Lectures On Fourier Series

S. Kesavan

Institute of Mathematical Sciences

Chennai-600 113, INDIA

Third Annual Foundational School - Part IDecember 4–30, 2006

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 23

7 Fejer’s Theorem 27

8 Regularity 30

9 Pointwise Convergence 39

10 Termwise Integration 44

11 Termwise Differentiation 47

References 52

2

1 Introduction

With the invention of calculus by Newton (1642–1727) and Leibnitz (1646–1716), 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 (1685–1731), D. Bernoulli(1700–1782), L. Euler (1707–1783) and d’Alembert (1717–1783).

By the middle of the eighteenth century, d’Alembert, 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 sinnπx

ccos

nπat

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 sinnπx

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 functionson a finite interval by a series of sine functions.

The French mathematician d’Alembert gave an elegant solution in theform

y(x, t) = v(at+ x) − v(at− x), (1.3)

and he believed that he had solved the problem completely.

3

At that time, the word function had a very restricted meaning and wasunderstood 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 Bernoulli’s 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) wasan analytic formula and hence a function, while the left-hand side, f , couldbe any graph. So Euler believed that d’Alembert’s solution was valid whenf was any graph while Bernoulli’s solution was applicable only to a veryrestricted class of functions.

J. B. Fourier (1768–1830) 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 (1805–1859) 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 Cantor’s 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 bya 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 ∪ cosnt, sinnt | n ∈ N on

4

[−π, π]. Given any point t ∈ [−π, π], the constant function does not vanishat t. Further if t1 and t2 are distinct points in [−π, π] 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 +N

∑

n=1

(an cosnt + bn sinnt) (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’ and‘separates points’. By the Stone-Weierstrass theorem (which generalizes theWeierstrass’ approximation theorem), every periodic continuous function on[−π, π] 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) =

N∑

n=−N

cn exp(inx). (1.5)

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

an = cn + c−n; bn = i(cn − c−n)

or, equivalently,

cn =an − ibn

2; c−n =

an + ibn2

.

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

1

2π

∫ π

−π

exp(inx) dx =

1 if n = 0

0 if n 6= 0.(1.6)

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

cm =1

2π

∫ π

−π

f(x) exp(−imx) dx. (1.7)

This gives us, for any positive integer m,

am =1

π

∫ π

−π

f(x) cosmxdx

bm =1

π

∫ π

−π

f(x) sinmxdx.

(1.8)

5

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

We now generalize this to define the trigonometric series

∞∑

n=−∞exp(inx),

or, equivalently,a0

2+

∞∑

n=1

(an cosnx+ bn sinnx) .

Given a 2π-periodic function f on [−π, π], 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 6= j, i, j ∈ I.

(2.1)

Example 2.1. The standard basis ei1≤i≤n in Rn, where ei has 1 in theith-coordinate and zero elsewhere, is orthonormal in R

n with the usual inner-product and the euclidean norm.

6

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 √

2 sin nπx is orthonormal in L2(0, 1).

Example 2.4. The sequence 1√2π ∪ sinnx√

π, cos nx√

π| n ∈ N is orthonormal in

L2(−π, π).

Proposition 2.1. Let H 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 balls

Bn = B(xn;√

2/4) can contain at most one element of an orthonormal set.Since the xn form a dense set, every member of H 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 Hilbertspace H. Then, for any x ∈ H, we have

n∑

i=1

|(x, ei)|2 ≤ ‖x‖2. (2.2)

Proof. We have ‖x−∑ni=1(x, ei)ei‖2 ≥ 0. Expanding this, we get, using the

fact that the ei are orthonormal,

‖x‖2 +

n∑

i=1

|(x, ei)|2 − 2

n∑

i=1

|(x, ei)|2 ≥ 0

which proves (2.2).

Theorem 2.1 (Bessel’s Inequality). If ei is an orthonormal set in aHilbert Space H, then

∑

i

|(x, ei)|2 ≤ ‖x‖2. (2.3)

7

Proof. Since H is separable, ei is at most countable. The result, in thefinite 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 elementsx1, . . . , 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 −∑k

i=1(xk+1, ei)ei‖.

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

Thus, if H 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 respectto 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 =

n∑

i=1

(x, en)en.

8

Then, for m > n,

‖yn − ym‖2 =

m∑

i=n+1

|(x, ei)|2

and the right-hand side tends to zero, by Bessel’s 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)ei‖2 = ‖x‖2 −∞

∑

i=1

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

Proof. Given any n, set yn =∑n

i=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− yn‖2 = ‖x‖2 +

n∑

i=1

|(x, ei)|2 − 2

n∑

i=1

|(x, ei)|2

= ‖x‖2 −n

∑

i=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, then

x =∑

j

(x, ej)ej. (2.5)

9

(iv) (Parseval’s Identity) If x ∈ H, then

‖x‖2 =∑

i

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

Proof. (i)⇒ (ii) If (x, ei) = 0 for all i and x 6= 0, then ei ∪

x‖x‖

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‖ = 1and (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→∞

n∑

i=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 eiis 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

‖x‖2 =∞

∑

i=1

|xi|2 =∞

∑

i=1

|(x, ei)|2.

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

10

trigonometric polynomials. It then follows, a fortiori, that they can also beapproximated in the L2-norm. Thus, the trignometric polynomials are densein L2(−π, π) and so, by the preceding corollary, the functions

1√2π

∪

cosnt√π,sinnt√

π| n ∈ N

form a complete orthonormal set in L2(−π, π). In particular, iff ∈ L2(−π, π), we have, by Parseval’s identity,

∫ π

−π

|f |2 dx =1

2π

(∫ π

−π

f(t) dt

)2

+

+∞

∑

n=1

[

1

π

(∫ π

−π

f(t) cosnt dt

)2

+1

π

(∫ π

−π

f(t) sinnt dt

)2]

which yields1

π

∫ π

−π

|f |2 dx =a2

0

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 series∑∞

i=1(x, ei)ei and the quantities (x, ei) are called the Fourier coefficients.

3 Variations on the Theme

Let f be a 2π-periodic function defined on [−π, π]. The Fourier coefficientsof f are given by the formulas (1.8). These formulas also make sense iff ∈ L1(−π, π). 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 theequivalence 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(π) 6= h(−π), we can redefine it so that h(π) = h(−π) and thenextend it periodically as a 2π-periodic function over R. We can also declarethe function to be undefined at −π or π. Thus we are forced to consider theFourier series of functions with jump discontinuities.

11

If a function belongs to L1(0, π) then we can extend it either as an oddfunction or as an even function to L1(−π, π). In the former case, all thecoefficients an will be zero and we get a series only involving the functionssinnt. 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 cosnt, is called the Fourier cosineseries of the function.

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

a0

2+

∞∑

n=1

(an cosnt+ bn sin nt) (3.1)

be the Fourier series of a function. Set

an = dn cosφn, bn = dn sinφn.

In other words,

dn =√

a2n + b2n, φn = cos−1(an/dn).

Then the series (3.1) can be rewritten as

a0

2+

∑

n

dn cos(nt− φn). (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(−π, π) we deduce that

limn→∞

∫ π

−π

f(t) cosnt dt = limn→∞

∫ π

−π

f(t) sinnt dt = 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.

12

Theorem 4.1 (Generalized Riemann-Lebesgue Lemma). Letf ∈ L1(a, b), where −∞ ≤ a < b ≤ +∞. Let h be a bounded measurablefunction defined on R, such that

limc→±∞

1

c

∫ c

0

h(t) dt = 0. (4.1)

Then

limω→∞

∫ b

a

f(t)h(ωt) 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

χ[c,d]h(ωt) dt =

∫ d

c

h(ωt) dt =1

ω

∫ dω

cω

h(t) dt

=1

ω

∫ dω

0

h(t) dt− 1

ω

∫ cω

0

h(t) dt

and, by hypothesis, both integrals tend to zero as ω → ∞. 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, if f ∈ L1(0,∞), then find a step function g such that∫ ∞0

|f − g| dx < ε2M

, for a given ε > 0. Now,∣

∣

∣

∣

∫ ∞

0

f(t)h(ωt) dt

∣

∣

∣

∣

≤∫ ∞

0

|f(t) − g(t)|h(ωt) dt+∣

∣

∣

∣

∫ ∞

0

g(t)h(ωt) dt

∣

∣

∣

∣

≤ ε

2+

∣

∣

∣

∣

∫ ∞

0

g(t)h(ωt) dt

∣

∣

∣

∣

and for ω large enough the second term can also be made to be less thanε2. A similar argument holds for

∫ 0

−∞ f(t)h(ωt) dt and this completes theproof.

Corollary 4.1. If f ∈ L1(a, b) then

limn→∞

∫ b

a

f(t) cosnt dt = limn→∞

∫ b

a

f(t) sinnt dt = 0. (4.3)

13

Proof.

∣

∣

∣

∣

1

c

∫ c

0

cos t dt

∣

∣

∣

∣

=

∣

∣

∣

∣

1

c(sin c)

∣

∣

∣

∣

≤ 1

|c|∣

∣

∣

∣

1

c

∫ c

0

sin t dt

∣

∣

∣

∣

=

∣

∣

∣

∣

1

c(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 αn, we know that αn → 0 as n → ∞. We now ask if a

trigonometric series

a0

2+

∞∑

n=1

(an cosnt+ bn sin nt)

is convergent, whether an → 0 and bn → 0 as n→ ∞.

Theorem 4.2 (Cantor-Lebesgue Theorem). If a trigonometric seriesa0

2+

∑∞n=1(an cosnt+ 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 and

φn = cos−1(an/dn). Since the series converges, it follows that for all t ∈ E,

dn cos(nt− φn) → 0

as n → ∞. Assume that dn does not converge to zero. Then, there existsε > 0 and a subsequence nk such that

dnk≥ ε > 0,

for all k. Then, it follows that cos(nkt − φnk) → 0 as k → ∞ for all t ∈ E.

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

∫

E

cos2(nkt− φnk) dt→ 0. (4.4)

14

Now cos2(nkt− φnk) = 1

2[1 + cos 2(nkt− φnk

)]. But then χE ∈ L1(R) and so

∫

E

cos 2(nkt− φnk) dt =

∫

R

χE(t) cos 2(nkt− φnk) dt

= cos 2φnk

∫

R

χE(t) cos 2nkt dt+

+ sin 2φnk

∫

R

χE(t) sin 2nkt dt

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

∫

E

cos2(nkt− φnk) dt→ µ(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(−π, π) with Fourier series given by

f(t) ∼ a0

2+

∞∑

n=1

(an cosnt + bn sinnt) (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 ∈ [−π, π]?

• If it converges at t ∈ [−π, π], does it converge to f(t)?

• If it converges at all t ∈ [−π, π], 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) =a0

2+

n∑

k=1

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

15

Equivalently, in exponential form,

sn(t) =

n∑

k=−n

ck exp(ikt).

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

sn(t) =

n∑

k=−n

1

2π

∫ π

−π

f(x) exp(ik(t− x)) dx

=1

2π

∫ π

−π

f(x)Dn(t− x) dx

where

Dn(t) =n

∑

k=−n

exp(ikt). (5.3)

The 2π-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(−π, π) which is 2π-periodic. Then

sn(t) =1

2π

∫ π

−π

f(x)Dn(t− x) dx (5.4a)

=1

2π

∫ π

−π

f(t− x)Dn(x) dx (5.4b)

=1

2π

∫ π

−π

f(x)Dn(x− t) dx (5.4c)

=1

2π

∫ π

−π

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

=1

2π

∫ π

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

2π

∫ π+t

−π+t

f(t− y)Dn(y) dy.

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

16

(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 2π. Finally, we split the integral in (5.4d) as the sum of integralsover [−π, 0] and [0, π]. Now,

∫ 0

−π

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

∫ π

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. Let n ≥ 0 be an integer. Then

Dn(t) =

sin(n+ 12)t

sin t2

t 6= 2kπ, k ∈ N ∪ 0.

2n + 1 t = 2kπ, k ∈ N ∪ 0.(5.5)

Further,1

2π

∫ π

−π

Dn(t) dt = 1. (5.6)

Proof. When n = 2kπ, clearly Dn(t) = 2n+1. Assume n 6= 2kπ, 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 of Dn(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

Φ(ω, t) =

sinωt

t, t 6= 0

ω, t = 0(5.7)

17

where ω and t are real numbers. The associated discrete Fourier kernel isgiven by

Φn(t) =

sin(n + 12)t

tt 6= 0

n+ 12

t = 0,(5.8)

where t ∈ R and n is a non-negative integer.

Remark 5.1. Clearly Dn is 2π-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

n∑

k=0

Dk(t). (5.9)

Proposition 5.3. Let n ≥ 0 be an integer. Then

Kn(t) =1

n+ 1

1 − cos(n+ 1)t

1 − cos t(5.10)

1

2π

∫ π

−π

Kn(t) dt = 1. (5.11)

Further, Kn ≥ 0 and if 0 < δ ≤ |t| ≤ π, we have

Kn(t) ≤ 2

(n + 1)(1 − cos δ). (5.12)

Proof. As before, observe that

(n+ 1)Kn(t)(

exp(it) − 1)(

exp(−it) − 1)

=(

exp(−it) − 1)

·m

∑

k=0

(

exp(

i(k + 1)t)

− exp(−ikt))

= 2 − exp(

i(n+ 1)t)

−− exp

(

−i(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).

18

We now derive some estimates for integrals of the Dirichlet and Fourierkernels. First we need a technical result.

Lemma 5.1. Let Ak be a sequence of real numbers such that A2k−1 > 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) + . . .+ (Ak−1 + Ak)

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

(A1 + A2) + (A3 + A4) + . . .+ (Ak−2 + Ak−1) + 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. Let 0 ≤ a < b ≤ π. Let n ≥ 0 be an integer. Then

∣

∣

∣

∣

∫ b

a

sin(n+ 1/2)t

sin(t/2)dt

∣

∣

∣

∣

≤ 4π. (5.14)

Proof. Let

Ak =

∫ kπ/(n+1/2)

(k−1)π/(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 (k−1)π and kπ. 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 ∈[

(k−1)πn+1/2

, kπn+1/2

)

or a ∈[

nπn+1/2

, π]

.

If a is in the interior of any of these intervals (or, if a ∈(

nπn+1/2

, π]

, when

k = n+ 1), we have

∫ a

(k−1)π/(n+1/2)

sin(n+ 1/2)t

sin(t/2)dt

19

is either greater than 0 (because the integrand is greater than 0, when k isodd) or is less than 0 (k even) and

∣

∣

∣

∣

∫ a

(k−1)π/(n+1/2)

sin(n+ 1/2)t

sin(t/2)dt

∣

∣

∣

∣

< |Ak|.

Thus∫ a

0

sin(n+ 1/2)t

sin(t/2)dt = A1 + . . .+ Ak−1 +

∫ a

(k−1)π/(n+1/2)

sin(n+ 1/2)t

sin(t/2)dt

and it follows that

0 <

∫ a

0

sin(n+ 1/2)t

sin(t/2)dt < A1.

Now, in the interval [0, π/(n+1/2)], the integrand is positive and decreasingwith maximum value (2n+ 1) at t = 0. Thus

A1 < (2n + 1)π

(n+ 1/2)= 2π.

Now∣

∣

∣

∣

∫ b

a

sin(n + 1/2)t

sin(t/2)dt

∣

∣

∣

∣

=

∣

∣

∣

∣

∫ b

0

sin(n+ 1/2)t

sin(t/2)dt−

∫ a

0

sin(n+ 1/2)t

sin(t/2)dt

∣

∣

∣

∣

≤ 4π

Proposition 5.5. Let 0 ≤ a < b. Then

∣

∣

∣

∣

∫ b

a

sinωt

tdt

∣

∣

∣

∣

≤ 2π (5.15)

for ω > 0.

Proof. Define

Ak =

∫ kπ/ω

(k−1)π/ω

sinωt

tdt =

∫ kπ

(k−1)π

sin t

tdt.

20

Then, again, Ak has alternating signs and decreases in absolute value. Asin the previous lemma, we get

0 <

∫ a

0

sinωt

tdt < A1.

Since∣

∣

sin tt

∣

∣ ≤ 1, we get A1 ≤ π and so∣

∣

∣

∣

∫ b

a

sinωt

tdt

∣

∣

∣

∣

=

∣

∣

∣

∣

∫ b

0

sinωt

tdt−

∫ a

0

sinωt

tdt

∣

∣

∣

∣

≤ 2π.

The discrete Fourier kernel is a very good approximation of the Dirichletkernel.

Proposition 5.6. Let f ∈ L1(0, π). Let 0 < r ≤ π. Then

limn→∞

∫ r

0

f(t)sin(n + 1

2)t

sin t2

dt = limn→∞

∫ r

0

f(t)sin(n+ 1

2)t

t2

dt (5.16)

whenever either limit exists.

Proof. By L’Hospital’s rule, we get

limt→0

(

1

sin t− 1

t

)

= 0.

Define

g(t) =

1

sin(t/2)− 1

t/2t 6= 0

0 t = 0.

Then g is continuous. Thus it is bounded in [0, r] for 0 < r ≤ π and thefunction f(t)g(t)χ[0,r](t) is integrable. So are the functions

φ(t) = f(t)g(t) cos(t/2)χ[0,r](t)

andψ(t) = f(t)g(t) sin(t/2)χ[0,r](t).

Thus,∫ r

0

f(t) sin(n+1

2)t

[

1

sin(t/2)− 1

t/2

]

dt =

∫ π

0

φ(t) sinnt dt+

∫ π

0

ψ(t) cosnt dt

and, by the Riemann-Lebesgue lemma (cf. Corollary 4.1), both the integralson the right-hand side tend to zero as n→ ∞ and the result follows.

21

Both the functionsDn(t) and Φn(t) enjoy the Riemann-Lebesgue property.More precisely, we have the following result.

Proposition 5.7. Let f ∈ L1(0, π) and let 0 < r ≤ π. Then

limn→∞

∫ π

r

f(t)Dn(t) dt = limr→∞

∫ π

r

f(t)Φn(t) dt = 0. (5.17)

Proof. Since r > 0, the functions

φ(t) = f(t) sin(t/2)g(t)χ[r,π](t)

ψ(t) = f(t) cos(t/2)g(t)χ[r,π](t)

where g(t) = 1sin(t/2)

or 1t/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(−π, π). Let 0 <r < π. Then, for x ∈ [−π, π],

limn→∞

(∫ −r

−π

+

∫ π

r

)

(

f(x− t)Dn(t))

dt = 0.

Proof. Fix x ∈ [−π, π]. Then, by the preceding proposition, we have∫ π

r

f(x− t)Dn(t) dt→ 0 as n→ ∞.

Now,∫ −r

−π

f(x− t)Dn(t) dt =

∫ π

r

f(x+ t)Dn(t) dt

using the evenness of Dn and so once again this integral also tends to zero.This completes the proof.

Note that the Fourier coefficients depend on the values of a function fthroughout the interval [−π, π]. However, if f and g are in L1(−π, π) andfor some t ∈ [−π, π] and r > 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, andonly if, the Fourier series of g 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.

22

6 Fourier Series of Continuous Functions

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

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

Proposition 6.1. We have

limn→∞

∫ π

−π

|Dn(t)| dt = +∞. (6.1)

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

∫ π

−π

|Dn(t)| dt ≥ 4

∫ π

0

| sin(n + 12)t|

tdt

= 4

∫ (n+ 1

2)π

0

| sin t|t

dt

> 4n

∑

k=1

∫ kπ

(k−1)π

| sin t|t

dt

> 4

n∑

k=1

∫ kπ

(k−1)π

| sin t|kπ

dt

=8

π

n∑

k=1

1

k

from which (6.1) follows immediately.

Proposition 6.2. Let V = Cper[−π, π], the space of continuous 2π-periodicfunctions with the usual sup-norm (denoted ‖ · ‖∞) and define φn : V → R

byφn(f) = sn(f)(0)

where sn(f) is the nth-partial sum of the Fourier series of f. Then φn is acontinuous linear functional on V and

‖φn‖ =1

2π

∫ π

−π

|Dn(t)| dt. (6.2)

23

Proof. On one hand,

φn(f) =1

2π

∫ π

−π

f(t)Dn(t) dt

(cf. (5.4d)). Thus,

|φn(f)| ≤ ‖f‖∞1

2π

∫ π

−π

|Dn(t)| dt

and so

‖φn‖ ≤ 1

2π

∫ π

−π

|Dn(t)| dt.

Now, let En = t ∈ [−π, π] | Dn(t) ≥ 0. Define

fm(t) =1 −md(t, En)

1 +md(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[−π, π], (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 ∈ Ec

n. By the dominated convergence theorem, it nowfollows that

φn(fm) → 1

2π

∫ π

−π

|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, Baire’s theorem also states that a complete metric spacecannot be the countable union of nowhere dense sets.

One of the important consequences of Baire’s theorem is the Banach-Steinhaus theorem also known as the uniform boundedness principle.

24

Theorem 6.2 (Banach-Steinhaus). Let X be a Banach space and Y anormed linear space and Λαα∈A a collection of bounded linear transforma-tions from X into Y , where α ranges over some indexing set A. Then eitherthere exists M > 0 such that

‖Λα‖ ≤M, for all α ∈ A (6.3)

or,supα∈A

‖Λαx‖ = ∞ (6.4)

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

(Recall that a Gδ-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 = N

and set Λn = φn, again defined in above mentioned proposition. Since, byProposition 6.1 and 6.2, we have ‖φn‖ → ∞ as n → ∞, it follows from theBanach-Steinhaus theorem, that there exists a dense Gδ-set (of continuous2π-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 [−π, π] in the same manner.

By another application of Baire’s theorem, we can strengthen this further.Let Ex be the dense Gδ-set of continuous 2π-periodic functions in V such

that the Fourier series of these functions diverge at x. Let xi be a countableset of points in [−π, π] and let

E = ∩ni=1Exi

⊂ V. (6.5)

Then, by Baire’s theorem E is also a dense Gδ-set. (Each Exiis the countable

intersection 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 Gδ-set in (−π, π) for each f . If we choose the xi above so that xi isdense (take all rationals, for instance in (−π, π)) then we have the followingresult.

25

Proposition 6.3. The set E ⊂ V is a dense Gδ-set such that for all f ∈ E,the set Qf ⊂ (−π, π) where its Fourier series diverges, is a dense Gδ-set in(−π, π).

Proposition 6.4. In a complete metric space, which has no isolated points,no countable dense set can be a Gδ.

Proof. Let E = x1, . . . , xn, . . . be a countable dense set. Assume E isa Gδ. Thus E = ∩∞

n=1Wn, Wn open and dense. Then, by hypothesis, Wn \∪n

i=1xi = Vn is also open and dense. But ∩∞n=1Vn = ∅, contradicting Baire’s

theorem.

Thus, there exists uncountably many 2π-periodic continuous functions on[−π, π] whose Fourier series diverge on a dense Gδ-set of (−π, π).

Having answered our first general question negatively, let us now prove apositive result.

Proposition 6.5. Let f be a 2π-periodic function on [−π, π] 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 [−π, π].

Proof. Choose 0 < r < π such that

1

π

∫ r

−r

|t/2|| sin(t/2)| dt <

ε

2K.

This is possible since∣

∣

∣

t/2sin(t/2)

∣

∣

∣is a bounded continuous function, and hence

integrable on [−π, π]. If C is an upper bound for this function, we need onlychoose r such that 2rC < επ

2K. Let x ∈ [−π, π].

sn(x) =1

2π

∫ π

−π

f(x− t)Dn(t) dt.

Thus

|sn(x) − f(x)| =1

2π

∣

∣

∣

∣

∫ π

−π

(

f(x− t) − f(x))

Dn(t) dt

∣

∣

∣

∣

.

26

If r is chosen as above, then, by the localization principle (Theorem 5.1), wehave, for n large enough

1

2π

∣

∣

∣

∣

(∫ −r

−π

+

∫ π

r

)

(

f(x− t) − f(x))

Dn(t) dt

∣

∣

∣

∣

<ε

2.

On the other hand,∣

∣

∣

∣

1

2π

∫ r

−r

(

f(x− t) − f(x))

Dn(t) dt

∣

∣

∣

∣

≤ K

2π

∫ r

−r

|t|| sin(n+ 1/2)t|| sin t/2| dt

≤ K

π

∫ r

−r

|t|/2| sin(t/2)| dt <

ε

2.

Thus for n large we have

|sn(x) − f(x)| < ε

which completes the proof.

Corollary 6.1. If f ∈ C1[−π, π] is 2π-periodic, then the Fourier series of fconverges to f on [−π, π].

In a later section, we will relax these conditions on f and study thepointwise convergence of Fourier series.

Remark 6.1. The convergences in Proposition 6.5 and Corollary 6.1 aboveare, in fact, uniform over [−π, π], cf. Exercise 32.

Remark 6.2. If f is Lipschitz continuous in a neighbourhood of t ∈ [−π, π],then we can show, using identical arguments, that the Fourier series of fconverges to f(t) at t.

7 Fejer’s Theorem

In the previous section, we saw that there exist an uncountable number ofcontinuous 2π-periodic functions whose Fourier series diverge over a denseset of points. Nevertheless, as we have observed earlier, such functions can beapproximated uniformly by means of trigonometric polynomials over [−π, π],thanks to the Stone-Weierstrass theorem. Indeed, we can now prove thefollowing result.

27

Theorem 7.1 (Fejer). Let f be a continuous 2π-periodic function on[−π, π]. Let snn≥0 be the sequence of partial sums of its Fourier series.Define

σn(x) =s0(x) + . . .+ sn(x)

n+ 1.

Then σn → f uniformly over [−π, π].

Proof. It is immediate, from the definition of the Fejer kernel Kn, to see that

σn(x) =1

2π

∫ π

−π

f(x− t)Kn(t) dt.

Thus

σn(x) − f(x) =1

2π

∫ π

−π

(

f(x− t) − f(x))

Kn(t) dt,

in view of (5.11). Now, since f is continuous on [−π, π], it is bounded anduniformly continuous. Let |f(x)| ≤ M for all x ∈ [−π, π] and, given ε > 0,let δ > 0 such that |x− y| < δ implies |f(x) − f(y)| < ε/2.

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

∣

∣

∣

∣

∫ δ

−δ

(

f(x− t) − f(x))

Kn(t) dt

∣

∣

∣

∣

≤ ε

2

∫ π

−π

Kn(t) dt = πε,

again by (5.11). On the other hand,∣

∣

∣

∣

(∫ −δ

−π

+

∫ π

δ

)

(

f(x− t) − f(x))

Kn(t) dt

∣

∣

∣

∣

≤ 2M · ε

4M· 2π = πε

if n ≥ N . It now follows that for all x ∈ [−π, π], we have

|σn(x) − f(x)| < ε

for n ≥ N which completes the proof.

Corollary 7.1. Let f and g be two 2π-periodic and continuous functions on[−π, π]. If they have the same Fourier series, then f ≡ g.

Proof. If the two functions have the same Fourier series, then the σn will bethe same for both functions and we know that σn → f and σn → g uniformlyon [−π, π]. Hence the result.

28

Given a series∑

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

σn =s1 + . . .+ sn

n,

sk being the partial sums of the series. Thus the Fourier series of a continuous2π-periodic function is always Cesaro summable to the function.

Thus if f is a continuous 2π-periodic function whose Fourier series is givenby

a0

2+

∞∑

n=1

(an cosnt + bn sinnt),

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

σn(x) =a0

2+

n∑

k=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 ∈ [−π, π]. Then g is2π-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

a0

2+

∞∑

k=1

ak cos kt.

Then

σn(t) =a0

2+

n∑

k=1

(

1 − k

n+ 1

)

ak cos kt.

Thus, given ε > 0, there exists N such that, for all n ≥ N ,

∣

∣

∣

∣

∣

f(cos t) − a0

2−

n∑

k=1

(

1 − k

n+ 1

)

ak cos kt

∣

∣

∣

∣

∣

< ε

for all t ∈ [−π, π]. This is the same as

∣

∣

∣

∣

∣

f(t) − a0

2−

n∑

k=1

(

1 − k

n + 1

)

ak cos(

k(cos−1 t))

∣

∣

∣

∣

∣

< ε

29

for all t ∈ [−1, 1]. Now it only remains to show that

Pk(t) = cos(k cos−1 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

cosns = 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 Pn−1(cos s) − Pn−2(cos s).

It then follows thatPn(t) = 2tPn−1(t) − Pn−2(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 Regularity

In 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+) = limt→a

f ′(t)

exists. Then f ′ is bounded in a subinterval (a, a+δ) (where δ > 0) and so, bythe mean-value theorem, f is uniformly continuous on (a, a+ δ). Thus it canbe continuously extended to [a, a + δ]. Consequently f(a+) = limt→a 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+ δ, we have

f(t) = f(a+) + f ′(a + θ(t− a))(t− a)

30

where 0 < θ < 1. Thus

limt→a

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) = limt↓a

f(t) − f(a)

t− a.

We also define, f ′(b−) = limt→b f′(t).

Definition 8.1. We say that a function f : [a, b] → R is piecewise 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 <. . . < xn = b be any partition of [a, b]. Define

V (P; f) =

n∑

i=1

|f(xi) − f(xi−1)|

andV (f ; a, b) = sup

PV (P; f)

the supremum being taken over all possible partitions of [a, b]. The quantityV (f ; a, b) is called the total variation of f over the interval [a, b].

Definition 8.2. A real (or complex) valued function defined on [a, b] is saidto be of bounded variation on [a, b] if V (f ; a, b) <∞.

Example 8.1. Any monotonic function defined on [a, b] is of bounded varia-tion. In this case

V (f ; a, b) = |f(b) − f(a)|.

31

Example 8.2. If f is uniformly Lipschitz continuous, then it is of boundedvariation. For,

∑

i

|f(xi) − f(xi−1)| ≤ L∑

i

(xi − xi−1) = L(b− a) <∞

Example 8.3. Define

f(x) =

x2 sin 1x2 , 0 < x < 1

0, x = 0.

Then, f is not of bounded variation. To see this, choose a partition of [0, 1]as follows:

0, 1 ∪

√

2

π(2j + 1)| 0 ≤ j ≤ n

.

|f(xj) − f(xj−1)| =2

π(2j + 1)+

2

π(2j − 1)=

2

π

4j

4j2 − 1.

≥ 2

π

4j

4j2=

2

πj.

Thus for all n, V (f ; 0, 1) ≥ 2π

∑nj=1

1j

and so V (f ; 0, 1) = ∞.

Given a real number r, define r+ = maxr, 0 and r− = −minr, 0.Then r = r+ − r− and |r| = r+ + r−. Thus, if f : [a, b] → R, and P = a =x0 < . . . < xn = b is any partition, set

p(P; f) =

n∑

i=1

(

f(xi) − f(xi−1))+

and

n(P; f) =

n∑

i=1

(

f(xi) − f(xi−1))−.

Then V (P; f) = p(P; f) + n(P; f) and

f(b) − f(a) = p(P; f) − n(P; f).

DefineP (f ; a, b) = sup

Pp(P; f) and N(f ; a, b) = sup

Pn(P; f);

where, again, the supremum are taken over all possible partitions of [a, b].

32

Proposition 8.1. Let f be of bounded variation on [a, b]. Then

V (f ; a, b) = P (f ; a, b) +N(f ; a, b) (8.2)

f(b) − f(a) = P (f ; a, b) −N(f ; a, b). (8.3)

Proof. We know that, for any partition P,

p(P; f) = n(P; f) + f(b) − f(a)

≤ N(f ; a, b) + f(b) − f(a).

Then, taking the supremum over all possible partitions, we get

P (f ; a, b) ≤ N(f ; a, b) + f(b) − f(a). (8.4)

Similarly n(P; f) = p(P; f) + f(a) − f(b) yields

N(f ; a, b) ≤ P (f ; a, b) + f(a) − f(b) (8.5)

Relations (8.4) and (8.5) yield (8.3), since V (f ; a, b) < ∞ implies thatP (f ; a, b) <∞ and N(f ; a, b) <∞. Now

V (P; f) = p(P; f) + n(P; f)

gives usV (f ; a, b) ≤ P (f ; a, b) +N(f ; a, b). (8.6)

On the other hand,

V (f ; a, b) ≥ p(P; f) + n(P; f)

= 2p(P; f) −(

f(b) − f(a))

= 2p(P; f) +N(f ; a, b) − P (f ; a, b)

using (8.3). Again, taking the supremum over all partitions, we get

V (f ; a, b) ≥ 2P (f ; a, b) +N(f ; a, b) − P (f ; a, b)

= P (f ; a, b) +N(f ; a, b) (8.7)

Relations (8.6) and (8.7) yield (8.2).

33

Proposition 8.2. If f and g are of bounded variation on [a, b], then f + gis of bounded variation and

V (f + g; a, b) ≤ V (f ; a, b) + V (g; a, b). (8.8)

Proof. If P is any partition,

V (P; f + g) ≤ V (P; f) + V (P; g)

by the triangle inequality and the result follows.

We now have an important characterization of functions of bounded vari-ation.

Theorem 8.1. A function f : [a, b] → R is of bounded variation if, and onlyif, f is the difference of two monotonic increasing functions.

Proof. If f is the difference of two monotonic increasing functions, sinceeach of these is of bounded variation (cf. Example 8.1), it follows from thepreceding proposition that f is of bounded variation.

Conversely, if f is of bounded variation on [a, b], by Proposition 8.1, wehave, for any x ∈ [a, b],

f(x) − f(a) = P (f ; a, x) −N(f ; a, x).

Let g(x) = P (f ; a, x) and h(x) = N(f ; a, x)−f(a). Clearly both functions aremonotonic increasing and f(x) = g(x)−h(x). This completes the proof.

From the properties of monotonic functions, we can now deduce the fol-lowing result.

Corollary 8.1. If f : [a, b] → R is of bounded variation, then f is dif-ferentiable a.e. Further, for all x ∈ (a, b), f(x+) and f(x−) exist. So dof(a+) and f(b−). The function f has at most a countable number of jumpdiscontinuities.

The fundamental theorem of Lebesgue integration states that if f is in-tegrable on [a, b] and if

F (x) = c +

∫ x

a

f(t) dt,

then F ′ = f a.e. on [a, b]. We now ask the question as to when a functioncan be expressed as an indefinite integral of an integrable function.

34

Proposition 8.3. Let f be integrable on [a, b]. Then set

F (x) =

∫ x

a

f(t) dt.

Then F is of bounded variation on [a, b].

Proof. If P = a = x0 < x1 < . . . < xn = b is any partition of [a, b], then

n∑

i=1

|F (xi) − F (xi−1)| ≤n

∑

i=1

∫ xi

xi−1

|f(t)| dt =

∫ b

a

|f(t)| dt <∞.

Thus

V (F ; a, b) ≤∫ b

a

|f(t)| dt <∞.

Hence the result.

Thus a function must be at least of bounded variation on [a, b] for it tobe expressed as an indefinite integral of an integrable function. But this isnot enough.

Proposition 8.4. Let f : [a, b] → R be integrable. Given ε > 0, there existsδ > 0 such that if E ⊂ [a, b] is a measurable set with µ(E) < δ, then

∣

∣

∣

∣

∫

E

f(x) dx

∣

∣

∣

∣

< ε.

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

∣

∣

∣

∫

E

f dx

∣

∣

∣

∣

≤ Mµ(E),

where µ(E) is the Lebesgue measure of E.In the general case, define

fn(x) =

|f(x)|, if |f(x)| ≤ n

n, 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→∞

∫ b

a

fn dx =

∫ b

a

|f | dx <∞.

35

Given ε > 0, choose N such that

∫ b

a

(|f | − fN) dx < ε/2.

Now choose δ > 0 such that µ(E) < δ implies that

∫

E

fN dx < ε/2.

Then |∫

Ef dx| ≤

∫

E|f | dx ≤

∫ b

a(|f |− fN) dx+

∫

EfN dx < ε. This completes

the proof.

Definition 8.3. A function f : [a, b] → R is said to be absolutely contin-uous on [a, b] if for every ε > 0, there exists δ > 0 such that whenever wehave a finite collection of disjoint intervals (xi, x

′i)n

i=1 satisfying

n∑

i=1

(x′i − xi) < δ

we haven

∑

i=1

|f(x′i) − f(xi)| < ε.

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(x′i) − f(xi)| ≤ L∑

(x′i − xi) < Lδ.

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.

36

Proof. Let δ correspond to ε = 1 in the definition of absolute continuity. LetK be the integral part of 1 + (b − a)/δ, where [a, b] is the given interval.Given any partition P of [a, b], we can refine it to a partition consisting of Ksets of sub-intervals each of total length less than δ. 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 µ(E) = c − a. Let ε, η > 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)| < ηh.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 thanδ, where δ corresponds to ε 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,n

∑

k=0

|xk+1 − yk| < δ.

Now,n

∑

k=1

|f(yk) − f(xk)| < η∑

(yk − xk) < η(c− a).

Alson

∑

k=1

|f(xk+1) − f(yk)| < ε

by absolute continuity. Thus

|f(c) − f(a)| ≤ ε+ η(c− a)

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

37

Theorem 8.2. A function can be expressed as an indefinite integral of anintegrable 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 ′ = F ′1 − F ′

2 a.e. andF ′

1 ≥ 0, F ′2 ≥ 0. Thus,

∫ b

a

|F ′| dx ≤∫ b

a

|F ′1| dx+

∫ b

a

|F ′2| dx

=

∫ b

a

F ′1 dx+

∫ b

a

F ′2 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) +

∫ x

a

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

∫ b

a

f(t)g′(t) dt = f(b)g(b) − f(a)g(a) −∫ b

a

f ′(t)g(t) dt. (8.9)

38

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

∫ b

a

∫ x

a

f ′(x)g′(y) dy dx.

It is a routine verification to check that Fubini’s theorem applies and so

∫ b

a

∫ x

a

f ′(x)g′(y) dy dx =

∫ b

a

∫ b

y

f ′(x)g′(y) dx dy.

The left-hand side gives us

∫ b

a

(∫ x

a

g′(y) dy

)

f ′(x) dx =

∫ b

a

(

g(x) − g(a))

f ′(x) dx

=

∫ b

a

g(x)f ′(x) dx− g(a)[

f(b) − f(a)]

by repeated use of Theorem 8.2. The right-hand side gives

∫ b

a

(∫ b

y

f ′(x) dx

)

g′(y) dy =

∫ b

a

(

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. Let f ∈ L1(0, π) and assume that f ′(0+) exists. Then

limn→∞

1

π

∫ π

0

f(t)sin(n+ 1

2)t

tdt =

1

2f(0+). (9.1)

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

∫ π

0

f(t)sin(n+ 1

2)t

tdt =

∫ π

0

(f(t)−f(0+))sin(n + 1

2)t

tdt+f(0+)

∫ π

0

sin(n + 12)t

tdt.

39

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

∫ (n+ 1

2)π

0

sin t

tdt

which converges to π/2 as n→ ∞. Thus,

limn→∞

1

πf(0+)

∫ π

o

sin(n+ 12)t

tdt =

1

2f(0+).

Hence we need to show that the first term tends to zero as n→ ∞. Let ε > 0be an arbitrarily small number. Choose 0 < r < π such that for 0 < t ≤ r,

∣

∣

∣

∣

f(t) − f(0+)

t− f ′(0+)

∣

∣

∣

∣

< ε

( cf. (8.1)). Then,

∫ π

0

(f(t) − f(0+))sin(n+ 1

2)t

tdt =

∫ r

0(f(t) − f(0+))

sin(n+ 1

2)t

tdt

+∫ π

r(f(t) − f(0+))

sin(n+ 1

2)t

tdt.

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

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

∫ r

0

(

f(t) − f(0+)

t− f ′(0+)

)

sin(n+1

2)t dt+ f ′(0+)

∫ r

0

sin(n+1

2)t dt.

But,

∣

∣

∣

∣

∫ r

0

(

f(t) − f(0+)

t− f ′(0+)

)

sin(n +1

2)t dt

∣

∣

∣

∣

< rε < πε

and∫ r

0

sin(n+1

2)t dt =

∫ r

0

sint

2cosnt dt+

∫ r

0

cost

2sinnt dt

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

40

Theorem 9.1. (Dirichlet) Let f ∈ L1(−π, π) be a piecewise smooth func-tion. Then its Fourier series converges to 1

2(f(t+) + f(t−)) at all points

t ∈ [−π, π]. 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

2π

∫ π

0

(f(t+ x) + f(t− x))Dn(x) dx.

Now, by (9.1), we have

limn→∞

1

π

∫ π

0

f(t+ x)sin(n+ 1

2)x

xdx =

1

2f(t+).

Thus,

limn→∞

1

2π

∫ π

0

f(t+ x)sin(n+ 1

2)x

x2

dx =1

2f(t+).

Hence, by Proposition 5.6, we get

limn→∞

1

2π

∫ π

0

f(t+ x)sin(n+ 1

2)x

sin x2

dx =1

2f(t+).

Similarly,

limn→∞

1

2π

∫ π

0

f(t− x)sin(n+ 1

2)x

sin x2

dx =1

2f(t−)

from which the result follows.

Example 9.1. Consider the function

f(t) =

−1, −π ≤ t < −π/20, −π/2 ≤ t ≤ π/21, π/2 < t ≤ π.

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

bn =1

π

∫ π

−π

f(t) sinnt dt =2

π

∫ π

π

2

sin nt dt.

41

Thus,

bn =

2nπ, n odd

0, n = 4k, k ∈ N

− 4nπ, n = 4k + 2, k ∈ N

At t = π/2, the series must thus converge to 1/2. Thus,

1

2=

2

π

[

1 − 1

3+

1

5− 1

7+ · · ·

]

which yields the well known Gregory series

π

4= 1 − 1

3+

1

5− 1

7+ · · ·

In order to prove the next result, we recall a version of the mean valuetheorem for integrals.

Proposition 9.2. Let g : [a, b] → R be continuous and let f : [a, b] → R

be non-negative and monotonic increasing. Then, there exists c ∈ [a, b] suchthat

∫ b

a

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

∫ b

c

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 be

∫ 1

−1

t2 dt =2

3.

The right-hand side is

f(1)

∫ 1

c

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).

42

We are now in a position to prove a convergence theorem for a larger classof functions. We need a preliminary result analogous to Proposition 9.1.

Proposition 9.3. Let f be of bounded variation on [0, π]. Then

limn→∞

1

π

∫ π

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,

∫ π

0

f(t)sin(n + 1

2)t

tdt =

∫ π

0(f(t) − f(0+))

sin(n+ 1

2)t

tdt

+ f(0+)∫ π

0

sin(n+ 1

2)t

tdt.

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

Let ε > 0 be an arbitrarily small number. Choose 0 < r ≤ π such that

|f(t) − f(0+)| < ε/4, 0 < t ≤ r. (9.4)

Then, splitting the first integral over [0, r] and [r, π], we see that the integralon [r, π] tends to zero by the Riemann-Lebesgue property for Φn (cf. Propo-sition 5.7) since f(t) − f(0+) is integrable over that interval. Thus, for nsufficiently large, we have

∣

∣

∣

∣

1

π

∫ π

r

(f(t) − f(0+))sin(n+ 1

2)t

tdt

∣

∣

∣

∣

<ε

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∣

∣

∣

∣

1

π

∫ r

0

(f(t) − f(0+))sin(n + 1

2)t

tdt

∣

∣

∣

∣

<ε

2.

This completes the proof.

43

Theorem 9.2. (Jordan) Let f be a function of bounded variation on [−π, π].Then, its Fourier series at any point t in this interval converges to 1

2(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. Jordan’s result was proved in 1881. In 1904, Fejer showedthat if f ∈ L1(−π, π), and if f(t+) and f(t−) exist at t ∈ [−π, π], 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(−π, π) converges to thevalue of that function a.e..This was extended by Hunt in 1968 to all functionsin Lp(−π, π) for 1 < p <∞.

10 Termwise Integration

In general, when we have a series f(x) =∑

n fn(x), we can integrate theseries 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, uniformconvergence 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. Let f ∈ L1(−π, π) be extended periodically over R and havethe Fourier series

a0

2+

∞∑

n=1

(an cosnt + bn sinnt).

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

a0

2x+

∞∑

n=1

(

an

nsin x− bn

ncosx

)

+ C

where

C =∞

∑

n=1

bnn,

44

converges to∫ x

0f(t) dt.

(ii) This convergence is uniform if f ∈ L2(−π, π).

Proof. Let

F (x) =

∫ x

0

(

f(t) − a0

2

)

dt.

Since f is integrable, it follows that F is absolutely continuous. It is also2π-periodic. For,

F (x+ 2π) =∫ x

0

(

f(t) − a0

2

)

dt+∫ x+2π

x

(

f(t) − a0

2

)

dt

= F (x) +∫ π

−π

(

f(t) − a0

2

)

dt

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

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

cn = 1π

∫ π

−πF (t) cosnt dt

= 1nπF (t) sinnt

∣

∣

π

−π− 1

nπ

∫ π

−π

(

f(t) − a0

2

)

sinnt dt.

The boundary terms vanish by periodicity. So does the integral of a0

2sinnt.

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

F (t) =c02

+∞

∑

n=1

(

an

nsin nt− bn

ncosnt

)

.

Evaluating this at t = 0, we get

c02

=∞

∑

n=1

bnn.

45

This proves (i).If f ∈ L2(−π, π), then the series

∑

n a2n and

∑

n b2n are convergent, by

virtue of Parseval’s 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(−π, π), 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(−π, π).

Remark 10.2. If f is continuous and 2π-periodic, then it is in L2(−π, π) andso, for any continuous 2π-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

sinnt

log n. (10.1)

The sequence (logn)−1 is non-negative and monotonically decreases tozero. Further,for t 6= 2nπ,

n∑

k=1

sin kt =cos t

2− cos(n+ 1

2)t

2 sin t2

and so∣

∣

∣

∣

∣

n∑

k=1

sin kt

∣

∣

∣

∣

∣

≤ 1

| sin t2| ≤ 1

sin a2

46

for t ∈ [a, 2π−a] for any a ∈ (0, π). Thus, by Dirichlet’s test, the given seriesin (10.1) converges uniformly in [a, 2π − a] for any a ∈ (0, π). 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 that∑

n bn/nis convergent. However, there is no such condition on

∑

n an/n. Indeed,Stromberg (1981) has shown that the series

∞∑

n=2

cosnt

logn

is a Fourier series!

11 Termwise Differentiation

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

f(t) = 2

∞∑

n−1

(−1)n+1

nsin nt

and, by Jordan’s theorem, the series converges to f(t) at all points in (−π, π)and to zero at −π and π. The termwise derivative of the series is

2

∞∑

n=1

(−1)n+1 cosnt

which is divergent everywhere since cosnt 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.

47

Theorem 11.1. Let f be a continuous and piecewise smooth 2π-periodicfunction on [−π, π] with Fourier series

a0

2+

∞∑

n=1

(an cosnt + bn sinnt).

If its derivative f ′ is also piecewise smooth, then

f ′(t+) + f ′(t−)

2=

∞∑

n=1

(nbn cosnt− nan sin nt). (11.1)

Proof. Let f ′ have the Fourier series expansion

c02

+∞

∑

n=1

(cn cosnt + dn sin nt).

Since f is 2π-periodic, we have

c0 =1

π

∫ π

−π

f ′(t) dt = 0.

Again, for n ≥ 1,

cn = 1π

∫ π

−πf ′(t) cosnt dt

= 1πf(t) cosnt

∣

∣

π

−π+ n

π

∫ π

−πf(t) sinnt dt,

by the absolute continuity of f , which is piecewise smooth. The boundaryterms cancel out by the 2π-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 (k−1)

are all continuous and f (k) is piecewise smooth, then nkan and nkbn tendto zero as n → ∞. Also, in this case, since f (k) ∈ L2(−π, π), we have, byParseval’s identity, that

∞∑

n=1

n2ka2n <∞, and

∞∑

n=1

n2kb2n <∞.

48

We conclude with a nice appliation of Fourier series to a problem in thecalculus of variations.

One of the earliest known problems in the calculus of variations is knownas the classical isoperimetric problem and can be traced to Virgil’s 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 ≥ 4πA, (11.2)

with equality if, and only if, the curve is a circle. For the circle, we do indeedhave equality since L = 2πr and A = πr2, where r is its radius. On the otherhand, for any curve of given length L, the maximum possible value for theenclosed area is L2/4π, which is achieved for the circle. Thus, this establishesthe circle as the optimal solution. The uniqueness follows from the ‘only if’part 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 =2π

Ls

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

x(t) = a0

2+

∑∞n=1(an cos nt+ bn sin nt)

y(t) = c02

+∑∞

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

(11.3)

49

Thus,x′(t) =

∑∞n=1(nbn cosnt− nan sinnt)

y′(t) =∑∞

n=1(ndn cos nt− ncn sinnt).(11.4)

Since(

dx

ds

)2

+

(

dy

ds

)2

= 1,

we get that(

dx

dt

)2

+

(

dy

dt

)2

=

(

L

2π

)2

.

Then, by Parseval’s identity, it follows that

2

(

L

2π

)2

=1

π

∫ 2π

0

[

(

dx

dt

)2

+

(

dy

dt

)2]

dt =∞

∑

n=1

n2(a2n + b2n + c2n + d2

n).

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

A =

∫

C

x dy =

∫ 2π

0

x(t)dy

dt(t) dt = π

∞∑

n=1

n(andn − bncn). (11.6)

From (11.5) and (11.6), we get

L2 − 4πA = 2π2

∞∑

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 all

n ≥ 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 t

from which we get(

x(t) − a0

2

)2

+(

y(t) − c02

)2

= a21 + b21

50

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

The isoperimetric inequality exists in all dimensions. The n-dimensionalball 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 ≥ 36πV 2

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

n is a bounded domain, then let us denote its n-dimensional(Lebesgue) measure by |Ω|n and the ‘(n−1)-dimensional surface measure’ ofthe boundary ∂Ω (which has to be suitably defined) by |∂Ω|n−1. The classicalisoperimetric inequality now reads as

|∂Ω|n−1 ≥ nω1

n

n |Ω|1−1

n

n

where

ωn =π

n

2

Γ(n2

+ 1)

is the volume of the unit ball in Rn. Equality occurs in the isoperimetricinequality if, and only if, Ω is a ball.

51

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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