Chapter 10: Fourier Series - University of Louisville …lee/RealAnalysis/IntroRealAnal... · · 2016-07-18CHAPTER 10 FourierSeries ... huge area of research when Joseph Fourier

CHAPTER 10

Fourier Series

In the late eighteenth century, it was well-known that complicated func-tions could sometimes be approximated by a sequence of polynomials. Someof the leading mathematicians at that time, including such luminaries as DanielBernoulli, Euler and d’Alembert began studying the possibility of using sequencesof trigonometric functions for approximation. In 1807, this idea opened into ahuge area of research when Joseph Fourier used series of sines and cosines tosolve several outstanding partial differential equations of physics.1

In particular, he used series of the form1X

n=0an cosnx +bn sinnx

to approximate his solutions. Series of this form are called trigonometric series,and the ones derived from Fourier’s methods are called Fourier series. Much ofthe mathematical research done in the nineteenth and early twentieth centurywas devoted to understanding the convergence of Fourier series. This chapterpresents nothing more than the tip of that huge iceberg.

1. Trigonometric Polynomials

DEFINITION 10.1. A function of the form

p(x) =nX

k=0Æk coskx +Øk sinkx(112)

is called a trigonometric polynomial. The largest value of k such that |Æk |+Øk | 6=0 is the degree of the polynomial. Denote by T the set of all trigonometricpolynomials.

Evidently, all functions in T are 2º-periodic and T is closed under additionand multiplication by real numbers. Indeed, it is a real vector space, in the senseof linear algebra and the set {sinnx : n 2N}[ {cosnx : n 2!} is a basis for T .

The following theorem can be proved using integration by parts or trigono-metric identities.

THEOREM 10.2. If m,n 2Z, thenZº

°ºsinmx cosnx d x = 0,(113)

1Fourier’s methods can be seen in most books on partial differential equations, such as [3]. Forexample, see solutions of the heat and wave equations using the method of separation of variables.

10-1

10-2 CHAPTER 10. FOURIER SERIES

Zº

°ºsinmx sinnx d x =

8><

>:

0, m 6= n

0, m = 0 or n = 0

º m = n 6= 0

(114)

and

Zº

°ºcosmx cosnx d x =

8><

>:

0, m 6= n

2º m = n = 0

º m = n 6= 0

.(115)

If p(x) is as in (112), then Theorem 10.2 shows

2ºÆ0 =Zº

°ºp(x)d x,

and for n > 0,

ºÆn =Zº

°ºp(x)cosnx d x, ºØn =

Zº

°ºp(x)cosnx d x.

Combining these, it follows that if

an = 1º

Zº

°ºp(x)cosnx d x and bn = 1

º

Zº

°ºp(x)sinnx d x

for n 2!, then

p(x) = a0

2+

1X

n=1an cosnx +bn sinnx.(116)

(Remember that all but a finite number of the an and bn are 0!)At this point, the logical question is whether this same method can be used

to represent a more general 2º-periodic function. For any function f , integrableon [°º,º], the coefficients can be defined as above; i.e., for n 2!,

an = 1º

Zº

°ºf (x)cosnx d x and bn = 1

º

Zº

°ºf (x)sinnx d x.(117)

The numbers an and bn are called the Fourier coefficients of f . The problem iswhether and in what sense an equation such as (116) might be true. This turnsout to be a very deep and difficult question with no short answer.2 Because wedon’t know whether equality in the sense of (116) is true, the usual practice is towrite

f (x) ª a0

2+

1X

n=1an cosnx +bn sinnx,(118)

indicating that the series on the right is calculated from the function on the leftusing (117). The series is called the Fourier series for f .

2Many people, including me, would argue that the study of Fourier series has been themost important area of mathematical research over the past two centuries. Huge mathematicaldisciplines, including set theory, measure theory and harmonic analysis trace their lineage back tobasic questions about Fourier series. Even after centuries of study, research in this area continuesunabated.

July 18, 2016 http://math.louisville.edu/ªlee/ira

2. THE RIEMANN LEBESGUE LEMMA 10-3

FIGURE 10.1. This shows f (x) = |x|, s1(x) and s3(x), where sn(x)is the nth partial sum of the Fourier series for f .

EXAMPLE 10.1. Let f (x) = |x|. Since f is an even functions and sinnx is odd,

bn = 1º

Zº

°º|x|sinnx d x = 0

for all n 2N. On the other hand,

an = 1º

Zº

°º|x|cosnx d x =

8<

:º, n = 02(cosnº°1)

n2º, n 2N

for n 2Z. Therefore,

|x|ª º

2° 4cos x

º° 4cos3x

9º° 4cos5x

25º° 4cos7x

49º° 4cos9x

81º+·· ·

(See Figure 10.1.)

There are at least two fundamental questions arising from (118): Does theFourier series of f converge to f ? Can f be recovered from its Fourier series,even if the Fourier series does not converge to f ? These are often called theconvergence and representation questions, respectively. The next few sectionswill give some partial answers.

2. The Riemann Lebesgue Lemma

We learned early in our study of series that the first and simplest convergencetest is to check whether the terms go to zero. For Fourier series, this is always thecase.

THEOREM 10.3 (Riemann-Lebesgue Lemma). If f is a function such thatRb

a fexists, then

limÆ!1

Zb

af (t )cosÆt d t = 0 and lim

Æ!1

Zb

af (t )sinÆt d t = 0.



PROOF. Since the two limits have similar proofs, only the first will be proved.Let "> 0 and P be a generic partition of [a,b] satisfying

0 <Zb

af °D

°f ,P

¢< "

2.

For mi = glb{ f (x) : xi°1 < x < xi }, define a function g on [a,b] by g (x) = mi whenxi°1 ∑ x < xi and g (b) = mn . Note that

Rba g =D

°f ,P

¢, so

0 <Zb

a( f ° g ) < "

2.(119)

Choose

Æ> 4"

nX

i=1|mi |.(120)

Since f ∏ g ,ØØØØZb

af (t )cosÆt d t

ØØØØ=ØØØØZb

a( f (t )° g (t ))cosÆt d t +

Zb

ag (t )cosÆt d t

ØØØØ

∑ØØØØZb

a( f (t )° g (t ))cosÆt d t

ØØØØ+ØØØØZb

ag (t )cosÆt d t

ØØØØ

∑Zb

a( f ° g )+

ØØØØØ1Æ

nX

i=1mi (sin(Æxi )° sin(Æxi°1))

ØØØØØ

∑Zb

a( f ° g )+ 2

Æ

nX

i=1|mi |

Use (119) and (120).

< "

2+ "

2= "

⇤COROLLARY 10.4. If f is integrable on [°º,º] with an and bn the Fourier

coefficients of f , then an ! 0 and bn ! 0.

3. The Dirichlet Kernel

Suppose f is integrable on [°º,º] and 2º-periodic on R, so the Fourier seriesof f exists. The partial sums of the Fourier series are written as sn( f , x), or moresimply sn(x) when there is only one function in sight. To be more precise,

sn( f , x) = a0

2+

nX

k=1(ak coskx +bk sinkx) .

Notice sn is a trigonometric polynomial of degree at most n.We begin with the following calculation.


3. THE DIRICHLET KERNEL 10-5

-p -p

2

p

2p

-3

3

6

9

12

15

FIGURE 10.2. The Dirichlet kernel Dn(s) for n = 1,4,7.

sn(x) = a0

2+

nX

k=1(ak coskx +bk sinkx)

= 12º

Zº

°ºf (t )d t +

nX

k=1

1º

Zº

°º

°f (t )coskt coskx + f (t )sinkt sinkx

¢d t

= 12º

Zº

°ºf (t )

√

1+nX

k=12(coskt coskx + sinkt sinkx)

!

d t

= 12º

Zº

°ºf (t )

√

1+nX

k=12cosk(x ° t )

!

d t

Substitute s = x ° t and use the assumption that f is 2º-periodic.

= 12º

Zº

°ºf (x ° s)

√

1+2nX

k=1cosks

!

d s(121)

The sequence of trigonometric polynomials from within the integral,

Dn(s) = 1+2nX

k=1cosks,(122)

is called the Dirichlet kernel. Its properties will prove useful for determining thepointwise convergence of Fourier series.

THEOREM 10.5. The Dirichlet kernel has the following properties.

(a) Dn(s) is an even 2º-periodic function for each n 2N.(b) Dn(0) = 2n +1 for each n 2N.(c) |Dn(s)|∑ 2n +1 for each n 2N and all s.



(d)1

2º

Zº

°ºDn(s)d s = 1 for each n 2N.

(e) Dn(s) = sin(n +1/2)s

sin s/2for each n 2N and s/2 not an integer multiple

of º.

PROOF. Properties (a)–(d) follow from the definition of the kernel.The proof of property (e) uses some trigonometric manipulation. Suppose

n 2N and s 6= mº for any m 2Z.

Dn(s) = 1+2nX

k=1cosks

Use the facts that the cosine is even and the sine is odd.

=nX

k=°ncosks +

cos s2

sin s2

nX

k=°nsinks

= 1sin s

2

nX

k=°n

≥sin

s

2cosks +cos

s

2sinks

¥

= 1sin s

2

nX

k=°nsin(k + 1

2)s

This is a telescoping sum.

=sin(n + 1

2 )s

sin s2

⇤According to (121),

sn( f , x) = 12º

Zº

°ºf (x ° t )Dn(t )d t .

This is similar to a situation we’ve seen before within the proof of the Weier-strass approximation theorem, Theorem 9.13. The integral given above is aconvolution integral similar to that used in the proof of Theorem 9.13, althoughthe Dirichlet kernel isn’t a convolution kernel in the sense of Lemma 9.14 becauseit doesn’t satisfy conditions (a) and (c) of that lemma. (See Figure 10.3.)

4. Dini’s Test for Pointwise Convergence

THEOREM 10.6 (Dini’s Test). Let f :R!R be a 2º-periodic function integrableon [°º,º] with Fourier series given by (118). If there is a ±> 0 and s 2R such that

Z±

0

ØØØØf (x + t )+ f (x ° t )°2s

t

ØØØØ d t <1,

thena0

2+

1X

k=1(ak coskx +bk coskx) = s.


4. DINI’S TEST FOR POINTWISE CONVERGENCE 10-7

FIGURE 10.3. This graph shows D50(t) and the envelope y =±1/sin(t/2). As n gets larger, the Dn(t) fills the envelope more com-pletely.

PROOF. Since Dn is even,

sn(x) = 12º

Zº

°ºf (x ° t )Dn(t )d t

= 12º

Z0

°ºf (x ° t )Dn(t )d t + 1

2º

Zº

0f (x ° t )Dn(t )d t

= 12º

Zº

0

°f (x + t )+ f (x ° t )

¢Dn(t )d t .

By Theorem 10.5(d) and (e),

sn(x)° s = 12º

Zº

0

°f (x + t )+ f (x ° t )°2s

¢Dn(t )d t

= 12º

Zº

0

f (x + t )+ f (x ° t )°2s

t· t

sin t2

· sin(n + 12

)t d t .

Since t/sin t2 is bounded on (0,º), Theorem 10.3 shows sn(x)° s ! 0. Now use

Corollary 8.11 to finish the proof. ⇤

EXAMPLE 10.2. Suppose f (x) = x for °º < x ∑ º and is 2º-periodic on R.Since f is odd, an = 0 for all n. Integration by parts gives bn = (°1)n+12/n forn 2N. Therefore,

f ª1X

n=1(°1)n+1 2

nsinnx.

For x 2 (°º,º), let 0 < ±< min{º° x,º+ x}. (This is just the distance from x toclosest endpoint of (°º,º).) Using Dini’s test, we see

Z±

0

ØØØØf (x + t )+ f (x ° t )°2x

t

ØØØØ d t =Z±

0

ØØØØx + t +x ° t °2x

t

ØØØØ d t = 0 <1,



-p -p

2

p

2p 2 p

-p

-p

2

p

2

p

FIGURE 10.4. This plot shows the function of Example 10.2 ands8(x) for that function.

so1X

n=1(°1)n+1 2

nsinnx = x for °º< x <º.(123)

In particular, when x =º/2, (123) gives another way to derive (105). When x =º,the series converges to 0, which is the middle of the “jump” for f .

This behavior of converging to the middle of a jump discontinuity is typical.To see this, denote the one-sided limits of f at x by

f (x°) = limt"x

f (t ) and f (x+) = limt#x

f (t ),

and suppose f has a jump discontinuity at x with

s = f (x°)+ f (x+)2

.

Guided by Dini’s test, consider

Z±

0

ØØØØf (x + t )+ f (x ° t )°2s

t

ØØØØ d t

=Z±

0

ØØØØf (x + t )+ f (x ° t )° f (x°)° f (x+)

t

ØØØØ d t

∑Z±

0

ØØØØf (x + t )° f (x+)

t

ØØØØ d t +Z±

0

ØØØØf (x ° t )° f (x°)

t

ØØØØ d t

If both of the integrals on the right are finite, then the integral on the left is alsofinite. This amounts to a proof of the following corollary.


5. GIBBS PHENOMENON 10-9

COROLLARY 10.7. Suppose f :R!R is 2º-periodic and integrable on [°º,º].If both one-sided limits exist at x and there is a ±> 0 such that both

Z±

0

ØØØØf (x + t )° f (x+)

t

ØØØØ d t <1 andZ±

0

ØØØØf (x ° t )° f (x°)

t

ØØØØ d t <1,

then the Fourier series of f converges to

f (x°)+ f (x+)2

.

The Dini test given above provides a powerful condition sufficient to ensurethe pointwise convergence of a Fourier series. There is a plethora of ever moreabstruse conditions that can be proved in a similar fashion to show pointwiseconvergence.

The problem is complicated by the fact that there are continuous functionswith Fourier series divergent at a point and integrable functions with Fourierseries diverging everywhere [13]. In Section 6, a continuous function whoseFourier series diverges is constructed.

5. Gibbs Phenomenon

For x 2 [°º,º) define

f (x) =( |x|

x , 0 < |x| <º0, x = 0,º

(124)

and extend f 2º-periodically to all of R. This function is often called a squarewave. A straightforward calculation gives

f ª 4º

1X

k=1

sin(2k °1)x

2k °1.

Corollary 10.7 shows sn(x) ! f (x) everywhere. This convergence cannot be uni-form because all the partial sums are continuous and f is discontinuous at everyinteger multiple of º. A plot of s19(x) is shown in Figure 10.5. Notice the higherpeaks in the oscillation of sn(x) just before and after the jump discontinuitiesof f . This behavior is not unique to f , as it can also be seen in Figure 10.4. If afunction is discontinuous at some point, the partial sums of its Fourier serieswill always have such higher peaks near that point. This behavior is called Gibbsphenomenon.3

Instead of doing a general analysis of Gibbs phenomenon, we’ll only analyzethe simple case shown in the square wave f . It’s basically a calculus exercise.

3It is named after the American mathematical physicist, J. W. Gibbs, who pointed it out in 1899.He was not the first to notice the phenomenon, as the British mathematician Henry Wilbrahamhad published a little-noticed paper on it it 1848. Gibbs’ interest in the phenomenon was sparkedby investigations of the American experimental physicist A. A. Michelson who wrote a letter toNature disputing the possibility that a Fourier series could converge to a discontinuous function.The ensuing imbroglio is recounted in a marvelous book by Paul J. Nahin [16].



FIGURE 10.5. This is a plot of s19( f , x), where f is defined by (124).

FIGURE 10.6. This is a plot of s9( f , x), where f is defined by (124).

To locate the peaks in the graph, differentiate the partial sums.

s02n°1(x) = d

d x

4º

nX

k=1

sin(2k °1)x

2k °1= 4º

nX

k=1cos(2k °1)x

It is left as Exercise 10.10 to show this has a closed form.

s02n°1(x) = 2º

sin2nx

sin x

Looking at the numerator, we see s02n°1(x) has critical numbers at x = kº/2nfor k 2Z. In the interval (0,º), s2n°1(kº/2n) is a relative maximum for odd k anda relative minimum for even k. (See Figure 10.6.) The value s2n°1(º/2n) is theheight of the left-most peak. What is the behavior of these maxima?

From Figure 10.7 it appears they have an asymptotic limit near 1.179. Toprove this, consider the following calculation.

s2n°1

≥f ,

º

2n

¥= 4º

nX

k=1

sin°(2k °1) º

2n

¢

2k °1

= 2º

nX

k=1

sin≥

(2k°1)º2n

¥

(2k°1)º2n

º

n


6. A DIVERGENT FOURIER SERIES 10-11

The last sum is a midpoint Riemann sum for the function sin xx on the interval

[0,º] using a regular partition of n subintervals. Example 9.16 shows

2º

Zº

0

sin x

xd x º 1.17898.

Since f (0+)° f (0°) = 2, this is an overshoot of a bit less than 9%. There is asimilar undershoot on the other side. It turns out this is typical behavior at pointsof discontinuity [21].

6. A Divergent Fourier Series

This is an advanced section that can be omitted.Eighteenth century mathematicians, including Fourier, Cauchy, Euler, Weier-

strass, Lagrange and Riemann, had computed the Fourier series for many func-tions. They believed from these examples that the Fourier series of a continuousfunction must converge to that function. Fourier went far beyond this claim in hisimportant 1822 book Théorie Analytique de la Chaleur. Cauchy even published aflawed proof of this “fact.”

Finally, in 1873, Paul du Bois-Reymond, settled the question by giving theconstruction of a continuous function whose Fourier series diverges at a point[17]. It was finally shown in 1966 by Lennart Carleson [7] that the Fourier series ofa continuous function converges to that function everywhere with the exceptionof a set of measure zero.

The problems around the convergence of Fourier series motivated a hugeamount of research that is still going on today. In this section, we look at the tip ofthat iceberg by presenting a continuous function F (t ) whose Fourier series failsto converge when t = 0.

6.1. The Conjugate Dirichlet Kernel.

FIGURE 10.7. This is a plot of sn( f ,º/2n) for n = 1,2, · · · ,100. Thedots come in pairs because s2n°1( f ,º/2n) = s2n( f ,º/2n).



LEMMA 10.8. If m,n 2! and 0 < |t | <º for k 2Z, then

nX

k=msinkt =

cos(m ° 12 )t °cos(m + 1

2 )t

2sin t2

.

PROOF.nX

k=mkt = 1

sin t2

nX

k=msin

12

sinkt

= 1

sin t2

nX

k=m

µcos

µk ° 1

2

∂t °cos

µk + 1

2

∂t

∂

=cos

°m ° 1

2

¢t °cos

°n + 1

2

¢t

2sin t2

⇤

DEFINITION 10.9. The conjugate Dirichlet kernel4 is

D̃n (t ) =nX

k=1sinkt , n 2N.(125)

We’ll not have much use for the conjugate Dirichlet kernel, except as a conve-nient way to refer to sums of the form (125).

Lemma 10.8 immediately gives the following bound.

COROLLARY 10.10. If 0 < |t | <º, then

ØØD̃n (t )ØØ∑ 1ØØsin t

2

ØØ .

6.2. A Sawtooth Wave. If the function f (x) = (º°x)/2 on [0,2º) is extended2º-periodically to R, then the graph of the resulting function is often referred toas a “sawtooth wave”. It has a particularly nice Fourier series:

º°x

2ª

1X

k=1

sinkx

k.

According to Corollary 10.7

1X

k=1

sinkx

k=

(0, x = 2nº, n 2Zf (x), otherwise

.

We’re interested in various partial sums of this series.

LEMMA 10.11. If m,n 2! with m ∑ n and 0 < |t | < 2º, thenØØØØØ

nX

k=m

sinkt

k

ØØØØØ∑1

mØØsin t

2

ØØ .

4In this case, the word “conjugate” does not refer to the complex conjugate, but to the har-monic conjugate. They are related by Dn (t )+ i D̃n (t ) = 1+2

Pnk=1 eki t .


6. A DIVERGENT FOURIER SERIES 10-13

PROOF.ØØØØØ

nX

k=m

sinkt

k

ØØØØØ=ØØØØØ

nX

k=m

°D̃k (t )° D̃k°1 (t )

¢ 1k

ØØØØØ

Use summation by parts.

=ØØØØØ

nX

k=mD̃k (t )

µ1k° 1

k +1

∂+ D̃n (t )

n +1° D̃n°1 (t )

m

ØØØØØ

∑ØØØØØ

nX

k=mD̃k (t )

µ1k° 1

k +1

∂ØØØØØ+ØØØØ

D̃n (t )n +1

ØØØØ+ØØØØ

D̃n°1 (t )m

ØØØØ

Apply Corollary 10.10.

∑ 1

2sin t2

√nX

k=m

µ1k° 1

k +1

∂1

n +1+ 1

m

!

= 1

2sin t2

µ1m

° 1n +1

+ 1n +1

+ 1m

∂

= 1

m sin t2

.

⇤

PROPOSITION 10.12. If n 2N and 0 < |t | <º, thenØØØØØ

nX

k=1

sinkt

k

ØØØØØ∑ 1+º.

PROOF.ØØØØØ

nX

k=1

sinkt

k

ØØØØØ=

ØØØØØØ

X

1∑k∑ 1t

sinkt

k+

X

1t <k∑n

sinkt

k

ØØØØØØ

∑

ØØØØØØ

X

1∑k∑ 1t

sinkt

k

ØØØØØØ+

ØØØØØØ

X

1t <k∑n

sinkt

k

ØØØØØØ

∑X

1∑k∑ 1t

kt

k+ 1

1t sin t

2

∑X

1∑k∑ 1t

t + 11t

2º

t2

∑ 1+º

⇤



6.3. A Continuous Function with a Divergent Fourier Series. The goal inthis subsection is to construct a continuous function F such that limsup sn(F,0) =1. This implies sn(F,0) does not converge.

EXAMPLE 10.3. The first step in the construction is to define a sequence oftrigonometric polynomials that form the building blocks for F .

fn(t ) = 1n+ cos t

n °1+ cos2t

n °2+·· · cos(n °1)t

1

° cos(n +1)t

1° cos(n +2)t

2° · · ·° cos2nt

n

Note that,(

sm( fn ,0) = fn(0) = 0, m ∏ 2n

sm( fn ,0) > 0, 1 ∑ m < 2n.(126)

Rearrange the sum in the definition of fn to see

fn(t ) =nX

k=1

cos(n °k)t °cos(n +k)t

k

= 2sinntnX

k=1

sinkt

k.

This closed form for fn combined with Proposition 10.12 implies the sequence offunctions fn is uniformly bounded:

ØØ fn(t )ØØ=

ØØØØØ2sinntnX

k=1

sinkt

k

ØØØØØ∑ 2+2º.(127)

At last, the main function can be defined.

F (t ) =1X

n=1

f2n3 (t )

n2

The Weierstrass M-Test along with (127) implies F is uniformly convergentand therefore continuous on R. Using (126), consider

s2m3 (F,0) = 12º

Zº

°ºF (t )D2m3 (t )d t

= 12º

Zº

°º

1X

n=1

f2n3 (t )

n2 D2m3 (t )d t

The uniform convergence allows the sum and integration to be reordered.

=1X

n=1

1n2

12º

Zº

°ºf2n3 (t )D2m3 (t )d t

=1X

n=1

1n2 s2m3

°f2n3 ,0

¢


7. THE FEJÉR KERNEL 10-15

Use (126).

=1X

n=m

1n2 s2m3

°f2n3 ,0

¢

> 1m2 s2m3

°f2m3 ,0

¢

= 1m2

2m3X

k=1

1k

> 1m2 ln2m3

= m ln2

This implies, limsup sn(F,0) ∏ limm!1 m ln2 =1, so sn(F,0) does not converge.

7. The Fejér Kernel

Since pointwise convergence of the partial sums seems complicated, whynot change the rules of the game? Instead of looking at the sequence of partialsums, consider a rolling average instead:

æn( f , x) = 1n +1

nX

k=0sn( f , x).

The trigonometric polynomialsæn( f , x) are called the Cesàro means of the partialsums. If limn!1æn( f , x) exists, then the Fourier series for f is said to be (C ,1)summable at x. The idea is that this averaging will “smooth out” the partialsums, making them more nicely behaved. It is not hard to show that if sn( f , x)converges at some x, then æn( f , x) will converge to the same thing. But there aresequences for which æn( f , x) converges and sn( f , x) does not. (See Exercises 3.21and 4.4.25.)

As with sn(x), we’ll simply write æn(x) instead of æn( f , x), when it is clearwhich function is being considered.

We start with a calculation.

æn(x) = 1n +1

nX

k=0sk (x)

= 1n +1

nX

k=0

12º

Zº

°ºf (x ° t )Dk (t )d t

= 12º

Zº

°ºf (x ° t )

1n +1

nX

k=0Dk (t )d t(*)

= 12º

Zº

°ºf (x ° t )

1n +1

nX

k=0

sin(k +1/2)t

sin t/2d t

= 12º

Zº

°ºf (x ° t )

1

(n +1)sin2 t/2

nX

k=0sin t/2 sin(k +1/2)t d t



-p - p2

p2

p

2

4

6

8

10

FIGURE 10.8. A plot of K5(t ), K8(t ) and K10(t ).

Use the identity 2sin A sinB = cos(A°B)°cos(A+B).

= 12º

Zº

°ºf (x ° t )

1/2

(n +1)sin2 t/2

nX

k=0(coskt °cos(k +1)t ) d t

The sum telescopes.

= 12º

Zº

°ºf (x ° t )

1/2

(n +1)sin2 t/2(1°cos(n +1)t ) d t

Use the identity 2sin2 A = 1°cos2A.

= 12º

Zº

°ºf (x ° t )

1(n +1)

√sin n+1

2 t

sin t2

!2

d t(**)

The Fejér kernel is the sequence of functions highlighted above; i.e.,

Kn(t ) = 1(n +1)

√sin n+1

2 t

sin t2

!2

, n 2N.(128)

Comparing the lines labeled (*) and (**) in the previous calculation, we see an-other form for the Fejér kernel is

Kn(t ) = 1n +1

nX

k=0Dk (t ).(129)


7. THE FEJÉR KERNEL 10-17

Once again, we’re confronted with a convolution integral containing a kernel:

æn(x) = 12º

Zº

°ºf (x ° t )Kn(t )d t .

THEOREM 10.13. The Fejér kernel has the following properties.5

(a) Kn(t ) is an even 2º-periodic function for each n 2N.(b) Kn(0) = n +1 for each n 2!.(c) Kn(t ) ∏ 0 for each n 2N.(d) 1

2º

Rº°ºKn(t )d t = 1 for each n 2!.

(e) If 0 < ±<º, then Kn ‚ 0 on [°º,±][ [±,º].(f) If 0 < ±<º, then

R±°ºKn(t )d t ! 0 and

Rº± Kn(t )d t ! 0.

PROOF. Theorem 10.5 and (129) imply (a), (b) and (d). Equation (128) implies(c).

Let ± be as in (e). In light of (a), it suffices to prove (e) for the interval [±,º].Noting that sin t/2 is decreasing on [±,º], it follows that for ±∑ t ∑º,

Kn(t ) = 1(n +1)

√sin n+1

2 t

sin t2

!2

∑ 1(n +1)

√1

sin t2

!2

∑ 1(n +1)

1

sin2 ±2

! 0

It follows that Kn ‚ 0 on [±,º] and (e) has been proved.Theorem 9.16 and (e) imply (f). ⇤

THEOREM 10.14 (Fejér). If f :R!R is 2º-periodic, integrable on [°º,º] andcontinuous at x, then æn(x) ! f (x).

PROOF. Since f is 2º-periodic andRº°º f (t )d t exists, so does

Rº°º

R( f (x ° t )°

f (x))d t . Theorem 8.3 gives an M > 0 so | f (x ° t )° f (x)| < M for all t .Let "> 0 and choose ±> 0 such that

ØØ f (x)° f (y)ØØ< "/3 whenever

ØØx ° yØØ< ±.

By Theorem 10.13(f), there is an N 2N so that whenever n ∏ N ,

12º

Z±

°ºKn(t )d t < "

3Mand

12º

Zº

±Kn(t )d t < "

3M.

5Compare this theorem with Lemma 9.14.



We start calculating.

|æn(x)° f (x)| =ØØØØ

12º

Zº

°ºf (x ° t )Kn(t )d t ° 1

2º

Zº

°ºf (x)Kn(t )d t

ØØØØ

= 12º

ØØØØZº

°º( f (x ° t )° f (x))Kn(t )d t

ØØØØ

= 12º

ØØØØZ°±

°º( f (x ° t )° f (x))Kn(t )d t +

Z±

°±( f (x ° t )° f (x))Kn(t )d t

+Zº

±( f (x ° t )° f (x))Kn(t )d t

ØØØØ

∑ØØØØ

12º

Z°±

°º( f (x ° t )° f (x))Kn(t )d t

ØØØØ+ØØØØ

12º

Z±

°±( f (x ° t )° f (x))Kn(t )d t

ØØØØ

+ØØØØ

12º

Zº

±( f (x ° t )° f (x))Kn(t )d t

ØØØØ

< M

2º

Z°±

°ºKn(t )d t + 1

2º

Z±

°±| f (x ° t )° f (x)|Kn(t )d t + M

2º

Zº

±Kn(t )d t

< M"

3M+ "

31

2º

Z±

°±Kn(t )d t +M

"

3M< "

This shows æn(x) ! f (x). ⇤

Theorem 10.14 gives a partial solution to the representation problem.

COROLLARY 10.15. Suppose f and g are continuous and 2º-periodic on R. Iff and g have the same Fourier coefficients, then they are equal.

PROOF. By assumption, æn( f , t) = æn(g , t) for all n and Theorem 10.14 im-plies

0 =æn( f , t )°æn(g , t ) ! f ° g .

⇤

In the case of continuous functions, the convergence is uniform, rather thanpointwise.

THEOREM 10.16 (Fejér). If f : R! R is 2º-periodic and continuous, thenæn(x) ‚ f (x).

PROOF. By Exercise 6.31, f is uniformly continuous. This can be used to showthe calculation within the proof of Theorem 10.14 does not depend on x. Thedetails are left as Exercise 10.12. ⇤

A perspicacious reader will have noticed the similarity between Theorem10.16 and the Weierstrass approximation theorem, Theorem 9.13. In fact, theWeierstrass approximation theorem can be proved from Theorem 10.16 usingpower series and Theorem 9.24. (See Exercise 10.13.)


8. EXERCISES 10-19

-p -p

2

p

2p 2 p

-p

-p

2

p

2

p

-p -p

2

p

2p 2 p

-p

-p

2

p

2

p

FIGURE 10.9. These plots illustrate the functions of Example 10.4. Onthe left are shown f (x), s8(x) and æ8(x). On the right are shown f (x),æ3(x), æ5(x) and æ20(x). Compare this with Figure 10.4.

EXAMPLE 10.4. As in Example 10.2, let f (x) = x for °º< x ∑º and extend f tobe periodic on Rwith period 2º. Figure 10.9 shows the difference between the Fe-jér and classical methods of summation. Notice that the Fejér sums remain muchmore smoothly affixed to the function and do not show Gibbs phenomenon.

8. Exercises

10.1. Prove Theorem 10.2.

10.2. Suppose f : R! R is 2º-periodic and integrable on [°º,º]. If b °a = 2º,then

Rba f =

Rº°º f .

10.3. Let f : [°º,º) be a function with Fourier coefficients an and bn . If f is odd,then an = 0 for all n 2!. If f is even, then bn = 0 for all n 2N.

10.4. If f (x) = sgn(x) on [°º,º), then find the Fourier series for f .

10.5. Is1X

n=1sinnx the Fourier series of some function?

10.6. Let f (x) = x2 when °º∑ x <º and extend f to be 2º-periodic on all of R.Use Dini’s Test to show sn( f , x) ! f (x) everywhere.

10.7. Use Exercise 10.6 to proveº2

6=

1X

n=1

1n2 .

10.8. Suppose f is integrable on [°º,º]. If f is even, then the Fourier series for fhas no sine terms. If f is odd, then the Fourier series for f has no cosine terms.

10.9. If f : R! R is periodic with period p and continuous on [0, p], then f isuniformly continuous.



10.10. ProvenX

k=1cos(2k °1)t = sin2nt

2sin t

for t 6= kº, k 2Z and n 2N. (Hint: 2Pn

k=1 cos(2k °1)t = D2n(t )°Dn(2t ).)

10.11. The function g (t) = t/sin(t/2) is undefined whenever t = 2nº for somen 2Z. Show that it can be redefined on the set {2nº : n 2Z} to be periodic anduniformly continuous on R.

10.12. Prove Theorem 10.16.

10.13. Prove the Weierstrass approximation theorem using Fourier series andTaylor’s theorem.

10.14. If f (x) = x for °º ∑ x < º, can the Fourier series of f be integratedterm-by-term on (°º,º) to obtain the correct answer?

10.15. If f (x) = x for °º ∑ x < º, can the Fourier series of f be differentiatedterm-by-term on (°º,º) to obtain the correct answer?


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