Notas de Aula - Lectures on Fourier Series - S. Kesavan

7/29/2019 Notas de Aula - Lectures on Fourier Series - S. Kesavan

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Lectures On Fourier Series

S. Kesavan

Institute of Mathematical SciencesChennai-600 113, INDIA

Third Annual Foundational School - Part IDecember 430, 2006


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Contents

1 Introduction 3

2 Orthonormal Sets 6

3 Variations on the Theme 11

4 The Riemann-Lebesgue Lemma 12

5 The Dirichlet, Fourier and Fejer Kernels 15

6 Fourier Series of Continuous Functions 237 Fejers Theorem 27

8 Regularity 30

9 Pointwise Convergence 39

10 Termwise Integration 44

11 Termwise Differentiation 47

References 52

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1 Introduction

With the invention of calculus by Newton (16421727) and Leibnitz (16461716), there was a surge of activity in various topics of mathematical physics,notably in the study of boundary value problems associated to vibrations ofstrings stretched between points and vibration of bars or columns of air asso-ciated with mathematical theories of musical vibrations. Early contributorsto the theory of vibrating strings include B. Taylor (16851731), D. Bernoulli(17001782), L. Euler (17071783) and dAlembert (17171783).

By the middle of the eighteenth century, dAlembert, Bernoulli and Eulerhad advanced the theory of vibrating strings to the stage where the partialdifferential eqaution (now called the wave equation)

2y

t2= a2

2y

x2

was known and a solution of the boundary value problem had been foundfrom the general solution of that equation. The concept of fundamentalmodes of vibration led them to the notion of superposition of solutions andBernoulli proposed a solution of the form

y(x, t) =

n=1

bn sinnx

c

cosnat

c

(1.1)

(where c is the length of the string). The initial position of the string f(x)will then be represented by

f(x) =n=1

bn sinnx

c(0 x c). (1.2)

Later, Euler gave the formulas for the coefficients bn.But the general concept of a function had not been clarified and a lengthy

controversy took place over the question of representing arbitrary functions

on a finite interval by a series of sine functions.The French mathematician dAlembert gave an elegant solution in the

formy(x, t) = v(at + x) v(at x), (1.3)

and he believed that he had solved the problem completely.

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At that time, the word function had a very restricted meaning and was

understood as something given by an analytic expression. Euler thought thatthe initial position of a plucked string need not always be a function, butsome form where different parts could be expressed by different functions. Inother words, function and graph meant different things. For every function,we can draw its graph but every graph that can be drawn need not comefrom a function.

Euler strongly objected to Bernoullis claim that every solution to theproblem of a plucked string could be represented in the form (1.1), on twocounts. First of all the right-hand side of (1.1) was a periodic function, whilethe left-hand side was arbitrary. Further, the right-hand side of (1.2) was

an analytic formula and hence a function, while the left-hand side, f, couldbe any graph. So Euler believed that dAlemberts solution was valid whenf was any graph while Bernoullis solution was applicable only to a veryrestricted class of functions.

J. B. Fourier (17681830) presented many instructive examples of expan-sions of functions in trigonometric series in connection with boundary valueproblems associated to the conduction of heat. His book Theorie Analytiquede la Chaleur (1822) is a classic. Fourier never justified the convergence ofhis series expansions and this was objected to by his contemporaries La-grange, Legendre and Laplace. Fourier asserted that any periodic functioncould be written as a trigonometric series.

Dirichlet (18051859) firmly established in 1829 (nearly seventy yearsafter the controversy started), sufficient conditions on a function f so thatits Fourier series converges to its value at a point.

Since then a lot of ideas and theories grew out of a need to understandwhat these series meant. Amongst them are Cantors theory of infinite sets,the rigorous notion of a function, the theories of integration due to Riemannand Lebesgue and the theories of summability of series.

In mathematical analysis, we always try to find approximation of objectsby simpler objects. For example, we approximate real numbers by rationals.By truncating the Taylor series of a function, we approximate the function by

a polynomial. However, for a function to admit a Taylor series, it has to beinfinitely differentiable (but this is not sufficient!) in some interval but this isquite restrictive. Indeed Weierstrass approximation theorem states that anycontinuous function defined on a finite closed interval can be approximateduniformly by a polynomial.

Now, consider the set of all functions {1} {cos nt, sin nt | n N} on

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[

, ]. Given any point t

[

, ], the constant function does not vanish

at t. Further if t1 and t2 are distinct points in [, ] then we can alwaysfind a function in the above set such that it takes different values at t1 andt2. Thus, the set of all trigonometric polynomials, viz. functions of the form

f(t) = a0 +Nn=1

(an cos nt + bn sin nt) (1.4)

form an algebra (i.e. the set is closed under pointwise addition, multipli-cation and scalar multiplication) and it does not vanish at any point andseparates points. By the Stone-Weierstrass theorem (which generalizes theWeierstrass approximation theorem), every periodic continuous function on

[, ] can be approximated uniformly by trigonometric polynomials.A trigonometric polynomial of the form (1.4) can also be written in ex-

ponential form:

f(x) =Nn=N

cn exp(inx). (1.5)

It is easy to see that a0 = c0, b0 = 0 and that

an = cn + cn; bn = i(cn cn)or, equivalently,

cn =an

ibn

2 ; cn =an + ibn

2 .

If n is a non-zero integer, then exp(inx) is the derivative of exp(inx)/in,which also has period 2. Thus

1

2

exp(inx) dx =

1 if n = 0

0 if n = 0. (1.6)

Mulitplying (1.5) by exp(imx) and integrating over [, ], we get, in viewof (1.6),

cm =1

2

f(x)exp(imx) dx. (1.7)This gives us, for any positive integer m,

am =1

f(x)cos mxdx

bm =1

f(x)sin mxdx.

(1.8)

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It is common practice to replace a0 by a0/2, so that (1.8) is valid for a0 as

well.We now generalize this to define the trigonometric series

n=

exp(inx),

or, equivalently,a02

+n=1

(an cos nx + bn sin nx) .

Given a 2-periodic function f on [

, ], we define an (n

0) and bn(n 1) by (1.8) and the resulting series is called the Fourier series of thefunction f. The an and bn are called the Fourier coefficients of f.

The basic question now is when does the Fourier series of a functionconverge? If it converges, does it converge to the value of f at the givenpoint? In other words, to what extent does the Fourier series of a functionrepresent the function itself?

In the sequel we will try and answer some of these questions.To begin with, we will look at an abstract situation suggested by the

relations (1.6).

2 Orthonormal Sets

Let H be a Hilbert Space (over R or C). We denote the inner product in Hby (, ) and the norm it generates by .Definition 2.1. Let S = {ui | i I}, where I is an indexing set, be acollection of elements in H. The set S is said to be orthonormal if

ui = 1 for all i I(ui, uj) = 0 for all i

= j, i, j

I.

(2.1)

Example 2.1. The standard basis {ei}1in in Rn, where ei has 1 in theith-coordinate and zero elsewhere, is orthonormal in Rn with the usual inner-product and the euclidean norm.

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Example 2.2. Consider the space of square summable sequences 2, i.e.

2 =

x = (xi) |

i=1

|xi|2 <

.

Then again, the set of sequences {en}n=1 where en has 1 as the nth entry andzero at all other places, is orthonormal in 2.

Example 2.3. The sequence {2sin nx} is orthonormal in L2(0, 1).Example 2.4. The sequence { 1

2} { sinnx

, cosnx

| n N} is orthonormal inL2(, ).

Proposition 2.1. LetH be a separable Hilbert space. Then any orthonormalset is at most countable.

Proof. Let {xn} be a countable dense set in H. If u and v are elementsin an orthonormal set, we have u v = 2. Thus each of the ballsBn = B(xn;

2/4) can contain at most one element of an orthonormal set.

Since the {xn} form a dense set, every member ofH must belong to one suchball. Hence the result.

Henceforth, we will assume that H is a separable Hilbert space over R.

Proposition 2.2. Let

{e1, . . . , en

}be a finite orthonormal set in a Hilbert

space H. Then, for any x H, we haveni=1

|(x, ei)|2 x2. (2.2)

Proof. We have x ni=1(x, ei)ei2 0. Expanding this, we get, using thefact that the {ei} are orthonormal,

x2 +ni=1

|(x, ei)|2 2ni=1

|(x, ei)|2 0

which proves (2.2).

Theorem 2.1 (Bessels Inequality). If {ei} is an orthonormal set in aHilbert Space H, then

i

|(x, ei)|2 x2. (2.3)

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Proof. Since H is separable,

{ei

}is at most countable. The result, in the

finite case, has already been shown. If{ei} is countably infinite, then for eachn, we have that (2.2) is valid. Thus, since the result is true for all partialsums, it is true for the series as well and we get (2.3).

Corollary 2.1. If {en} is an orthonormal sequence in H, then for everyx H, (x, en) 0 as n .

It is immediate to see that the elements of an orthonormal set are lin-early independent. Further, given a set of linearly independent elements{x1, . . . , xn} in H, we can produce an orthonormal set {e1, . . . , en} such thatthe linear spans of {x1, . . . , xk} and {e1, . . . , ek} coincide for all 1 k n.Indeed, set e1 = x1/x1. Define

e2 =x2 (x2, e1)e1

x1 (x2, e1)e1 .

(Notice, by the linear independence of x1 and x2, the vector x2 (x2, e1)e1cannot be zero.) It is easy to see that e2 = 1 and that (e1, e2) = 0. Ingeneral, assume that we have constructed e1, . . . , ek such that

(i) each ei (1 i k) is a linear combination of x1, . . . , xi and(ii) the {ei} are orthonormal.

Now define

ek+1 =xk+1

ki=1(xk+1, ei)ei

xk+1 ki=1(xk+1, ei)ei

.

Thus, inductively we obtain {e1, . . . , en}. This procedure is called the Gram-Schmidt orthogonalization procedure.

Thus, ifH is a finite dimensional space, we can construct an orthonormalbasis for H.

Henceforth, we will assume that H is infinite dimensional and separable.

Definition 2.2. An orthonormal set is complete if it is maximal with respect

to the partial ordering on orthonormal sets in H induced by inclusion.Let {en} be an orthonormal sequence in an infinite dimensional (separa-

ble) Hilbert space. Let x H. Define

yn =ni=1

(x, en)en.

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Then, for m > n,

yn ym2 =mi=n+1

|(x, ei)|2

and the right-hand side tends to zero, by Bessels inequality. Thus, {yn} is aCauchy sequence and so converges in H. We define the limit to be

i=1

(x, ei)ei.

Proposition 2.3. The vector x

j=1(x, ej)ej is orthogonal to each ei.

Further,

x i=1

(x, ei)ei2 = x2 i=1

|(x, ei)|2. (2.4)

Proof. Given any n, set yn =ni=1(x, ej)ej. Then yn

i=1(x, ei)ei in H.

Now, if 1 i n, clearly (x yn, ei) = 0. Fix i, and the above relationholds for all n i. Thus,

(x j

(x, ej)ej, ei) = 0.

Now,

x yn2 = x2 +ni=1

|(x, ei)|2 2ni=1

|(x, ei)|2

= x2 ni=1

|(x, ei)|2.

Thus, passing to the limit as n , we get (2.4).Theorem 2.2. Let H be a Hilbert space and {ei} an orthonormal set in H.The following are equivalent.

(i) {ei} is complete.(ii) If x H such that (x, ei) = 0 for all i, then x = 0.

(iii) If x H, thenx =

j

(x, ej)ej. (2.5)

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(iv) (Parsevals Identity) If x

H, then

x2 =i

|(x, ei)|2. (2.6)

Proof. (i) (ii) If (x, ei) = 0 for all i and x = 0, then {ei} xx

will be

an orthonormal set contradicting the maximality of{ei}.(ii) (iii) By Proposition 2.3, x j(x, ej)ej is orthogonal to each ei andso, by (ii) we get (2.5).(iii) (iv) This is an immediate consequence of (2.4).(iv)

(i) If

{ei

}were not maximal, there exists e

H such that

e

= 1

and (e, ei) = 0 for all i. This contradicts (2.6).

Corollary 2.2. An orthonormal set {ei} in a Hilbert space H is complete,if, and only if, the linear span of the {ei}, i.e. the space of all (finite) linearcombinations of the {ei}, is dense in H.Proof. If{ei} is complete, then by (2.5) we get that each x H is such that

x = limn

ni=1

(x, ei)ei.

Thus the linear span of {ei} is dense in H. Conversely, if the linear span isdense, then, if x is orthogonal to all the ei, it follows that x = 0. Thus {ei}is complete.

Remark 2.1. In view of (2.5), a complete orthonormal set is also called anorthonormal basis.

Example 2.5. Consider the sequence {en} in 2 (cf. Example 2.2). Thissequence is complete in 2 since

x2

=

i=1 |xi|

2

=

i=1 |(x, ei)|

2

.

Example 2.6. Consider the Hilbert space L2(, ). It is known that contin-uous functions with compact support are dense in this space. Such functionsare periodic (they vanish at and ) and we saw in 1 that, by virtueof the Stone-Weierstrass theorem, they can be uniformly approximated by

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trigonometric polynomials. It then follows, a fortiori, that they can also be

approximated in the L2-norm. Thus, the trignometric polynomials are densein L2(, ) and so, by the preceding corollary, the functions

12

cos nt

,sin nt

| n N

form a complete orthonormal set in L2(, ). In particular, iff L2(, ), we have, by Parsevals identity,

|f|2 dx = 12

f(t) dt2

+

+n=1

1

f(t)cos ntdt

2+

1

f(t)sin ntdt

2

which yields1

|f|2 dx = a20

2+

n=1

|an|2 + |bn|2 (2.7)where an, bn are the Fourier coefficients of f given by (1.8).

By analogy with the above example, if{ei} is an orthonormal basis for aseparable Hilbert space H, we say that the Fourier series of x is the seriesi=1(x, ei)ei and the quantities (x, ei) are called the Fourier coefficients.

3 Variations on the Theme

Let f be a 2-periodic function defined on [, ]. The Fourier coefficientsof f are given by the formulas (1.8). These formulas also make sense iff L1(, ). By changing the values of a function at a finite numberof points, we do not change the values of the Fourier coefficients. (Indeed,recall that the spaces Lp are only equivalence classes of functions, under the

equivalence relation given by f g if f = g a.e.; thus, it is meaningless totalk of the value of a function in Lp at a particular point). Given a functionh such that h() = h(), we can redefine it so that h() = h() and thenextend it periodically as a 2-periodic function over R. We can also declarethe function to be undefined at or . Thus we are forced to consider theFourier series of functions with jump discontinuities.

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If a function belongs to L1(0, ) then we can extend it either as an odd

function or as an even function to L1(, ). In the former case, all thecoefficients an will be zero and we get a series only involving the functionssin nt. This is called the Fourier sine series of the given function. Similarly,in the latter case, only the coefficients an will be non-zero and the resultingseries, which involves only the functions cos nt, is called the Fourier cosineseries of the function.

We can also rewrite the Fourier series of a function in the amplitude-phaseform. Indeed, let

a02

+

n=1(an cos nt + bn sin nt) (3.1)

be the Fourier series of a function. Set

an = dn cos n, bn = dn sin n.

In other words,

dn =

a2n + b2n, n = cos

1(an/dn).

Then the series (3.1) can be rewritten as

a0

2+ n

dn cos(nt

n). (3.2)

4 The Riemann-Lebesgue Lemma

We have seen earlier that if{en} were an orthonormal sequence in a Hilbertspace H, then, for any x H, we have

limn

(x, en) = 0.

In particular, if f

L2(

, ) we deduce that

limn

f(t)cos ntdt = limn

f(t)sin ntdt = 0.

This result is one of the forms of what is called the Riemann-Lebesgue lemma.We now prove a very useful generalization of this.

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Theorem 4.1 (Generalized Riemann-Lebesgue Lemma). Let

f L1(a, b), where a < b +. Let h be a bounded measurablefunction defined onR, such that

limc

1

c

c0

h(t) dt = 0. (4.1)

Then

lim

ba

f(t)h(t) dt = 0. (4.2)

Proof. We extend f by zero outside (a, b) so that we can consider f L1(R).Let us consider an interval [c, d] (0, ). Then

0

[c,d]h(t) dt =

d

c

h(t) dt =1

d

c

h(t) dt

=1

d0

h(t) dt 1

c0

h(t) dt

and, by hypothesis, both integrals tend to zero as . The result nowfollows, by linearity, to all step functions made up of characteristic functionsof intervals. However such functions are dense in L1(0, ). (Indeed contin-uous functions with compact support are dense in L1(0, ); such functionsare uniformly continuous and by partitioning the interval containing the sup-

port, we can approximate continuous functions with compact support by stepfunctions uniformly and hence, a fortiori, in L1(0, ).)

Let |h| M. Thus, iff L1(0, ), then find a step function g such that0

|f g| dx < 2M

, for a given > 0. Now,0

f(t)h(t) dt

0

|f(t) g(t)|h(t) dt +0

g(t)h(t) dt

2+

0

g(t)h(t) dt

and for large enough the second term can also be made to be less than2 . A similar argument holds for

0 f(t)h(t) dt and this completes theproof.

Corollary 4.1. If f L1(a, b) then

limn

ba

f(t)cos ntdt = limn

ba

f(t)sin ntdt = 0. (4.3)

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Proof. 1cc0

cos t dt

=1c (sin c)

1|c|1cc0

sin t dt

=1c (1 cos c)

2|c|and both tend to zero as |c| .

We now give an immediate application of this result. Given a conver-gent series

n=1 n, we know that n 0 as n . We now ask if a

trigonometric series

a02

+n=1

(an cos nt + bn sin nt)

is convergent, whether an 0 and bn 0 as n .Theorem 4.2 (Cantor-Lebesgue Theorem). If a trigonometric seriesa02

+n=1(an cos nt + bn sin nt) converges on a set E whose (Lebesgue) mea-

sure is positive, then an 0 and bn 0.Proof. Without loss of generality, we may assume that E has finite measure.

We rewrite the trigonometric series as in (3.2) where dn =

a2n + b2n andn = cos

1(an/dn). Since the series converges, it follows that for all t E,

dn cos(nt n) 0

as n . Assume that {dn} does not converge to zero. Then, there exists > 0 and a subsequence {nk} such that

dnk > 0,

for all k. Then, it follows that cos(nkt

nk)

0 as k

for all t

E.

Since E is of finite measure, it follows, from the dominated convergencetheorem, that

E

cos2(nkt nk) dt 0. (4.4)

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Now cos2(nkt

nk) =

12

[1 + cos 2(nkt

nk)]. But then E

L1(R) and so

E

cos2(nkt nk) dt =R

E(t)cos2(nkt nk) dt

= cos 2nk

R

E(t)cos2nkt dt +

+sin2nk

R

E(t)sin2nkt dt

and both the integrals on the right-hand side tend to zero by Corollary 4.1.It then follows that

E

cos2(nkt nk) dt (E)2 > 0

which contradicts (4.4). This shows that dn 0 and so an 0 and bn 0.

5 The Dirichlet, Fourier and Fejer Kernels

Let f L1(, ) with Fourier series given by

f(t) a02

+ n=1

(an cos nt + bn sin nt) (5.1)

where an, bn are the Fourier coefficients defined as in (1.8). We wish toexamine the convergence of this series. In particular, we will be interested,in the sequel, to answers (positive or negative) to the following questions.

Does the Fourier series converge at all points t [, ]? If it converges at t [, ], does it converge to f(t)?

If it converges at all t

[

, ], is the convergence uniform?

To discuss the convergence, pointwise or uniform, of the Fourier series, weneed to discuss the convergence of the sequence {sn} of partial sums. Wehave

sn(t) =a02

+nk=1

(ak cos kt + bk sin kt). (5.2)

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Equivalently, in exponential form,

sn(t) =nk=n

ck exp(ikt).

Using the expression for the ck (cf. (1.7)) we get

sn(t) =nk=n

1

2

f(x) exp(ik(t x)) dx

=1

2

f(x)Dn(t x) dx

where

Dn(t) =nk=n

exp(ikt). (5.3)

The 2-periodic function Dn(t) is called the Dirichlet Kernel.

Proposition 5.1. Let {sn} be the sequence of partial sums of the Fourierseries of f L1(, ) which is 2-periodic. Then

sn(t) =1

2

f(x)Dn(t

x) dx (5.4a)

=1

2

f(t x)Dn(x) dx (5.4b)

=1

2

f(x)Dn(x t) dx (5.4c)

=1

2

f(t + x)Dn(x) dx (5.4d)

=1

2

0

(f(t + x) + f(t x))Dn(x) dx (5.4e)

Proof. We have already established (5.4a). By a change of variable t x = ywe get

sn(t) =1

2

+t+t

f(t y)Dn(y) dy.

By the 2-periodicity, it follows that the integral does not change as long asthe length of the interval of integration is 2. This proves (5.4b). The relation

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(5.4c) follows from (5.4a) since Dn is easily seen to be an even function.

Relation (5.4d) follows from (5.4c), again by a change of variable y = x tand the fact that the integrals do not change as long as the length of theinterval is 2. Finally, we split the integral in (5.4d) as the sum of integralsover [, 0] and [0, ]. Now,0

f(t + x)Dn(x) dx =

0

f(t y)Dn(y) dy

using the change of variable y = x and the evenness of Dn. This proves(5.4e).

Proposition 5.2. Letn 0 be an integer. Then

Dn(t) =

sin(n + 12

)t

sin t2

t = 2k, k N {0}.2n + 1 t = 2k, k N {0}.

(5.5)

Further,1

2

Dn(t) dt = 1. (5.6)

Proof. When n = 2k, clearly Dn(t) = 2n+1. Assume n

= 2k, k

N

{0

}.

Then(exp(it) 1)Dn(t) = exp(i(n + 1)t) exp(int)

Multiplying both sides by exp(it/2) we immediately deduce (5.5). Therelation (5.6) immediately follows from the definition ofDn(t) (cf. (5.3)) andthe relations (1.6).

We now introduce two other functions which, together with the Dirichletkernel, will play an important role in the study of the convergence of Fourierseries.

Definition 5.1. The continuous Fourier kernel is given by

(, t) =

sin t

t, t = 0

, t = 0(5.7)

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where and t are real numbers. The associated discrete Fourier kernel is

given by

n(t) =

sin(n + 12

)t

tt = 0

n + 12

t = 0,(5.8)

where t R and n is a non-negative integer.Remark 5.1. Clearly Dn is 2-periodic. It is easy to see that

sin(n+1/2)tsin(t/2)

2n + 1 as t . Thus Dn is continuous. Similarly, it is easy to see that thecontinuous and discrete Fourier kernels are also continuous.

Definition 5.2. The Fejer kernel is defined by

Kn(t) =1

n + 1

nk=0

Dk(t). (5.9)

Proposition 5.3. Letn 0 be an integer. Then

Kn(t) =1

n + 1

1 cos(n + 1)t1 cos t (5.10)

1

2

Kn(t) dt = 1. (5.11)

Further, Kn 0 and if 0 < |t| , we have

Kn(t) 2(n + 1)(1 cos ) . (5.12)

Proof. As before, observe that

(n + 1)Kn(t)

exp(it) 1exp(it) 1 = exp(it) 1m

k=0

exp

i(k + 1)t

exp(ikt)

= 2 expi(n + 1)t expi(n + 1)t

from which we deduce (5.10). The relation (5.11) follows directly from thedefinition (cf. (5.9)) and the relation (5.6). That Kn is non-negative followsimmediately from (5.10). So does relation (5.12).

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We now derive some estimates for integrals of the Dirichlet and Fourier

kernels. First we need a technical result.

Lemma 5.1. Let{Ak} be a sequence of real numbers such that A2k1 > 0and A2k < 0 and such that |Ak+1| < |Ak| for all k N. Then, for everyk N, we have

0 < A1 + . . . + Ak < A1. (5.13)

Proof. Let k be odd. Then

A1 + (A2 + A3) + (A4 + A5) + . . . + (Ak1 + Ak)

is such that each term in parentheses is negative. Thus the sum is less thanA1. Again

(A1 + A2) + (A3 + A4) + . . . + (Ak2 + Ak1) + Ak

is such that each term in parentheses is greater than 0 and Ak > 0. Thusthe sum is greater than 0. This proves (5.13) in the case k is odd. The proofwhen k is even is similar.

Proposition 5.4. Let0 a < b . Let n 0 be an integer. Then

ba

sin(n + 1/2)t

sin(t/2) dt 4. (5.14)

Proof. Let

Ak =

k/(n+1/2)(k1)/(n+1/2)

sin(n + 1/2)t

sin(t/2)dt, 1 k n + 1.

Then in each such interval, the numerator of the integrand varies like sin tbetween (k1) and k. On the other hand sin(t/2) is positive and increases.Thus clearly A1 > 0 and Ak alternates in sign and decreases in absolute value.

Now let a (k1)n+1/2 , kn+1/2 or a nn+1/2 , .If a is in the interior of any of these intervals (or, if a

nn+1/2

,

, when

k = n + 1), we have a(k1)/(n+1/2)

sin(n + 1/2)t

sin(t/2)dt

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is either greater than 0 (because the integrand is greater than 0, when k is

odd) or is less than 0 (k even) anda(k1)/(n+1/2)

sin(n + 1/2)t

sin(t/2)dt

< |Ak|.Thusa

0

sin(n + 1/2)t

sin(t/2)dt = A1 + . . . + Ak1 +

a(k1)/(n+1/2)

sin(n + 1/2)t

sin(t/2)dt

and it follows that

0 0.

Proof. Define

Ak =

k/(k1)/

sin t

tdt =

k(k1)

sin t

tdt.

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Then, again,

{Ak

}has alternating signs and decreases in absolute value. As

in the previous lemma, we get

0 0, the functions

(t) = f(t) sin(t/2)g(t)[r,](t)

(t) = f(t) cos(t/2)g(t)[r,](t)

where g(t) = 1

sin(t/2)

or 1

t/2

, are integrable and the result follows, once again,

from the Riemann-Lebesgue lemma (Corollary 4.1).

Theorem 5.1 (The localisation principle). Let f L1(, ). Let 0 0, we have f g in (t r, t + r) it followsfrom the above theorem that the Fourier series of f will converge at t if, and

only if, the Fourier series ofg converges at t and in this case the sums of theFourier series are the same. Thus the behaviour of the Fourier series at apoint t depends only on the values of the function in a neighbourhood of t.

This is in strong contrast with the behaviour of power series. If two powerseries coincide in an open interval, then they are identical throughout theircommon domain of convergence.

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6 Fourier Series of Continuous Functions

A basic question that can be asked is the following: does the Fourier seriesof a continuous 2-periodic function, f, converge to f(t) at every point t [, ]?

Unfortunately, the answer is No! and we will study this now.

Proposition 6.1. We have

limn

|Dn(t)| dt = +. (6.1)

Proof. For t R, we have | sin t| |t| and so

|Dn(t)| dt 40

| sin(n + 12

)t|t

dt

= 4

(n+ 12)

0

| sin t|t

dt

> 4nk=1

k(k1)

| sin t|t

dt

> 4n

k=1

k

(k1)

| sin t|k

dt

=8

nk=1

1

k

from which (6.1) follows immediately.

Proposition 6.2. LetV = Cper[, ], the space of continuous 2-periodicfunctions with the usual sup-norm (denoted ) and define n : V Rby

n(f) = sn(f)(0)

where sn(f) is the nth-partial sum of the Fourier series of f. Then n is a

continuous linear functional on V and

n = 12

|Dn(t)| dt. (6.2)

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Proof. On one hand,

n(f) =1

2

f(t)Dn(t) dt

(cf. (5.4d)). Thus,

|n(f)| f 12

|Dn(t)| dt

and so

n

1

2

|Dn(t)

|dt.

Now, let En = {t [, ] | Dn(t) 0}. Define

fm(t) =1 m d(t, En)1 + m d(t, En)

where d(t, A) = inf {|t s| | s A} is the distance of t from a set A.Since d(t, A) is a continuous function (in fact, |d(t, A) d(s, A)| |t s|)fm Cper[, ], (it is periodic since Dn is even and so En is a symmetricset about the origin). Also fm 1 and fm(t) 1 if t En whilefm(t)

1 if t

Ecn. By the dominated convergence theorem, it now

follows that

n(fm) 12

|Dn(t)| dt

from which (6.2) follows.

Let us now recall a few results from topology and functional analysis.

Theorem 6.1 (Baire). If X is a complete metric space, the intersection ofevery countable collection of dense open sets of X is dense in X.

Equivalently, Baires theorem also states that a complete metric space

cannot be the countable union of nowhere dense sets.One of the important consequences of Baires theorem is the Banach-

Steinhaus theorem also known as the uniform boundedness principle.

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Theorem 6.2 (Banach-Steinhaus). Let X be a Banach space and Y a

normed linear space and {}A a collection of bounded linear transforma-tions from X into Y, where ranges over some indexing set A. Then eitherthere exists M > 0 such that

M, for all A (6.3)

or,supA

x = (6.4)

for all x belonging to some dense G-set in X.

(Recall that a G-set is a set which is the countable intersection of opensets).

Now consider X = V as defined in Proposition 6.2 and Y = R. Let A = Nand set n = n, again defined in above mentioned proposition. Since, byProposition 6.1 and 6.2, we have n as n , it follows from theBanach-Steinhaus theorem, that there exists a dense G-set (of continuous2-periodic functions) in V such that the Fourier series of all these functionsdiverge at t = 0. We could have very well dealt with any other point in theinterval [, ] in the same manner.

By another application of Baires theorem, we can strengthen this further.

Let Ex be the dense G-set of continuous 2-periodic functions in V suchthat the Fourier series of these functions diverge at x. Let {xi} be a countableset of points in [, ] and let

E = ni=1Exi V. (6.5)

Then, by Baires theorem E is also a dense G-set. (Each Exi is the countableintersection of dense open sets and so, the same is true for E). Thus for eachf E, the Fourier series of f diverges at xi for all 1 i . Define

s(f; x) = supn

|sn(f)(x)|.

Then s(f, ) is a lower semi-continuous function. Hence {x | s(f; x) = }is a G-set in (, ) for each f. If we choose the xi above so that {xi} isdense (take all rationals, for instance in (, )) then we have the followingresult.

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Proposition 6.3. The set E

V is a dense G-set such that for allf

E,

the set Qf (, ) where its Fourier series diverges, is a dense G-set in(, ).Proposition 6.4. In a complete metric space, which has no isolated points,no countable dense set can be a G.

Proof. Let E = {x1, . . . , xn, . . .} be a countable dense set. Assume E isa G. Thus E = n=1Wn, Wn open and dense. Then, by hypothesis, Wn \ni=1{xi} = Vn is also open and dense. But n=1Vn = , contradicting Bairestheorem.

Thus, there exists uncountably many 2-periodic continuous functions on[, ] whose Fourier series diverge on a dense G-set of (, ).Having answered our first general question negatively, let us now prove a

positive result.

Proposition 6.5. Let f be a 2-periodic function on [, ] which is uni-formly Lipschitz continuous, i.e. there exists K > 0 such that

|f(x) f(y) K|x y|

for all x, y. Then, the Fourier series of f converges to f on [, ].Proof. Choose 0 < r < such that

1

rr

|t/2|| sin(t/2)| dt 0 such that |x y| < implies |f(x) f(y)| < /2.

Now, choose N large enough such that, for n N, Kn(t) /4M forall < |t| . This is possible because of the estimate (5.12). Thus, sinceKn 0,

f(x t) f(x)Kn(t) dt

2Kn(t) dt = ,

again by (5.11). On the other hand,

+

f(x t) f(x)Kn(t) dt

2M 4M 2 = if n N. It now follows that for all x [, ], we have

|n(x) f(x)| < for n

N which completes the proof.

Corollary 7.1. Letf and g be two 2-periodic and continuous functions on[, ]. If they have the same Fourier series, then f g.Proof. If the two functions have the same Fourier series, then the n will bethe same for both functions and we know that n f and n g uniformlyon [, ]. Hence the result.

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Given a series n an, we say that it is Cesaro summable or (C, 1)summable to a if n a as n , where

n =s1 + . . . + sn

n,

sk being the partial sums of the series. Thus the Fourier series of a continuous2-periodic function is always Cesaro summable to the function.

Thus iff is a continuous 2-periodic function whose Fourier series is givenby

a02

+

n=1(an cos nt + bn sin nt),

then f is uniformly approximated over [, ] by the trigonometric polyno-mials

n(x) =a02

+nk=1

1 k

n + 1

ak cos kt + bk sin kt

. (7.1)

Starting from this, we can deduce Weierstrass approximation theorem. In-deed, let f C[1, 1]. Define g(t) = f(cos t), for t [, ]. Then g is2-periodic and continuous. Further, g is an even function and hence itsFourier series will only consist of cosine terms. Let the Fourier series of g begiven by

a02

+ k=1

ak cos kt.

Then

n(t) =a02

+nk=1

1 k

n + 1

ak cos kt.

Thus, given > 0, there exists N such that, for all n N,f(cos t) a0

2n

k=1

1 kn + 1

ak cos kt

<

for all t [, ]. This is the same asf(t) a02 nk=1

1 k

n + 1

ak cos

k(cos1 t)

<

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for all t

[

1, 1]. Now it only remains to show that

Pk(t) = cos(k cos1 t)

is a polynomial in t for every non-negative integer k. Indeed, P0(t) 1 andP1(t) t. Assume that Pk(t) is a polynomial in t, of degree k, for every1 k n 1, n 2. Then

cos ns = cos ns + cos(n 2)s cos(n 2)s= cos

(n 1)s + s + cos(n 1)s s cos(n 2)s

= 2 cos(n 1)s cos s cos(n 2)s= 2 cos s Pn1(cos s) Pn2(cos s).

It then follows thatPn(t) = 2tPn1(t) Pn2(t) (7.2)

and so Pn(t) is a polynomial of degree n in t.The polynomials {Pn} defined recursively via (7.2) where P0 1 and

P1(t) = t, are called the Chebychev polynomials and play an importantrole in numerical analysis, especially in numerical quadrature.

8 RegularityIn this section, we will briefly discuss various regularity assumptions to bemade on functions when discussing the pointwise convergence of Fourier se-ries.

Let f : [a, b] R be a given function which is differentiable on (a, b).Assume that

f(a+) = limta

f(t)

exists. Then f is bounded in a subinterval (a, a+) (where > 0) and so, bythe mean-value theorem, f is uniformly continuous on (a, a + ). Thus it can

be continuously extended to [a, a + ]. Consequently f(a+) = limta f(t),exists.

Again by the mean-value theorem, applied to f where f(a) = f(a+) andf(t) = f(t) for a < t a + , we have

f(t) = f(a+) + f(a + (t a))(t a)

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where 0 < < 1. Thus

limta

f(t) f(a+)t a = f

(a+). (8.1)

Notice that f(a+) is different from the right-sided derivative of f at a (if itexists), which is given by

D+f(a) = limta

f(t) f(a)t a .

We also define, f(b) = limtb f(t).

Definition 8.1. We say that a functionf : [a, b] R ispiecewise smoothif there exist a finite number of points

a = a0 < a1 < .. . < an = b

such that f is continuously differentiable in each subinterval (ak, ak+1), 0 k n 1 and f(c+) and f(c) exist at all points c [a, b] except at a,where f(a+) exists, and at b, where f(b) exists.

Let f : [a, b] R (or C) be a given function. Let P= {a = x0 < x1 0, there exists > 0 such that if E [a, b] is a measurable set with (E) < , then

E

f(x) dx < .

Proof. If |f| M on [a, b], the result is trivially true, sinceE

f dx

M(E),where (E) is the Lebesgue measure of E.

In the general case, define

fn(x) = |f(x)|, if |f(x)| nn, if |f(x)| > n.

Then fn is bounded and fn |f| pointwise. Further {fn} is an increasingsequence of non-negative functions. By the monotone convergence theorem,

limn

ba

fn dx =

ba

|f| dx < .

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Given > 0, choose N such thatba

(|f| fN) dx < /2.

Now choose > 0 such that (E) < implies thatE

fNdx < /2.

Then |E

f dx| E

|f| dx b

a(|f| fN) dx +

E

fNdx < . This completesthe proof.

Definition 8.3. A function f : [a, b] R is said to be absolutely contin-uous on [a, b] if for every > 0, there exists > 0 such that whenever wehave a finite collection of disjoint intervals {(xi, xi)}ni=1 satisfying

ni=1

(xi xi) <

we haven

i=1

|f(xi) f(xi)| < .

Clearly, any absolutely continuous function is uniformly continuous on[a, b].

Example 8.4. Any uniformly Lipschitz continuous function is absolutely con-tinuous, since

|f(xi) f(xi)| L

(xi xi) < L.

Example 8.5. Any indefinite integral of an integrable function is absolutelycontinuous by virtue of Proposition 8.4.

We will show presently that a function can be written as an indefiniteintegral if, and only if, it is absolutely continuous.

Proposition 8.5. An absolutely continuous function is of bounded variation.

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Proof. Let correspond to = 1 in the definition of absolute continuity. Let

K be the integral part of 1 + (b a)/, where [a, b] is the given interval.Given any partition Pof [a, b], we can refine it to a partition consisting ofKsets of sub-intervals each of total length less than . Thus V(P; f) K andso V(f; a, b) K.

Consequently, any absolutely continuous function is differentiable a.e.. on[a, b].

Proposition 8.6. Let f : [a, b] R be absolutely continuous. Assume thatf = 0 a.e. in [a, b]. Then f is a constant function.

Proof. Let c

(a, b). Let

E = {x (a, c) | f(x) = 0}Then (E) = c a. Let , > 0 be arbitrary. If x E, there exists asufficiently small h > 0 such that [x, x+h] [a, c] and |f(x+h)f(x)| < h.By the Vitali covering lemma, there exists a finite disjoint collection of suchintervals which cover all of E except possibly a subset of measure less than, where corresponds to in the definition of absolute continuity. We labelthese intervals [xk, yk], with xk increasing. Thus

y0 = a

x1 < y1

x2 < y2

. . .

yn = c = xn+1.

Then,nk=0

|xk+1 yk| < .

Now,nk=1

|f(yk) f(xk)| <

(yk xk) < (c a).

Alsonk=1 |f(xk+1) f(yk)| <

by absolute continuity. Thus

|f(c) f(a)| + (c a)or f(c) = f(a) for all c (a, b). Hence the result.

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Theorem 8.2. A function can be expressed as an indefinite integral of an

integrable function if, and only if, it is absolutely continuous. The derivativeof this function (which exists a.e.) is equal a.e. to the integrand.

Proof. If f were an indefinite integral, it is absolutely continuous (cf. Exam-ple 8.5).

Conversely, let F : [a, b] R be absolutely continuous. Then it is ofbounded variation and

F(x) = F1(x) F2(x)

where Fi(x), i = 1, 2 are monotonic increasing. Thus F = F1

F2 a.e. and

F1 0, F2 0. Thus,ba

|F| dx ba

|F1| dx +ba

|F2| dx

=

ba

F1 dx +ba

F2 dx

F1(b) F1(a) + F2(b) F2(a) < .

Thus F is integrable. Now let G(x) =

x

aF(t) dt. Then G, being an indefi-

nite integral of an integrable function, is absolutely continuous and G = F

a.e. Thus G F is absolutely continuous and (G F) = 0 a.e.. Thus, byProposition 8.6, (G F)(x) = (G F)(a) for all x [a, b]. Hence,

F(x) = F(a) +

xa

F(t) dt.

This completes the proof.

Absolutely continuous functions share many properties of continuouslydifferentiable functions. In particular, we have the following result.

Theorem 8.3 (Integration by parts). Let f, g be absolutely continuouson [a, b]. Then

ba

f(t)g(t) dt = f(b)g(b) f(a)g(a) ba

f(t)g(t) dt. (8.9)

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Proof. Consider the function f(x)g(y) on [a, b]

[a, b]. Consider the integralb

a

xa

f(x)g(y) dy dx.

It is a routine verification to check that Fubinis theorem applies and soba

xa

f(x)g(y) dydx =ba

by

f(x)g(y) dxdy.

The left-hand side gives us

ba

xa

g(y) dy

f(x) dx =ba

g(x) g(a)f(x) dx

=

ba

g(x)f(x) dx g(a)f(b) f(a)by repeated use of Theorem 8.2. The right-hand side givesba

by

f(x) dx

g(y) dy =ba

f(b) f(y)g(y) dy

= f(b)g(b) g(a) b

a

f(y)g(y) dy.

Equating the two, we deduce (8.9). This completes the proof.

9 Pointwise Convergence

In this section, we will prove the convergence theorems of Dirichlet and Jor-dan.

Proposition 9.1. Letf L1(0, ) and assume that f(0+) exists. Then

limn 1

0

f(t) sin(n +1

2)tt

dt = 12

f(0+). (9.1)

Proof. Adding and subtracting f(0+) in the integrand, we get0

f(t)sin(n + 1

2)t

tdt =

0

(f(t)f(0+))sin(n +12

)t

tdt+f(0+)

0

sin(n + 12

)t

tdt.

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The second integral on the right-hand side becomes (after a change of vari-

able), (n+ 12)

0

sin t

tdt

which converges to /2 as n . Thus,

limn

1

f(0+)

o

sin(n + 12

)t

tdt =

1

2f(0+).

Hence we need to show that the first term tends to zero as n . Let > 0be an arbitrarily small number. Choose 0 < r < such that for 0 < t r,f(t) f(0+)t f(0+)

< ( cf. (8.1)). Then,

0

(f(t) f(0+))sin(n +12

)t

tdt =

r0

(f(t) f(0+))sin(n+ 12 )tt

dt

+r

(f(t) f(0+))sin(n+ 12 )tt

dt.

The second term tends to zero, by the Riemann-Lebesgue property for n

(cf. Proposition 5.7), since f(t) f(0+) L1(0, ). Now, the first termabove can be written asr

0

f(t) f(0+)

t f(0+)

sin(n +

1

2)t dt + f(0+)

r0

sin(n +1

2)t dt.

But,

r0

f(t) f(0+)

t f(0+)

sin(n +

1

2)t dt

< r <

and r0

sin(n +1

2)t dt =

r0

sint

2cos nt dt +

r0

cost

2sin nt dt

and both these integrals tend to zero, as n , by the Riemann-Lebesguelemma (cf. Corollary 4.1). This completes the proof.

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Theorem 9.1. (Dirichlet) Let f

L1(

, ) be a piecewise smooth func-

tion. Then its Fourier series converges to 12(f(t+) + f(t)) at all pointst [, ]. In particular, if f is continuous at a point t, then its Fourierseries converges to f(t) at that point.

Proof. Recall that (cf. (5.4e))

sn(t) =1

2

0

(f(t + x) + f(t x))Dn(x) dx.

Now, by (9.1), we have

limn 1

0

f(t + x) sin(n +1

2)xx

dx = 12

f(t+).

Thus,

limn

1

2

0

f(t + x)sin(n + 1

2)x

x2

dx =1

2f(t+).

Hence, by Proposition 5.6, we get

limn

1

2

0

f(t + x)sin(n + 1

2)x

sin x2

dx =1

2f(t+).

Similarly,

limn

1

2

0

f(t x)sin(n +12

)x

sin x2

dx =1

2f(t)

from which the result follows.

Example 9.1. Consider the function

f(t) =

1, t < /20, /2 t /21, /2 < t .

This function is piecewise smooth and is odd. Thus, its Fourier series consistsonly of sine functions. Now,

bn =1

f(t)sin nt dt =2

2

sin nt dt.

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Thus,

bn =

2n , n odd0, n = 4k, k N

4n

, n = 4k + 2, k NAt t = /2, the series must thus converge to 1/2. Thus,

1

2=

2

1 1

3+

1

5 1

7+

which yields the well known Gregory series

4 = 1 1

3 +

1

5 1

7 + In order to prove the next result, we recall a version of the mean value

theorem for integrals.

Proposition 9.2. Let g : [a, b] R be continuous and let f : [a, b] Rbe non-negative and monotonic increasing. Then, there exists c [a, b] suchthat b

a

f(t)g(t) dt = f(b)

bc

g(t) dt. (9.2)

Remark 9.1. The result is false if f is not non-negative. To see this, takef(t) = g(t) = t on [1, 1]. Then the left-hand side will be11

t2 dt =2

3.

The right-hand side is

f(1)

1c

t dt =1 c2

2

and we can never have c

[

1, 1] such that the two are equal.

Remark 9.2. In the mean value theorem of differential calculus, the pointc will be in the interior of the interval. In case of mean value theorems forintegrals, this need not be necessarily the case. For instance, if f 1 on[a, b] and if g is strictly positive on that interval, then we cannot have (9.2)with c (a, b).

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We are now in a position to prove a convergence theorem for a larger class

of functions. We need a preliminary result analogous to Proposition 9.1.Proposition 9.3. Letf be of bounded variation on [0, ]. Then

limn

1

0

f(t)sin(n + 1

2)t

tdt =

1

2f(0+). (9.3)

Proof. Since f is of bounded variation, it can be written as the difference oftwo monotonic increasing functions. Hence we may assume, without loss ofgenerality, that f is monotonic increasing. Now,

0 f(t)

sin(n + 12

)t

t dt =

0(f(t) f(0+))sin(n+ 12 )t

tdt

+ f(0+)0

sin(n+ 12)t

tdt.

As already shown in the proof of Proposition 9.1, the second integral con-verges to /2. Thus, as before, it is enough to show that the first integraltends to zero as n .

Let > 0 be an arbitrarily small number. Choose 0 < r such that|f(t) f(0+)| < /4, 0 < t r. (9.4)

Then, splitting the first integral over [0, r] and [r, ], we see that the integralon [r, ] tends to zero by the Riemann-Lebesgue property for n (cf. Propo-

sition 5.7) since f(t) f(0+) is integrable over that interval. Thus, for nsufficiently large, we have 1r

(f(t) f(0+))sin(n +12

)t

tdt

< 2 .On the interval [0, r], we redefine the function f as f(0+) at t = 0, so thatthe function f(t) f(0+) remains non-negative and monotonic increasingwithout altering the value of the integral over that interval. Hence, by themean value theorem (cf. Proposition 9.2), there exists c [0, r] such that

r

0

(f(t)

f(0+))sin(n + 1

2)t

tdt = (f(r)

f(0+))

r

c

sin(n + 12

)t

tdt.

Thus, by Proposition 5.5 and by (9.4), it follows that 1r0

(f(t) f(0+))sin(n +12

)t

tdt

< 2 .This completes the proof.

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Theorem 9.2. (Jordan) Letf be a function of bounded variation on [

, ].

Then, its Fourier series at any pointt in this interval converges to 12(f(t+) +f(t)). In case f is continuous at t, then the series converges to f(t).Proof. The proof is identical to that of Theorem 9.1, except that we appealto Proposition 9.2 in place of Proposition 9.1.

Remark 9.3. Dirichlet proved his theorem on the convergence of Fourierseries in 1829. Jordans result was proved in 1881. In 1904, Fejer showedthat if f L1(, ), and if f(t+) and f(t) exist at t [, ], thenthe Fourier series is (C, 1)-summable at t to (f(t+) + f(t))/2. One of thegreatest triumphs in the history of Fourier series is the result of Carlesson

(1966) that the Fourier series of any function in L2

(, ) converges to thevalue of that function a.e..This was extended by Hunt in 1968 to all functionsin Lp(, ) for 1 < p < .

10 Termwise Integration

In general, when we have a series f(x) =n fn(x), we can integrate the

series term-by-term if the series is uniformly convergent. However, we haveseen that Fourier series (under appropriate hypotheses) converge to (f(t+) +f(t

))/2 at a discontinuity. Thus, if the function is discontinuous, uniform

convergence is ruled out and the above principle does not apply. However,Fourier series are special and enjoy special properties. Vis-a-vis integration,we have the following result.

Theorem 10.1. Letf L1(, ) be extended periodically overR and havethe Fourier series

a02

+n=1

(an cos nt + bn sin nt).

Then,(i) the series obtained by termwise integration, viz.

a02

x +n=1

ann

sin x bnn

cos x

+ C

where

C =n=1

bnn

,

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converges to x0 f(t) dt.(ii) This convergence is uniform if f L2(, ).Proof. Let

F(x) =

x0

f(t) a0

2

dt.

Since f is integrable, it follows that F is absolutely continuous. It is also2-periodic. For,

F(x + 2) =x0

f(t) a0

2

dt +

x+2x

f(t) a0

2

dt

= F(x) + f(t) a0

2 dt

since, by 2-periodicity, the integration can be done over any interval oflength 2 without affecting its value. But the last integral vanishes by thedefinition of a0. Thus F(x + 2) = F(x).

Since it is absolutely continuous, it is continuous and of bounded vaiationand so its Fourier series converges to its value at each point. Let

F(t) =c02

+n=1

(cn cos nt + dn sin nt).

We can also use integration by parts (available for absolutely continuousfunctions (cf. (8.9)) to obtain

cn =1

F(t)cos nt dt

= 1n

F(t)sin nt

1n

f(t) a02

sin nt dt.

The boundary terms vanish by periodicity. So does the integral of a02

sin nt.Thus we deduce that cn = bn/n. Similarly, dn = an/n. Thus,

F(t) =

c0

2 +

n=1

ann sin nt

bn

n cos nt

.

Evaluating this at t = 0, we get

c02

=n=1

bnn

.

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This proves (i).

If f L2(, ), then the series n a2n and n b2n are convergent, byvirtue of Parsevals identity (cf. (2.7)). Hence, by the Cauchy-Schwarz in-equality, the series

n an/n and

n bn/n are absolutely convergent. Thus,

by the Weierstrass M-test, the above series for F(t) is uniformly conver-gent.

Remark 10.1. Since we have shown that

c02

=n=1

bnn

,

it follows that for any f L1(, ), the Fourier sine coefficients bn aresuch that

n bn/n is convergent. The absolute convergence of this series,

however, is true under the additional hypothesis that f L2(, ).Remark 10.2. Iff is continuous and 2-periodic, then it is in L2(, ) andso, for any continuous 2-periodic function, the termwise integrated Fourierseries is uniformly convergent.

We now give an application of the observation made in Remark 10.1 bypresenting an example of a uniformly convergent trigonometric series whichcannot be a Fourier series.

Example 10.1. Consider the trigonometric series

n=2

sin nt

log n. (10.1)

The sequence {(log n)1} is non-negative and monotonically decreases tozero. Further,for t = 2n,

n

k=1

sin kt =cos t

2 cos(n + 1

2)t

2sin t

2

and so nk=1

sin kt

1| sin t2|

1

sin a2

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for t

[a, 2

a] for any a

(0, ). Thus, by Dirichlets test, the given series

in (10.1) converges uniformly in [a, 2 a] for any a (0, ). However, theseries

n=2

bnn

=n=2

1

n log n

is divergent and so the given series cannot be a Fourier series.

Remark 10.3. Note that in a Fourier series, it is necessary thatn bn/n

is convergent. However, there is no such condition onn an/n. Indeed,

Stromberg (1981) has shown that the series

n=2

cos ntlog n

is a Fourier series!

11 Termwise Differentiation

Consider the Fourier series of the 2-periodic extension of the function f(t) =t on [, ]. Of course, such an extension has jump discontinuities at all oddmultiples of . Nevertheless,

f(t) = 2n1

(1)n+1n

sin nt

and, by Jordans theorem, the series converges to f(t) at all points in (, )and to zero at and . The termwise derivative of the series is

2n=1

(1)n+1 cos nt

which is divergent everywhere since cos nt does not converge to zero, as n , for any t.Thus, while any Fourier series may be integrated termwise meaningfully,

extra hypotheses are needed for termwise differentiation.

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Theorem 11.1. Let f be a continuous and piecewise smooth 2-periodic

function on [, ] with Fourier seriesa02

+n=1

(an cos nt + bn sin nt).

If its derivative f is also piecewise smooth, then

f(t+) + f(t)2

=n=1

(nbn cos nt nan sin nt). (11.1)

Proof. Let f have the Fourier series expansion

c02

+n=1

(cn cos nt + dn sin nt).

Since f is 2-periodic, we have

c0 =1

f(t) dt = 0.

Again, for n 1,cn =

1

f(t)cos nt dt

= 1

f(t)cos nt

+ n

f(t)sin nt dt,

by the absolute continuity of f, which is piecewise smooth. The boundaryterms cancel out by the 2-periodicity and so we get cn = nbn. Similarly, dn =nan. The relation (11.1) follows from the Dirichlet convergence theorem(Theorem 9.1) for piecewise smooth functions.

Remark 11.1. By the Riemann-Lebesgue lemma, cn and dn tend to zero asn . Thus, nan and nbn tend to zero. More generally, the smoother thefunction, the faster its Fourier coefficients tend to zero. If f, f, f, f(k1)are all continuous and f

(k)

is piecewise smooth, then nk

an and nk

bn tendto zero as n . Also, in this case, since f(k) L2(, ), we have, byParsevals identity, that

n=1

n2ka2n < , andn=1

n2kb2n < .

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We conclude with a nice appliation of Fourier series to a problem in the

calculus of variations.One of the earliest known problems in the calculus of variations is known

as the classical isoperimetric problem and can be traced to Virgils Aeneid,which describes the problem of Dido, future queen of Carthage. In mathe-matical terms, the problem can be stated as follows.

Of all simple closed curves in the plane with a fixed length, which oneencloses the maximum area?Or, equivalently:

Of all simple closed curves in the plane enclosing a fixed area, whichone has the least length?

Given the existence of such a curve, it is possible to give a fairly elemen-tary geometric argument to identify it as the circle. However, the existenceof an optimal solution requires extra proof.

We can deal with the existence and the uniqueness in one stroke if weprove the classical isoperimetric inequality: if L is the length of a simpleclosed plane curve and if A is the enclosed area, then

L2 4A, (11.2)with equality if, and only if, the curve is a circle. For the circle, we do indeedhave equality since L = 2r and A = r2, where r is its radius. On the otherhand, for any curve of given length L, the maximum possible value for theenclosed area is L2/4, which is achieved for the circle. Thus, this establishesthe circle as the optimal solution. The uniqueness follows from the only ifpart of the proof.

Let us assume that a simple closed curve C is parametrized by the equa-tions x = x(s), y = y(s), where s is the arc length, which varies over theinterval [0, L]. We assume that the functions x(s) and y(s) verify the hy-potheses of Theorem 11.1. We reparametrize the equations using the param-eter

t =2

Ls

so that t [0, 2] and x and y are 2-periodic smooth functions in t withpiecewise smooth derivaives. Let

x(t) = a02

+n=1(an cos nt + bn sin nt)

y(t) = c02

+n=1(cn cos nt + dn sin nt).

(11.3)

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Thus,

x(t) =n=1(nbn cos nt nan sin nt)

y(t) =n=1(ndn cos nt ncn sin nt).

(11.4)

Since dx

ds

2+

dy

ds

2= 1,

we get that dx

dt

2+

dy

dt

2=

L

2

2.

Then, by Parsevals identity, it follows that

2

L

2

2=

1

20

dx

dt

2+

dy

dt

2dt =

n=1

n2(a2n + b2n + c

2n + d

2n).

(11.5)Now, the enclosed area A is given by

A =

C

x dy =

20

x(t)dy

dt(t) dt =

n=1

n(andn bncn). (11.6)

From (11.5) and (11.6), we get

L2 4A = 22 n=1

(nan dn)2 + (nbn + cn)2 + (n2 1)(c2n + d2n)

which is non-negative, thus proving (11.2).

If equality occurs in (11.2), then the above relation shows that for alln 1,

nan = dn and nbn = cn.Also, it shows that, for n > 1, cn = dn = 0, and hence it follows thatan = bn = 0 as well for those n. Thus,

x(t) =

a0

2 + a1 cos t + b1 sin t

y(t) = c02

b1 cos t + a1 sin tfrom which we get

x(t) a02

2+

y(t) c02

2= a21 + b

21

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and so the curve has to be a circle, thus proving the uniqueness of the optimal

solution.The isoperimetric inequality exists in all dimensions. The n-dimensional

ball is the unique domain with given (n 1)-dimensional surface area andmaximizing the enclosed (n-dimensional) volume amongst all possible do-mains.

In three dimensions, the analogue of (11.2) reads as

S3 36V2

where S is the surface area and V is the enclosed volume.If

Rn is a bounded domain, then let us denote its n-dimensional

(Lebesgue) measure by ||n and the (n 1)-dimensional surface measure ofthe boundary (which has to be suitably defined) by ||n1. The classicalisoperimetric inequality now reads as

||n1 n1

nn ||1

1

nn

where

n =

n

2

(n2

+ 1)

is the volume of the unit ball in Rn. Equality occurs in the isoperimetric

inequality if, and only if, is a ball.

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References

[1] Bachman, G., Narici, L. and Beckenstein, E. Fourier and WaveletAnalysis, Springer.

[2] Churchill, R. V. Fourier Series and Boundary Value Problems,McGraw-Hill.

[3] Royden, H. L. Real Analysis, Macmillan.

[4] Rudin, W. Principles of Mathematical Analysis, McGraw-Hill.

[5] Rudin, W. Real and Complex Analysis, Tata McGraw-Hill.

[6] Simmons, G. F. Introduction to Topology and Modern Analysis,McGraw-Hill.

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Notas de Aula - Lectures on Fourier Series - S. Kesavan