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

Подождите немного. Документ загружается.

14.6 Ces

aro Summation of Fourier Series 493

This proof has a strong resemblance to our proof of the Weierstrass Approxi-

mation Theorem. The common underlying technique here is controlling the integral

of a product by splitting the integral into two parts, where one factor of the product

is well-behaved on each part. For another example, look at the proof of one of the

inequalities in Theorem 14.4.3 (3).

In fact, it is possible to prove the Weierstrass Approximation Theorem using

Fej

er’s Theorem. One proof is outlined in the Exercises. Another will be given in

Section 14.9.

14.6.5. EXAMPLE. Consider the function h introduced in Example 14.4.6,

h(x) =

(

1 for 0 ≤ x ≤ π

−1 for −π < x < 0.

The sequenceS

h willexhibit Gibbs’s phenomenon (see Exercise 14.5.A) as shown

in Figure 14.5. We will compute the Ces

aro means for this function using the F

ejer

kernel. Since h is an odd function, the approximants S

h and σ

h are also all odd.

Also, h(π − x) = h(x), and so S

h and σ

h also have this symmetry. Consider a

point x in [0, π/2].

h(x) =

−π

h(x + t)K

(t) dt

= −

−x

−π

(t) dt +

π−x

−x

(t) dt −

π−x

(t) dt

−x

(t) dt −

π+x

π−x

(t) dt

Here we have exploited the fact that K

is even to cancel the integral from x to

π − x with the integral from x − π to −x.

–1

–2 2

FIGURE 14.5. The graphs of h, S

h, and σ

h on [−π, π].

By Proposition 14.6.3, the ﬁrst integral converges to 1 for x in (0, π/2]; while

the second term tends to 0. On the other hand, σ

h(0) = 0 for any n. Rewrite

h(x) as 2

(t) − K

(π −t) dt. Since K

is positive on [0, π], it follows

494 Fourier Series and Approximation

that σ

h(x) is monotone increasing on [0, π/2], then decreases symmetrically back

down to 0 at π. Also, 0 < σ

h(x) < 1 here because

−x

(t) dt < 1. Likewise,

by symmetry, σ

h(x) converges to −1 on (−π, 0) and σ

h(−π) = 0 for all n.

From the monotonicity, we can deduce that this convergence is uniform on

intervals [ε, π − ε] ∪ [ε − π, −ε] for ε > 0. Since the function h has jump discon-

tinuities at the points 0 or ±π, continuous functions cannot converge uniformly to

it near these points.

Exercises for Section 14.6

A. Show that if f is an absolutely integrable function with lim

n→∞

f(θ) = a, then

lim

n→∞

f(θ) = a.

B. Show that if f is a piecewise continuous function with a jump discontinuity at θ, then

lim

n→∞

f(θ) =

f(θ

) + f(θ

−

)

HINT: Write f as the sum of a continuous function g and a piecewise C

function h.

Use Exercise A on h.

C. Show that kσ

∞

≤ kfk

∞

D. Let

j≥0

be an inﬁnite series. Deﬁne s

j=0

and σ

n−1

j=0

(a) If lim

n→∞

= L, show that lim

n→∞

= L.

(b) Show by example that the converse of (a) is false.

n→∞

= 0 and lim

n→∞

= L, then

lim

n→∞

= L. HINT: Verify that s

− σ

N+1

j=1

E. Suppose that f is a 2π-periodic function and that |A

| + |B

| ≤ C/n for n ≥ 1 and

some constant C.

(a) Find a bound for S

f(θ) − σ

f(θ) =

k=1

n + 1

coskθ +

n + 1

sinkθ.

Hence show that kS

∞

≤ kfk

∞

+ C.

(b) Apply this to obtain a uniform bound for S

f for the function f used in our exam-

ple of Gibbs’s phenomenon.

F. Prove Weierstrass’s Approximation Theorem for a continuous function f on [0, π] as

follows:

(a) Set g(θ) = f(|θ|) for θ ∈ [−π, π]. This is an even, continuous, 2π-periodic func-

tion. Use Fej

er’s Theorem to approximate g within ε/2 by a trig polynomial.

(b) Use the fact that the Taylor polynomials for cosnθ converge uniformly on [0, π] to

approximate the trigonometric polynomial by actual polynomials within ε/2.

14.7 Best Approximation by Trig Polynomials 495

14.7. Best Approximation by Trig Polynomials

We return to the theme of Chapter 10, uniform approximation. Here we are

interested in approximating 2π-periodic functions by (ﬁnite) linear combinations

of trigonometric functions. In this section, we obtain some reasonable estimates.

Later we will establish the Jackson and Bernstein Theorems, which yield optimal

estimates. It will turn out that there are close connections with approximation by

polynomials which we also explore.

The degree of a trigonometric polynomial q(x) = a

k=1

coskx +

sinkx is n if |a

|+ |b

| > 0 and a

= b

= 0 for all k > n. We let TP

denote

the subspace of C[−π, π] consisting of all trigonometric polynomials of degree at

most n.

14.7.1. DEFINITION. The error of approximation to a 2π-periodic function

f by trigonometric polynomials of degree n is

(f) = inf{kf − qk

∞

: q ∈ TP

}

For example, for any 2π-periodic function f ∈ C[−π, π], both S

f and σ

are in TP

. The subspace TP

has dimension 2n + 1 because it is spanned by the

linearly independent functions {1, coskx, sinkx : 1 ≤ k ≤ n}. It follows from the

compactness argument of Theorem 7.6.5 that there is a best approximation in TP

to any function f. That is, given f in C[−π, π], there is a trig polynomial p in TP

so that

kf − pk

∞

= inf{kf − qk

∞

: q ∈ TP

} =

(f).

In a certain sense, the functions S

f and σ

f are natural approximants to f

in TP

. Theorem 14.1.2 shows that S

f is the best L

norm approximant to f in

. However, it has some undesirable wildness when it comes to the uniform

norm. F

ejer’s Theorem (Theorem 14.6.4) suggests that σ

f is a reasonably good

approximant in the uniform norm. The following theorem gives bounds on how

close S

f is to the best approximation in TP

. It says that S

f can be a relatively

bad approximation for large n. Nevertheless, the degree of approximation by S

is sufﬁciently good to yield reasonable approximations if the Fourier series decays

at a sufﬁcient rate.

On the other hand, while σ

f is in general a superior approximant, consider

f(x) = sin nx. Then S

f = f is the best approximation with

(f) = 0, while

f(x) =

n+1

sinnx is a rather poor estimate. In spite of this, good general results

can be obtained from the obvious estimate

(f) ≤ kf − σ

∞

The reason that S

works in the following theorem is the fact that S

p = p for all

p in TP

14.7.2. THEOREM. If f : R → R is a continuous 2π-periodic function, then

kf − S

∞

≤ (3 + log n)

(f).

496 Fourier Series and Approximation

PROOF. By Theorem 14.4.2,

f(x) =

−π

f(x + t)D

(t) dt.

Hence by Theorem 14.4.3 (3), we have

∞

≤ kf k

∞

−π

(t)

dt ≤ (2 + logn)kfk

∞

Observe that if p ∈ TP

, then S

p = p. In particular, if we let p ∈ TP

be the best

approximation to f, then

kf − S

∞

≤ k(f − p) − S

(f − p)k

∞

≤ kf − pk

∞

+ kS

(f − p)k

∞

≤ kf − pk

∞

+ (2 + log n)kf − pk

∞

≤ (3 + log n)kf − pk

∞

As kf − pk

∞

(f) for our choice of p, we obtain the desired estimate. ¥

We can now apply this estimate to show that the Dirichlet–Jordan Theorem

actually yields uniform convergence when the piecewise Lipschitz function is con-

tinuous.

14.7.3. THEOREM. If f is a 2π-periodic Lipschitz function with Lipschitz con-

stant L, then for n ≥ 2

kf − σ

∞

≤

(1 + 2 log n)L

Consequently,

kf − S

∞

≤

2π(1 + logn)

In particular, S

f converges to f uniformly on [−π, π].

PROOF. We need a decent estimate for the F

ejer kernel. For our purposes, the

following is enough:

(t) =

2π(n + 1)

sin

n+1

sin

≤ min

n + 1

2π

2(n + 1)t

Indeed, K

(t) ≤ K

(0) =

n+1

2π

yields the ﬁrst upper bound. And the inequality

|sint/2| ≥ (2/π)|t/2| = |t|/π on [−π, π] is enough to show that

2π(n + 1)

sin

n+1

sin

≤

2π(n + 1)

|t|/π

2(n + 1)t

The ﬁrst bound is better for small |t|, and the second becomes an improvement at

the point δ = π/(n + 1).

14.7 Best Approximation by Trig Polynomials 497

This proof follows the proof of F

ejer’s Theorem using the additional informa-

tion contained in the Lipschitz condition to sharpen the error estimate. Looking

back at that proof, we obtain an estimate by splitting [−π, π] into two pieces. We

use the Lipschitz estimate |f(x + t) − f(x)| ≤ L|t|, and we take δ as found previ-

ously:

|σ

f(x) − f(x)| ≤

−π

|f(x + t) − f (x)|K

(t) dt

≤

−δ

L|t|

n + 1

2π

dt +

−δ

−π

L|t|

2(n + 1)t

(n + 1)L

π/n+1

t dt +

πL

(n + 1)

π/n+1

−1

(n + 1)L

2(n + 1)

πL

(n + 1)

log(n + 1)

Lπ

2(n + 1)

1 + 2 log(n + 1)

Finally, a little calculus shows that f (x) = (1 + 2logx)/x is decreasing for x >

√

e. So for n ≥ 2 we obtain

kσ

f − fk

∞

≤

Lπ

(1 + 2 log n).

Now apply Theorem 14.7.2 to obtain

kf − S

∞

≤ (3 + log n)

(f)

≤ (3 + log n) kf − σ

∞

≤ (3 + log n)

Lπ

(1 + 2 log n) ≤

Lπ

4(1 + log n)

Now lim

n→∞

2πL(1 + logn)

= 0. Therefore, S

f converges uniformly to f. ¥

Exercises for Section 14.7

A. Show that if f is a Lipschitz function on [−π, π] with Lipschitz constant L, then the

Fourier coefﬁcients satisfy |A

| ≤

and |B

| ≤

for n ≥ 1.

HINT: Split the integral into n pieces and replace each integral by

+π/n

−π/n

f(x) − f(c

)

sinnx dx. Then estimate each piece.

B. For a 2π-periodic continuous function f, deﬁne approximants P

(f)(θ) =

k=1

coskθ+B

sinkθ+

n−1

k=1

(1−

)

n+k

cos(n+k)θ+B

n+k

sin(n+k)θ

(a) Show that P

(f) = 2σ

2n−1

(f) − σ

n−1

(f).

(b) Hence deduce that kP

(f)k

∞

≤ 3kfk

∞

, then P

(p) = p.

498 Fourier Series and Approximation

(d) Hence show that kf − P

(f)k

∞

≤ 4

(f).

HINT: Follow the method of Theorem 14.7.2.

C. Use the previous exercise to obtain a lower bound

|sinθ|

> C/n.

HINT: See Exercise 13.6.B. Show:

|sinθ|

(0)

≥

∞

k=n

− 1

πn

D. Let 0 < α < 1. Recall that a 2π-periodic function f is of class Lipα if there is a

constant L so that |f(x) − f(y)| ≤ L|x − y|

for all −π ≤ y ≤ x ≤ π.

(a) If f ∈ Lipα, show that there is a constant C such that kf − σ

∞

≤ Cn

−α

(b) Hence show that S

f converges uniformly to f.

HINT: Follow the proof of Theorem 14.7.3 using the new estimate.

E. Let f be a 2π-periodic function with Fourier coefﬁcients A

and B

(a) Prove that (A

+ B

)

1/2

≤ ω(f;

HINT: Show that A

2π

−π

f(t) − f(t +

)

cos(nt) dt.

(b) If f ∈ Lipα, prove that there is a constant C so that (A

+ B

)

1/2

≤ Cn

−α

and f

(p)

∈ Lipα, show that (A

+ B

)

1/2

≤ Cn

−p−α

F. Let f and g be 2π-periodic with Fourier series f ∼ A

n≥1

cosnθ + B

sinnθ

and g ∼ C

n≥1

cosnθ + D

sinnθ. Suppose that f is absolutely integrable and

g is Lipschitz. Prove that

2π

−π

f(θ)g(θ) dθ = A

n≥1

+ B

HINT: Use Theorem14.7.3.

G. (a) Find the Fourier series of g(x) =

(

−x − π for −π ≤ x < 0

−x + π for 0 ≤ x ≤ π

(b) For n ≥ 1, deﬁne g

(x) = sin(2nx) S

g(x). Find the Fourier series for g

HINT: 2 sinA sinB = cos(A − B) − cos(A + B)

∞

≤ π + 2. HINT: Use Exercise 14.6.E.

(d) Show that S

(0) > logn.

HINT:

k=1

is an upper Riemann sum for

n+1

dx.

(e) Hence prove that kg

− S

∞

≥

logn

π + 2

). This shows that the logn

term is needed in Theorem 14.7.2.

(f) Show that h(x) =

∞

n=1

2·3

(x) is a continuous function such that S

h(0)

diverges. NOTE: This is difﬁcult.

14.8. Connections with Polynomial Approximation

We will now connect trigonometric polynomial approximation with polyno-

mial approximation on an interval by exploiting an important connection between

trig polynomials and the Chebychev polynomials of Section 10.7.

14.8 Connections with Polynomial Approximation 499

The idea is to relate each function f in C[−1, 1] to an even function Φf in

C[−π, π] in such a way that the set P

of polynomials of degree n is carried into

. This map is deﬁned by

Φf(θ) = f(cos θ) for − π ≤ θ ≤ π.

Notice immediately that since cosθ takes values in [−1, 1], the right-hand side is

always deﬁned. Also,

Φf(−θ) = f(cos(−θ)) = f(cosθ) = Φf(θ).

Thus Φf is even.

A crucial property of the map Φ is that it is linear. Observe that for f, g in

C[−1, 1] and α, β in R,

Φ(αf + βg)(θ) = (αf + βg)(cosθ) = αf(cosθ) + βg(cosθ)

= αΦf(θ) + βΦg(θ) = (αΦf + βΦg)(θ).

So Φ(αf + βg) = αΦf + βΦg, which is linearity.

Recall from Deﬁnition 10.7.1 that the Chebychev polynomials are deﬁned on

the interval [−1, 1] by T

(x) = cos(n cos

−1

x). Since T

is a polynomial of degree

n, every polynomial can be expressed as a linear combination of the T

s, and, in

particular, P

is spanned by {T

, . . . , T

}. Also recall that there is an inner product

on C[−1, 1] given by

hf, gi

−1

f(x)g(x)

√

1 − x

for f, g ∈ C[−1, 1].

The Chebychev polynomials form an orthonormal set by Lemma 10.7.5.

Notice that

ΦT

(θ) = cos(n cos

−1

(cosθ)) = cos(nθ).

It follows that Φ maps P

onto the span of {1, cosθ, . . . , cosnθ}, which consists

of the even trig polynomials in TP

This establishes the ﬁrst parts of the following theorem. Let E[−π, π] denote

the closed subspace of C[−π, π] consisting of all even continuous functions on

[−π, π].

14.8.1. THEOREM. The map Φ of C[−1, 1] into E[−π, π] satisﬁes the follow-

ing:

(1) Φ is linear, one-to-one, and onto.

(2) ΦT

(θ) = cos(nθ) for all n ≥ 0.

(3) Φ(P

) = E[−π, π] ∩ TP

(4) kΦf − Φgk

∞

= kf − gk

∞

for all f, g ∈ C[−1, 1].

(5) E

(f) =

(Φf) for all f ∈ C[−1, 1].

(6) hf, gi

= hΦf, Φgi for all f, g ∈ C[−1, 1].

500 Fourier Series and Approximation

PROOF. For (1), linearity has already been established. To see that Φ is one-to-one,

suppose that Φf = Φg for functions f, g in C[−1, 1]. Then f(cosθ) = g(cosθ) for

all θ in [−π, π]. Since cos maps [−π, π] onto [−1, 1], it follows that f (x) = g(x)

for all x in [−1, 1]. Thus f = g as required.

To show that Φ is surjective, we construct the inverse map from E[−π, π] to

C[−π, π]. For each even function g in E[−π, π], deﬁne a function Ψg in C[−1, 1]

Ψg(x) = g(cos

−1

x).

(Notice that cos

−1

takes all values in [0, π]. So Ψg depends only on g(θ) for θ in

[0, π]. This is ﬁne because g is even.) Compute

ΦΨg(θ) = Ψg(cosθ) = g(cos

−1

(cosθ)) = g(|θ|) = g(θ).

So Φ maps Ψg back onto g, showing that Φ maps onto E[−π, π].

We proved (2) and (3) in the discussion before the theorem. Now consider (4).

Note that

kΦfk

∞

= sup

θ∈[−π,π]

|f(cosθ)| = sup

x∈[−1,1]

|f(x)| = kfk

∞

Hence by linearity,

kΦf − Φgk

∞

= kΦ(f − g)k

∞

= kf − gk

∞

Applying this, we obtain

(f) = inf{kf − pk

∞

: p ∈ P

}

= inf{kΦf − Φpk

∞

: p ∈ P

}

= inf{kΦf − qk

∞

: q ∈ TP

∩ E[−π, π]}.

However, since Φf is even, the trig polynomial in TP

closest to Φf is also even.

To see this, suppose that r ∈ TP

satisﬁes kΦf − rk =

(Φf), and let

q(θ) =

r(θ) + r(−θ)

You should verify that if r(θ) = a

k=1

coskθ + b

sinθ, then q(θ) =

k=1

coskθ. Then q belongs to TP

∩ E[−π, π] and

|Φf(θ) − q(θ)| =

Φf(θ) + Φf(−θ)

−

r(θ) + r(−θ)

≤

Φf(θ) − r(θ)

Φf(−θ) − r(−θ)

≤

(Φf) +

(Φf) =

(Φf).

Hence kΦf −qk

∞

(Φf). Putting this information into the preceding inequal-

ity, we obtain E

(f) =

(Φf).

14.8 Connections with Polynomial Approximation 501

Finally, to prove (6), we make the substitution x = cosθ in the integral.

hf, gi

−1

f(x)g(x)

√

1 − x

f(cosθ)g(cos θ)

−sinθ dθ

sinθ

2π

−π

f(cosθ)g(cos θ) dθ = hΦf, Φgi

In summary, this theorem shows that C[−1, 1] with the inner product h·, ·i

and E[−π, π] with the inner product h·, ·i are, as inner product spaces, the same.

Approximation questions for Fourier series are well studied, and this allows a

transference to polynomial approximation. The reason we obtain estimates more

readily in the Fourier series case is that the periodicity allows us to obtain nice

integral formulas for our approximations.

Part (6) of this theorem shows that the Chebychev series for f ∈ C[−1, 1]

correspond to the Fourier series of Φf as f ∼

∞

k=0

, where

= hf, 1i

and a

= 2hf, T

for n ≥ 1.

Then deﬁne two series corresponding to the Dirichlet and Ces

aro series.

f =

k=0

and Σ

f =

k=0

1 −

n + 1

Now it is just a matter of reinterpreting the Fourier series results for polynomials

using Theorem 14.8.1.

14.8.2. THEOREM. Let f be a continuous function on [−1, 1] with Chebychev

series

∞

k=0

. Then (Σ

∞

n=1

converges uniformly to f on [−1, 1]. If f is

Lipschitz, then (C

∞

n=1

also converges uniformly to f. In any event,

kf − C

∞

≤ (3 + log n)E

(f).

PROOF. The map Φ converts the problem of approximating f by polynomials of

degree n to the problem of approximating Φf by trig polynomials of degree n.

Property (6) of Theorem 14.8.1 shows that ΦC

f = S

Φf and ΦΣ

f = σ

Φf.

So the fact that Σ

f converge uniformly to f on [−1, 1] is a restatement of F

ejer’s

Theorem (Theorem 14.6.4).

If f has a Lipschitz constant L, then

|Φf(α) − Φf(β)| = |f(cosα) − f(cos β)|

≤ L|cosα − cos β| ≤ L|α − β|.

The last step follows from the Mean Value Theorem since the derivative of cosθ is

sinθ, which is bounded by 1. Thus Φf is Lipschitz with the same constant. (Warn-

ing: This step is not reversible.) By Theorem 14.7.3, the sequence S

Φf converges

uniformly to Φf, whence C

f converges uniformly to f by Theorem 14.8.1(4).

502 Fourier Series and Approximation

Theorem 14.7.2 and Theorem 14.8.1(4) provide the estimate

kf − C

∞

= kΦf − S

∞

≤ (3 + log n)

(Φf) = (3 + logn)E

(f).

14.8.3. EXAMPLE. Let us try to approximate f(x) = |x| on [−1, 1]. We con-

vert this to the function

g(θ) = Φf(θ) = |cosθ| for − π ≤ θ ≤ π.

This is an even function and thus has a cosine series. Also, g(π − θ) = g(θ). The

functions cos2nθ have this symmetry, but

cos

(2n + 1)(π − θ)

= −cos(2n + 1)θ.

So A

2n+1

= 0 for n ≥ 0. Compute

2π

−π

|cosθ|dθ =

and for n ≥ 1,

−π

|cosθ|cos2nθ dθ

π/2

cosθ cos2nθ dθ

π/2

cos(2n − 1)θ + cos(2n + 1)θ dθ

sin(2n − 1)θ

2n − 1

sin(2n + 1)θ

2n + 1

π/2

(−1)

n−1

2n − 1

(−1)

2n + 1

(−1)

n−1

π(4n

− 1)

These coefﬁcients are absolutely summable since they behave like the series

1/n

. Thus S

g converges uniformly to g by the Weierstrass M-test (see Exer-

cise 13.4.D). Hence by Theorem 14.8.1, the sequence of polynomials

f(x) =

−

k=1

(−1)

− 1

(x)

converges to |x| uniformly on [−1, 1].