Fourier Series (Mathematical Association of America Textbooks)

C L ~\

~

S I{

l) () ~\ 1

R I

~

0. The partial sums s, of this series are alternately I and 0. So the series does not converge. However, its divergence is different in character from that of the series L n or L 1/n. In these latter cases the partial sums diverge to oo, whereas in the former case the partial sums oscillate between I and 0 and hence on an average take the value I/2. We say that a series L x, is (C, I) sum mabie or Ces(uv summab/e if the sequence a, = ~(so + s 1 + .. · + s,_ 1). formed by taking the averages of the first n partial sums of the series, converges. Every convergent series is (C, I) sununable~ a series may be (C, I) summable but not convergent. for example this is the case when x, = (- I )11 • In 1904. Fejer proved that the Fourier series of every continuous function is (C. I) summable and, in this new sense, converges uniformly to the function. This theorem is extremely useful and gave an impetus to the study of sununability of series. (Fejer's Theorem is proved in Chapter 2, Section 2.)

A HISTORY OF FOURIER SERIES

9

11. H. lebesgue It had already been noted, by Cantor among others, that changing a function at ··a few" points does not alter its Fourier series, because the value of the integral (8) defining the Fourier coefficients is not affected. So it is not proper to ask whether the sum of the Fourier series of f is equal to f at e\'et)' point; rather we should ask whether the two are equal everywhere except on those sets that are irrelevant in integration. This problem led to a critical examination of the Riemann integral. Just as Riemann was motivated by problems in Fourier series to define a new concept of integration, the same motivation led Lebesgue to define a new integral that is more flexible. The notions of sets of measure zero and al11wst e\'erywhere equality offimctions now changed the meaning of function even more. We regard two functions f and g as identical if they differ only on a set of measure zero. Thus, Dirichlet's function (the characteristic function of the rationals) is equal to 0 ··almost everywhere." So, from a formula to a graph to a rule, a function now became an equi,•alence class. The Lebesgue integral is indispensable in analysis. Many basic spaces of functions are defined using this integral. We will talk about the L 11 spaces LP

= {f

:

j I!I oo} . I P
0. there exists ro such that for r > ro.

1

-lJ

Q,.(t)dt

-;r

In this case Qr

*f

converges to

+

1"lJ

Q,.(t)dt <e.

f uniformly as r

( 1.31)

___,. I.

Exercise 1.3.2. Show that the Poisson Kernel defined by ( 1.26) has the following propertics: (i) P(r. cp) > 0,

Cii) P(r. cp) = P(r. -cp), (iii) P(r. cp) is a monotonically decreasing function of cp in [0. rr ],

= P(r. 0) = (I + r)/2rr( I - r), min_rr:::tp:::rr P(r. cp) = P(r. rr) = (I - r)/2rr( I + r).

(iv) maX-rr:::tp:::rr P(r, cp) (v) (vi) f~rr P(r. cp)dcp = I.

(vii) for each e, D> 0, there exists 0 < ro < I such that, for ro < r < I. we have -lJ

1

P(r. cp)dcp +

-TT

111' P(r. cp)dcp < e. lJ

Hint: To prove (vii) proceed as follows. Since P(r. cp) is an even function of cp you need to prove P(r. cp)dcp < ef2. Since P(r. cp) is monotonically decreasing.

J.;

max P (r. cp) •S:::tp:::rr

=

I - r2 ,. 'J.rr I - 2r cos 8 + rI

23

HEAT CONDUCTION AND FOURIER SERIES

Now for a fixed D > 0. . hm

r-+ I

I - ,.2

., = 0. I - 2r cos D+ r-

Use this to choose the ro that is required.

3

-1T

1T

FIGURE 1 The Poisson kernel

Theorem 1.3.3. (Poisson's Theorem) Let f be a continuousfimction on T and let u(r. 0) be defined by ( 1.25). Then as r

~

I. u(r, 0)

4>

J(O) uniformly.

Proof For 0 < r < 1. let Pr(cp) = P(r. cp) denote the Poisson kernel. Then u(r. 0) = ( Pr

* /)(0).

The family Pr is a Dirac family (by the preceding exercise). Use Theorem 1.3.1 now.

•

Let us sum up what we have achieved so far. We have shown that u(r. 0) defined by ( 1.25) satisfies the conditions ( 1.2), ( 1.3) and (1.4) of the Dirichlet problem and further (for every continuous function f) it satisfies the boundary condition (1.5) in that limr- 1 u(r, 0) = f(O) and the convergence is unifonn in 0. So we have found a solution to the Dirichlet problem. We will now show that the solution is unique, i.e., ifv(r,O) is another function that satisfies (1.2), (1.3) and (1.4) and if v(r. 0) converges to f .:. T(

sin (2n + I )u

0

du.

ll

(iii) Now write

Ia 0

} sin(2n + l)u ll

~ ~:!~~~~-! sin(2n + l)u

du = ~

kO

·=

- - - - - du.

_1:._· -r ln+l'"!

ll

(iv) Each of these integrals can be estimated from below by replacing u occm,-ing in the denominator by the larger quantity

k+1 2n + I

T(

2

After this the resulting integrals arc simple to evaluate.

CONVERGENCE OF FOURIER SERIES

35

(v) This gives 4

I

>-'"""2n

L"- rr 2 ~ k +I k=O

Use Exercise 2.2.7 now. (vi) At step (ii) above use the inequality of Exercise 2.2.6 (ii) instead, to estimate L11 from above. Make a similar change in step (iv). Get an upper bound for L11 •

Exercise 2.2.9. Estimate L11 in another way as follows:

.

(1) L 11

21TT

= -rr

sin(n + ~)u ? .

,;

-Sm 2

0

du.

(ii) Split the integral in two parts; one from 0 to I In and the other from I In to rr. By Exercise 2.2.1 (iii)

!

sin(n + )u -< 2 sin I

I

ll

+ -. 2

So

loo

lfn

sin(n +

!>u

I du 2.

2.3 Pointwise convergence of Fourier series By the Weierstrass approximation theorem every continuous function on T is a unifonn limit of exponential polynomials. (See Exercise 1.3.9; this follows also from Theorem 2.2.5). In other words iff is a continuous function on T. then for every E > 0 there exists an exponential polynomial N

PN(t) =

L ani"' n=-N

(2.1 0)

FOURIER SERIES

36 such that

1/(t)- PN(t)l

sup

(2.11)

<E.

-TT :::£1:::£TT

This can be used to prove: Theorem 2.3.1. (The Riemann-Lebesgue Lemma) Iff is a continuous .function on T.

then

lim

lnl-oo

f (n) = 0.

Proof We want to show that given an E > 0 we can find an N such that for alllnl > N we have 1j (n) 1 < E. Choose p N to satisfy (2.10) and (2.11 ). Note that for In I > N. fJ (n) = 0. Hence for In I > N we have A

f(n)

= f(n)A

= (f-

p(n)

p)(n).

But from (2.1 0) we get -

ll =

1111 I iTT . rr -TT [f(t)- p(t)]e- dt 2

I " -< -E.-7r') _rr

•

E.

Exercise 2.3.2. Let f be a continuous function on T. Show that

lim iTT f(t) sinnt = 0.

lim iTT f(t)cosnt =0.

n-ov

n-oo

-TT

-TT

Exercise 2.3.3. Let f be a continuous function on T. Show that

nl!':J.,

L

{(I) sin

(

(n + Dt) dt

= 0.

Hint: Use the two statements of Exercise 2.3.2 replacing .f(t) by f(t) respectively. Exercise 2.3...1. If show that

sin~

and f(t) cos~

f is continuously differentiable on [a, b], use integmtion by parts to

lim •\-+X

lb

f(t) sin tx dt

= 0.

ll

This gives another proof of the Riemann-Lebesgue Lemma for such functions. More generally. use this method to prove this lemma for piecewise C 1 functions. [Definition: A function f on [a. b] is called piecewise continuous if there exist a finite number of points

37

CONVERGENCE OF FOURIER SERIES

aj.a =ao 0 such that lfn(f; B)- S,(f: 8)1
l k,< lii.:S(k+ I )II

(k+l>n A

k such that for all n > n 0 •

Now let m > kno. Then for some n > no we will have kn < m < (k Lemma 2.3.18, lakn.lk+l>n(f; B)- S,(f; 8)1
k and ljl < m(k). then A

f,,(r(k)

+ j) =

1

2k pk(j).

(2.30)

for all j < 0.

(2.31)

and A

f,,(j) = 0

If n' > n + 1. then II'

I J+II' (t)

-

J+II (f

)I =

'"""' ~ ,..1k ' eir!k>tPk(t) k=ll+l II


0, c, =

I

(a,- ib,),

2

c_, =

I

2(a + ib,). 11

Exercise 3.1.3. The series (3.5) can be integrated term by tenn to give

lo

0

x

f(O)dO

00

lJ11

00

n

11=1

=-;:;- + L- + L l lo-•,.

-

11=1

(

-n sinnx--l cosnx ) .

a,

')II

ll

(3.7)

ODDS AND ENDS

53

Iff is sufficiently smooth, the series (3.5) can be differentiated term by term, and then 00

['{e)= "L, 0 there exists a finite sum g(8) = E:~=O a, cos n8 such that sup 1/(8)- g(fJ)I < e. OeT

(3.9)

This can be used to prove the Weierstruss appmximation theorem as follows: (i) Show, by induction, that there exists a polynomial T,, of degree n with real coefficients such that cosn8 = T,,(cose).

(3.10)

T,, is called the Tchebychev polynomial of degree n.

(ii) Let q; be any continuous function on [0, I]. Define a function f on T by JW) q;(l cos 81). Then f is an even continuous function. Use (3.9) and (3.10) to show that N

sup lq;(t)- L,a,T,,(t)l <e. O:::t:::l

11=0

This shows that every continuous function on [0. I] is a uniform limit of polynomials. (iii) By a change of variables show that every continuous function on an interval [a, b] is a uniform limit of polynomials. This is called the Weierstrass approximation theorem. (iv) Show that this result and the one we proved in Exercise 1.3.9 can be derived from each other.

Exercise 3.1.5. Use the Weierstrass approximation theorem to prove that if f and g are continuous functions on [a. b] such that

J.b x" f(x)dx = J.b x"g(x)dx for alln > 0, then f = g. (This is called the Hausd01f! moment theorem and is very useful in probability theory).

3.2 Functions with arbitrary periods So far. we have considered periodic functions with period 2rr, identified them with functions on T and obtained their Fourier series. A very minor modification is required to deal with functions of period 2L, where L is any real number. Iff is periodic with period 2L,

54

FOURIER SERIES

then the function g(x) = j(xLjrr) is periodic. with period 2rr. So. we can first derive the Fourier series for g and then translate everything back to f by a change of variables. All the results proved earlier remain valid with appropriate modifications.

Exercise 3.2.1. Let L be any positive number and let f be an integrable function on [-L. Ll with f(L) = J(-L). Then the Fourier series off can be written as

L A,ei":mfL, where . rr fL dt. A, = - I 1L j(t)e- 1111 2L -L A

(3.11)

In the same way. iff is even. then its cosine series can be written as I

00

-

11=1

nxrr

-Ao +"'"'A, cos--, "J ~ L where

{L

2

A,

= L lo

ntrr

.f(t) cosT dt;

(3.12)

and iff is odd. then its sine series can be written as 00

•

nxrr

LB,smL. II= I

where 21L

B, = -

L

. ntrr

[(t) sm- dt.

0

L

(3.13)

Now we come to an interesting bit: how to apply these ideas to any function on a bounded interval. Let J be any continuous (or piecewise C 1) function defined on the interval (0. L]. Extend f to [-L. L] as follows. Define {(0) = J(O+>·

f(-x)

=

f(x)

for 0 < x < L.

This gives a continuous (or piecewise C 1) function on [-L. L]. Now extend f to all of IR by putting

f<x + 2L) = f<x> for all x. This is an even continuous (or piecewise C 1) function defined on all of IR having period 2L. So we can expand f into a cosine series. Notice that our extended function is continuous

55

ODDS AND ENDS

at x = 0 and x = L. We could also have extended f to an odd function as follow~. Define

= 0.

j(O)

{( -x) = - j(x).

0

<X
0. (iii) B, (0)

= B, (I) for all

n > 0 except n

= 1.

(Note 81 (0) = -B, (1) = -1/2.) The Bemoulli numbers are the sequence B, defined as

B,

= n!B,(O).

(3.46)

We should warn the reader here that different books use different conventions. Some do not put the factor n! in (3.46) while others attach this factor to B,(x).

FOURIER SERIES

66 Exercise 3.6.2. Show that

L n! 1

B,(x) = -

II

(

Bkx

)

k

k=O

Use Exercise 3.6.1 to show that Bo

ll

= I. and for n > I

n-l

B,=--L n + I k=O

11-k

(3.47)

1.

(n + 1) Bk.

(3.48)

k

The last relation can be w1itten out as a sequence of linear equations: 1 + 2B, = 0.

+ 382 = 0. I + 48, + 682 + 483 = 0. 1 + 58, + 1082 + 1083 + 584 = 0. 1 + 3B,

From here we see that

B 1 = -1/2.

B2 = 1/6.

B6 = 1/42,

B?. = 0,

B1 = 0.

84 = -1/30.

Bs = 0,

Bs = -1/30, ....

Exercise 3.6.3. Show that t 1

e

-

1

=I:

B, 1,, n=O n!

(3.49)

(Hint: W1ite down the power se1ies for e' - 1. multiply the two series, and then equate coefficients.) From this we see that t

t

IJC

B

- + - = I + "'""' ___!!.. r". e1 - I 2 ~ n!

(3.50)

11=2

The left-hand side can be rewritten as r(e 1 + l)/2(e'- 1). This is an even function of r. Hence we must have for n ~ I.

B2n+l = 0

(3.51)

What has all this to do with Fourier series? We have seen that the functions B, (x), n > 2 are periodic on IR with period 1. The function 8 1(.t) has the Fourier series I ~sin 2rrnx B ,(x)=--~ . JT

"='

ll

0

<X

< I.

(3.52)

ODDS AND ENDS

67

(See (3.20)). Repeated integration shows that form > I, ")

B2m (x)

= ( -l)m-1

N

-.,

(2rr )-111

B2m+ I (.\:·) -_ (-I )m-1

')

"'""'cos ~rrnx • ~ n-111

11=1

L sm (2rr)-"'+l ")

N

-

•

11=1

_rrnx n-m+l

= 0 for all m >

(3.54)

1.)

I

oc

{(s)

")

---:.,=---~

.,

for 0 ~ x < I. (This shows again that B2m+l The Riemann ::.era function is the function

(3.53)

= "'""'---;. ~/l"

(3.55)

11=1

a meromorphic function defined on the complex plane. We will be concerned here only with its values at positive integers. We have seen formulas for {(2) and {(4) in (3.39) and (3.44 ). More generally we have the following.

Theorem 3.6.4. For e\'el')' positi\•e imeger m the \'Cdue of {(2m) is a rational multiple of rr 2"' gi\•en by the formula 2111 B., (-I (2rr) r (? ) _ _ _)"'-I __ ___ -_m ~ -Ill 2(2m )!

(3.56)

Exercise 3.6.5. Show that (3.57)

The result of Theorem 3.6.4 was proved by Euler. The corresponding question about {(2m+ I) was left open and has turned out to be very difficult. Not much is known about the values of the zeta function at odd integers. In 1978 R. Apery showed that {(3) is irrational. Very recently (2000) T. Rivoal has shown that among the values {{3), {(5), ... , {(2n + 1) at least log(n)/3 must be irrational, and that among the nine values {(5), {(7). ... , {(21) at least one is irrational. W. Zudilin has improved this to show that at least one among the four numbers {(5), {(7), {(9) and {(II) is irrational. These methods do not show which of them is irrational.

3.7 sinxjx The function sin x fx and the related integral Si (x) defined as Si(x)

=

lo

0

x sin t

-dt t

appear in several problems. The function sinxjx is even while Si(x), called the sine integral, is odd. Their graphs are shown in Figure 6.

FOURIER SERIES

68

:::::;;;;>"'""'=::::::::---.~:::::;;::;;-=........:=·.---==-

-7T

-27T

1T

21T

( a) sin(x) .l

-27T (b) Si(x)

FIGURE 6

Exercise 3.7.1. Show that the integral smr

dt

t

is divergent. (See the discussion in Exercise 2.2.8.)

Exercise 3.7.2. Show that the integral 00

1 o

sin r

--dr t

=

lim A-co

InA --dr sin r o

(3.58)

t

is convergent. (Split the integral into two parts

The first one is clearly finite. Integrate the second by parts. Alternately. represent the integral as an infinite sum with terms (11+1)7T

1

sinrjrdr.

111T

These tem1s alternate in sign and decrease in absolute value.) Thus we have an example of a function for which the Riemann integral (in the .. improper.. sense (3.58)) is finite but which is not Lebesgue integrable over JR. The main result of this section is the fonnula 00

1 0

sin r rr -dt=-. t 2

(3.59)

There are several ways to prove this. We give two proofs related to our discussion of Fourier series. In our discussion of the Dirichlet kernel (Proposition 2.2.2) we have seen that

"sin

Ino

((n

+ 1/2)1)

. sm (t /2)

dt =

T(

(3.60)

69

ODDS AND ENDS

for every positive integer n. Let I ----sin (1/2>' g(t) = { 1/2 0

0< 1

1

:5 rr,

= 0.

This is a continuous function (use the L'Hopital rule). Hence by the Riemann-Lebesgue Lemma (see Exercises 2.3.4 and 2.3.5) lim [" g(1) sin ((n

11-oo

lo

+ 1/2)1) d1

(3.61)

= 0.

From (3.60) and (3.61) it follows that . [," sin ((n IliD

11-00 0

Changing variables by putting x = (n .

+ 1/2)1) d1 = -. rr

+ 1/2)1, we get

[,(11+1/2)rr

hm

(3.62)

2

1

sinx

--dx

11-"'00 0

X

rr

= -. 2

(3.63)

This is the formula (3.59).

Exercise 3. 7.3. The integral I in (3.59) can be written as

L

oo 1(11+ I )rr /2

I=

sin 1 -d1.

11=0 111T /2

When n =2m, put 1 = mrr

+ x, to get

(1l+l>rrf 2

1

sin1

-d1=(-l)"'

11TT/2

When n

= 2m -

1

[,"'2 smx 0

1

dx.

mT! +X

I, put 1 = mrr - x to get 11 +l)rr/ 2

1
0. Show that for alln, lc,l 0 for alln.

ODDS AND ENDS

73

Exercise 3.9.4. Let f be a convex function on (0, 2rr ). Show that a, :=: 0 for n = I. 2, .... A doubly infinite sequence {c,} of complex numbers is said to be a positive definite sequence if (3.64) for all positive integers N. and for all complex numbers .:o, .: 1, lent to saying that for every N theN x N matrix

co

••• ,

.:N _ 1• This is equiva-

C-(N-1)

Cl

CO

C-1

co CN-l

(3.65)

co

CN-2

is positive (semi) definite. (This matrix has a special form; each of its diagonals is constant. Such a matrix is called a Toeplitz matrix.) Clearly, if this matrix is positive definite, then

c_, = c,. Exercise 3.9.5. Let (i)

Iff (t) >

f

co> 0,

E C(T) and let

lc,l


0.

O~r.s~N-l

This implies that the Fejer sums

aN(f; t)

L

N-l

=

(

I ') c,e i N 11

1-

Ill

(3.66)

11=-(N-J)

are nonnegative. Hence, by Fejer's Theorem (2.2.5), /(1) > 0. This statement is a part of a general theorem called Herglot.: 's Theorem which says that every positive definite sequence is the sequence of Fourier coefficients associated with a positive measure 11; i.e.,

c,

=

i

TT

e -ill I dp,(t).

(3.67)

-TT

(This can be proved along the lines of Exercise 3.9.5. Given c,,let /N(t) be the sequence of functions defined by the expression on the right-hand side of (3.66). Then {fN} is a family of nonnegative continuous functions with the property 1 2rr

-

Jj( /N(t) dt =co. -rr

14

FOURIER SERIES

This is a kind of compactness condition from which we can derive (3.67) using standard theorems of Functiona1 Analysis.)

Exercise 3.9.6. A linear operator A on the space C(T) is called positi\le if it maps nonnegative functions to nonnegative functions. (The elements of C(T) are complex-valued continuous functions on T). Let go, g1, g2 be the three functions

go(t)

= I,

g 1(r) = sin t,

Koro"kin 's Theorem is the statement: If {A,} is a sequence of positive linear operators on C(T) such that as n ~ oo, A,gi

~

gi

j = 0, 1, 2,

uniformly, for

then

A,f

~

f

f

uniformly for all

E

C(T).

A proof of this theorem is sketched below. Fill in the details. (i) It is enough to prove the assertion for all real-valued functions

f

(ii) Fix s E T and let g5 (1) = sin2 [(t - s)/2]. Note that g5 (t) = go(t) - g, (s)gl (t) -

g2(s)g2(1). (iii) Let M = sup Ifl. Let e > 0, and let D be such that 1/(1) - f(s)l < e whenever It- sl 4D 2 /rr 2 , whenever It- sl >D. Thus

.,

1/(1) - f(s)l

Mrr-

< e + 2D2 gs(t).

(iv) Put K = Mrr 2 /2D 2 , a constant that depends on written as

f

and e. The last inequality can be

an inequality between functions (in the pointwise sense). Since the operators A, are linear and order preserving, this gives

I(A,f)(t)- f(s)(A,go)(t)l < e(A,go)(t) + K(A,g5 )(1) for all

(3.68)

r.

(v) Choose t =sin (3.68) and let n ~ oo. Note (A,go)(s) ~ I, and by the observation in (ii), (A,gs)(s) ~ 0. The convergence is uniform overs. So from (3.68) we see that (A,f)(s) ~ f(s) unifonnly.

Exercise 3.9.7. Use Korovkin's Theorem to give another proof ofFejcr's Theorem. (Hint: The Fcjcr sums A,f = a,(j) define positive operators on C(T).)

ODDS AND ENDS

75

Exercise 3.9.8. There is a Korovkin Theorem for the space C[a, bj where [a. b) is any interval. The three functions go. g 1. g2 now are g,(r) = 1, g 1(/) = r. g2 (I) = r 2. Fonnulate and prove the assertion in this case.

3.10 A historical digression The series (3.39) is a famous one. The mathematician brothers Jakob and Johann Bernoulli had discovered in 1689 a simple argument to show that the harmonic series L I1n is divergent. This was contrary to the belief, held earlier. that a series L X 11 with positive tern1s decreasing to zero converges. Among other series they tried to sum was the series L II n'2 They could see that this series is convergent to a sum less than 2 but did not succeed in finding the sum. This problem became famous as the ··Basel Problem.. after the Swiss town Basel where the Bernoullis lived. Solving the Basel problem was one of the early triumphs of L. Euler (1707-1783) who went on to become one of the most prolific and versatile mathematicians of all time. Euler was a student of Jakob and Johann. and a friend of Johann's sons Daniel and Nicolaus (all mathematicians-there were more in the family). It is worthwhile to recall. very briefly. some of the stages in Euler's attack on this problem as they are illustrative of a great scientist's working... . In 1731 Euler proved the fornmla ., ""I {(2) = (log2)- + 2"'"' --.,. ~ 211 nll=l

using calculus and infinite series expansions of log( I -

x).

In this one can substitute

1

00

Joa2 o = "'"'-. ~ n211 ll=l

The factors 211 occurring here make the series converge very rapidly. Using this he found that {(2) =

1.644934 ....

This number did not look familiar. More calculations of this kind followed. The proof of (3.39), however, came through another route: Newton's theorem on roots of polynomials generalized to ..infinite degree polynomials." Let p(x) be a polynomial of degree n and suppose p(O) = I. Suppose we know the n numbers a 1• a 2 , ••• , a 11 are roots of p. Then we can write p(x)

=

(1 -_::__) (1- _::__) ···(1 -_::__). a1

a2

(3.69)

a11

Euler considered the real function •

smx f(x) = .r

")

x= I - -3!

5

x +· ·· . 5!

(3.70)

FOURIER SERIES

76 noticed f(O) = I, and f(x) = 0 precisely when x = mr, n he factored (3.70) as

f (X) = =

[(

E

Z. In analogy with (3.69)

1- ~ ) (1 + ; ) ] [ ( 1 - ~: ) (I + ~: ) ] ···

( .,)( ") r-

1--rr2

r-

1-~

4rr-

...

He then multiplied out this infinite product to w1ite I

f(x) = 1 - ( ? rr-

4 I I ) ., + -, + -., + · · · x- + (· · ·)x + · · · . 4rr9rr-

(3.71)

Now comparing coefficients of x 2 in (3.70) and (3.71) he saw that I 1 I - =?"'"" --:;. 3! rr- ~ n-

This is the remarkable fonnula (3.39). The answer confinned the numerical calculations made earlier by Euler and others. Several questions arise when one sees such a calculation. Can one do to an infinite series whatever one does to a polynomial? One knows all the real roots off (x ), but are there other roots? What is the meaning of an infinite product? The function g(.r) =ex f(x) has the same roots as f(x). Can both be represented by the same fonnula? Euler certainly was aware of all these problems and spent several years resolving them. This led to a more rigorous proof of the product expansion (3.33). These days infinite product expansions for entire functions (and meromorphic functions) are routinely used in analysis. (See the Weierstrass Fact01isarion Theorem and the Mittag-Leffler Theorem in L. V. Ahlfors, Complex Analysis.) Euler found great delight in coaxing out of equation (3.71) more and more complicated sums. The connection with Bernoulli numbers (introduced by Jakob Bernoulli in his famous work Ars Conjectandi) became apparent and he proved a general fonnula for {(2k) in tenns of these numbers. You may enjoy reading two books by W. Dunham. Chapters 8 and 9 of his book Journey Tluvugh Genius are titled The Bemoullis and the Harmonic Series, and The Extraordinary Sums of Leonhard Euler. The other book Euler: The Master of Us All is a short elementary account of some of Euler's work. Chapter 3 of A. Weil. Number Theory: An Approach Tluvugh His tot)' .from Hammurapi to Legendre is devoted to Euler. This book by a great twentieth century mathematician is an excellent historical introduction to a substantial part of mathematics. The proof of (3.39) that we have given is one of several known now. Like Euler's proof this uses a general method that leads to several other formulas. A more special proof given by Euler is outlined in the following exercise:

ODDS AND ENDS

77

Exercise 3.10.1. (i) Show that

[,.t

1 -(sin- 1 x) 2 =

2

sin- 1 t

J1 -

o

t2

dt.

(ii) Use the integral

to expand sin- 1 x in a power series. Substitute this in the integral in (i) and integrate term by term. Use the recurrence relation 1

11

2

+ 1 [,' t dt = - [ , --;::===::;;:: o J1 - t 2 n + 2 o J1 t +

11

ll

dt t2

starting with { I --;::=t::::::;;: 2

lo J1- r

dt

= I.

(iii) The result of this calculation is a power series expansion for the function ~(sin -I x )2 . Put x = 1 in this fonnula to obtain the sum (3.38).

Exercise 3.10.2. Show that oo

L

11=1

I -ll2

co (n - I )2

= 3 L --!11=1

[Hint: Use the power series for (sin- 1 x) 2 .] One of the reasons for interest in the zeta function is the intimate connection it has with prime numbers.

Exercise 3.1 0.3. For s > 1. let col

{(s)

= ~-. ~ ns n=l

Show that {(s) =

n p

I 1 - p-S •

where, in the product on the right. p varies over all prime numbers. Among other things this shows that there are infinitely many prime numbers. This fommla was discovered by Euler.

78

FOURIER SERIES

Let us demonstrate an interesting application of it that links prime factors, probability, and the number rr. A naive and intuitive idea of probability is adequate for our discussion. A natural number picked ..at random" is as likely to be even as odd. We express this by saying that the probability of a natural number being even (or odd) is 112. In the same way, the probability of a natural number picked at random being a multiple of k is 1I k (as such multiples occur at jumps of length k in the sequence of natural numbers). Pick up a natural number n at random and factor it into primes. What is the probability that no prime is repeated in this factoring? This happens if n is not a multiple of p 2 for any prime p. As we have seen, the probability of this for any given prime p is 1 - 1I p 2 • The probability of all these ..events" happening simultaneously is the product of these individual probabilities, i.e., nr(l - 11 p 2). We have seen that this product is equal to

1

6

{(2)

rr-

- - - -., . Exercise 3.10.4. Think of a plausible argument that shows that the probability that two natural numbers picked at random are coprime is 61rr 2 . Generalise this statement to k numbers. The fonnulas of Viete and Wallis (Exercises 3.7.7 and 3.5.3) were discovered in 1593 and 1655, respectively. Thus they preceded the work of Euler ( 1707-1783). Of course, the arguments ofViete and Wallis were different from the ones given here. Since both fonnulas give expressions for 2lrr. one may wonder whether there is a single expression that unites them. There is one, and it was found quite recently. The curious reader should see the paper T. Osler, The union of Vieta's and Wallis's product for rr, American Mathematical Monthly. Volume 106. October, 1999, pp. 774--776. A final tidbit. The symbol rr seems to have been first used by William Jones in 1706. For thirty years it was not used again till Euler used it in a treatise in 1736. It won general acceptance when Euler used it again in his famous book Introduction to the Analysis of the Infinite published in 1748. There are a few books devoted exclusively to the number rr. A recent one is J. Arndt and C. Haenel, rr Unleashed, Springer, 2001.

Convergence in L2 and L1

In Chapter 2 we saw that the Fourier series of a continuous function on T may not converge at every point. But under weaker notions of convergence (like Cesaro convergence) the series does converge. Another notion of convergence is that of convergence in the mean square or L2 convergence. However, this notion is in some sense even more natural than pointwise convergence for Fourier series. We will now study this and the related L 1 convergence. We assume that the reader is familiar with basic properties of Hilbert space and the spaces L 1 and L 2 • Some of them are quickly recalled below in the form of exercises.

4.1 L2 convergence of Fourier series Let 1t be a Hilbert space with inner product (... ) and norm II interest to us are: 1. The space /2 of all sequences of complex numbers x

II· Special examples of

= (x1. x2 •.. .) such that

The inner product in this space is defined as 00

L X11Y11

(x. y) =

11=1

and the associated norm is co

llxll2 =

(

L lx11l

) 1/2

2

11=1

79

FOURIER SERIES

80

We need also the space of doubly infinite sequences {x, l~-oo satisfying c-:

L

lx,l2 < oo.

11=-X

We denote this space by /2(2). It is easy to see that /2 and /2(2) are isomorphic Hilbert spaces. 2. For any bounded interval I on the real line, the space L2(1) consists of all Lebesgue 2 measurable functions f on I such that 1 1!1 < oo. The inner product on this space is defined as

J

If -

(f. g)=-

f(x)g(x)d x,

Ill

I

where Ill denotes the length of /. The associated norm is

II !112 = (The factor

1 {

2 ( 1/1 lr lf(x)l dx

) 1/2

1/111 is inserted to make some calculations simpler.)

3. The space L2(IR) consists of all Lebesgue measurable functions on the real line IR such that fi?.I/1 2 < oo. Here the inner product and the norm are given as

(f. g) =

{ f(x)g(x) dx.

liP.

1/2

11!112 = (£.1f(x)l

2

dx)

Of course. when we talk of measurable functions. we identify functions that are equal almost everywhere. The space L2[-rr. rr] can be identified with L2(T). We are interested also in the space L 1(/) which consists of all integrable functions on /,i.e., L1(l)

=

{t: f lf(x)ldx

When I is a bounded interval. we define. for

1

11!111=Ill

f

E


l dx. The spaces L 1(/) and L 1(!R} arc Banach spaces with this nonn. Neither of them is a Hilbert space. (You can prove this by showing that the nonn does not satisfy the "parallelogram law" which a Hilbert space nonn must satisfy).

CONVERGENCE IN L2 AND L1

81

Exercise 4.1.1. Show that the nonn and the inner product in any Hilbert space satisfy the Cauchy-Schwar~

inequality:

l(x. y)l

~

llxll IIYII·

Exercise 4.1.2. Use this inequality to show that for every bounded interval/,

Show that this inclusion is proper.

Exercise 4.1.3. Show that neither of L2(IR) and L 1(IR) is contained in the other. Exercise 4.1..1. Show that continuous functions are a dense subset of L2 and of L l· Functions of class C 00 also are dense in L2 and in L l· Recall that a sequence {e,} in a Hilbert space 1-l is called orthogonal if (e,, e111 ) = 0 for

n f= m, and orthonormal if in addition lie, II= I. Given any vector x in 1-llet (x, e,) =a,. The numbers a, are called the Fourier coefficients of x with respect to the orthonormal system {e, }.

Exercise -1.1.5. Show that in the Hilbert space L2(T) the sequence e,, defined as e,(x) = ei"x.

n E Z.

is an orthonormal sequence. Note that here the Fourier coefficients are

1 (f, e,) = 2rr

iTT

..

{(x)e-"1.1; dx = f(n). A

-:r

So the general results on orthononnal sets in Hilbert spaces are applicable here.

Exercise -1.1.6. (Bessel's inequality) Prove that if {e,} is any orthononnal system in a Hilbert space 1-l and a, the Fourier coefficients of a vector x with respect to this system, then

Therefore, the sequence {a,} belongs to the space l2. For f E L2(T) this gives

L lfl

2


l 11=-00

2

CONVERGENCE IN L2 AND L1

83

Thus the Fourier series L j (n )einx converges to f in the sense of convergence with respect to the norm of L2(T). In particular. this is true for all continuous functions on T. This explains our remark at the beginning of the chapter. Exercise 4.1.11. Let (a11 } be any sequence in l2. Let 1-l be a Hilbert space with an orthonormal basis {en}. Show that there exists a vector x in 1-l whose Fourier coefficients with respect to e" are a". Hint: Show that the series L a11 e11 converges in 11. In particular. given any sequence {a 11 }~_ 00 in l2(Z) we can find a function f in L2(T)

such that j(n)

=a

11 •

Exercise 4.1.12. (Riesz-Fischer Theorem) Show that the spaces L2(T) and l2 are isomorphic: the map that sends an element f of L2(T) to the sequence consisting of its

Fourier coefficients is an isomorphism. (This is a very important fact useful in several contexts. For example. in Quantum Mechanics the equivalence between the ..Wave Mechanics" and the .. Matrix Mechanics" approaches is based on this fact. See the classic book. J. von Neumann. fv/athematical Foundations of Quantum Mechanics.) Exercise 4.1.13. (Parseval's Relations) Show that if f. g E L2(T). then

-

I

2rr

ITT

f(xfg(x) dx

=

-rr

L (X)

A

-

f(n)g(n).

n=-oo

Prove a general version of this for any Hilbert space. Exercise 4.1.14. Use the Weierstrass approximation theorem (Exercise 3.1.4) and Exercise 4.1.4 to show that the polynomial functions are dense in the Hilbert space L2[ -1. I]. Note that a polynomial is a finite linear combination of the functions x". n > 0. Obtain

an orthonormal basis of L2 f-I. I] by applying the familiar Gram-Schmidt process to the functions x". The functions you get as a result are. up to constant multiples. Po(x)

= 1.

P,(x) = x.

3 .,

1

5 3

3

-

-

P2(x) =

2x-- 2'

PJ(X) =

?X - ?X.

I d" ., II P11 (X) = - - ( . C -1) . 2"n! dx"

These are called the Legendre Polynomials. Every function in L2[ -1. I] can be expanded with respect to the basis formed by these functions. This is called the Fourier-Legendre

FOURIER SERIES

84

series. (There are several other orthonom1al systems for the space L2 that are useful in the study of differential equations of physics).

Exercise 4.1.15. Let {e,} be an orthononnal system in a Hilbert space 11. Show that the following conditions are equivalent: (i) {e,} is a basis.

(ii) if x E 11 is such that (x. e,) (iii) llx f =

= 0 for alln. then x = 0.

2::,1 (x. e,) 12 for all x

E 11.

These results are very useful in different contexts. Here are some examples.

Exercise4.1.16. Let f E C 1(T). Recall that then inj(n) = ['(n). (See Exercise 2.3.11). Use tltis and the Cauchy-Schwarz and the Bessel inequalities to show that. for 0 < N < N' < oo,

ISN(f; 0)- SN'(f; 0)1 < L

lil

l11l>N

=

1 L 1]\n)l llll>N In I

c

< Nl/2

I

II! 112-

This shows that SN(f; 0) converges to f(O) uniformly at the rate I/ N 112 as N-..:;. oo. By the same argument. show that iff E Ck(T). then SN(f; 0) converges to {(0) unifmmly at the rate I 1Nk-l/ 2.

Exercise 4.1.17. In Exercise 2.3.12 we saw that iff E Ck(T). then j(n) = 0(1/nk). Show that iff E L 1(T) and j(n) = 0(1/nk) for some integer k > 2. then fisk - 2 times continuously differentiable. Show that no more can be concluded. Hint: L j(n)inei"x converges uniformly.

Exercise 4.1.18. A sequence {x, l~-oo is called rapidly decreasing if lnlk x, converges to zero as In I -..:;. oo. for all k > 0. Construct an example of such a sequence. Show that f E C 00 (T) if and only if the sequence j(n) is rapidly decreasing. Exercise 4.1.19. Let f have the Fourier series 00

ao 2

"'""' cos nO+ b, +~(a, II=

sin nO).

I

(See (3.5) in Chapter 3). Show that

I :rr

1T

1

2

_ lfl =

co

.,

.,

2 + Ll dt < 11/lloo· -rr

(4.1)

86

FOURIER SERIES

So if j;, ~ fin the space C(T). i.e.. if the sequence j;, of continuous functions converges uniformly to f. then .f, ~ fin L 1(T) also. i.e.• IIJ,, - !II 1 ~ 0. For}: g E L 1(T) define. as before. (f

* g)(t) = frr

f(t - x)g(x) dx.

-TT

A routine argument using Fubini's Theorem shows that f *g is well defined for almost all t. and is in L 1(T). It is perhaps worthwhile to go through this argument in detail. though it is somewhat dull. Let I. J be any two (finite or infinite) intervals and let I x ./ be their product. The theorems ofFubini and Tonnelli relate the double integral of a measurable function f(x. y) with respect to the product measure on I x J to iterated one-variable integrals. Fubini's Theorem. Let f E L1 (I x J), i.e., let {{

JlrxJ

lf(x • .v>l dx dy < oo.

Then for almost all x E I.

i i.e., the jimction J:r:(Y) the jimction

lf(x. y)j dy < oo.

= f(x. y) is an integrable function of y for almost all x.

i

Furthe1;

f(x. y)dy

is an integrable fimction of x, and

ffxl

f(x, y)dxdy =

f [i

l

f(x. y)dy dx.

A corresponding statement with the roles ofx andy interchanged is true.

Tonelli's Theorem. Let

f

be a IWilllegath•e measurable function on I x J. If one ~f the

integrals

f [i i [!

f(x,y)dy]dx, f(x. )') dx] dy

is finite, then so is the otlte1; and f E L 1(I x J).

87

CONVERGENCE IN L2 AND L1 Hence (by Fubini's Tlteorem)

flxJ

f(.r,y)d.rdy =

1[.l

f(x.y)dy]d.r

= .£[1t<x.y)dx]dy. (Fubini's Theorem says that if f(x, y) is an integrable function on I x J. then its integral can be evaluated in either of the three ways-as an integral with respect to the area measure or as an iterated integral in two different ways. Tonelli's Theorem says that iff is nonnegative. then even the hypothesis of integrability is unnecessary. The three integrals are either all finite or all infinite.) Let us return to convolutions now. Theorem 4.2.1. Let f, g, E L l(T). Then the integral defining 0 there exists a continuous function g such that II f - g II 1 < e. If F, denotes the Fejcr kernel, defined by (2.8), then by Theorem 2.2.5 and the inequality (4.1) there exists N such that II g - g * F,, II 1 < £ for alln > N. Hence, we have,

II f

-

f * F, II I

+ II g - g * F,, II I + II g * F, < 2£ + II (g - !> * F,, II I . II f


N.

•

Corollary 4.2.3. (The Riemann-Lebesgue Lemma) Iff E L 1(T), then A

lim f(n) lnl-oo

= 0.

Proof Use Theorem 4.2.2 and the idea of the proof of Theorem 2.3.1.

•

Let co denote the space of sequences converging to 0. The Riemann-Lebesgue Lemma says that the map f ~ {j (n)} is a map from L 1(T) into co. From results on continuous functions obtained earlier, we know that this map is one-to-one. This map has a particularly pleasing behaviour towards convolutions:

Theorem 4.2.4. Let f and g be in L 1(T). Then -

A

(f*g)(n) =2rrf(n)g(n) fora/ln.

CONVERGENCE IN L2 AND L1

89

Pmof Once again. using Fubini 's Theorem we can write

(f- * g)(n) = - 1 2:rr

I

(f

* g)(t)e-

.

1111

dt

dt d.t 2~ 11 rcr= 2~ I I f(t- .r)e_;,,, __,,g(x)e-;"·' dt dx

=

.t)g(.t)e_;.,

1 = 2:rr

I

· I

f (t)e _,, d t

g (x )e

·

_,,x d x

= 2:rr j(n)g(n). (All integrals above are over the interval [-:rr, :rr ].)

Remark. Some books define f

•

* g with a different normalisation as

1 0 for alln.

Proposition 4.2.8. Suppose the sequence {a, }~ 0 is conl'ex and bounded. Then (i) {a,} is monotonically decreasing and convergent,

(ii) limn D.a,

= 0,

(iii) L~o(n + l)D. 2a, = ao -lima,. Proof If D.a, > 0 for any n, then by the given convexity, D.a, > D.a, > 0 for allm > n. But then {a,} cannot be bounded. So D.a, < 0 for alln. This proves (i). Note that II

L

D.ak =a,+ I

-

ao.

k=O

Thus the series L( -D.ak) is convergent and its terms are positive and decreasing. This proves (ii). We have N

L Ofor al/n,

(ii) a, =a_,, (iii) a, ~ 0 as n ____:;. oo, (iv) {a, }~ 0 is com'e.Y.

Then there exists a nonnegative fimction f E L 1( T) such that a, =

f (n) for a/ln.

Proof Let F, be the Fejer kernel and 00

[(t) = 2rr

L 0 since all the terms involved are positive. For each integer k. we have 00

i =

2rr

L