Remark LINEARITY O F FOURIER COEFFICIENTS AND SERIES

Since the integral operators .; ( (cos or sin n are linear , the Fourier cosine (for example) coefficients of + for 9E any scalars a and b, are given by

J�'II" ) -

nt t) dt en af bg, f, L2[-1r, 1r],

dt + -b 1 '11"'II" g(t) cos nt dt = aan + ba� , where an and a� are the Fourier cosine coefficients of f and respectively. It immediately follows that the Fourier series for af + bg is a (Fourier series for J) + b (Fourier series for g). 1r 1

�

-'11"'II"

/(t) cos nt

1r

g,

X

x

This observation can sometimes save some computation . 0 Example 4 . 1 .4 SQUARE WAVE I I I

f (t) = {

O,

- 1r < t < °

1, ° < t <

1r

Show that the Fourier series of

- l) t .

. 1 2 " sin (2n + 2 2 -1 EN -; n

IS

L.J

n

4. Fourier Series

149

Sol ution . Forget Example 4. 1 .2 where we computed the Fourier series of the step function for the moment. Let us use the fact that we know the Fourier series of the square wave 9 of Example 4 . 1 .3. We can use that and linearity to get the Fourier series for Since = l ( 1 + its Fourier + series is given by l l (Fourier series of or

f

f.

g ),

f

g)

!2 + �71" L...tn EN sin 2(2nn--1 1) t . "

0

Example 4 . 1 . 5 AN EVEN SAWTO OTH Sh ow that the Fourier series for is t - ;. ( cos + CO;23t + CO;25 t + · · l on

[ -71", 71"] t Solutio n . Since f is even, its sine coefficients b n are 0 for every n E N . The cosine coefficients are an = -71"2 1'"0 t cos nt dt. f (t) = I t I

Integrating by parts , we obtain

2 2 cos n 7l" - 1 ) an = 7I" n (

=

{

-4 -2 ' n odd, 7I" n n even, 0,

a o = ( 2 / 71" ) fo'" t dt = 71". The Fourier series for I t I is therefore cos 3t + cos 5t + . . 4 -. . 0 - - - ( cos t + -2 71" 32 52 ) Remark PERIOD 2p INSTEAD OF 271" Of course, there is nothing special about [-71", 71"] . If [-p, p], p > 0, is any closed interval and if f E L2[- p, p], the Fourier coefficients for f are given by 1 P n7l"t (4. 10) a n = - j f(t) cos - dt, n E N U {O} p P and

71"

-P

P n7l"t bn = -p1 j f(t) sm. dt, n E N . P -P The Fourier series for f is ao + (an cos n 7l"t + bn sin n 7l"t . L P P ) nEN and

2

{ t,-6 - t

(4. 1 1 )

0

(4 . 1 2)

Example 4 . 1 . 6 AN ODD SAWTO OTH Sh ow that the Fou rier series for the sawtooth < < -3' ' = -3 � � 3 , 3 3 ". is 2 4 (sin ". t ..!.. sin t + ..!.. sin !1l!:! on )

f (t)

[- 6 , 6]

".2

6

- 32

-6 t t 6 - t, 5: t 5: 6, 52 6

6

.

.

•

.

150

4. Fou rier Series

2

A saw-tooth Sol utio n . Since f is odd, we need only compute the sine coefficients bn since f (t) sin [( rt) /6] is even, we can double the expression in equation (4.11) and integrate from 0 t o 6: 26 [ 6 f (t) sm. mrt dt bn 10 fi 1 [3 t sm. mrt dt + 1 [ 6 (6 - t) sm. n7l"t dt, n E N . a la fi a 13 fi With w = 7I"t/6, 12 1 '1r (71" - w) sin nw dw. 12 1 '1r / 2 w sin nw dw + 2" bn = 2" 71" a 71" / 2 j

m

Integrating by parts and simplifying gives

'Ir

24 . n 7l" bn = 22 7I" n sm -2 '

24 ( . 7I"t 1 . 371"t 1 . 571"t 71" 2 sm "6 - 3 2 sm (; + 5 2 sm (; - . . .) . 0 Example 4. 1 . 7 A PARABOLIC CUP Sh ow that the Fourier series for f (t) = t 2 on [-71", 71"] . + 4 " L.me N ( _ l)Rn cos n t so the Fourier series for the sawtooth is

IS

Solution .

"3 'lr 2

2

•

271"2 and an = -I j'lr t 2 cos nt dt for n E N . ao = -71"I j'lr t 2 dt = 3 71" - 'Ir

- 'Ir

4. Fou rier Series

=

151

Integrating the latter by parts yields

-2 an n 1'" t sin nt dt . 7r

_ ...

Another integration by parts yields

2 1'" cos nt dt 4 (- 1t t n ... n n Since 1 is even , each bn is O. Thus, the Fourier series for I (t) is " (- 1 r cos nt 7r 2 +4 L.J 3 n2 . nEN an

=

2

-2- cos t L ... n

7r

-

27r

=

_ ...

2

0

Exercises 4 . 1

1 . EXP ONENTIAL FORM

Ln EZ cn e i nt OF FOURIER SERIES

(a) Verify the relationships of equation (4.4) . (b) FINITE FOURIER TRANSFORM Show that if 1 E L1 even or odd, then

=

[

]

- 7r , 7r

is

I( t ) e - int dt , n E Z , ... is real or purely imaginary, respectively, for all n E Z. j(n)

� I:.

2

2. AMPLITUDE-PHASE FORMULATION

Show that the series

� + L an cos nt + L bn sin nt nEN n EN

may be written in AMPLITUDE-PHASE form

!'f +

L dn cos (nt + ipn ) . nEN The coefficient dn is called the AMPLITUDE and i.pn the PHASE (PHASE ANGLE) of the nth component . Express an and bn in terms of Cn and ipn and vice versa. 3. SHIFTS AND MULTIPLES Let 1 E L'2 [- 7r , 7r] . (a) SCALE CHANGE T

=

D ILATION Given the Fourier series

+ L an cos nt + L bn sin nt nEN nEN

for the function 1 and some constant C series for 1 (

ct)?

# 0, what is the Fourier

152

4. Fourier Series

)

I 2, Ln EN dn (nt if'n) I (t a) � + L dn cos (nt + [if'n + na]) . n EN

ao/2

(b TIME SHIFT If the function has the Fourier series + cos + of Exercise show that the Fourier se­ ries + is given by

4.

I (t) = {

Use the linearity of Fourier coefficients as in Example pute the Fourier series of

io,

O� � � �. O ,

4.1.4 to com­

5 . Find the Fourier series of the following functions. (a (b (c (d

6. 7.

) ) ) )

I (t) = 0 for -1r ::; t < 0 , I (t) = t for 0 ::; t ::; 1r. I (t) = t for - 1r ::; t ::; 1r. I (t) = t 2 for 0 ::; t ::; 21r. I (t) = 112 (1r 2t - t3 ) , -1r ::; t ::; 1r. IF ( note: IF; cf. 4.6.2) the series converges pointwise, show that ;� 1 ;3 + 5\ - . . . . =

-

Find the exponential Fourier series expansions of:

( a) I (t) = at/21r, 0 < t < 21r, where (b ) I (t) = t,
- 1r

a

is constant.

DERIVATIVES AND INTEGRALS OF PERIODIC FUNCTIO NS Let the function be p-periodic.

I

) ) If I

I

I' be pe­ (x) = f: I(t) dt

(a Assuming that is differentiable , must the derivative riodic? (b is integrable over all closed intervals is F periodic? What if

f: +P I(t ) dt =

8 . Find the Fourier series for

O?

[a, b],

I (t) = { 01 ,, � << tt ::;< tl '. 2

9 . Find the Fourier series for the triangular wave

2n - 1 ::; t ::; 2n, , n E N . I (t) = { t2n -2nt - t, 2n ::; t ::; 2n + 1 , -

-

2'

1 0 . PARSEVAL'S IDENTITY Energies of waves are related to the squares of their amplitudes. The Parseval identity equation (3.9) of 3 .3 . 1 ) is useful in evaluating these. Let

( I E L2[-1r, 1r). ( a) Show that � r:.,. I / (t) 1 2 dt = a2� + Ln EN (a� + b�) , where an and bn are the Fourier coefficients of I of equations (4.6) and (4 . 7) .

4. Fourier Series

(b) With

i (n)

1 ( a) , show that 2 2� f:'7r I/ (t) 1 2 dt = Ln EZ Ii (n) 1 .

as in Exercise

(c) Use the result of Exercise 5(c) to show that 7r4 /90

11.

153

= L n EN 1/n4.

Does there exist an I E L; [- 7r, 7r] whose Fourier coefficients are : (a) (b)

an = bn = lifo with a o = O? an = bn = l/ n with ao = O?

(t) = (-t)) 12.

12. Any function I can be written as the sum of an even function Ie -I +I /2 and an odd function 10 = For any I E L2 [-7r, 7r] , show that

(f (t)

(-t))

� f:'7r P (t) dt

=

(t) (f (t)

� f:'7r {; (t) dt + � D7r {; (t) dt.

Does this relationship among squares look familiar? Can you explain it? 1 3 . PRODUCTS Let I, 9 E L; [-7r, 7r] have Fourier cosine and sine coeffi­ cients for I and for g , respectively. Show that

(an , bn )

(cn, dn)

Since we have considered Fourier series only for functions from L; [-7r, 7r] , the Fourier series for the product Ig falls outside our purview-it may not be square-summable-for now. The product of two L; functions is an Ll function, however, and we consider Fourier series for Ll functions in Section 4.4.

t,

n 1) t, . . . is an orthonormal

14. Show that sin sin 3t , sin 5t , . . . , sin (2 + basis for L; [O, 7r/2] .

1 5 . PERIODICITY If there exists a smallest period Po > 0 for a function Which of the I, then Po is called the FUNDAMENTAL PERIOD of functions below is periodic? If it is periodic, what is the fundamental period? (a) sin 7rt (b) sin tl5 (c) sin 3tl5 (d) sin 3t + cos 5t (e) sin tl5 + sin tl3 (f) I (t) = sin lit , t # 0 , and 1 (0) = 0 (g) e it + e i 7r t .

f.

16 .

EXISTENCE O F FUNDAMENTAL PERIOD Let I be a nonconstant con­ tinuous function on R. Show that if I is periodic with period then I has a fundamental period.

p

154

4. Fourier Series

Hints/ Answers

dn 'Pn ), an = dn 'Pn , dn Ja� b� 'Pn = (an/dn). n sinnn-t · " EN cos(n2n -1?1)t - " 5 A nswers. ( a) ". - ;-2 wn wn EN ( - 1) (2 (b) 2 (sin t - t sin 2t + i sin 3t - � sin 4t + . l " E Z - {O} ( -ni ) " ei nt . 6(b) A nswer: . wn 8. See 4 . 2 . 1 ( c). 9 wn EN cos(( 22nn--1 1)�".t . 11. Use 3 .3.4. 1 2. It looks like the Pythagorean relation: 11/11 2 = II g l1 2 + II h ll 2 . If M denotes the subspace of even functions, show that M 1. is the subspace of odd functions. Conclude that L5[-1r, 1r] = M � M 1. . 14. For any 1 E L5[O, 1r/2] orthogonal to each of these functions, extend 1 so that it is even with respect to /2 (i.e . , symmetric about the line t 1r /2) on [0, 1r]. Then extend this function to an even function on [ - 1r, 1r] and compute its Fourier coefficients. 15. Answers. (c) 101r/3. (d) 21r. (e) 301r. (f) not periodic. (g) Note that e it and e i". t are linearly independent . 1 6 . If 1 doesn't have a fundamental period, let (Pn) be a decreasing se­ quence of positive numbers decreasing to 0 such that 1 (t + Pn) = 1 (t) for each n and all t. Then 1 (t + mpn) = 1 (t) for all integers m, n � 1. For a given s, let mn be the largest integer such that mnPn :::; Then I - mnPn l :::; Pn. Since 1 is continuous and Pn 0, conclude that 1 (t + s ) = 1 (t) for a fixed t and arbitrary s . This contradicts the nonconstancy of f. 2.

cos Given the amplitude-phase series (the and the sin Conversely, = + and cos - 1 This is easy to remember by drawing a right triangle.

bn = -dn 'Pn . 4"

·

·

•

·

Z

=-1 " 2 ".

1r

=

s

s.

4.2

-+

Fourier Sine Series and Cosine Series

(4 . 10)-(4.12), if a square-summable function 1 is [- p, p], its Fourier series is given by n1rt + bn sm. n1rt ) , aO L..J ( an cos "2 + '" (4.13) P P nEN

As noted in equations defined on an interval

4. Fourier Series

155

mrt dt, n E N U {O} , (4 . 14) an = p-I jP f(t) cos P -P and mrt dt, n E N . (4. 15) bn = p-I j-P f(t) . P P If f is defined only on [0, p], we can extend f to [-p, p] as an even or as an odd function or, depending on f, neither (see Example 4 . 2 . 1 (c )) . If we consider the ODD EXTENSION fo-taking fo on [-p, O] to be the negative of what f is on [0, p] and taking its 2p-periodic extension-then all the Fourier cosine coefficients an are 0, n E N u {O} , and equation (4. 13) is a series consisting only of sines, a FOURIER SINE SERIES for f: where

Sill

f

Likewise , if had been extended as an even function, then all the sine coefficients drop out, and we get a FOURIER COSINE SERIES representation for We illustrate these HALF-RANGE or HALF-INTERVAL EXPANSIONS next .

f.

Example

4.2.1

SINE AND COSINE SERIES Expand the fun ction

f (t) = { 0,I, � << tt << �1,, 2

.

a s (a) a Fourier sine series; (b) a Fourier cosine series; and (c) a Fou rier serzes.

{,

Solution . (a) Fou rier sine series. The odd extension of

f is

t _ 12 ' fo (t) = -1, 1 , 0 <�
IS

-1 < < -

2'

even ,

1 11 2 mr ( 1 - cos -) . bn = 1- I fo(t) sin mrt dt = 2 0 / 2 sin mrt dt = mr 2

We get the following pattern

b1 = -11"2 , b2 = -11"2 , b3 = -311"2 , b4 = 0, . . . .

156

4. Fourier Series

Therefore, the Fourier sine series

(

I is

-- --

)

2 sin 271"t sin 371"t sin 571"t 2 sin 671"t . . . 2 . sm 7l"t + . + + + + 71" 2 3 5 6

I t E �) , t E ((-�, - I, - V U (�, I).

(b) Fourier cosin e series. The even extension of is �

Je

_

(t) -

{ 0,

1,

f; / 2 dt = 1 and 1 1 / 2 cos n7l"t dt = -2 sin 11 n7l" , E N. an = 2 /e (t) cos n7l"t dt = 2 2 n n7l" o 0 The Fourier cosine series for I is therefore The sine coefficients bn are all 0 ; a o = 2 fol le (t) dt = 2

!2 + !71" ( cos1 7I"t _ cos3371"t + cos5571"t _ . . .) .

(c) Fourier series. The periodic extension of I happens to be neither even nor odd with period 2p = 1, so 2n7l"t d 1 1 / 2 cos 2n7l"t dt , n E Nu {O} , 1 1 / 2 I(t) cos -t=2 1 0 - 1/2 1 1 / 2 sin 2n7l"t dt , n E N . 1 11 / 2 2n7l"t I(t) sin -- dt = 2 bn = / 1 2 - 1/2 1 0

1 an = 1/2 and

This yields the Fourier series

(.

)

sin 671"t sin 107l"t sin 1471"t . . . . 0 2 1 + - sm 2 7I"t + -- + + 3 5 7 + 71" Lemma 4.2.2 shows that if we integrate a 2p-periodic function over any interval of length 2p, we get the same value as over any other such interval. If is continuous, it is easy to prove this by differentiating:

I

lx + 2P /(t) dt = I (x + 2p) - I (x) = 0 for any x , x from which it follows that f: + 2P I(t) dt is constant. More generally, d dx

4.2.2 INTEGRAL SHIFT Iff is 2p-periodic and int egrable over [-p, p] , then for any real numbers a and b,

I b /(t) dt = I b+2P /(t) dt and jP I(t) dt = la + 2P I(t) dt . a +2p a a -P

4. Fourier Series

157

Proof. For any a and b, with s = t +2p and the fact that f ( s - 2p) = 1 ( s ) , b I(t) dt = b+2P f(s - 2p) ds = b+2P I(s) ds , Ia+2p Ia+2p Ia which proves the first equality. Now replace a by a - 2p and b by a to get then let

a = -po

b+2P ra+2P la+2p = la ;

0

Exercises 4 . 2

( (t) = t

1. Find a) the Fourier cosine series and (b) the Fourier sine series for 1 on [0, .

2.

'11"]

Find the Fourier (a) cosine series for I (t) = sin t , 0 0 t p. (b) cosine series for f (c) sine series for = a constant, for (d) sine and cosine series for 1 = 0

(t) = t 2 , I (t) c,

3. HALF-RANGE COEFFICIENTS general form of the coefficients (a)

(b)

0 � t � p.

(t) t, < t < 2. If f i s defined o n [O, p], determine the

an in its cosine series expansion. bn in the sine series expansion .

4. SHIFT BY

'II"

Suppose you know that the Fourier series for f E

L2[- , '11"] is ao/2+ E n eN an cos nH E n eN bn sin nt. Find the Fourier series for 9 (t) = 1 (t - '11" ) . EXP ONENTIAL SERIES For any constants a and p, find the exponen­ 'II"

5.

� t � '11" . � �

tial Fourier series (Exercise 4.1- 1) for (a)

(b) (c)

I (t) = at/2'11" , 0 < t < 2'11" . I (t) = t, < t < '11" . I (t) = at/2p, 0 < t < 2p. - 'II"

L2[-'II" , '11"] , ao Ibn I � ao I an I bn '11") , I I � n b for

6. SIZE OF FOURIER COEFFICIENTS For nonnegative f E show that the Fourier coefficients satisfy and � for all n E If 1 is odd and f � 0 in (0, then n . ..

= 1, 2,

N.

.

1

158

4. Fourier Series

f (t) = cos at where a is not an integer. ( a) Show that the Fourier series for f is

7. Let

Substituting t

= 7r and dividing by sin a7r,

2 a ( 2 \ + �12 + 2 1 2 2 + . . .)

cot 7ra = With w

we obtain

a

".

a -

a

-

.

= 7ra we get _n + _1n_ ) , cot w - w + " ( L-nEN w - ". w + ". 1

.!.

-

_

which is the partial fraction expansion of the cotangent. (b ) Use the expansion in a) to show that

(

_si n1_w -_ w + " ( _ I) n ( _ L-w - n ". + w +n ". ) . nEN 1

.!.

__ 1_

_

Hints/ Answers 1

(

1r. . a) 2

n n _ .i". " L--n E N cos(( 2n2- -1)21 )t . ( b ) 2 " L--n EN ( _ 1) + 1 s i nnn t .

2. Answers: ( a) - .i ( COS 2t + cos 4 t + cos 6 t + . . . ) 2-n1t · Note - ". _ ". L-- n EN cos n 1 ". ". 3 2 5 35 4 that this returns I sin t I . (b) � - � 3 ". 2 ( cos ".Pt - 4 cos 2".tP + cos 3". t _ . . .) . (c ) 4".c ( sin ". t + sin 3". t + 5 sin 5 ".t ). P 3 P P (d ) cosine series·. 1 - ". 2 ( cos 2 + 32 cos 3 ".t + 52 cos 5 ".t + . . . ) . 2 2 3. (a) a n = foP f(t) cos u t dt, n = 0, 1, . . (b ) bn = foP f(t) sin u t dt, P P P n N P 4. The cosine coefficients are ( - 1 r an the sine coefficients are (-1 r bn. 5 . (a) Use 4.2.2 to assert that Cn = 21". !�". f (t) e - i n t dt . in ". /p in ( b ) i I:nEZ - { O } (-�t e t . ( c ) Cn = (1/2p) !:' p f (t) e - t dt. 6 . For the second statement, use the fact that I sin nt l ::; n i sin t l . �

- �

1

E

.

£

J(

1

..§..

1 9

1

".t

P

_

.

.

.

...L

...L

.

;

.i "

.

£

J(

4. Fourier Series

4.3

159

Smo othness Not long ago many thought that the mathematical world was created out of analytic functions. It was the Fourier series which disclosed a terra incognita in a second hemisphere. -E. B . van Vleck , 1 9 1 4

1

For I E L [ - 'II" , '11"] , what, if anything, does the nth partial sum

a; + L an cos nt + L bn sin nt nEN n EN

( 4. 16)

of the Fourier series for I converge to pointwise? In many cases-wave propagation , for one-it is important to know how good the approxima­ tion of the wave by its harmonics is at specific points, i.e . , to know about pointwise convergence. The problem of pointwise convergence of Fourier series is an old one . Fourier first used such series in his Th eone de la propaga tion de la ch aleur dans les solides (reproduced with comments in Grattan-Guiness 1972) , a manuscript submitted to the Institut de France on 2 1 December 1807 , to a committee consisting of Laplace, Lagrange , Lacroix, and Monge. (Whew! Poisson was only the committee's clerk .) Lagrange questioned the conver­ gence of such series. Fourier offered the following trigonometric series for I (t) = 1 on [- 1 , 1] 'll"t 1 3'11"t 1 5'11"t 1 ht cos "2 - 3" cos 2 + 5" cos 2 - 7 cos 2 + . . .

(

4 . 17)

How could this converge for all t? A series of real numbers converges ab­ solutely if and only if it converges unconditionally (see 3.6. 1 and Exercise 3.6-3) . Since the series of absolute values of the coefficients of the series of equation ( 4 . 17) , 1 1 1 1 + - + - + - + · · · ----. 00 3 5 the terms cannot be rearranged-if the series converges at all, it can only do so conditionally. Not only that, in Example 4 . 1 . 3 we computed the Fourier series of the square wave

'

7

I (t) = as

±

{

'"

'II" nL.J

-1, 1,

EN

-'II" < t < 0 , 0 < t < '11" ,

sin (2n - l ) t . 2n - 1

160

4. Fourier Series

does not even con verge conditionally. Though the others would probably have consented to the publication of Fourier's article, Lagrange was adamant. Simeon Poisson tersely (5 pages) summarized the committee's findings in 1808 which were as follows: Re­ jected on the grounds that it contained nothing new or interesting. Fourier went back to his desk . He submitted a revised version, M emoire sur la prop­ agation de la chaleur, to essentially the same group of j udges at the end of 1 8 1 1 as a candidate for the Grand prix de matMmatiques for 1 8 12. Though they awarded him the prize , their reservations were such that the article was not published in the Memoires de l 'A cademie des Sciences. Fourier resented this bitterly. In 1824 Fourier became Secretary of the Aca demie and published his 1 8 1 1 paper , practically unchanged, in the Memoires. Nevertheless, the problems about convergence of his series were serious . Poisson published an erroneous argument about their convergence in 1 820 . Fourier kept trying, and Cauchy published a flawed proof in 1 826 . In 1829 the 23-year old Prussian mathematician of French origin, Peter Gustav Lejeune Dirichlet pointed out Cauchy's mistake (a rather courageous act , considering Cauchy ' s influence and crotchetiness) ; Dirichlet gave sufficient conditions for pointwise convergence of Fourier series. After Gauss's death in 1855 , Dirichlet succeeded Gauss as professor of higher mathematics at Gottingen. It is essentially Dirichlet's proof that we give in 4.6.2. It asserts that the Fourier series of a piecewise smooth integrable function converges at each point to + 2 Hence , the Fourier series converges to at points of continuity and to the average of the limit Ing values at a jump discontinuity. We look at Jordan's ( 1 881) improvement of Dirichlet 's pointwise convergence theorem in 4.6.6: If the piecewise smoothness condition on is weakened to bounded varia­ tion (Definition 4.3 .4) , its Fourier series still converges to t + at every point t. There is no hope of weakening the smoothness condition to mere continuity. As we show in Section 4.9, there are continuous functions whose Fourier series diverge at certain points. Indeed, there are continu­ ous functions on [- '11' , '11'] whose Fourier series diverge on a dense subset of [- '11' , '11'] . Of course, since such functions cannot even be of bounded varia­ tion, they are pretty wild. The key to convergence of the Fourier series of the integrable function is the smoothness of The smoother is (i.e . , the more continuous and its derivatives are ) , the better the convergence of its Fourier series to If is merely continuous at a point, its Fourier series may diverge, but (4.6.2) if is continuous and piecewise smooth, the series converges uniformly and absolutely to (4.7.4) . The convergence is also faster if the function is smoother. The essential content of 4 . 1 2.3 is that if is continuously differentiable k times, then its Fourier coefficients go to 0 faster and than

f

t

f (t + ) f (t - ) f (t) f

f.

f

(f (t - ) f (t + ))

f f f. f

f

f

li n k.

an

f

bn

161

4. Fourier Series

Before demonstrating the benefits of smoothness in regard t o conver­ gence, let us look at the other side of the coin first, at the deleterious effect of discontinuity on pointwise convergence. Dirichlet invented the function 1 Q of Example 4 .3 . 1 , which is as discontinuous as a function can be and its Fourier series has correspondingly atrocious pointwise convergence behav­ ior . As a comment on Dirichlet ' s bold ingenuity, at the time of his invention entities like 1 Q were not even considered to be functions. It forced math­ ematicians to confront the question: What is a function? EXaIllple 4 . 3 . 1 AWFUL POINTWISE CONVERGENCE

(

The Dirichlet FUNCTION not the Dirichlet kerne0 1Q

�atio.nal, E Lr2 [-1I" , 11"] (t) = { 0, tt IS�s IrratIOnal, I,

has a Fourier series that does not converge pointwise to the function at any rational number. Since we can disregard the set Q (of rational numbers , which has Lebesgue measure 0,

an = -11"1 1 " 1 Q ( t ) cos t dt = -11"1 1Q _

n

"

n [ - " , "l

l

)

td

cos n t

=O

,

n

E N,

and O. Likewise , 1 Q (t) sin n 0 for all n. The Fourier .;. series is therefore identically 0 and does not converge to 1 Q (t) at any rational number t. 0

ao =

bn = J�"

t dt =

[1914] n cnx

As E. B . van Vleck put it, " . . the power series obtained by com­ bining terms of the form define the most civilized members of math­ ematical society-the so-called analytic functions-which are most orderly in their behavior , being continuous throughout their domains, possessing derivatives of all orders." Fourier series can represent some of the most pathological elements of mathematical society, even those schizophrenic functions that are continuous everywhere but differentiable nowhere. Weier­ strass proved that functions defined by series of the form

f

f (t) = n=L bn sin (an t) , 0 < b < I , a E N, ab > I , 00

O

(

(4.18)

are continuous everywhere by Weierstrass's M-test, the series converges uniformly but differentiable nowhere this is not so easy to prove . To see how wildly these functions oscillate, consider the graphs of some partial sums of

Sn

)

£; 00

( l )k 2

(

sin

)

(5kt) .

( 4. 19)

1 62

4. Fourier Series

86

(t)

As we discuss in Section 4.1 1 , convergent trigonometric series

�o + L an cos nt + L bn sin nt neN neN and Fourier series are not the same thing; there , we give an example of a convergent trigonometric series that is not the Fourier series of any function in [ -11", 11"] . But the series of equation (4.19) is a Fourier series (by 4.7. 1 , any uniformly convergent trigonometric series is a Fourier series) . Even so, wild as the limit is, the Fourier series represents it deftly ; the Taylor series cannot touch it. For the sake of the pointwise convergence theorems to follow, we need to elaborate on the theme of smoothness. Specifically, we need to discuss piecewise continuity, one-sided derivatives, and functions of bounded vari­ ation.

Ll

4. Fourier Series

163

Definition 4 . 3 . 2 PIECEWISE CONTINUITY A N D SMOOTHNESS

I

[a, b]

A real-valued function defined on the closed interval except for possibly a finite number of points is PIECEWISE (SECTIONALLY ) CONTIN­ UOUS if it is continuous at all but a finite number of points and the right­ and left-hand limits and exist at every point except the endpoints where only and must exist ; in other words , is continuous except for a finite number of finite jump discontinuities . We de­ note the space of all such functions by If is composed of a finite number of differentiable components and f' and f' exist at every point E [ a , except the end points where only and must exist , we say that I is PIECEWISE SMOOTH; in other words, is piecewise 0 smooth if f ' E a,

c

I (c+ ) I (c - ) c E [a, b], I (a + ) I (b - ) I, PC[a, b]. I (c+) +(c-) b] I' (a ) f' (b-) I PC [ b].

Step functions are piecewise continuous-indeed, they are piecewise smooth. In fact , any function that is continuous except for a finite number of jump discontinuities is piecewise continuous. The definition of piecewise conti­ nuity rules out highly oscillatory functions such as sin ( l ft) for '# 0, 1 (0) 0 , on any interval containing 0 ; f is not piecewise continu­ ous on because it does not approach a limit as ---+ 0 from either side. The function 9 1 ft for 0, 9 (0) = 0 on [0 , 1] is not piece­ wise continuous for the same reason: 9 does not exist . By breaking the closed interval a , up into an appropriate finite number of subintervals , it is easy to see that c for any � 1 . Since the step functions are dense in (Example 2.4.5) , it follows that is dense in [a , as well. It is simple to show (basically, you connect the j umps with steep , but not vertical, straight lines) that piecewise smooth functions are dense in , too.

t

= [ - 11', 11']

t f. (0 + ) [ b] PC [a, b] L; [a, b] Lp [a, b] L; b] L; [a b],

I (t) = t

(t) =

p

PC [a, b]

Definition 4.3.3 ONE-SIDED DERIVATIVES

Let I be defined in an interval containing TIVE of at a point c is

I: (c) I

.fI

Jr

c. A RIGHT-HANDED DERIVA­

( ) - hl....' O+ I ( c + h)h - I ( c ) . C

_

un

(4.20)

Ii. (c) . Remark A SUBTLE DISTIN CTION f' (c + ) f. I: (c) If I is piecewise smooth on the closed interval [a, b], then I' is continuous at all but a finite number of points of [a , b] , and f' (c + ) and f' (c - ) exist at every point c E ( b) . Note the distinction between the right-handed derivative I: ( c) and f' ( c+): f' (c + h) . I: (c) f. I' ( c+ ) = lim .... O+

A similar definition applies to LEFT-HANDED DERIVATIVES

a,

h

0

164

4. Fou rier Series

I, as we discuss in Exercise 1 , · I(c + h) - / (c+) ' 1' ( c+ ) - 11m h-+O+ h 0 for i .e . , with I (c+) in the numerator instead of I (c) . If I (t) t < 0 and I (t) 1 for t > 0 , then f' (0+) and f' (0- ) both exist ; neither fl (0) nor I: (0) exists because I is not even defined at O. We For piecewise smooth

=

=

say a little more about this in Exercise 2.

Dirichlet's 1829 pointwise convergence theorem 4.6.2 shows that the Fourier series of a piecewise smooth periodic function converges to I at every point Dirichlet believed that weaker conditions would suffice , but was unable to prove it. Jordan took up the cause and focused his attention on functions that were not smooth-functions such as the nondif­ ferentiable function of equation 4 . 18 , for example . He noted that the wild oscillation that destroyed differentiability also meant infinite arc length . He defined and investigated functions of "bounded variation" Definition 4.3.4 , then improved Dirichlet 's result in 1 8 8 1 . He showed that Fourier series of functions of bounded variation converge to at every point 4.6.6 .

+ (t+))

I

t.

�(I (t - )

( )

(

)

t ( l (t - ) + /(t+))

I t( )

Definition 4.3.4 BOUNDED VARIATION

Let h be a complex-valued function defined on the ( finite ) closed interval [a, b] and let P {t o , h , . . , tn}, t o = a < t l < . . . < tn b, be a partition of [a, b]. The VARIATIO N OF h WITH RESPECT TO P is n-l Vp (I) = L I h (t ; + d - h (t ; )I · =

(

)

=

.

;=0

h over [a, b] is V (h, a, b) V (h) = sup p Vp

The TOTAL VARIATIO N of

=

[a, b], BV [a, b]

as P ranges over all partitions of and we allow the possibility that V = 00. If < 00 , then we say that is of BOUNDED or FINITE VARIATION on The set of all real-valued functions of bounded variation on is easily verified to form a real vector space. 0

(h)

V (h) [a, b]. [a, b]

(

h

[a, b]

)

Clearly, any monotonic function on a closed interval is of bounded variation there-indeed, = Since the monotone func­ tion I = on (0, 1 is of unbounded variation , it is essential that the domain be a closed interval to conclude that a monotone function be of bounded variation.

(t) lit

]

V (I)

II (b) - I (a) l .

4. Fourier Series

165

What connection is there between continuity and bounded variation? The uniformly continuous function I (t) =

{ 0,

t cos h t E (0 , 1] , t = 0,

is not of bounded variation (Exercise 3) , so not even uniform continuity implies bounded variation. But suppose I is smoother. Suppose that I is continuous on and has a bounded derivative on the open interval then I must be of bounded variation (Exercise 6(a) ) . Conversely, how smooth must a function of bounded variation be? By 4.3.6 and the following theorem due to Lebesgue [Royden p. or Natanson pp. pretty smooth.

[a, b]

(a, b);

1968, 96,

2 1 1-212]:

1961,

4.3.5 MONOTONE ==> DIFFERENTIABLE A .E . A ny monotone function I on a closed interval is differentiable a. e., and the derivative I' is integra ble.

Not only are monotone functions of bounded variation, there is even the following result in the converse direction (the proof is outlined in Exercise

7):

1961, 218] [a, b] f V a,

4.3.6 T HE JORDAN DECOMPOSITION THEOREM [Natanson p. A ny function I 01 bounded variation on the closed interval can be writ­ ten as the difference of two increasing functions, namely (t) = (I, t ) ­ (I, t) - I (t)) .

(V a,

Consequently, a function of bounded variation must be differentiable a.e. and have an integrable derivative. Thus, the nondifferentiable functions of equation are of unbounded variation. Since differentiable a.e. implies continuous a.e . , monotone functions are continuous a.e .-hence Riemann integrable , therefore also Lebesgue integrable . Some other facts about func­ tions of bounded variation are given in the Exercises.

(4.18)

Exercises 4 . 3

1.

THE VALUE I' (a + ) (a) (b)

2.

Suppose that I is differentiable in the open

(a, b), and that limt -+a+ f' (t) exists. Show that I (a + ) exists. f (a + u) - / (a + ) . f' (t) = 'Ulim f' ( a+ ) = t lim -+a+ -+O+ U

interval

ONE-SIDED DERIVATIVES Suppose that I is defined on the open interval is differentiable in and that limt -+ a + I' (t) exists. The right-handed derivative I: depends on the of equation existence of I and is generally different from

(a, b),

(a),

(a, b), (a)

(4.20) + f' (a ) .

166

4. Fourier Series

f' (a + ).

(a) I: (0)

(a) Show that if I: exists, then so does The converse is not true by the Remark after Definition 4.3.3, a case where exists, but does not. 2 (b) For I (t) = t sin ( l/t) for t E and = show that I i s differentiable everywhere (even at but that limt -+ o+ f' (t) does not exist . Therefore I is not piecewise smooth.

f' (0 + )

[- 71", 71"] {O} ,

1 (0) 0, 0)

{

3. UNIFORM CONTINUITY # BOUNDED VARIATION Show that

0, t = 0, I (t) = t cos � 0 < t _< 1, 2t '

is continuous-actually, uniformly continuous since but not of bounded variation on

[0, 1] .

4. PRODUCTS For I, 9 E BV [a, b], show that

[0, 1] is closed­

Ig E BV [a, b] .

5. LIPSCHITZ CONDITION AT A POINT c A condition in between differ­

entiability and continuity-i.e. , stronger than continuity but weaker than differentiability-is the Lipschitz condition . Let be a closed interval. The function � R is said to satisfy a LIPSCHITZ CONDITION OF ORDER 1 at E if there is some neighborhood r) n r > and some constant K such that - r,

[a, b] I : [a, b] c [a, b] c + [a, b] , 0, II (t) - I ( c) I � K It - cl for all t E (c - r, c + r ) n [a, b] ,

(c

or its equivalent formulation

I/ (c + u) - l (c) 1 � K l u i for l u i < r, and c + u E [a, b] with the obvious adjustments at the endpoints.

(a) If I is continuous on the closed interval and f' and I' exist at E show that I satisfies a Lipschitz con­ dition of order 1 at (b) Show that

(c- )

c (a, b), c.

differentiable at and

c

(c+ )

[a, b],

::}

Lipschitz condition at

Lipschitz condition at

c

::}

continuous at

c

c

and that neither of the implications is reversible .

6 . UNIFORM LIPSCHITZ CONDITION Let I be an interval. The function I I � R is said to satisfy a UNIFORM LIPSCHITZ CONDITIO N on I if there is some constant K such that :

x I (y) 1 � K I x - y l

II ( ) -

for all

x, y E I.

4 . Fourier Series

167

f

(a) UNIFORM LIPSCHITZ CONDITION =} B V Show that if satisfies a uniform Lipschitz condition, then is of bounded variation on

( b)

f

I.

B OUNDED DERIVATIVE =} LIPSCHITZ If f is continuous on and has a bounded derivative on the closed interval show that satisfies a uniform Lipschitz condition on In particular, the function ( t = t sin for t E 11" 11" and 1 0) = 0 of Exercise 2(b) i s o f bounded variation o n 0 c) PIECEWISE SMOOTH =} BV Show that any piecewise smooth function on the closed interval is of bounded variation.

I

(

[a, b]

(a, b), [a, b]. [- , ] - {O} [ , 1].

f ) 2 (1/t)

(

7. THE VARIATION FUN CTION

[a, b] Let [a, b] be a closed interval. For I, 9 E

BV [a, b] show that: ( a) The variation (of Definition 4.3.4) V (I) is 0 if and only if I is constant. ( b ) For a < e < b, V (I, a, b) = V (I, a, e) + V (I, e, b) . (c) For x, y E [a, b], let V (I, x, y) denote the variation of f on [x, y] with V (I, a, a) = O. Show that for any a ::; x ::; y ::; b, V (I, a, y) = V (I, a, x) + V (I, x, y). Conclude that V (I, a, x) is an increasing function of x. ( d ) Show that p (x) = V (I, a, x) - f (x) is increasing. Since I (x) = V (I, a, x) - (V (I, a, x) - f (x)), conclude that f can be written as the difference of two increasing functions and therefore that functions of bounded variation are differentiable a.e .

8 . BOUNDS O N FOURIER COEFFICIENTS BY TOTAL VARIATION A mea­ sure of the rapidity of convergence of a series is how quickly its terms go to Show that:

O.

( a)

The Fourier sine and cosine coefficients

LH-1I", 1I"] satisfy 1 (I, [ l en I ::; -V n 1l"

])

- 11", 11"

en of f E BV [ n

,

E

]

- 11" , 11"

C

N.

( b) These estimates cannot be improved. f

9 . ABSOLUTE CONTINUITY What functions may be recovered as the integral of their own derivatives (to within an arbi­ trary constant) ? They are precisely the a solutely continu ous func­ tions. A real-valued function f defined on the closed interval is ABSOLUTELY CONTINUOUS on if for every t > 0 there ex­ ists 8 > 0 such that for any n E N , for all E such that < < and < < 8, we

f: f'(t) dt

[a, b] a l b l ::; a 2 b2 ::; . . . ::; an bn ,

b

[a, b]

ai , bi [a, b] L:?= l (bi - a i )

168

4. Fou rier Series

2:7=1 I I (bi ) - I (a i ) 1

have < f. Since n could be 1 , absolutely contin­ uous functions must be uniformly continuous. We denote the set of absolutely continuous functions on by Prove the following:

[a, b] AC [a, b].

(a) PRODUCTS The product of absolutely continuous functions is absolutely continuous. (b) AC => U C Absolute continuity implies uniform continuity. (c) If satisfies a uniform Lipschitz condition, then is absolutely continuous. (d) If I has a bounded derivative on then is absolutely con­ tinuous on a , (e) AC => BV An absolutely continuous function is of bounded variation. (Note that this implies the existence of continuous functions that are not absolutely continuous such as x sin ( l /x) on By the remarks about functions of bounded variation after it follows that absolutely continuous functions have integrable derivatives: If is absolutely continuous on then is differentiable a.e. , and f' E The next result is in the converse direction . (f) INTEGRALS OF FUNCTIONS ARE ABSOLUTELY CONTINUOUS If then 9 (x) is absolutely continuous. (g) INTEGRATION BY PARTS [Stromberg 1 9 8 1 , p. By (e) and (f) of this exercise, absolutely continuous functions are dif­ ferentiable a.e . and have integrable derivatives. If and 9 are absolutely continuous on then

I

I

[a, b],

[ b].

(0, a).) 4 . 3.6

I

I

I E L'i [a, b],

Ll

I

[a, b],

LHa, b].

= f: I(t) dt

6.90, 323] I

[a, b], f: I (t) g ' (t) dt + f: f' ( t) 9 (t) dt = I ( b) 9 (b) - I ( a) 9 (a) .

Hints

(f (tn))

tn

f' (a + ) b.

a. a

1 . (a) . Show that is Cauchy whenever --+ Since exists, has a bounded derivative in r) for some < r < By Example 2.7. 1(d) it follows that is uniformly continuous in ( a , r) . Hence , by the discussion in Section 2.7, maps Cauchy sequences into Cauchy sequences. By the uniform continuity, if --+ and --+ limn then limn = (b) Define let --+ and apply the mean value theorem to on r] for some r >

I

I

(a,

I

I (tn) = I (Yn). I (a) I (a + ), tn a, 0. l(tn) [a, (a) . If I: (a + ) exists, show that I (a + ) I (a). Yn a,

2.

=

tn

a

4. Fourier Series

3. Consider the partitions [0 , 1] .

169

Pn = {O, 1/2n , 1 / (2n - 1) , . . . , 1/3, 1 / 2 , 1 } of

6 . (a) . Use the mean value theorem. (b) Use (a) o n each of the subin­ tervals. (c) Let I satisfy a uniform Lipschitz condition on b] . For . .< E [ , b] , [{

[a, Xn a ( I (X k + 1 ) - I (X k )) � (X k + 1 - X k ) . Xl < X2 < 7. (d) . For X < y , show that p ( y) - p ( x ) = V (I, X, y) - (f ( y) - I (x)) . Note that I ( y) - / (x) � V (j, a, x). 8 . (a) . an = t D" I(t) cos nt dt = n1" f:'" I (t) d (sin nt) . Integrating by parts , we get an = �; f:'" sin nt dl (t) . (b) Consider Example 4 . 1 .3. .

9 . (f) . Make use of the following property of the integral [Nat anson 196 1 , p. 148] : If I E b] then, for any > 0, there exists 6 > ° such that for any measurable set E C b] of measure less than 6, I (t) dt E l I

L'1 [a,

f

4.4

[a,

< f.

f

The Riemann-Leb esgue Lemma

We enlarge the class of functions for which we compute Fourier series from to in this section. For I E the products (t) sin nt l and If (t) cos nt l are each less than or equal to l l (t) l , so the integrals that define the Fourier coefficients still converge; therefore it makes sense to consider the Fourier series of such functions. One reason to consider the theory for functions instead of functions is that there are more functions: As noted in Exercise 1 , the space is a proper subset of b] for all > 1 ; indeed, generally, as p increases , the spaces decrease in size. We cannot really go beyond for if rf: 11"] , what sense could we make of the Fourier coefficients

L2 L1

L1

LH-1I", 1I"],

L2

p

II

L1 Lp [a, b] Lda, Lp [a, b] L 1 , I Ld-1I",

an = -11"1 1" f (t) cos nt dt? Another reason to enlarge the theory to L1 functions is that it is fairly easy to do. As noted in the Remark before Example 4.1 .4, the Fourier coefficients of a linear combination of L; functions are the corresponding _"

linear combination of the Fourier coefficients of the original functions. Since this result depends only on the linearity of the integral, it remains true for functions. Also, for functions the product Ig is in 71"] , since

L1

L2

2 41g = (f + g)

L1 [-7I",

-

(j - g) 2 E L1[-1I", 11"] ,

so we can now speak of the Fourier series of the product of two another convenience.

L2 functions,

170

4. Fourier Series

L1

Although we consider functions from now on , we do not consider con­ vergence in the norm. Rather, we consider various kinds of convergence: pointwise , uniform, (e, 1), and Abel convergence. Lagrange , who was so troubled by the erratic behavior of Fourier co­ efficients , would perhaps have been mollified by the Riemann-Leb esgue lemma 3 .3.2; it demonstrates at least that the Fourier coefficients of any go to 0, i.e . , E

L1

f L;[-7I", 71"] 11" an = 71"1 1 f ( t ) cos n t dt -

- 11"

11"

---+

O, and bn = -71"1 1-11" f (t) sin nt dt

---+

O. (4.2 1 )

Intuitively, as the oscillation of the sinusoids increases , it puts as much of the function's area above the axis as below ; we mention two more instances where the increasing oscillations of a kernel function drive an integral to 0 in 4.5.4 and Exercise 2( a) . For our development of the theory of Fourier series for functions, we generalize the Riemann-Lebesgue lemma to functions fE b] , -00 � � b � 00 , and to more general kernel functions than sinusoids; the key to 4.4 . 1 is the density of the step functions in mentioned in 2.4.5( d).

L'i LHa,

a

L'i [a, b] ,

4.4.1 THE GENERALIZED RIEMANN-LEBESGUE LEMMA For any f E -00 � ::; 00 , and any bounded measurable fun ction defined on R it follows that

Li:[a, b],

then

h

�a b

c

l c h (t ) dt

lim ! .... ± oo c 0

limoo

w-+

-+

0

( the averaging condition) ,

l b f (t) h (wt) dt = O . a

Remarks Sinusoids are bounded, measurable (indeed, continuous) func­ tions, and, using sin t (t) for example,

=h �I l c sin t dt l = I � (1 - cos C)I ::;

I� I

-+

It therefore follows from 4.4. 1 that for any f E lim

W -+ OO

0 as c -+ ±oo.

Li:[a, b],

l b f (t) coswt dt = lim lb f (t) sinwt dt = O. W -+ OO

a

a

L1

(4.22)

In particular, it follows that Fourier coefficients of functions ap­ proach Note , too, that oscillation is not the driving force for the conclusion of 4.4. 1 ; rather the averaging condition is. Clearly,

O.

h (t) = {

�:t , IItt II

> ::;

1, 1,

4. Fourier Series

171

h( )

satisfies the conditions of 4.4. 1 . As w - 00 , the graph of y = wt becomes a horizontal line the t-axis with a peak of 1 at t = 0 , it does not oscillate at all.

(

Proof.

b

)

I(

< a,

If a or is finite, define t ) to be 0 for t > b or t respectively, so that we can restrict attention to E L� (R) . Suppose that l [ c,d] is the characteristic function of C [0 , (0 ) . Then, letting x = wt ,

[e, d]

l1 ( )

100 l[c,d] (t ) h (wt ) dt

l

I

d

dw

-wI

o

By hypothesis ,

1

dw

() Thus, 1000 l[c, d] ( t ) h ( wt ) dt any step function g , J uW

0

1

h x dx - O and �

-

h wt dt

0 as w

l0

-

h ( x ) dx - -wl

cW

1

c

0

W

()

h x dx .

()

h x dx - O as w - oo.

00 . It follows by linearity that for

r oo 9 (t ) h (wt ) dt = O.

W -+ OO 10 lim

As noted in 2.4.5, the step functions are dense in L'i (R) . Hence, if C is a bound for h , then for any E L'i (R) and { > 0 , there exists a step function {/2C. For sufficiently large t - 9 t dt 9 such that {/2, so w , 1J000 9 t ) h wt dt

I l

(

10(00 I I) ( ) < l

I

( ) 1 � II I gil l <

1 1 00 I (t ) h (wt ) dt l

<

100 II ( )

l

( ) l l h (wt ) 1 dt 1 1 00 9 (t ) h (wt ) dt � 2C C + 2 = c. A similar argument applies to I� oo I ( t ) h ( wt ) dt . Consider a more general question . If a trigonometric series a 0 + L an cos nt + L bn sin nt 2 n eN n eN converges, must an , bn O? As this need not be a Fourier series ( see o +

€

t -9 t

€

0

-

Section 4.1 1 ) , you do not know how the coefficients were obtained. Cantor showed that if the series above converges for all t in a closed interval, then - O. Lebesgue then generalit;ed Cantor's result to sets of positive measure . Since we refer to it in the proof of 4.4.3, we state Lebesgue's dominated convergence theorem here for reference.

an, bn

172

4. Fourier Series

1961,

161;

p. 4.4.2 DOMINATED CONVERGENCE THEOREM (Natanson is a sequence Rudin p . 305) If of measurable functions defined on the m easurable set E that is domin at ed on E by the integra ble function ---+ a. e. for every and 9 in the sense that a. e. on E, then is integrable, a n d

1976,

I

(In) l in (t) 1 � g (t)

n E N,

In (t) I (t)

n JfE In (t) dt JfE I (t) dt. 4.4.3 THE CANTOR-LEBESGUE THEOREM If a trigonometric series a o /2+ Ln EN an cos nt+ L n EN bn sin nt converges on a set E of positive Lebesgu e measure, then a n , bn ---+ O . lim

=

Jl.

Jl. [a, b]: Jl.

Proof. Let denote Lebesgue measure . We can assume that E is of finite measure , and even that E is a subset of a closed interval If (E) = 00, w e can choose the first positive integer m such that (E n [- m , mn > 0, and use E n m instead of E i n the argument . A s i n Exercise we can rewrite the series in amplitude-phase form as

[-m, ]

4.1-2,

�o + L: dn cos (nt - !f'n ) , where dn = Ja� + b� and 'Pn cos - 1 (an/dn) . nEN For t E E , dn cos (nt - !f'n) ---+ O. I f dn -f+ 0 then there exists f. > 0 such that dnk � f. > 0 for some subsequence of (dn). Hence cos (n k t - !f'nk) goes to 0, and therefore so does cos 2 (n k t - 'Pnk) ' Since I cos 2 (n k t - !f'nk) I � 1 on E, by the Lebesgue dominated convergence theorem 4.4.2, it follows that =

we can pass to the limit under the integral:

2 2 k JfE cos (n k t - 'Pnk) dt = JfE limk cos (n k t - 'Pnk ) dt

lim But

L cos 2 (n k t - !f'nk) dt

Now consider the integral

=

-

=

O.

L [1 + cos 2 (n k t - 'Pnk)] dt t [Jl.(E) + L 2 (n k t - 'Pnk ) dt] . t

L cos 2 (nkt - 'Pnk ) dt. Let I = 1 on E , and

f = 0 on [a, b] n CE; clearly, I E L'i [a, b]. Now

4. Fourier Series

173

0

As the latter two integrals go to by the Riemann-Leb esgue lemma 4.4. 1 , i t follows that the limit is p. (E) 1 2 , which contradicts the previous equality. o

Exercises 4 . 4

1. Lp GETS SMALLER AS P GETS BIGGER Show that (a) L q [a, b] is a proper subset of L p [a, b] for all 00 � > p � 1 , the exact opposite of what happens for the lp spaces. Does this remain true for infinite intervals? (b) There are functions I, g E L2[-7I", lr], such that Ig fi. L2[-7I", 71"]. 2. RIEMANN-LEBESGUE LEMMA VARIATIONS Let [a, b] be a closed in­ terval, and let I E Ll [a, b]. (a) Divide [a, b] into n subintervals of length (b - a) In, and define gn = ±1 on alternate subintervals . Show that for continuous I, limn co f: I (t) gn(t) dt = O. (You can use 4.4. 1 for this, but it is easy to give a more revealing elementary argument .) (b) OTHER ORTHONORMAL BASES Show that if ( gn) is a uniformly bounded (i .e . , for some M, I gn (t)1 � M for all t, and every ) countable orthonormal basis for L2[a, b], a, b finite, then nlim � oo ta I (t) gn(t) dt = O. 3 . ALTERNATIVE PRO OF OF RIEMANN-LEBESGUE Let I E Ll ( R) . (a) Show that lim6 o f�co I I (t + 6) - I (t)1 dt = O. (b) With g (w) = f�oo e iwt l (t) dt , without using 4.4. 1 , show that limw 9 (w ) = O. q

.....

n

.....

.....

o

Hints

L [a, b], split [a, b] into {t E [a, b] : I / (t)1 < I } U {t E : I / (t)1 � qPI}, and use the fact that I / (tW I / (tW on the latter . Since I / I � max { I , lf n , it follows that I /P I is integrable . Since q p � 1 , q p + for some O. Let 9 (t) = II (p + ) on

1 . (a) . For I E

[a, b]

>

>

c

c

>

�

c

(0, 1], 9 (0) = O. Then 9 E Lp [0, 1] but 9 fi. L q [0, 1]. The inclusion L q Lp for q � p does not hold on infinite intervals: Consider I (t) C

lit on [1 , 00) .

=

174

4. Fourier Series

2. (a) . Use the uniform continuity of f, and consider Riemann sums.

-g

3 . (b) . Since (w) = it follows that

f�oo eiw (t +1r /w ) I (t) dt = f�oo e iwt I (t - 7r/w) dt,

2g (w) = Therefore,

4.5

f�oo eiwt [I (t) I (t -

-

7r/w)]

dt.

Ig (w) 1 � � f�oo I I (t) - f (t - 7r/w ) 1 dt.

The Dirichlet and Fourier Kernels

This section paves the way for the pointwise convergence theorem 4.6.2 by providing certain ways to express the nth partial sum of a Fourier series (4.5. 1 ) . The device by which we accomplish this is the simpler way provided by equation (4.24) to express a sum of cosines as a ratio of two sines . In practice, Fourier series are always truncated to approximate a function as a finite sum of harmonics (t) = sin kt . cos k t Through the use of the nth DIRICHLET KERNEL

Bn

+ L � = l bk

ao/2+ L� = l a k

(4.23)

Bn

(see also equation (4.25) ) , we can theoretically calculate (t) in just one integration (4.5. 1 ) . Clearly, for n 1= 0, is an even 27r-periodic function . As illustrated in the figure below , (t) becomes more and more highly concentrated at t = 0 as n increases, becoming like a pulse 2n 1 units high on a shrinking pedestal of width 47r / (2n 1) . The shrinking base and increasing height are such that the area under the curve remains constant (Example 4.5.3) . (To sketch it, use the representation of equation (4.24) , and treat 1/ [sin (t/2)] as the envelope.)

Dn

Dn

+

+

The Dirichlet kernel for

n

=

3 and

n

=

7

4. Fourier Series

175

, Dn =

For t = 0 , ±21r, ±47r, . . . (t) 2n + 1 by substitution. To calculate # 0, we can avoid the summation by showing that for any n E N,

Dn (t) for t

sin ( n + t ) t , t # 0, ±21r, ±47r , . . . . sin (t/2) (4.24) To see this, add the equalities below for k = 1 , 2, . . . , n : _

1 + 2 cos t + 2 cos 2t + · · · + 2 cos nt -

sin

(k �) +

( �)

t - sin k -

t = 2 cos k t sin

�.

The sum on the left telescopes t o yield just the end terms:

which implies that

For t # 0, ±21r, ±47r, . . . , we can divide both sides by sin (t/2) to get equation (4 .24) . (An alternate derivation is suggested in Exercise 1 .) By I 'Hopital 's rule, for any 0, ±21r, ±47r, . . . ,

p=

. sin (n + l/2) t hm . sm ( t /2)

t -+ p

{

= 2n + 1 = Dn (p) .

Hence , the function defined as sin (n + l/2) t , t # 0, ±27r, ±47r, . . . , sin (t/2) otherwise, 2n + 1 ,

n E NU

{O} ,

(4.25)

Dn

is continuous and agrees with (t) as defined in equation (4.23) for all t. When we write in the sequel, we do so with the understanding that it has the value 2n + 1 when t = 0, ±21r, ±47r, . . .. Various expressions for the nth partial sum of a Fourier series are given in 4.5 . 1. As we shall see, as n -- 00, (t) behaves like Dirac 's 6 function in the integrals below , sieving out the value of the function f (t) when the argument of (t - ) is 0, i.e . , when = t; in other words, (t) -- f

Si:�:t;��)t

Dn

x

Dn x

Dn

Sn

(t) .

4.5 . 1 THE nTH PARTIAL SUM USIN G For any n E N U { O } , the n th partial sum of the Fourier series for f E L1 [-1r, 7r] is given by any of the convolution integrals below:

Sn

176

4. Fourier Series

-211'1 1_�� f(x) Dn (t - x) dx = -211'1 1_� f(x) Dn (x - t) dx, Sn (t) = (b) -211'1 1_�� f (t - x) Dn (x) dx = -211'1 1_� f {t + x) Dn (x) dx, � (c) 1 ior [f (t + x) + f (t - x)]Dn (x) dx. 211' Proof. The statements are all trivial for = 0, so we assume i- 0 in the arguments below . (a) The nth partial sum of the Fourier series of f is � Sn (t) = 211'1 1_ f (x) dx + � L:� = l cos kt I: f (x) coskx dx "" nk = l sin kt 1 � f (x) sin kx dx + 2:.11' L...-; [cos kt cos kX + sin kt sin kx] dx 1;: 1 � f (x) ( 21 + {; _� ) ) n ( 1 ;:1 1_� f (x) 2 + {; cos k (t - x) dx -211'1 1_�� f (x) Dn (t - x) dx -211'1 1_�� f (x) Dn (x - t) dx because Dn i s even. (b) With w = t - x, the penultimate equation becomes Sn (t) = -2111' I-t + � f (t - w) Dn (w) dw. t� Since f and Dn are each of period 211' , by 4.2.2 , the value of the integral remains the same as long as the range of integration is 211'; hence Sn (t) = 211'1 1_� f (t - w)Dn (w) dw. � With w = x - t, the equation Sn (t) = � 1 � f (x) Dn (t - x) dx becomes 211' _ � -t � Sn (t) = 211'1 1-� -t f (w + t) Dn (w) dw = 211'1 1_ � f (w + t) Dn (w) dw. � (c) We split the first integral 21 J::� f (t - x ) Dn (x) dx in (b) to get � 1 10 Sn (t) 211' _ f (t - w) Dn (w) dw + 211'1 ior f (t - w) Dn (w) dw. � (a)

n

=

n

4. Fourier Series

Now replace w by to get

177

-u in the first integral , and use the fact that Dn is even

sn (x) = 211"1 Jr [f (x + u) + f (x - u)] Dn (u) duo o We need the estimates of 4.5.2 to prove Dirichlet ' s theorem 4.6.2 about 0

the pointwise convergence of Fourier series.

4.5.2 Two UNIFO RM BOUNDS ON INTEGRALS (a) THE DIRICHLET KERNEL For any 0 :S :S b :S {O} , b sin n t t < 411". . a sm

NU

( + l /2) d ( t/2)

l

a

11" ,

a n d any n E

-

(b) THE CONTINU OUS FO URIER KERNEL For all real w and t , the qu a n­ tity cp (t , w ) = sin wt t for t 0, cp (O, w ) w , is called the CONTIN U O U S FOURIER KERNEL. We write sin wt t sometimes, but with t h e understan d­ ing that sin wt t is defined to be w when t O. For any 0 :S <

/

=P

/

= =

/

lb -sin wt dt a

t

<

a b,

11" .

Proof. (a) If > 0 and AI , . . . . . are numbers that alternate in sign and decrease in absolute value, then dominates the partial sums: For any k = 1 , 2 , . . . , n ,

Al

. , An ,

Al

= 1 , . . . , n , the areas l k1r /(n + l / 2) ( n + 1/2) t . { positive, A k = k l n l sinsm . (t/2) dt negative, ( - ) 1r/( + / 2) sin ( n + 1/2) t under the various lobes of Dn (t ) 0 sin (t/2) For k

IS

7

3

if k is odd, if k is even, < t <

11",

consti-

178

4. Fourier Series

tute a set of numbers of alternating sign and decreasing absolute value , since sin is increasing. Since is contained in the rectangle of base [0 , (n and height 2n it follows that � 1r . For any [0 , 1r , either there exists some � n such that

(t/2) 1r / + 1/2)] aE ]

Al + 1, Al 2 k a E [(k - 1) 1r/ (n + 1/2) , h/ (n + 1/2)] , or a E [n1r / (n + 1/2) , 1r] ( take the interval on the right if a belongs to two of them) . As the latter argument is quite similar, we argue only the former case. If k is odd, A k > 0, and A k < 0, so for some E [0, 1] , 10a Dn(t) dt = Al +A2 + · · +Ak - l + cA k � Al +A2 + · ·+Ak - l � Al � 2 11". Consequently, for any ° � a � b � 1r, and any n E N u {O}, c

If k is even,

br

Ak

sin u duo A k = jk I 'II" -(-) u

The alternate in sign and decrease in absolute value-indeed, by a geometric argument as in a , It follows that � k , for each = ( 'II" Sl�. t for any n , 11", and therefore that

( ) I Ak I 1/ k E N. n dt < J E k = l Ak < Al o 1 '11" sin wt sin t 1 6 -< t d t - 0 t dt < 1r .

a

0

Example 4.5.3 CONSTANT AREA UNDER DIRICHLET KERN EL O N [0 , 11"] Sh ow that

1 '11" sin (.n + 1/2) t dt = 1 '11" Dn (t) dt = 11", for all n E NU {O} .

( t-/2) 0 Solution. Since fo'll" cos n t dt = ° for every n , o

sm

1o '11" Dn (t) dt = 1'11" (1 + 2 t cos kt ) dt = 1r. 0

k= l

0

4. Fourier Series

179

Dn

(t) = S i :�:t;Wt jumps up to 2n + 1 at t = 0, The Dirichlet kernel then stays at about constant amplitude between t = 7r / (n + 1 / 2 ) and t = 7r while becoming increasingly more oscillatory. We might expect that the oscillations of are enough to make

Dn

1,..,./.(n + l/ 2)

f (t)

Dn (t) dt

--->

0 as n ---> 00

f E Ll [- 7r, 7r] , and indeed this is what happens. 4.5.4 T HE RIEMANN -LEBESGUE PROPERTY O F Dn L e t Dn ( t ) th e Dirichlet kernel. For any f E Ll [ - 7r, 7r) , and r E (0, 7rJ , '" lim ! n f (t) Dn (t) dt = O.

for any

den ote

Proof. Since r > 0, then sin (t/2) � sin r/2 > 0 for all r S; t S; 7r. It follows r

f?j

E Ll [r, 7r) . Hence, by the Riemann-Lebesgue lemma 4.4. 1 , as that . sm t 2 ) n � oo ,

Definition 4.5.5 DISCRETE FO URIER KERNEL

Suppose we replace the sin (t/2) in the denominator of the Dirichlet kernel (t) = [sin (n + 1/2) t) / sin (t /2) by t /2, i.e . , consider

Dn

sin (n + l/2) t sin (n + l/2) t instead of t/2 sin (t/2) . . N U {O} , we call the contmuous function � n (t) =

sin (n + 1/2) t t/2 for t '# 0, �n (0) = 2n + 1, the DIS CRETE FOURIER KERNEL of order n . sin (n + 1/2) t . sin (n + 1/2) t . IS we mean that If we WrIte mstead of t/2 /2 defined to be 2n + 1 at = O. 0 For n E

t

t

�n ,

�n 2; �7, �7

The discrete Fourier kernel provides a very good approximation to the Dirichlet kernel for It I < and are sketched below. Eventually, (t) = sin the discrete Fourier kernel / (t/2) decreases more rapidly than but and are essentially indistinguishable until about = 2. (To sketch treat 1/ (t/2) as the envelope.)

Dn

D7,

�7 D7 �n (t),

D7 (7.5t)

t

4 . Fourier Series

The Dirichlet kernel

The Fourier kernel

Dr (t)

and

D7 (t)

7 (t)

r ( t)

=

=

�in 7 . 5 t sm(t/2}

sin 7 . 5t t/2

4. Fourier Series

181

AnotherFourier observation concerning the behavior similarityasbetween thein integral Dirichlets . The and discrete kernels concerns their kernels essentialby content of 4.Fourier 5 . 6 and kernel 4. 5 . 9 isas that you can replace the Dirichlet kernel the discrete n Let 1 E L1: [0, 7r] 4.5.6 sin (t/2) CAN BE REPLACED BY t/2 as n and r E (0, 71"]. Then, assuming that the limits exist, limn for I (t) sins�: �:;)2)t dt limn for I (t) sin (nt;21 /2) t dt. (4. 2 6) �i; -+ 00 .

=

sin(

In addition, the discrete Fourier kernel

-+ 00

/ 2)t

LEBESGUE PROPERTY : 11

has the ''RIEMA NN­

o.

l� 1" 1 (t) sin ( nt721 /2) t dt Proof. By two applications of l' H 6pital's rule, lim..... o (_ sin1 t - �)t = Therefore, 9 (t) � / is continuous everywhere if we define it to beis the0 atproduct t = O. Itoffollows that 9 isfunction integrableandanda bounded boundedintegrable on [0, 71"] . function, Since Ig an integrable Igwithis integrable on [0, 71"] . Thus, by the Riemann-Lebesgue lemma 4.4 .1 h (t) sin ( n + 1 /2) t , l� 1 " [I (t) Cin (lt/2) - t;2 ) ] sin (n + 1 /2)t dt 0 , or, assuming "that the limits exist, + 1 /2) dt = limn 1" 1 (t) sin (n + 1/2) t dt. limn 1 1 (t) sinsm�n (t/2) t/2 By an identical" argument + l/2) dt = limn " l (t) sin(n + 1 /2)t dt. limn 1r l (t) sinsm�n (t/2) t/2 1r By 4. 5 . 4 , however, limn Ir" 1 (t) :(;(i/N dt 0, and the desired result follows. For what 1 do the limits of equation (4. 2 6) of 4. 5 . 6 exis t? What is the limit? For certain 1 (piecewise smooth, for one type) the limi t is 7r 1 (0 + ) . This fact is the essence ofthese the pointwise convergence theorems 4. 6 . 2 andon 4.integrals. 6 . 6 . BeforeIn dealing we consider questions, we consider some variants with infinite series, wesomething allow absolutely convergent and nonabsolutely convergent ones. We do similar for integrals for more flexibility. =

o.

t

=

s i n( /2) - t 2

=

=

0

0

0

si

i

)

=

182

4. Fourier Series

-+ , pv , a < b 1a -6 1-+6a 1-+ a 6 1a b we say thatconvergence. f is integrable on [a, b], we mean that f: I f (t)1 dt < i. e-If. , absolute If f is integrable, f+ (t) = max:(f (t) , 0), and f (t) = min (f (t) , 0), then f = f+ -f - , and f: f(t) dt f: f+(t) dt­ f: F(t) dt. If we abandon absolute convergence and demand only that f be integrable on [a, c] for all c < b and that li c-+ f: exist, then we represent the limit as fa-+ b • Similarly, b and -+ = lim d = lim 1 1-+ba c-+ a c 1-+ a 6 c-+ a , d--+ 1c b kernel sin tit is not The distinction is important. For example, the Fourier in L'i (R+) since (4.48) ) r oo I sint t l dt ( eqUation of Section 4.9 ' Jo but the DIRICHLET INTEGRAL - 00 sin t 1o t dt 11"/2 (4. 5 . 8 (b )) . If f is integrable on [a, b], then 1a b f(t) dt = 1a -+ b f(t) dt = 1-+ba f(t) dt. For c E ( a, b) , the CAU CHY PRINCIPAL VALUE of f: is defined to be ( -€ PV { b lim r + 1 ) . Ja €-+ o Ja c+6 € By this convention, even though neither f'::' r 3 dt nor r:o t-3 dt exists, °O -€ PV 1 t-3 dt = lim ( 1 r 3 dt + J oo r3 dt ) = lim ( -� + -4) = 0, €-+o 2f 2f €_ o - 00 - 00 € whereas if we take 2f in the second integral, we get -1 --1 ) ' hm. ( e-O 2f 2 + 2 (2f) 2 sob, then the symmetric approach is crucial. Likewise, if a Cl < C2 . . . < Cn ::; Definition 4.5.7 IMPRO PER INTEGRALS

,

,

- 00

::;

::;

00 .

00 ,

=

-

m

6

= 00

=

=

= - 00

::;

<

4. Fourier Series

183

Last, if lima_ oo I� exists, we designate it to be the CAU CHY PRINCIPAL of I�oo ' and write °O l PV 1 = lim - 00 If I� oo and 1�000 exist,� i.e ., if f is absolutely integrable on (-00, 0] and then I oo = I oo ' and 1000 = 10..... 00 If the integrals I� oo' and 10- 00 exi[0, s00),t, then 00 00 PV 1 = ..... - 00 1-+- 00 A consequence of taking Cauchy principal values is that if function-f(t) t or f (t) = sint, say-then for all 0, If�(t)f is(t)anydt =odd0, t dt nor I:�oo sin t dt exists. If f is so PVI�oo f (t) dt = neither I�C: integrable, f E L'i ( b) , then I ..... b f, t+ f, I:: f, PVI: f exist and are equal to I: f. nelWenext.consider integrals of the Dirichlet kernel and continuous Fourier ker­ 4.5.8 (a) For any r E (0, 11"] , limn iro sinsm�n +(t/2)1/2) t dt = limn 10 sinsm�n +(t/2)1/2) t dt = 11". (b) THE DIRICHLET INTEGRAL For any r 0, l r sinwt dt -- 1 ..... 00 sint dt - 1I"/2 . 1m _ w oo 0 t o t Proof. (a) By Example 4.5.3, dt 11" for every n. Hence, for any r E (0, 11"] and any n , 10'" Si:�:�;Wt 1/2)t dt 11" = ior sin(sinn +(t/2) sin(n + 1 /2) t 1'" sin (n + 1/2) t ior sin (t/2) dt + r sin (t / 2 ) dt. with f (t)the=continuous By(b)4.5.4Consider 1, I: � n 00. Fourier kernel (t, w) defined to be w at t = 0, and