Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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14.4 Pointwise Convergence of Fourier Series 483

For the integral from δ to π, we have

f(θ + t) − f (θ

)

(t) dt =

f(θ + t) − f (θ

)

2π sint/2

sin(n +

)t dt

g(t) sin(n +

)t dt,

where

g(t) =

f(θ + t) − f (θ

)

2π sint/2

As g is piecewisecontinuous on [δ, π], we can apply the Riemann-Lebesgue Lemma

to this last integral. Thus, for all n sufﬁciently large,

f(θ + t) − f (θ

)

(t) dt

g(t) sin(n +

)t dt

Combining these two estimates, we have

f(θ + t) − f (θ

)

(t) dt

< ε

for all n sufﬁciently large, and thus the limit is zero. ¥

14.4.6. EXAMPLE. Let h be the following variant on the Heaviside step func-

tion:

h(x) =

(

−1 if −π < x < 0

1 if 0 ≤ x ≤ π.

Evidently, this function is piecewise C

. Let us compute its Fourier series. Since h

is odd, it has only sine terms.

−π

h(x) sinnx dx

sinnx dx

−2

nπ

cosnx

(

0 if n is even

nπ

if n is odd

Thus h ∼

∞

k=0

(2k + 1)π

sin(2k + 1)x.

The Dirichlet–Jordan Theorem tells us that this series converges to 1 for all

0 < x < π and to −1 for −π < x < 0. At the points of discontinuity, it converges

to the average 0. This latter fact is clear since sin kπ = 0 for all integers k. Let us

plug in a few points. For example, take x = π/2. Since sin(2k + 1)π/2 = (−1)

1 = h

∞

k=0

(2k + 1)π

(−1)

484 Fourier Series and Approximation

Therefore,

1 −

−

+ ··· =

Similarly, plugging in x = 1, we obtain

sin1 +

sin3 +

sin5 + ··· =

Both of these series converge exceedingly slowly. So they have no real computa-

tional value.

We will consider this function again in Example 14.6.5; see Figure 14.5 for a

graph of h and S

Exercises for Section 14.4

A. (a) Compute the Fourier series for f(x) = x for −π ≤ x ≤ π.

(b) Find the sum of the series.

B. (a) Find the Fourier series for f(θ) = sinh(θ) for |θ| ≤ π.

(b) Find a constant c so that the Fourier series for f(θ) − cθ converges uniformly on

[−π, π].

C. Show that

∞

n=2

(−1)

− 1

sinnx is the Fourier series of a piecewise C

function.

HINT: Use Exercise A to subtract a multiple of x to leave the uniformly convergent

Fourier series of a C

function.

D. Sum the Dirichlet kernel using complex exponentials.

E. Sum the series

∞

n=1

sinnθ

HINT: Exercise A and Exercise 13.3.D

F. (a) Show that the function h(x) = cos(x/2) for −π ≤ x ≤ π is continuous and

piecewise C

(b) Find the Fourier series for h. HINT: A

4(−1)

n−1

π(4n

− 1)

result.

HINT:

− 1

2n − 1

−

2n + 1

G. Prove Dini’s Test: If f is a piecewise continuous 2π-periodic function such that

f(θ

+ t) + f(θ

− t)

− s

dt < ∞,

then lim

n→∞

f(θ

) = s.

HINT: Study the proof of the Dirichlet–Jordan Theorem. Look for a place where this

integral condition may be used in lieu of the Lipshitz condition.

14.5 Gibbs’s Phenomenon 485

14.5. Gibbs’s Phenomenon

In this section, we show that pointwise convergence of Fourier series, which we

established in the previous section, is not good enough for many applications. In

particular, a sequence of functions can converge pointwise without “looking like”

their limit.

We have seen in Section 8.1 that pointwise convergence of functions allows

surprisingly bad behaviour. Such a phenomenon arises for the functions S

f near

any jump discontinuity of f. This was ﬁrst discovered by an English mathemati-

cian, Wilbraham, in 1848. Around the turn of the century, it was rediscovered by

Michelson and then explained by Gibbs, a (now) famous American physicist, in a

letter to the journal Nature. For a discussion of this history, see [43]. Put simply,

whenever f has a jump discontinuity, the graphs of S

f overshoot f near the dis-

continuity and increasing n does not reduce the error, it only pushes the overshoot

nearer to the discontinuity. See Figure 14.3, for example.

As an example, we demonstrate the phenomenon for the 2π-periodic function

given by

f(x) =

(

x if x ∈ (−π, π),

0 if x = ±π.

By the Dirichlet–Jordan Theorem (Theorem 14.4.5), lim

n→∞

f(x) = f(x), for

all x ∈ R. Nonetheless, S

f(x) always overshoots f(x) at some point near the

discontinuity by about 9% of the gap (which is 2π in this case).

–2

–2 2

–2

–2 2

FIGURE 14.3. The graphs of S

f and S

f, each plotted with f.

486 Fourier Series and Approximation

14.5.1. THEOREM. Let A =

sin(x)

dx ≈ 1.178979744. For the func-

tion f deﬁned previously, we have

lim

n→∞

π(1 −

)

= Aπ and lim

n→∞

− π(1 −

)

= −Aπ.

PROOF. Note that f is an odd function and hence has a sine series. An integration

by parts argument (see Exercise 14.4.A) shows that

f(x) ∼ 2

∞

k=1

(−1)

k+1

sinkx.

Thus,

π(1 −

)

= 2

k=1

(−1)

k+1

sin

kπ −

kπ

= 2

k=1

(−1)

k+1

sinkπ cos

kπ

− coskπ sin

kπ

and since sinkπ = 0 and cos kπ = (−1)

= 2

k=1

sin

kπ

k=1

2sin(kπ/n)

kπ/n

Remembering the formula for Riemann sums, we observe that this is the Riemann

sum for the integral of the function 2 sinx/x on the interval [0, π] using n+1 evenly

spaced points for the partition. Since lim

x→0

sinx/x = 1, this function is bounded and

continuous on [0, π]. Therefore, the Riemann sums converge to the integral. Thus,

we have

lim

n→∞

π(1 −

)

2sinx

dx = πA

and, similarly,

lim

n→∞

− π(1 −

)

= −

2sinx

dx = −πA.

It remains to estimate the integral A. The function (sinx)/x does not have a

closed form integral. However, we can get good mileage out of the Taylor series

for sinx because it converges so rapidly. Indeed, we obtain that

sinx

∞

k=0

(−1)

(2k + 1)!

14.5 Gibbs’s Phenomenon 487

for all real x. Since this converges uniformly on [0, π], we may integrate term by

term by Theorem 8.3.1. Therefore,

A =

sin(x)

dx =

∞

k=0

(−1)

(2k + 1)!

∞

k=0

(−1)

(2k + 1)!

dx =

∞

k=0

(−1)

(2k + 1)!

2k+1

(2k + 1)

= 2

∞

k=0

(−1)

(2k + 1)!(2k + 1)

= 2 −

300

−

17640

+ . . . .

Note that this is an alternating series in which the terms decrease monotonely

to 0. Therefore,

1.17357 ≈ 2 −

300

−

17640

< A

< 2 −

300

−

17640

1632960

≈ 1.17938.

This is enough for our purposes. However, using Maple, we found the integral to

be approximately 1.178979744. ¥

Gibbs’s phenomenon depends on the ﬁnite Fourier series being a best L

ap-

proximation; see [44] for a related example using using piecewise linear approx-

imations, much like Section 10.8 except that the L

norm replaces the uniform

norm.

Exercises for Section 14.5

A. (a) Suppose that h is a C

function on [−π, π] with h(π) = h(−π) but possibly

(π) 6= h

(−π). Show that S

h converges uniformly to h.

(b) Suppose that g is a C

function on [−π, π] but g(π) 6= g(−π). Subtract a multiple

of the function f used in this section from g to obtain a function h as in part (a).

B. (a) Follow through the proof of Gibbs’s phenomenon and show that

lim

n→∞

π −

sinx

dx.

(b) Let t

nπ

(n−1)π

sinx

dx. Show that t

alternates in sign, |t

n+1

| < |t

| and

lim

n→∞

= 0.

a>0

sinx

dx.

(d) Establish the existence of the improper Riemann integral

∞

sinx

dx = lim

a→∞

sinx

dx.

488 Fourier Series and Approximation

C. (a) Use the formula for D

as a sum of cosines to show that

π−a

2πD

(x) dx = π − a + S

f(a),

where 0 < a < π and f(x) = x for −π ≤ x ≤ π.

(b) Hence show that

f(a) − a| =

(n+

)(π−a)

2sinx

(2n + 1) sin

2n+1

dx − π

(n+

)

2sin x

(2n + 1) sin

2n+1

−

2sinx

dx =

g(x) sin(n +

)x dx

(for a certain continuous function g) tends to 0 as n goes to +∞.

(d) Use the Dirichlet–Jordan Theorem to deduce that

∞

sinx

dx =

14.6. Ces

aro Summation of Fourier Series

Although it is natural to try to recombine the harmonics of a function f sim-

ply by adding the ﬁrst number of terms, we have seen that such functions are not

good approximations to f in a number of ways. In this section we consider a new

sequence of approximations, built from the Fourier coefﬁcients of f, that converge

uniformly to the function f for all continuous functions f . The results of this sec-

tion were found about 1900 by Fej

er, a Hungarian mathematician, at the age of 19.

In order to obtain this better behaviour, we replace the sequence of functions

f with their averages, known as Ces

aro means:

f(x) =

n + 1

k=0

f(x).

This is deﬁned whenever the Fourier coefﬁcients of f are deﬁned, which includes

all absolutely integrable functions. Our primary interest will be for continuous

functions. This new sequence of functions also has an associated kernel that is

much better behavedthan the Dirichlet kernel. Itshares manyof the good properties

of the Poisson kernel, and computing σ

f does not require an inﬁnite sum. Indeed,

f(x) =

n + 1

k=0

j=0

cosjx + B

sinjx

= A

j=0

1 −

n + 1

´³

cosjx + B

sinjx

Our ﬁrst result is to turn this summation into an integral formula, using the

following sequence of functions.

14.6 Ces

aro Summation of Fourier Series 489

14.6.1. DEFINITION. The sequence of functions K

given by

(t) =











n + 1

2π

if t = 2mπ, m ∈ Z,

2π(n + 1)

sin

n+1

sint/2

if t 6= 2mπ,

for n = 1, 2, . . ., is called the Fej

er kernel.

14.6.2. THEOREM. If f : R → R is piecewise continuous and 2π-periodic,

then

f(x) =

−π

f(x + t)K

(t) dt.

PROOF. Using Theorem 14.4.2, we have

f(x) =

n + 1

k=0

−π

f(t + x)

sin(k +

2π sint/2

−π

f(t + x)

2π(n + 1) sint/2

k=0

sin(k +

)t dt.

So all we need to establish is the trigonometric identity

k=0

sin(k +

)t sin t/2 =

sin

n + 1

as we can then divide through by sin

t/2 and substitute into the preceding integral

to obtain the equality in the statement of the theorem.

One method for computing this sum is to use complex exponentials to obtain a

geometric series as in Lemma 13.10.4. Here we will give a real variable argument

using trig identities. Applying the identity 2sinA sinB = cos(A−B)−cos(A+B)

with A = k +

and B = k/2 to the left-hand side of the identity gives

k=0

sin(k +

)t sin t/2 =

k=0

coskt − cos(k + 1)t

1 − cos(n + 1)t

sin

n + 1

where we have used the half-angle formula in the last step. ¥

You should compare Figure 14.4 with the graphs of the Poisson and Dirichlet

kernels, Figures 13.2 and 14.2. The key difference between the kernels K

and

is that the K

are positive. Moreover, for t /∈ 2πZ, K

(t) → 0 as n goes to

inﬁnity, unlike D

(t). However, K

(2πm) → ∞ as n goes to inﬁnity, exactly like

. It is helpful to think of the K

as functions that become more and more like

spikes (i.e., large at zero and small elsewhere), as n goes to inﬁnity. All of these

properties are shared with the Poisson kernel. This parallel allows us to obtain

490 Fourier Series and Approximation

similar convergence results. As before, we collect the important properties of the

in a proposition.

14.6.3. PROPERTIES OF THE FEJ

ER KERNEL. (1) For each n, K

a positive, continuous, 2π-periodic, even function.

(2)

−π

(t) dt =

−π

(t)|dt = 1.

(3) For δ ∈ (0, π), lim

n→∞

−δ

−π

= 0.

(4) Moreover, if δ ∈ (0, π), then K

converges uniformly to zero on

[−π, −δ] ∪ [δ, π].

PROOF. It is evident from the formula that K

is positive, even, 2π-periodic, and

continuous except possibly at multiples of 2π. Because of the periodicity, it sufﬁces

to check continuity at 0. This is a simple consequence of the well-known limit

lim

t→0

sinat

= a. Hence

lim

t→0

(t) = lim

t→0

2π(n + 1)

sin

n+1

sint/2

= lim

t→0

2π(n + 1)

sin

n+1

sint/2

2π(n + 1)

n + 1

2π

For (2), taking f = 1 in Theorem 14.6.2, we have

−π

(t) dt = σ

f(0) = 1.

For (3), we could use the analogous fact for the D

and the averaging property.

However, (3) is an immediate consequence of (4). For (4), we let ε > 0 and observe

that |sint/2| ≥ sin δ/2 for t such that δ ≤ |t| ≤ π. Thus,

(t)| ≤

2(n + 1)

|sinδ/2|

for all t ∈ [−π, −δ] ∪ [δ, π].

As δ is ﬁxed, if we choose any N ≥ ε/(2 sin δ/2), then for all n ≥ N ,

(t)| ≤ ε for all t ∈ [−π, −δ] ∪ [δ, π].

That is, K

converges uniformly to zero on [−π, −δ] ∪ [δ, π]. ¥

We can now prove the main result of this section.

14.6.4. FEJ

ER’S THEOREM.

If f is continuous and 2π-periodic, then σ

f converges uniformly to f.

14.6 Ces

aro Summation of Fourier Series 491

–2 2

FIGURE 14.4. The graphs of K

, K

, and K

492 Fourier Series and Approximation

PROOF. Conceptually, this proof is much like the proof of Poisson’s Theorem. We

write σ

f(x) as an integral from −π to π and then split the integral into two parts:

the interval [−δ, δ], and second, the rest, namely [−π, −δ] ∪[δ, π]. On the ﬁrst part,

we control the integral using the fact that f is continuous; on the second, we control

the integral by using the fact that K

converges uniformly to zero.

Let M = kfk

∞

= max{|f(x)| : x ∈ [−π, π]} and let ε > 0. As f is con-

tinuous on the compact set [−π, π], it is uniformly continuous by Theorem 5.5.9.

Hence there is some δ > 0 so that

|f(x) − f(y)| <

whenever |x − y| < δ.

With this δ ﬁxed, we apply Theorem 14.6.3 (4), to conclude that there is an integer

N so that

(x) <

8πM

for all x ∈ [−π, −δ] ∪ [δ, π] and n ≥ N.

Using Theorem 14.6.2 and Theorem 14.6.3 (2), we have

|σ

f(x) − f(x)| =

−π

f(x + t)K

(t) dt − f(x)

−π

(t) dt

−π

f(x + t) − f (x)

(t) dt

≤

−π

|f(x + t) − f (x)|K

(t) dt.

Now, we split this integral into two parts, as promised above. Let I

= [−δ, δ] and

= [−π, −δ] ∪ [δ, π]. If t ∈ I

, then |f(x + t) − f (x)| < ε/2, and so

|f(x + t) − f (x)|K

(t) dt ≤

(t) dt

≤

−π

(t) dt =

If t ∈ I

, then |K

(t)| ≤ ε/(8πM) and so

|f(x + t) − f (x)|K

(t) dt ≤

2MK

(t) dt

≤ 2M

8πM

dt <

4π

−π

dt =

Adding these two results, we have

|σ

f(x) − f(x)| ≤

−π

|f(x + t) − f (x)|K

(t) dt ≤

= ε

for all x ∈ [−π, π] and all n ≥ N. Thus, by the deﬁnition of uniform convergence,

f converges uniformly to f. ¥