The Lebesgue-Stieltjes Integral: A Practical Introduction

1

If(x)IP

[a,b]

ax < 00.

Now, IIfll p = IIgllp for any f,g E AP[a, b] such that f = g a.e., so we know that II . lip is at best a seminorm for AP[a, b]. This problem can be easily remedied by using equivalence classes. A more serious concern is that AP[a, b] may not even be a vector space. In particular, if f, g E AP[a, b], it is not clear that f + g E AP[a, b]. Moreover, it is not obvious that II . lip will satisfy the triangle inequality. As it turns out, the sets AP[a, b] are vector spaces and II· lip is a seminorm on them for 1

1

1I[f]lIp = { If(x)IP ax } [a,b] Theorem 8.5.1 (Holder's Inequality) Let F E Il[a, b] and G E Lq[a, b], where 1 < P < Then FG E LI[a, b] and

00

and IIp+ IIq = 1.

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _...,..-_--,'.,..1- - - - - - - - - - - - - - - - - - - - - . "

144

8. The Lebesgue Spaces IY

Theorem 8.5.2 Let 1 < P < and

00

(Minkowski's Inequality)

and suppose that F, G E V[a, b]. Then F + G E V[a, b]

The proofs of these inequalities can be found in most texts on functional analysis, e.g., [38]. In the Holder inequality, the product is the pointwise product of functions, i.e., if F = rf], G = [g], then FG = rfg], where (fg)(x) = f(x)g(x). That V[a, b] is a vector space and II . lip defines a norm on it follows from Minkowski's inequality. As with the space (L1[a, b], II . III), the normed vector spaces (V[a, b], II . lip) are complete. This result is a generalized version of the classical Riesz-Fischer theorem. Theorem 8.5.3 The normed vector spaces (V[a, b], P < 00.

II . lip) are Banach spaces for 1

<

A detailed proof of this result can be found in [17] and [18]. The proof for the case 1 < P < 00 is similar to that for the case p = 1. Essentially, the civilized behavior of the Lebesgue integral (as manifested in the monotone convergence theorem) is responsible for completeness. The Lebesgue integral thus yields an entire family of Banach spaces. To simplify notation, we shall refer to the Banach space (V[a, b], II . lip) simply as V[a, b] unless there is some ambiguity regarding the norm. These Banach spaces are collectively referred to as the Lebesgue or V spaces. We also follow the common (and convenient) practice of blurring the distinction between AP[a, b] and V[a, b] by treating elements of V[a, b] as functions. We trust the reader to make the correct technical interpretation ana to remember in this context that "f = g" means f = g a.e. Suppose that f E L 2 [a, bJ. The constant function g = 1 is also in L 2 [a, b], and therefore Holder's inequality implies that the function f ·1 = f is in L1[a, b]. In addition, we have that

IIflli

<

IIfll2111112 = (b - a)1/2l1fIl2'

This observation shows that L 2 [a, b] C L1[a, b]. The calculation works" because p = q = 2 in Holder's inequality and g = 1 is /I

----------------r-----,-,

T ' " " i-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-• • • •-

•..

8.5. The Lebesgue V

145

integrable on any interval of finite length. We can repeat this argument for the general P > 1 because g E Lq[a, b] for any q. Thus, V[a, b] C LI[a, b] for all P > 1. The next result indicates that if 1
E

VZ [a, b]. Then

IIfll p l < IIfll p z(b - a)IIPI-IIPz, and consequently f E VI [a, b]. • Proof If PI = Pz, the result is trivial. Suppose that 1
Jra,b]

IIf(x)IPIl

k

ax ==

[ If(x)IPZ ax < Jra,b]

00,

since f E VZ[a, b], and therefore If(x)IPI E Lk[a, b]. Holder's inequality with P = k, q = m, and g = 1 E Lm[a, b] yields the inequality

and

Consequently,

J

(lIfll p PI < (lIfll p zYI (b - ai-PIIPZ,

o

and the inequality follows.

Two positive numbers P and q are called conjugate exponents if IIp+ IIq = 1. Holder's inequality suggests that the Banach spaces V[a, b] and Lq[a, b] are related if P and q are conjugate exponents. In fact, the spaces are intimately related. A linear functional is an

, i

146

8. The Lebesgue Spaces IY

operator 1 from a normed vector space (X, 11·11) to the scalar field or C) of the vector space such that for any fI, fz E X

OR

l(fi + fz) = l(fi) + l(fz), and for any scalar ex

°

The functional 1 is bounded if there is a number c > such that 11 (f) I < c IIf II for all f EX. The norm of a bounded functional 1 is defined as the smallest number c such that 11 (f) I < c IIf II for all f E X and denoted by 111 II. Since ex = is a legitimate choice for a scalar, we see that

°

1(0 . fI)

= 1(0) = 01(fl) = 0,

Le., 1(0) = 0. The norm of 1 is thus given by

11111

= sup 11(f) I [EX

(any choice of c >

°satisfies 101

f#O

<

IIfll

cIlOIl)·

Exam.ple 8-5-1:

Let 1(f) =

1

k(x)f(x) dx,

[a,b]

where k E LZ[a, b], and let X = L2[a, b]. Now, 1 is evidently a linear functional, and Holder's inequality implies that for any f E LZ[a, b],

11 (f) I <

1

[a,b]

•

Ik(x)f(x) I dx = IIkflll < IIkllzllfllz;

thus,

11111 < IIkllz·

°

In fact, 11111 = IIkll z. To see this, note first that if IIkllz = 0, then k = a.e., and therefore kf = a.e. for any f E LZ[a, b]; hence, 1(f) = for all f E LZ[a, b] and 11111 = 0. If IIkll z -I- 0, choose f = k/llkll z E LZ[a, b].

°

°

8.5. The Lebesgue If

147

Now,

11er)I ==

1

[a,b]

1 == IIkll z

Since IIfllz

k(x) k(x)

ax

1

ax == IIkll z.

IIkllz

[a,b]

k

Z

(X)

== 1 for this choice, we see that 11111 = IIkll z.

The above example is typical of bounded linear functionals on the V[ a, b] spaces for 1 < P < 00. The following representation theorem expresses the situation: Theorem 8.5.5 Let p and q be conjugate exponents with 1 < P < 00, and suppose that 1 : V[a, b] ~ 1R is a bounded linear functional. Then there exists a unique element g E Lq[a, b] such that

ler) = for all f

E

1

f(x)g(x) ax,

[a,b]

V[a, b]. Moreover;

11111 == IIgll q . The proof of this fundamental result can be found in [37]. In the more general context of functional analysis, the dual space of a normed space (X, II •II) is defined to be the set of all bounded linear functionals on X. The above theorem indicates that Lq[a, b] is the dual space of V[ a, b] when p and q are conjugate exponents with 1 < P < 00. Conjugate exponents and dual spaces bring to the fore two exceptional cases. Each Banach space V[a, b] is paired with its dual space Lq[a, b], where lip + llq == I, 1 < P < 00. The conjugate exponent ofp = 2 is q = 2, and so LZ[a, b] is its own dual space. The space LZ[a, b] is clearly the only member of the V[a, b] spaces with this property, and it turns out that LZ[a, b] has several other special properties not shared by the other V[a, b] spaces. The space LZ[a, b] is special because it forms what is known as a Hilbert space. We postpone further discussion about the special properties of LZ[a, b] until the next chapter.

148

8. The Lebesgue Spaces L!

The other exceptional case is the space L1[a, b]. There is no conjugate exponent for p = I, yet there are certainly bounded linear functionals defined on Ll[a, b]. What then is the dual space of L1[a, b]? If Ik(x)1 < M < 00 for all x E [a, b], then the functional J defined by J(f)

= (

Jra,b]

k(x)f(x) ax

is a bounded linear functional, since IJ(f) I < M (

Jra,b]

If(x) I ax = Mllflll.

In fact, the condition that Ik(x)1 < M < 00 for all x E [a, b] can be relaxed to Ik(x)\ < M < 00 a.e. in view of the Lebesgue integral. This example serves as a guide to what might be expected as a dual space for Ll[a, b]. Let f : [a, b] -+ ~ be a measurable function. A number J-L is an essential upper bound for IfI if If(x) I < J-L a.e. If IfI has an essential upper bound, then it can be shown that there exists a least such bound. This leastbound is denotedby ess sup IfI. Let A 00 [a, b] denote the set of all measurable functions on [a, b] such that ess sup IfI < 00 and define the function II . II 00 by

IIflloo = ess sup Ifl· As with the AP[a, b] spaces, A 00 [a, b] is a vector space and II . 1100 is a seminorm on it. Let LOO[a, b] denote the set of equivalence classes of A 00 [a, b] modulo equality a.e. Then (LOO[a, b], 11·1100) is a normed vector space. The following theorem summarizes some of the properties of LOO[a, b]: Theorem 8.5.6 (i) (LOO[a, b], II . 1100) is a Banach space. (ii) The dual space of L1[a, b] is LOO[a, b]. (iii) Iff E L1[a, b] andg E LOO[a, b], thenfg E L1[a, b] and

IIfglll

<

IIflllllglloo

(an extension of the Holder inequality). (iv) Iff E LOO[a, b], then f E V[a, b] for all 1 < p <

00.

8.5. The Lebesgue V

149

The proofs of parts (i) and (ii) can be found in [33]. The proofs of parts (iii) and (iv) are left as exercises. Up and q are conjugate exponents with p > I, then the dual space of V[a, b] is Lq[a, b] and vice versa. In the language of functional analysis, the V spaces are reflexive for p > 1. In contrast, the dual space of V'O[a, b] is not L 1 [a, b]. The dual space of V'O[a, b] turns out to be a space of measures. More details concerning the dual space of V'O[a, b] can be found in [34]. The space L 1 [a, b] is thus unusual among the V[a, b] spaces in that it is not reflexive. We have t4us far been concerned with V spaces where the interval of integration is bounded. The definitions for V spaces can be extended to include unbounded intervals of integration. If I c ~ is any interval, the space V(l) consists of the set of functions f : I -+ R (Le., equivalence classes) such that

IIfll p =

Ii

If(x)IP

ax}

lip

<

00.

All the results stated for the V[a, b] spaces carry over to the general V(l) spaces with the notable exception of the inclusion results. The proof of Theorem 8.5.4 relies crucially on the fact that the interval is bounded. If the interval is not bounded, then results such as L 2 (l) c L 1 (I) are not valid. For example, consider the functions fI and f2 defined by fi(x) hex)

I =I

=

1/,JX., if 0 < x < I, 0, if x > I, 0,

llx,

ifO<x 1.

Now, fI E L 1 (O, 00), but f2 E L 1 (O,OO); alternatively, fI E £2(0,00) but fz E L 2(O,oo). Hence L2 (O, 00) does not contain L 1 (O, 00), and L 1 (O, 00) does not contain L2 (O, 00). Exercises 8-5: 1. Suppose that the function r positive on the interval [a, b].

[0, 1] -+ ~ is continuous and

150

8. The Lebesgue Spaces If

(a) Prove that for 1

(

r(x)lf(x)IP

( 1[a,b]

)

1~

ax

is a norm on V[ a, b].

(b) Prove that the norm rll . lip is equivalent to the norm II . lip. 2. Let 1 < PI < pz and let I be a bounded interval. Prove that IIfll pI < 1 + IIfllp 2 and from this deduce that V2(I) C VI (I).

3. The Fredholm integral operator K is defined by (Kf)(x)

= ( k(x,~)f(~)~, 1[0,1]

where k : [0,1] X [0,1] -+ ~ is a given function. Suppose that k is bounded in the square [0, 1] X [0,1]. Prove that the operator K maps functions in £z[O, 1] to functions in £z[O, 1]. 4. Prove parts (iii) and (iv) of Theorem 8.5.6.

5. Iff

E

£<'°[0,1], prove that

IIfll1 < IIfllz < ... < IIflln < IIflln+1 < ... < IIflloo.

8.6

Separable Spaces

A Banach space (X, II . II) is separable if X contains a countable set

that is dense in X in the II . II norm. The Banach space (~, II . lie), for example, is a separable space. To establish this clairp. we note that the rational numbers form a countable set that is dense in the real numbers. Separability is a desirable property, because the existence of a countable dense set simplifies problems concerning the representation of functions (e.g., bases for spaces) and the approximation of functions. In the Hilbert space theory discussed in Chapter 9, separability is a key ingredient in establishing the existence of an orthonormal basis (cf. Theorem 9.2.4). Fortunately, the V spaces are separable.

8.6. Separable Spaces

Theorem 8.6.1 The V spaces are separable for 1

151

00.

The proof of this result can be found in [17]. We limit ourselves here to a brief informal discussion of the ideas underlying the proof and focus on sets that are dense in the V spaces. These sets are ofinterest in their own right for the purpose of approximation. The construction of the Lebesgue-Stieltjes integral in Chapter 4 brings to the fore one class of functions that must be dense in any V space, viz. the set of functions consisting of x-summable step functions (Section 4.5). This density relationship is used to prove the Riemann-Lebesgue theorem (Theorem 9.3.4) in the next chapter. Unfortunately, this set is not countable. There are, of course, other sets dense in the V spaces. The next result (which we state without proof) provides asample of such a set. Theorem 8.6.2 Let ego (~) denote the set offunctions f : ~ -+ ~ with derivatives of all orders and with supports offinite length. The set Cgo~) is dense in each V(~) space for 1

O. Now, f has a support of finite length, and hence there is some number fJ > 0 such that f = 0 outside the interval [-fJ, fJ]. Certainly f (restricted to the interval [-fJ, fJ]) is in the space C[ -fJ, fJ], and thus there is a polynomial PE such that E

sup IPE(X) - f(x) I < (2fJ)l/p

xE[-P,Pl

(8.1)

Moreover, we know from Exercises 8-3-2 that the set PQ [ -fJ, fJ] of polynomials with rational coefficients is dense in P[ - fJ, fJ], and hence we can find a polynomial with rational coefficients satisfying inequality (8.1). In fact, we can find a polynomial qE with rational coefficients such that inequality (8.1) is satisfied and qE(-fJ) = qE(fJ) = O. Now the polynomial qE canbe extended to a functiongE = qE' X[-P,Pl

152

8. The Lebesgue Spaces £P

defined on ~. Here X[-P,P] is the characteristic function defined by

X_

ex) =

[ P,P]

{I,

if x E [-,8, ,8], 0, otherwise.

Thus, IIgE - flip = { ( IgE(X) - f(x)I Pdx}l/P =

JJR

<

{i

IgE(X) - f(x)I Pdx}l/P

[-P,P]

sup IgE(X) - f(x) I(2,8)l/P <

E.

XE[-P,P]

Now, the set of rational numbers is countable, and this can be used to establish that the set PQ t-,8,,8] is also countable. If we define the set r = {g : g = q . X[-~,P] for some q E PQ [ -,8,,8] and some ,8 = 1,2, ...}, then it can be shown that r is a countable set in Cgo~). The above inequality indicates that r is dense in Cgo(R), and hence it is dense in V(~). Thus, V(~) contains a countable dense set.

8.7

Complex LP Spaces

In this section we pause to make a modest generalization ofV spaces to include complex-valued functions of a real variable. A fuller treatment of these spaces can be found in [17]. Let I c ~ be some interval and let f : I ~
We wish to construct spaces analogous to AP(!) and V(l). The natural generalization of the norm function is the function II· lip defined by

lIfllp =

{i

If(t)[P dt} liP,

__

._~-----------------_.

..- .... -

8.7. Complex V Spaces

153

where for 1 = Ref - ilmf (the complex conjugate), Ifl 2 = fl. The next theorem shows that IflP is integrable if and only if IRefl P and IImfl P are both integrable. The proof rests on the inequalities IRefl < If!' IImfl < If!' and If I < IRefl + IImf!' and is left as an exercise. Theorem 8.7.1 Let f : I ---+ C be a complex-valued function defined on an interval I c ~ and suppose f = Ref + Hmf, where Ref and Imf are real-valued measurable functions. Then, for 1

l

lf (t), Pdt

<

00

if and only if

1

IRe f(t)I P dt <

00

and lIImf(t)IP dt <

00.

Let A~(I) denote the set of functions f : I -+ C such that IIfllp < 00. The above result asserts that f E A~(I) if and only if Ref E AP(I) and Imf E AP(I). For 1 < P < 00 it can be shown that the function II . lip satisfies the Holder and Minkowski inequalities. In particular, the sets A~(I) are complex vector spaces and II . lip is a seminorm. As with the AP(!) sets, we can form a normed vector space by partitioning A~(I) into equivalence classes. Two functions belong to the same equivalence class if If - gl = 0 a.e. Let L~(I) denote the resulting set of equivalence classes. The spaces L~(I) share the same properties as the V(l) spaces. For example, if 1

I, then the dual space of L~(l) is the space Lt(I). If J is a bounded linear functional from L~(l) to C, then there is a unique g E Lt(I) such that

J(f) =

1

f(fJk(t) dt

for all f E L~(I). The space LC'(!) can be defined in a manner analogous to that used to define Loo(I), and this space is the dual of Lt(I).

154

8. The Lebesgue Spaces l j

The Lt(!) spaces often arise in applications involving line integrals in the complex plane. Recall that if y is some curve in C represented parametrically by some piecewise smooth function z(t) = x(t) + iy(t) for t E [to, tI], then the line integral of a function F defined on y is given by

1

F(z) dz

y

=

i

tl

F(z(t))z'(t) dt.

to

The definition of a line integral can be extended using the Lebesgue integral in an obvious way, and thus line integrals can be defined for a more general class of functions. Note that a curve y may be parametrized any number of ways. For example, the variable t may be replaced by any smooth function g : [so, sI] -* [to, tIl to give a new parametrization z(s) = z(g(s)), provided that g'(s) ¥= 0 for all s E [so, sI]. If g'(s) > 0 in the interval [so, sI] then the parametrization is orientation preserving, and it is well known that the value of the line integral in the Riemann setting is invariant under smooth orientation preserving reparametrizations. In the Lebesgue setting, Theorem 6.2.1 ensures that this is also the case. Thus, provided that the orientation of y is specified, we can use the notation f y F(z)dz without ambiguity. In this context we may also use the notation Lt(y) unless a specific parametrization of y is required. A particularly interesting case is that in which y is a simple closed curve (no self-intersections) and F(z) is ho1omorphic (analytic) in the region enclosed by y but not necessarily on y itself. In the next section, we investigate a class of spaces known as Hardy spaces and show that the Lt(!) spaces are closely related to spaces of ho1omorphic functions.

8.8

The Hardy Spaces HP

Hardy spaces are complex normed vector spaces of functions ho1omorphic on a region of the complex plane. These spaces are closely related to the Lebesgue spaces and share the same properties. The functions in a Hardy space are quite different in nature from elements in an Lt(!) space. Aside from the fact that elements in the

8.8. The Hardy Spaces HP

155

latter set are equivalence classes, the underlying functions in an L~(I) space need not be continuous. In contrast, the functions in a Hardy space are infinitely differentiable. That these spaces should be so closely linked is remarkable. Hardy spaces play an important role in complex function theory and harmonic analysis. These spaces also arise in applications such as control theory. The discussion in this section presumes some knowledge of complex analysis, but it is primarily a descriptive account and no proofs are given. For a basic reference on complex analysis the reader is directed to [2], [10], or [39]. A much fuller discu$sion of Hardy spaces along with the proofs of various results in this section can be found in [20], [21], and [37]. Let 0 C C denote a region (Le., a nonempty, connected open set) and let A(O) denote the set of functions holomorphic on O. It is clear from the elementary properties of holomorphic functions that A(O) is a vector space. A candidate for a norm on A(O) is the function II . II 00 defined by

IIflloo = sup !f(z)l. zeQ

Although this function satisfies the conditions for a norm such as the triangle inequality, it is not finite for all f E A(O). There are functions in A(O) With singularities on the boundary ao of O. If the set A(O) is restricted to the functions f for which Ilflloo < 00, then the resulting space is a normed vector space. Let ROO(O) = if E A(n) :

IIflloo <

oo}.

The set ROO (0) is still a vector space, and now II . II 00 is a norm on it. In fact, the space (ROO(O), II . 1100) is a Banach space. The value of IIflioo for a given f E ROO(O) is less tractable than the analogous function defined fot e[a, b] (Example 8-1-5) owing to the two-dimensional nature of the region 0 and the fact that 0 is an open set. A fundamental result in complex analysis known as the maximum modulus principle, however, mitigates these problems. Let 0 c C be a bounded region with boundary ao and suppose that f E A(O) and also that f is continuous in the set Q = 0 u ao. Under these conditions, the maximum modulus theorem implies that the function IfI assumes its maximum value on the boundary ao and

----------------------_

......

156

8. The Lebesgue Spaces IJ'

that a nonconstant function in A(O) cannot have local maxima for IfI in o. The maximum modulus principle cannot be applied directly to all the functions in ROO (0), sinoe functions in this set need not be continuous on 0 U ao. Nonetheless, the norm IIflloo of a function in ROO(O) reflects the values of If(z) I as z approaches the boundary of the region, and this suggests that the norm can be calculated using a limiting argument. The Riemann mapping theorem implies that arbitrary regions such as S can be conformally mapped to a unit disk in the complex plane. We can thus limit our investigati~nsmostly to the study of functions holomorphic on the unit disk centred at the origin. Let D(O; r) = {z E C : Izl < r}, D(O; r) = {z E C : Izl < r}, aDr = {z E C : Izl = r}, and for simplicity, denote DCa; I), D(O; I), aD l by D, D, and aD, respectively. If a < r < I, then, taking 0 = D(O; r), the maximum modulus principle implies that sup If(z) I = sup If(z)l. zeD(O;r)

The function

ze8Dr

Moo is defined by Moo(r,f) = sup If(reUP)1,
and it is evident from the maximum modulus principle that for a fixed f E ROO (D) the function Moo is monotonic increasing in rand that sup If(z) I = Moo(r,f)· zeD(O;r)

We thus expect that for any f lim

r-+l-

E

HOO(D),

Mooer , f) = IIflloo,

and in fact, this relationship can be proved formally. The function Moo suggests a ,generalization for norms analogous to the IY norms. Let Mp be the function defined by Mp(r,f)

= { .2..1· 2n

[0,211"]

lip

If(rei
}

,

-----------------~---------------_._-_.-.

__

.

8.8. The Hardy Spaces HP

157

and let HP(D) be defined as HP(D) =

if E A(D) : sup Mp(r, f)

< oo}.

r< 1

It can be shown that for 1

the Mp functions satisfy H6lderand Minkowski-type inequalities. Specifically, if p and q are conjugate exponents with p > I, then for any functions f E HP(D), g E Hq(D), 00

MI(r,fg) < Mp(r,f)Mq(r, g),

and ifh

E HP(rJ),

then Mp(r,f + h) < Mp(r,f)

+ Mp(r, h).

If P > I, the function II . lip defin¢d by

IIfllp =

SlilP Mp(r, f) r<: 1

is therefore a norm on HP(D). (Note that we need not form equivalence classes because the functions involved are holomorphic, and thus iff = g a.e., thenf = g.) As with Moo(r, f), it can be shown that for all f E HP (D), lim Mp (r 1 f) = Ilfll p .

r-+I-

The normed vector spaces (HP(D), II . lip) are called Hardy or HP spaces. As with the Lebesgue spaces, we refer to the normed vector space (HP(D), II . lip) as HP(D). 'Iihe Hardy spaces share the same properties as the Lebesgue spaces!. The next theorem gives a sample of some of these properties. Theorem 8.8.1 (i) Holder's Inequality: If 1 < P < exponent ofp, then for any f fg E HI (D) and

IIfglh

and q is the conjugate E HP (D), g E Hq (D) we have that

<

00

IIfllp lIgll q .

(ii) Minkowski's Inequality: For any f, g E HP (D) with P > 1,

IIf +gllp < IIfll p + IIgllp · (iii) Completeness: The normed vector spaces (HP(D), Banach spaces. .

II . lip)

are

158

8. The Lebesgue Spaces l j

::s P2

(iv) Inclusion: For any 1
<

00,

HOO(D) c HP2(D) c HP2(D).

The two inequalities in the above. theorem are direct consequences ofthe corresponding inequalities involving the Mp functions. A proof of completeness can be found in [21]. The inclusion result is established using the same approach as used to prove a similar result in Theorem 8.5.4. lt is not a coincidence that the HP spaces mimic the IY spaces. These spaces are closely related through the l'boundaryvalues" ofthe functions in HP(D). Cauchy's integral formula is a remarkable manifestation that a holomorphic function is determined by its boundary values. Recall that for any f E A(b(O; r)), r > 0, the Cauchy integral formula states that fez)

=~ ( 2m

few) dw - z

)aDr w

for any z E D(O; r). Here, the circle aDr is oriented anticlockwise. The value of f on the boundary aDr thus determines the function f uniquely in the interior D(O; r). We cannot apply Cauchy's integral formula directly on the boundary of D because f need not be holomorphic on aD, but we can still aJj>ply the formula on any boundary aDr when 0 < r < 1. This suggests that any function f E HP(D) can be lIidentified" with a function i defined on aD a.e. by some limiting process as r -+ 1- . Let fr(z) :; f(rz).

-

.

If 0 < r < I, then fr(z) is holomorphic on D and the Cauchy integral formula implies that fr(z) =

~

1.

fr(~)~.

2m. aD ~ - z

Intuitively, we should be able to define a function f : aD -+ C by considering the function lim r--+ 1- fr. The next result shows that this approach does lead to an isometry from the HP(D) spaces to the Lt(aD) spaces. The notation for the norms for these spaces is the

8.8. The Hardy Spaces HP

159

same by convention. Here, we shall use the notation II . IIHP, and II . Ilv to distinguish the norm f~r WeD) from the norm for Lt(aD). Theorem 8.8.2 L..,et 1

IIfIlHP. IIf - fr IIv

E

HP(D). Then there is an element

= o. fez) =

~

1 f(~)~.

2m aD ~ - z

The above result indicates that there is an isometry from HP (D) to Lt(aD). These spaces are Iljot isometric, since the set Lt(aD) contains elements with no conesponding members in HP(D). Let ,!!p(aD) C Lt(aD) denote the set of elements If] E Lt(aD) such that f corresponds to some function f E HP(D). The complex normed vector spaces (1-lP(aD), II . IIv) and (HP(D), II . IIHP) are isometric. Now, it can be shown that the space (1-lP (aD) , II . IIv) is a closed subspace of the Banach space C~t(aD), II . Ilv), and this means that (1-lP(aD), II . IIv) is also a Banach space. Since (1-l,P(aD), II . IIv) is isometric to (HP(D), II . IIHP), the ]atter space must also be a Banach space. Properties ofthe HP(D) spaces such as completeness are generally proved by establishing the analogous results for the 1-lP(aD) spaces, and in this sense the Hardy spaces "inherit" the qualities of the IY spaces. The HP(D) spaces can be generalized to sets of functions holomorphic in arbitrary regions ot the complex plane via conformal transformations. Perhaps the most frequently encountered Hardy spaces aside from the HP (D) spaces are the spaces of functions holomorphic on a half-plane. Specifically, the Hardy spaces of functions analytic on the right half-plane no = {z E C : Rez > o} are of interest owing to (among other things) their connection with the Laplace transform. The Hardy $paces in the half-plane share most of the properties of the HP(D) spaces. Indeed these properties are sometimes established by using a Mobius transformation such as

w-l w+l

z=--

160

8. The Lebesgue Spaces IJ'

This transformation maps points ZED to points w E no; the circle aD is mapped to the imaginary axis. If g E A(D), then the function f defined by few) =g

(W-l) w+l

(8.2)

is in A(no), and in this manner we can IItransport" some of the properties of the HP(D) spaces. We return to this comment at the end of the section. Let Mp(x,n =

{!

r

p

If(x + iY)IP dy

,

and define the function II . lip by

IIfllp

= supMp(x,f)· x>l!l

The Hardy spaces HP(n o) are defined as HP(n o) =

if

E

A(n o) : IIfllp < oo}.

In contrast with the set D, the set no is not bounded in the complex plane, and this complicates matters because along any line Ix = {z E C : Re z = x, x > O} the function fx defined by fx(y)

= r(x + iy)

must be in IY(Ix ). Moreover, we cannot appeal directly to the maximum modulus principle to evaluate IIfllp because the region is unbounded and the function may have a singularity at the point at infinity. The bound for Mp ( x, f) for large x is as important as that for small x for the general f E A(n o). It turns out, however, that the requirement that IIfllp < 00 forces f to tend uniformly to zero as z tends toward the point at infinity along any path inside any fixed half-plane of the form no = {z E C : Rez > 0 > O}. It can thus be shown that Mp ( x, f) is a decreasing function of x and that

As with the HP(D) spaces, the boundary values of functions in HP(n o) can be mapped to elements in an IY space. The

8.9. Sobolev Spaces

Wk,p

161

Lebesgue space associated with HP(n o) is the space L~(lR), and the corresponding Cauchy (Poisson) integral for this case is x ( f(it) fx(Y) = 1C fRo x2 + (y _ t)2 dt.

For each x > 0 the function f" is in L~ (lR), and

IIf" - fllv -+ 0 as x -+ 0+. Moreover, we have that IIf IIHP = IIlllv. As remarked earlier, the sets D and no are related by a Mobius transformation, and we can expect that the sets of functions in one space can be used to generate the functions in the other space. Ifg E HP(D) , then the function f as defined in equation (8.2) is certainly in A(n o), but it is not clear whether f E HP(no) and whether all the functions in HP(n o) can be generated by functions in HP(D). The final result of this section gives a concrete connection between the two spaces (cf. [21 D. Theorem 8.8.3 Letg E A(D) and define f as in equation (B. 2). lfp > 1, thenf E HP(n o) if and only if there is a function G E HP(D) such that g(z)

= (1 -

z)2/P G(z).

Equivalently, the function g is in HP(D) ifand only if there is a function F E HP(n o) such that few)

8.9

= (1 + w)2/p F (w).

Sobolev Spaces

Wk,p

Sobolev spaces are another class of function spaces based on the Lebesgue integral. These function spaces have found widespread applications in differential equations and feature norms that not only IImeasure" the modulus of a function but also the modulus of its derivatives in an IY setting. In this section we give a briefglimpse

iii

162

8. The Lebesgue Spaces IJ'

of these important spaces and hopefully whet the reader's appetite for a more serious study. A detailed account of the theory can be found in [1 ]. Let Cl (1) denote the set offunctions f mapping the intervall C lR to lR such that f' exists and is continuous for all x E 1. The set Cl (1) forms a vector space, and the function II . Ih,oo defined by

IIflh,oo = sup If(x)1 + sup I['(x) I xeI

xeI

is a norm. The space (Cl(l), II . Ih,oo) can be shown to be complete. More generally, we can define functions II . Ih ,P by

and investigate the correspondingnormed vector spaces (Cl(l), IIflh ,P ). For 1

8.9. Sobolev Spaces

Wk,p

163

by k

IIfllk,P =

L

IIfCJ)lI p ,

j=O

then the Sobolev space (Wk,p(I), II . IIk,P) is defined as the completion of the space (ek(I), II . IIk ,P)'

Hilbert Spaces 2 andL CHAPTER

9.1

Hilbert Spaces

Hilbert spaces are a special class ofBanach spaces. Hilbert spaces are simpler than Banach spaces owing to an additional structure called an inner product. These spaces playa significant role in functional analysis and have found widespread use in applied mathematics. We shall see at the end of this section that the Lebesgue space £2 (and its complex relative H 2 ) is a Hilbert space. In this and the next section, we introduce some basic definitions and facts concerning Hilbert spaces of immediate interest to our discussion of the space £2. Further details and proofs of the results presented in these sections can be found in most books on functional analysis, e.g., [25]Let X be a real or complex vector space. An inner product is a scalar-valued function (., .) on X x X such that for any f, g, hEX and any scalar a the following conditions hold: (i) if, f) > 0; (ii) if, f) = 0 if and only if f = 0; (iii) if + g, h} = if, h} + (g, h); (iv) if, g) = (g,f); (v) (af,g)

= a if, g). 165

166

9. Hilbert Spaces and L 2

A vector space X equipped with an inner product (., .) is called an inner product space and denoted by (X, (., .)). Note that in general, (f,g) is a complex number; however, condition (i) indicates that (f,f) is always a real nonnegative number. Note also that conditions (iii) and (iv) imply that (f,g + h)

= (f,g) + (f, h).

In general, (f, pg) = p(f, g) # f3(f, g), and consequently the inner product is not in general a bilinear function. The special case arises frequently in applications that X is a real vector space and (., .) is realvalued. These spaces are referred to as real inner product spaces, and for these spaces the inner product is a bilinear function.

Example 9·1·1: en and Rn Let X = en and for any z = (Zl, Z2, ... ,zn), W = (WI, W2, ... , w n) en define (., .) by

E

n

(z, w)

= LZjWj. j=l

Then it is straightforward to verify that (.,.) is an inner product on en. Similarly, if X = Rn and for any x = (Xl, X2, ... ,xn), y = (Y1, Y2, ... ,Yn) E Rn the function (., .) is defined by n

(x, y)

= LXjYj, j=l

then (., .) defines an inner product on Rn. The definition of the inner product is modeled after the familiar inner product (dot product) defined for Rn.

Example 9·1·2: -e 2 Let -e 2 denote the set of complex sequences {an} such that the series L~=l Ian 12 is convergent. If addition and scalar multiplication are defined the same way as for the space -e 1 in Example 9-1-3, then-e 2 is a vector space. Suppose that a = {an}, b = {b n } E -e 2, and let Cn = max (an , b n). Then the series L~=l Icn l2 is convergent, and hence the series L~l anb n is absolutely convergent. An inner product on this

9.1. Hilbert Spaces.

167

vector space is defined by 00

(a, b)

=L

anbn .

n=l

Let (X, (., .)) be an inner product space and let II . II : X -+ R be the function defined by IIfll

= J (f,f)·

We will show that II ·11 as defined above is a norm hence justifying our notation. The,conditions for the inner product ensure that II· II meets the requirements for a norm except perhaps the triangle inequality. In order to establish the triangle inequality we need the following result, which is of interest in its own right: Theorem 9.1.1 (Schwarz's Inequality) Let (X, (., .)) be an inner product space. Iff, g

E

X, then

I(f, g) I < IIf illig II· The proof of this inequality is left as an exercise. Now, IIf + gll2 = if + g,f + g) = (f,f) + (f,g) + (f,g) + (g,g) = IIfll 2 + 2Re (f,g) + IIgll 2 2 2 < IIfll + 21(f,g)1 + IIgll , and Schwarz's inequality implies that 2 IIf + gll2 < IIfl1 + 211fllllgll

+ IIgll 2

= (lIfll + IIglli· Thus, IIf + gil < IIfll

+ IIgll,

and hence II . II defines a norm on X. Given any inner product space (X, (., .)) we can construct a normed vector space (X, II . II). The function II . II is called the norm induced by the inner product. The normed vector space mayor may not be complete. If (X, II . II) is a Banach space, then the inner

168

9. Hilbert Spaces and £2

product space (X, (., .)) is called a Hilbert space. A Hilbert space is thus an inner product space that is complete in the norm induced by the inner product. The inner product spaces (Rn, (., .)) and (en, (., .)) are examples of finite-dimensional Hilbert spaces. Although we do not show it here, the inner product space (£2, (., .)) of Example 9-1-2 is an infinite-dimensional Hilbert space. It is of interest to enquire whether a given normed space (X, II . II) can be identified with an inner product space (X, (., .)). In other words, given a norm 11·11 on X can an inner product (., .) be defined on X such that II . II corresponds to the norm induced by (., .)? Suppose that the norm II . II can be obtained from some inner product (., .). Then for any f,g EX,

IIf + gll2 + IIf - gll2 = if + g,f + g) + if -

g,f - g)

+ (f,g) + (g,f) + (g,g) +(f,f) + (f, -g) + (-g,f) + (g,g) 2 2 2 (lIfll + IIg1l ) ;

= (f,f) =

thus, if a norm II . II can be obtained by some inner product, then it must satisfy the equation

IIf + gll2 + IIf - gll2 = 2 (lIfll 2+ IIgI1 2). This condition is also sufficient. This equation is called the parallelogram equality. The name comes from an elementary relation in plane geometry. If x and yare two vectors in R2 that are not parallel, then they can be used to define a parallelogram. Here, the quantities IIxll and IIYII correspond to the lengths of the vectors x and Y respectively and hence the lengths of the sides of the parallelogram. The quantities IIx + YII and IIx - Y II correspond to the lengths of the diagonals. The parallelogram equality is useful for (among other things) showing that certain norms cannot be obtained from an inner product. Example 9-1-3: Consider the normed vector space (C[a, b], 11·1100) defined in Example 8-1-5. We shall use the parallelogram equality to show that the norm II . 1100 on C[a, b] cannot be obtained by an inner product. Suppose, for contradiction, that the norm II . II 00 can be obtained by an inner product. Then the parallelogram equality must be satisfied for any

_ _ _ _ _ _~ - - - - - - - - - - - - - - _ . -••_., •.

I

9.1. Hilbert Spaces

choice of f, g

E

169

C[a, b]. Let f and g be the functions defined by

x-a . -a

f(x) = I, g(x) = b

Now, IIflloo = I, IIglloo = I,

IIf+glloo= sup 1+ xE[a,b]

x-a =2, b- a

and

IIf - glloo = sup 1xE[a,b]

x-a = 1; b- a

consequently,

IIf+gll~ + IIf-gll~ = 4+ 1= 5, and

2 (lIfll~ + IIgll~) = 4. As the parallelogram equality is not satisfied for these functions, the normed vector space (C[a, b], 11·1100) cannot be obtained from an inner product space. The parallelogram equality can be used to show that the only lY space that might arise from an inner product space is L 2 • Theorem 9.1.2 The only Lebesgue norm II .lip that can be obtained from an inner product is the L 2 norm II . 112. Proof We first establish the result for lY[ -I, 1] spaces. A modest change of the proofleads to the result for general lY[a, b] spaces and lY(l) spaces where I is an unbounded interval. Suppose that the norm II . lip on lY[ -I, 1] can be obtained from an inner product. Then the parallelogram equality indicates that

IIf + gil; + IIf - gil; = 2(lIfll; + IIgll;) for all f, g

E

lY[ -I, 1]. Consider the functions f and g defined by

f(x)

= 1 + x,

i I

g(x)

=1-

x.

9.1. Hilbert Spaces

1 71

Now,

/ p+ 1 8 (P) = 1 - P + 1 - log(p + 1) = - log(p + 1) < 0, and thus 8 is a monotonic strictly decreasing function of p. Since 8(1) = 1 - 2 log 2 < 0, it follows that 8(P) < 0 for all p > 1 and consequently that Q/(P) < 0 for all p > 1. This means that Q is a monotonic strictly decreasing function of p, and therefore the equation Q(P) = 0 can have at most one solution. Therefore, the parallelogram equality is satisfied only if p = 2. • A slight modification of the functions f, g for the general interval [a, b] leads to the proof of the result for the V[a, b] spaces. If I is not a bounded interval, then a suitable restriction of the functions can be used to establish the result for the V(1) spaces. For example, suppose that 1= (-00,00). Then we can choose the functions 1 +x,

ifxE[-I,I], otherwise,

g(x) = { 1 - x, 0,

if x E [-1,1], otherwise,

[(x)

={

0,

and the proof then is exactly the same.

0

It is not difficult to see that for any interval I, £2(1) is a Hilbert space.

We know from Chapter 8 that this space is complete with respect to the II . 112 norm, and the inner product (', .) defined by (f, g) = [f(X)g(x)

ax

for all f, g E £2(1) induces this norm. The complex space £~(1) is also a Hilbert space with an inner product (', .) defined by (f,g)

= [f(t)g(t)dt.

The conjugate exponent p = q = 2 is special throughout the function spaces based on the Lebesgue integral. One can show, for example, that the only Hilbert space among the Hardy spaces is H 2 , and a similar statement can be made about the Sobolev spaces. These spaces have found widespread use and have many special properties not enjoyed by the other V (HP, Wk,P) spaces, p # 2, because they

---,---,~

I

170

9. Hilbert Spaces and £2

These functions are (in equivalence classes) in V[ -I, 1] for all p > 1. Now,

1 =1 + gll~ = 1 gll~ 1

IItll~ IIgll~

and

lit

lit -

=

[-1,1]

2P+I P

+1

,

2P+I

1

(l-xfax=

[-1,1]

-1

1 1

P+1

,

1

11

[-1,1] 2P+ I

+ x + (1 -

x)IP ax

=

,

=

[-1,1]

= 2P+ I

=

(1 +xfax

-1

11-x1Pax

=

1

1 = 1

11 +xlP ax

=

11 + x - (1- x)IP ax =

2P ax

-1

1

12xlP ax

-1

2P+I xP ax = - o p+l

1 1

The parallelogram equality implies that

{2P+ I)2IP+

(2P+

P I 2P+I )2I = 2 {( 2P+I )2/P + )2/P} ( p+l p+l p+l

,

Le.,

i/

(p + 1

p

-

3 = O.

Note that p = 2 is a solution to this equation. We will show that this is the only solution for p > 1. Let Q(P) = (p + 1)2/p

-

3.

Then Q'(P) =

2

p2(p+ 1)

(p + li/p s(p)

,

where S(P) = p - (p + 1) log(p + 1).

--.-..--.. - I

172

9. Hilbert Spaces and £2

are Hilbert spaces. In the next section we investigate some special properties of Hilbert spaces. For the remainder of this chapter we shall denote an inner product space (X, (', .)) simply by X, unless there is some danger of confusion, and the norm II . II on X will always be the norm induced by the inner product unless otherwise stipulated. Exercises 9-1: 1. Verify that the function (', .) defined in Example 9-1-2 satisfies

the conditions for an inner product. 2. Let X be an inner product space and suppose f, g E X with g # O. Prove that l(f,g/lIgll)I < IIfll and use this to prove the Schwarz inequality. 3. Let X be an inner product space. Prove that for any elements f,g, hEX, IIh - fll

2

+ IIh -

gll2

1

= -llf -

gll2

+ 211h -

1 -(f + g)1I 2.

2 2 This relation is known as the Appolonius identity.

9.2

Orthogonal Sets

The paradigm for a finite-dimensional inner product space is the space R n discussed in Example 9-1-1. In this space, the inner product can be used to measure the angle between two vectors, Le., (x, y) = IIxllllYII coscP, where cP is the angle between the vectors x and y. This geometrical idea can be extended to infinite-dimensional real inner product spaces, but there is no satisfactory extension to general inner product spaces. It turns out, however, that the magnitude of the angle between elements in an infinite-dimensional real vector space is of limited interest in the general theory with one important exception. Recall that for x, Y E Rn the relationship (x, y) = 0 means geometrically that the vector x is orthogonal to the vector y. This concept of orthogonality can be readily extended to general inner product spaces. Let (X, (', .)) be any inner product space. 'TWo elements f, g E X are said to be orthogonal if (f, g) = o.

i

I

9.2. Orthogonal Sets

173

This relationship is denoted by f 1.. g. As we shall see, orthogonality plays an important part in Hilbert space theory. Example 9-2-1:

Consider the inner product space £2[ -Jr, Jr], and let fn denote the functions defined by fn(x) = sin(nx) for n = 1,2, .... Now, ifn, fm) =

=

1

[-1l',1l']

L:

fn(x)fm(x) ax

sin(nx) sin(mx) dx.

Integration by parts indicates that 1 ]1l' +n j1l' cos(nx)cos(mx)ax ifn,fm)= [ --sin(nx)cos(mx) m -1l' m_1l'

= -nj1l' cos(nx) cos(mx) ax, m -1l'

and integration again by parts yields ifn, fm)

= !!- {[~ sin(mx) COS(nx)]1l' + !!-j1l' sin(nx) sin(mx) ax} m

=

m

-1l'

m_1l'

(:)2 ifn,fm).

If n # m, then 1 - (n/m)2 n = m, then (fn,fm)

=

llfn1l

# 0, and consequently 2

=

L:

2

sin (nx) dx

ifn,fm) = 0. If

= Jr.

We therefore have that fn 1.. fm unless m = n. Suppose that f and g are elements in the inner product space X. Then

If f 1.. g then If,g) = 0, and we thus have an extension of Pythagoras's theorem:

174

9. Hilbert Spaces and £2

We pause here to introduce two terms applicable to a general vector space. Let X be a vector space, and let Y and Z be subspaces of X. If for each f E X there exist elements g E Y and h E Z such that f = g + h, then we say that X is the vector sum of Y and Z, and denote this relationship by X = Y + Z. If, in addition, the elements g E Y and h E Z are determined uniquely for every f EX, then we say that X is the direct sum of Y and Z, and write X = Y ffi Z. Given a set S C X, where (X, (., .)) is an inner product space, another set S1. can be formed by taking all the elements of X that are orthogonal to every element of S, i.e., the set S1. = if EX: if, g) = o for all g E S}. The set S1. is called the orthogonal complement of S. Example 9-2-2: Let el, e2, e3 be three linearly independent vectors in lR3 such that (ej, ek) = 0 forj, k = 1,2,3 unlessj = k. If S = {v E lR3 : v = ael +be2 for some a, b E }R} (i.e., the span of tel, e2}), then S1. = {w E lR3 : W = ce3 for some C E lR}. Since the set {el, e2, e3} forms a basis for }R3, we have that }R3 = S ffi S1. . Example 9-2-3: Let.e 2 be the inner product space defined in Example 9-1-2, and let el E f} be the sequence {I, 0, 0, ...}. Let S be the subspace defined by S = {a E f} : a = ael for some a E C}, and define the set B by B = {b = {b n } E .e 2 : b I = OJ. Then (a, b) = 0 for all a E Sand all b E B, and consequently B C S1.. In fact, B = S1., for if there is a sequence c = {cn} with CI # 0, then (a, c) = aCI, and aCI is not zero for all a E C. Since any sequence d = {tin} can be written as {dl , 0, 0, ...} + {O, d2 , d3 , ••• }, and this representation is unique, we have that .e 2 = S ffi S1. . The inner product space]R3 and the set S ofExample 9-2-2 demonstrate two geometrical properties both of which extend to any finite dimensional inner product space. First, given any vector x E }R3 there is a unique vector v E S at a minimum distance from x, i.e., IIx - vII < IIx - vII for all v E S \ {v}. Second, any vector x E }R3 can be represented in the form x = v + w, where v E Sand W E S1. are uniquely determined, Le., R 3 = SffiS1.. Any subspace of}R3 has these properties, and one can enquire whether these properties persist in

----------.-...------------_._-_

.. ,....-

,

9.2. Orthogonal Sets

175

the infinite-dimensional case. The space £2 in Example 9-2-3 demonstrates the second property, and it is straightforward to show that the set S in this example also has the minimum distance" property. Unfortunately, these properties do not carry over to the general inner product space. Under the crucial assumption of completeness, however, these properties do extend to infinite-dimensional inner product spaces. II

Theorem 9.2.1 Let S be a closed subspace of the Hilbert space X and. let f EX. Then • there exists a unique g E S at a minimum distance from f. Note that if S is a closed subspace, then S is a Hilbert space in its own right. Theorem 9.2.2 (Projection Theorem) Let S be a closed subspace ofthe Hilbert space X. Then S..L is also a closed subspace in X, and. X = S E9 S..L. Moreover, iff E X is decomposed into the form f = g + h, where g E S and h E S..L then g is the unique element in S closest to f. The projection theorem provides a key to understanding bases in Hilbert spaces. The study of bases in general Banach spaces is of limited value, but there is a rich and useful theory for bases in Hilbert spaces. As our main focus will eventually be on the space L 2 , which is separable, we limit our general discussion to bases for separable Hilbert spaces. In Example 9-2-2, consider the set K = {ell e2, e3}' The set K..L consists of all the vectors x E lR3 such that (x, ek) = 0 for k = 1,2,3, buttheonlyvectorwiththispropertyisx = O;thusK..L = O.Now, the set K forms a basis for lR3 , and this is characterized by the condition K..L = O. Motivated by this observation, we can extend the concept of an orthogonal basis to general (separable) Hilbert spaces. Let N be a set of elements in the Hilbert space X. If N..L = 0, then N is called a total set1 inX. If M is a countable set inX and (a, b) = 0 for all a, b E M, a =1= b, then M is called an orthogonal set in X. If, in addition, lIall = 1 for all a E M, then M is called an orthonormal I

IThese sets are also called "complete" in the literature. We avoid this term, since "complete" in this context has no connection with "complete" as used in Banach space theory.

176

9. Hilbert Spaces and £2

set in X. For our purposes, we are interested primarily in the case where M has an infinite number of elements but is countable. Let M = {4>nl be an orthonormal set in the Hilbert space X, and let f E X. The numbers if, 4>n} are called the Fourier coefficients of f with respect to M, and the series L~=l if, 4>n}4>n is called the Fourier series of f (with respect to M). Now, for any mEN, m

o < IIf -

m

L if, 4>n}4>n 11 n=l

2

= if -

m

m

L if, 4>n}4>n, f - L if, 4>n}4>n} n=l n=l m

= if,f} - if, Lif, 4>n}4>n} - (Lif, 4>n)4>n,f}

n=l

n=l

m

m

+ (L if, 4>n}4>n, L n=l

if, 4>n)4>n}

n=l m

= IIfll

2

-

m

+ II L

2 L (f, 4>n) if, 4>n} n=l

2

n=l

m

=

if, 4>n}4>n 11

m

IIfll 2 - 2 L Iif, 4>n}1 2 + II Lif, 4>n}4>nIl 2 , n=l

n=l

and Pythagoras's equation implies that m

m

II Lif,4>n}4>nIl = L 1Iif,4>n}4>nIl 2 2

n=l

n=l m

= L

Iif, 4>n} 12 114>n 11 2

n=l m

= L Iif, 4>n}1 n=l

2

and therefore we have m

o < IIf -

L if, 4>n}4>nIl n=l

2

m

=

IIfll

2

-

L 1(f,4>n}1 n=l

2

.

,

9.2. Orthogonal Sets

177

Hence, m

L /(f, rPn)/2 < I/fl/ 2,

n=l

and since the sequence {am} = {L::=1 I (f, rPn) 12 } is a monotonic increasing sequence bounded above, we have that {am} converges to some a E JR and Bessel's inequality holds: 00

L I(f, rPn)/2 < I/fl/ 2. n=l

Bessel's inequality indicates that the Fourier series of f with respect to M is absolutely convergent (in the norm). This means that the Fourier series is convergent (in the norm) and that its sum is independent of the order in which terms are added. Thus, the series L~=l (f, rPn)rPn converges to some functionPMf EX. Now, iff E [M]..L (= M..L), all the Fourier coefficients off are zero, and thus P Mf = O. Alternatively, iff E [M], then it can be shown thatPMf = f. For the general f EX, the set [M] is a closed subspace in the Hilbert space X, and the projection theorem indicates that f = PMf + h, where PMf E [M] and h~M]..L. The function PMf is thus the Ilclosest" approximation in [M] to f. An orthonormal set M = {rPn} in the separable Hilbert space X is called an orthonormal basis of X if for every f EX, 00

f

= L(f, rPn)rPn. n=l

If M is an orthonormal basis, then m

lim "(f, rPn) rPn £...J

m-+oo

= f,

n=l

and using the derivation ofBessel's inequality (and continuity of the norm function) we have that

178

9. Hilbert Spaces and £2

We thus arrive at Parseval's formula 00

2

IIfll =

L I(f, rPn) 1 2

•

n=l

Suppose now that f E M.1. Since M is an orthonormal basis of X, f = L~l (f, rPn}rPn and (f, rPn) = for all rPn E M; consequently, by Parseval's formula we have that IIfll = 0, and the definition of a norm implies that f = O. In this manner we see that if M is an orthonormal basis, then M.1 = 0, i.e., M is a total orthonormal set. Yet another implication of M being an orthonormal basis is that [M] = X. This follows from the projection theorem, since [M].1 c M.1 = 0, and so

°

X

--

--.1

--

--

= [M] E9 [M] = [M] E9 0 = [M].

It is interesting that the above implications actually work the

other way as well. For example, if M is an orthonormal set in X such that Parseval's formula is satisfied for all f EX, then it can be shown that M is an orthonormal basis. In summary we have the following result:

Theorem 9.2.3 Let X be a separable Hilbert space and suppose that M = {rPnl is an orthonormal set in X. Then the following conditions are equivalent: (i) M is an orthonormal basis; (ii) IIf11 2 = L~l 1(f, rPn) 12 for all f E X; (iii) M is a total set; (iv) [M] = x. It can be shown that every separable Hilbert space has an orthonormal basis. Formally, we have the following result: Theorem 9.2.4 Let X be a separable Hilbert space. Then there exists an orthonormal basis for X. The proof of this result is based on a Gram-Schmidt process analogous to that used in linear algebra to derive orthonormal bases. Parseval's formula is remarkable in this context because it essentially identifies all separable Hilbert spaces with the space e2 . Now, e2 is a separable Hilbert space, and given an orthonormal basis M = {rPnl

i i

9.2. Orthogonal Sets

179

on a general separable Hilbert space X there is a mapping T : X -+ e2 defined by the Fourier coefficients if, cPn}. In other words, T maps f E X to the sequence {if, cPn}} E e2 . Parseval's formula shows that T is an isometry from X to e2 . On the other hand, given any sequence {an} E e2 , the function defined by L~I ancPn is in X, since X is a Hilbert space and the series is convergent. We thus have that any separable Hilbert space is isometric to the Hilbert space e2 . In fact, an even stronger result is available, viz., all separable complex Hilbert spaces are isomorphic to e2 . This means that at an Iialgebraic level" a complex separable Hilbert space is indistinguishable from the space e2 , i.e., at this level there is only one distinct space. A similar statement is true for real Hilbert spaces, where the space e2 is replaced by its real analogue. In passing we note that any Hilbert space X possesses an Ilor_ thogonal basis," though it may not be countable. The cardinality of the set is called the Hilbert dimension of the space. If X and Xare two Hilbert spaces both real or both complex with the same Hilbert dimension, then it can be shown that X and X are isomorphic. This is a generalization ofthe situation in finite-dimensional spaces, where, for example, all n-dimensional real Hilbert spaces are isomorphic to JRn. Exercises 9-2:

1. Show that for any integers m, n

1

cos(mx)cos(nx)dx =

[-Jr,Jr]

{](o, ,

ifm =1= n, if m = n,

and

1

sin(mx) cos(nx) ax = 0.

[-Jr,Jr]

2. The first three Legendre polYnomials are defined by Po(x) = I, PI (x) = x, and P2(X) = ~(3x2 - 1). (a) Show that the set P onL 2 [-I,I].

=

{Po, PI, P2} forms an orthogonal set

(b) Construct an orthonormal set M from the set p' and find the Fourier coefficients for the function f(x) = ex.

180

9. Hilbert Spaces and £2

3. The Rademacher functions are defined by rn(x) = sgn(sin(Zn.rrx)) for n

= I, Z, ... , where sgn denotes the signum function sgn C) x

={

-I, I,

ifx < 0, if x > 0.

(a) Show that the set M = {rnl is an orthogonal set on £2[0,1].

(b) Iff is definedbYf(x) = cos(ZJl'X), showthatf J.. rnforalln = I, Z, ... and deduce that M cannot form a total orthogonal set. 4. Let X be an inner product space and let M = {rPnl be an orthonormal basis of X. Show that for any f, g EX, 00

if, g}

= L if, rPk} (g, rPk). k=1

9.3

Classical Fourier Series

We saw in Section 9-1 that £2 is a Hilbert space, and we know from Theorem 8.6.1 that all IY spaces are separable. Theorem 9.2.4 thus implies that the space £2 must have an orthonormal basis. It turns out that the classical Fourier series (i.e., trigonometric series) lead to an orthonormal basis for £2[a, b]. In this section we study classical Fourier series and present some basic results with little detail. There are many specialized texts on the subject ofFourierseries such as [15] and [45], and we refer the reader to these works for most the details. A particularly lively account of the theory, history, and applications of Fourier series can be found in [Z4]. For convenience we focus primarily on the space £2[ -Jr, Jr] and note here that the results can be extended mutatis mutandis to the general closed interval. A classical Fourier series is a series of the form 1

Zao +

L (an cos(nx) + bnsin(nx)), 00

n=l

(9.1)

9.3. Classical Fourier Series

181

where the an's and bn's are constants. We know from Example 9-2-1 that for any nonzero integers m, n,

1

sin(mx) sin(nx) ax = {O,

[-Jr,Jr]

]"(,

~ffm =1= n,

1

m - n.

We also know from Exercises 9-2, No. I, that for all integers m, n,

1

cos(mx) cos(nx)

[-Jr,Jr]

ax =

{' 0

if m =1= n,

]"(,

if m

= n,

and

1

sin(mx) cos(nx) ax

= o.

[-Jr,Jr]

In addition, it is evident that for any integer n,

1

sin(nx) ax

=

[-Jr,Jr]

Let

1

cos(nx) ax

= o.

[-Jr,Jr]

= {rPn} and \II = {'lfrn} denote the sets of functions defined by rPn(x)

=

cos(nx) ~

,

'lfrn(x)

=

sin(nx) ~

,

for n = I, 2, .... Then the above relationships indicate that the set S = {11.J2]r} U U \II forms an orthonormal set in £2[-]"(,]"(]. The Fourier coefficients of a function [ E £2[_]"(,]"(] with respect to S are given by

1) .J2]r 11 (, .J2]r f, -

= -

(f, rPn) = (f, 'lfrn) =

[-Jr,Jr]

1 ~1

~

'\/ ]"(

[-Jr,Jr]

'\/ ]"(

[-Jr,Jr]

[(x)

ax ,

[(x) cos(nx) ax, [(x) sin(nx)

ax,

and the Fourier series is (9.2)

\

,

182

9. Hilbert Spaces and £2

which is equivalent to expression (9.1) with the familiar coefficient relations ao

.!.1 .!.1 = .!.1 =

an = bn

f(x) dx,

re

[-Jr,3l']

re

[-Jr,Jr]

re

[-Jr,Jr]

f(x) cos(nx) dx, f(x) sin(nx) dx.

Fourier series can be expressed in atidier, more symmetric, form using the relation einx = cos(nx) + i sin(nx). Using this relation, the series (9.1) can be written in the form 00

inx

'L...J " cn e

(9.3)

,

-00

where the

Cn

are complex numbers defined by Cn

The set of functions B

= -1

1

2re [-Jr,Jr]

= {f3n},

.

f(x)e znx dx.

where

f3n(x) =

einx

$'

forms an orthonormal set for L~[ -re, re], and for any f E L~[ -re, re] (and consequently for any f E L 2 [ -re, reD the corresponding Fourier series is 00

PBf =

L(f, f3n)f3n.

(9.4)

-00

Now, for any f E L 2 [-re, re], Bessel's inequality guarantees that the Fourier series (9.2) (and (9.3)) converges in the \I . 112 norm to some function Psf E L 2 [ -re, re]. The central question here is whether the set S forms an orthonormal basis for L 2 [-re, re], so that Psf = f a.e. In fact, it can be shown that S forms an orthonormal basis for L 2 [ -re, re]. If we combine this fact with Theorem 9.2.3, we have the following result:

9.3. Classical Fourier Series

183

Theorem 9.3.1 Let f E L 2 [-re, re] and let

k=n

=

L

if, 13k}13k·

k=-n

Then: • (i) IIsn - fll2 -+ 0 as n -+ 00; (ii) Parseval's relation is satisfied:

Ilfll~ = 1if, 1}1 2+

00

L (I if, rPn}1

2

+ 1if, 1/Ik}1 2 )

n=l

re

00

2 2 2 = _a 20 + re "Ca L....Jn + bn)

n=l 00

00

= L 1if, f3n}1 = 2re L Icn l2 . 2

-00

-00

From the last section we know that all separable Hilbert spaces are isomorphic to the Hilbert space £2 and Parseval's relation is a manifestation of this relationship. This observation leads to the following result: I

Theorem 9.3.2 CRiesz-Fischer) Let {c n } E £2. Then there is a unique function f E L 2 [-re , re] such that the Cn are the Fourier coefficients for f. Note that Ilunique" in the above theorem means that the sequence {c n } determines an equivalence class of functions modulo equality a.e. An immediate consequence ofTheorem 9.3.1 and the projection theorem (Theorem 9.2.2 ) is the following result: Theorem 9.3.3 Let f E L 2 [-re, re]. Then for any E > 0 there exists a positive integer n and a trigonometric polynomial ofdegree n, say On = L~-n qkf3k such that liOn - fl12 < E. Moreover; among the trigonometric polynomials of

184

9. Hilbert Spaces and £2

degree n, the closest approximation to f in the II . 112 norm is that for which the qk correspond to the Fourier coefficients. In other words, trigonometric polynomials can be used to approximate any function in L 2 [-re, re], and the Fourier coefficients provide the best approximation among such polynomials in the II . 112 norm. Theorem 9.3.1 guarantees that the Fourier series will converge in the II ·112 norm, but this is not the same as pointwise convergence, and an immediate question is whether or not a Fourier series for a function f E L 2 [ -re, re] converges pointwise to f. More explicitly, for a fixed x E [-re, re] does the sequence of numbers {sn(x)} converge, and if so does sn(x) -+ f(x) as n -+ oo? The answer to this question is complicated, and much of the research on Fourier series revolved around pointwise convergence. Some simple observations can be made. First, the orthonormal basis defining the Fourier series is manifestly periodic, so that if sn(x) converges for x = -re, it also converges for x = re, and s( -re) = sere). The existence of a Fourier series does not require thatf( -re) = f(re), so pointwise convergence will fail at an endpoint unless f satisfies this condition. More generally, we see that any two integrable functions f and g such that f = g a.e. produce the same Fourier coefficients, so that for a specific function f, we expect that the best generic situation would be that sn(x) -+ f(x) a.e. We shall not Ilplumb the depths" of the vast results concerning pointwise convergence of Fourier series; however, we will discuss two results of interest in their own right that make crucial use of the Lebesgue integral. Prima facie, it is not obvious that the coefficients an and bn in the Fourier series have limit zero as n -+ 00. Given that arguments such as x = 0 lead to series such as L~=o an, this is clearly a concern. An elegant result called the Riemann-Lebesgue theorem resolves this • concern and is of interest in its own right. We state the result here in a form more general than is required for the question at hand.

Theorem. 9.3.4 (Riemann-Lebesgue) Let I C JR be some interval and f E L 1 (f). If {An} is a sequence of real numbers such that An -+ 00 as n -+ 00, then

1

{(x) caS(AnX) ax ---> 0 and

as n -+

00.

1

{(x) sin(AnX) ax ---> 0

9.3. Classical Fourier Series

185

Proof The proof of this result is of particular interest because the convergence theorems of Chapter 5 are of little help, and we must return to the definition ofthe integral itself. We sketch here the proof for the cosine integral. Suppose first that I = [a, b] and that f : I ~ 1R is bounded on I. If M denotes an upper bound for IfI, then [

f(x) COS(AnX) dx < M

J[a,b]

cos(Anx)dx

[ J[a,b]

=M

l

b cOS(AnX)d<

M = -lsin(Anb) - sin(Ana)1

An

2M <-

- An .

Since An ~ 00, as n ~ 00, we have that ~a,b] f(x) COS(AnX) dx ~ O. This calculation shows in particular that the theorem holds for any step function having a support of finite length. Now, the Lebesgue integral is defined in terms of a sequence of a-summable functions, with a = x, and by definition these functions have supports of finite length. If f E L 1 (!), then we know that there is a sequence ()j of x-summable functions such that

1,lf(X) - OJ(x)1 d< --+ a asj

~ 00.

Choose any E > 0, and select aj sufficiently large so that [If(x) - ()j(X) I dx <

J1

E.

2

Now, 1,f(X) cOS(AnX)d< 1,(f(x) - OJ(x) + OJ(x)) cOS(AnX) d< < 1,(f(X) - OJ(X)) COS (AnX) d<

+

1,OJ(X)cOS(AnX)d<

186

9. Hilbert Spaces and £2

< IIf(X) - OJ(x)1 ax + 10j(X) COS(AnX) ax < ;

+

l 0j(X) COS(AnX) ax

.

From the above discussion we know that l0j(X) COS(AnX) ax -*

as n

~ 00

a

for any j, so that there is an integer N such that E

< 2

whenever n > N. Therefore,

I

E

[f(x) cos(Anx)dx < 2

E

+2 =

E

whenever n > N, and by definition this means that 1 [(x) COS(AnX) ax -*

as n

~

a

o

00.

The Riemann-Lebesgue theorem with I = [-re, re] and An = n shows that the Fourier coefficients tend to zero as n ~ 00. This theorem is also crucial in proving another notable result known as the Riemann localization theorem, which we shall not prove. Theorem. 9.3.5 (Riemann Localization) Let f E L 1 [ -re, re] and x E [-re, re]. Then for any fixed 8 with 0 < 8 < re, sn(x)

~

f(x) if and only if

•

lim ~ [ f(x + t) + f(x - t) sin((n + ~ )t) dt = n~oo 2re 1[0,0] t 2

o.

Here it is assumed that f has been extended periodically to a function on 1R so that f is defined for arguments x ± t that may not be in [-re, re]. What is interesting in the above theorem is that the number 8 can be arbitrarily small, and this means that the pointwise convergence of the Fourier series depends only on the values that f

if

9.3. Classical Fourier Series

187

assumes in a small neighborhood of x. Given that the Fourier coefficients in the series defining Sn depend on the values f assumes in the entire interval [-re, re], this result is remarkable. We can study Fourier series outside the comfortable space 2 L [-re, re]. Naturally, we lose the results that rely on L 2 [ -re, re] being a Hilbert space such as Parseval's relation, but in the larger space L 1 [ -re, re] the Fourier coefficients are still well-defined, and results such as the Riemann-Lebesgue theorem are still valid. We even have the following uniqueness result: •

Theorem. 9.3.6 Iff, gEL1 [ - re, re] have the same Fourier coefficients, then f = g a. e. What is needed, however, is some result that shows that the partial sums of the Fourier series for a function f E L 1 [ -re, re] converge in the 11·111 norm to f, but this is where things go wrong. Iff E L 1 [ -re, re], then in general we do not have that IIsn - fill ~ 0 as n ~ 00. It can be shown, however, that for 1 Un

=-

n

n-1

LSk, k=l

I\un - fllI ~ 0 as n ~ 00 for any f E L 1 [ -re, re]. The quantity Un is called the Cesaro mean of the partial sum sequence {sn}. This is a weaker convergence result, since Un effectively measures an averaged partial sum. The sequence {un} may converge even if {sn} diverges, and if {sn} converges to some s, then {un} also converges to some function s E L 1 [ -re, re]. In the space L 1 [ -re,n] this is the sharpest result we can get. In fact, it can be shown that there are functions in L1 [-re, re] such that {sn(x)} diverges a.e. Though most of the above results concerning pointwise convergence were established by the early twentieth century, it was not until the 1960's that Carleson [7] proved that is f E L 2 [ -re, re] then {sn(x)} ~ f(x) a.e. From this perspective, the space L 2 [-re, re] is the natural space in which to study Fourier series. Exercises 9-3: 1. Let f(x)

= x on the interval [-re, re].

(a) Determine the Fourier series for f.

188

9. Hilbert Spaces and £2

(b) Show that the series obtained by differentiating term by term the Fourier series in part lea) is a divergent series. 2. The complex Fourier series for a function f E £2[ -A, A], A > 0, is given by 00

PMf(x)

= LCneirurxlA, -00

where en

= ~ i>(x)e-inm
for n = 0, ±l, ±2, ... , and it can be shown that the set M _ irurx1 {e A} sUitably normalized forms an orthonormal basis for £2[-A,A].

(a) Find the complex Fourier series for the function g : ~ lR defined by

[-!,!]

g(x)

= { ~l,

if _1 < x < 0 2 ' if 0 < x <

!.

(b) Use Parseval's relation and the Fourier series in 2(a) to prove that 1

00

]"(2

~ (2n + 1)2 = 8' 3. Suppose thatf has a continuous derivative on the interval [-]"(, ]"(], and let an, bn denote the Fourier coefficients off. Use integration by parts to prove the Riemann-Lebesgue theorem for this special case.

9.4

The Sturm-Liouville Problem

Theorem 9.2.4 indicates that for any interval I the Hilbert space £2(1) must have an orthonormal basis. If I is bounded, then the trigonometric functions provide such a basis. Theorem 9.2.4 does not, however, preclude the existence of other orthonormal bases, and there are, in fact, any number of orthonormal bases available. If

i i

9.4. The Sturm-Liouville Problem

189

I is not bounded, then the trigonometric functions are not even in the space L 2 (1), and the classical Fourier expansions are no longer

valid. Nonetheless, Theorem 9.2.4 guarantees the existence of an orthonormal basis for any interval I. In practice, many of the orthonormal bases for L 2 (1) arise as solutions to boundary value problems of the Sturm-Liouville type. A regular Sturm-Liouville problem consists in determining a solution y to a differential equation of the form

;

(r(x):)

+ (q(x) + J..p(x))y = a

(9.5)

on some interval [a, b], satisfying boundary conditions of the form k1y(a)

+ k2y'(a) = 0,

lly(b)

+ l2y'(b) = O.

(9.6)

Here, r, q, andp are given functions, J... is a complex parameter, and the k/s and ~'s are constants such that + k~ =f:: 0, + l~ =f:: O. In addition, it is assumed for the r¢gular Sturm-Liouville problem that the functions r andp are nonzero in the interval [a, b]. Sturm-Liouville problems are important in theory and applications of differential equations. As a consequence, this class of boundary value problems has been studied exhaustively for some 150 years. Relatively accessible accounts of the theory can be found in [5], [9], [11], [23], and [40]. In addition, many applications-oriented texts such as [3], [8], and [26] contain short summaries of the theory and various applications. We do not attempt here to replicate the general theory in any detail or depth, but merely focus on the Sturm-Liouville problem as a Ilmachine" for producing orthogonal sets in L 2 • Equation (9.5) is often written in the abbreviated form

kr

[,y = -J...p(x)y,

lr

(9.7)

where the linear operator [, is defined by

Cy =

;

(r(x):)

+ q(x)y.

(9.8)

If Y E G2[a, b], r E G'[a, b], and q E G[a, b] then Ly E G[a, b], and in particular £y E L 2 [a, b]. For linear operators such as [, there exists another operator [,* called the (Hilbert) adjoint. The adjoint operator

190

9. Hilbert Spaces and L 2

satisfies the relation (9.9) for all Yl, Y2 E L 2[a, b], which alsc:> satisfy the boundary conditions (9.6). The peculiar form of the operator £ for the Sturm-Liouville problem ensures that the operator is self-adjoint, Le., £ = £*. This can be verified directly as follows: (£Yl, Y2) = {

ira,b]

WI, £Y2)

(Y2(X)£Yl (x) - Yl (X)£Y2(X))

ax

=

la'bl; ((r(x)

=

[rex) (Y2(X) ;Yl(X) - Yl(X~ ;Y2(X)) J:

=

o.

;Yl(X)) Y2!CX) -

(r(x) ;Y2(X)) Yl(X)) ax

The final equality follows from the condition thatYl andY2 satisfy the same boundary conditions (9.6). (The specific details can be found in [26]). As we shall see shortly, the self-adjointness of £ is the key to obtaining orthogonal sets. For any value A E C, the Sturm-Liouville problem always has one obvious solution, viz., the triVial solution Y = o. Generically, this is the only solution available for a given value of A, but there may be some values of A for which nontrivial solutions exist. These values are called eigenvalues, arid the associated nontrivial solutions are called eigenfunctions. The set of all eigenvalues for the problem is called the spectrum2
i

i

9.4. The Sturm-Liouville Problem

191

The regular Sturm-Liouville problem has fairly tractable spectral properties, as the next theorem illustrates.

Theorem 9.4.1 Supposethatr E C1 [a,b],p,q E C[a,b], andthatr(x) > Oandp(x) > 0 for all x E [a, b]. Then: (i) the spectrum for the Sturm-Liouville problem is an infinite but countable set; (ii) the eigenvalues are all real; (iii) to each eigenvalue there c(J)rresponds precisely one eigenfunction (up to a'constant factor), i.e., the eigenvalues are simple; (iv) the spectrum contains no finite accumulation points. Thus, under the conditions of the above theorem, the SturmLiouville problem always has an infinite (but countable) number of eigenfunctions. The spectrum is a countable set, so we can regard it as a sequence V.n} and impose the condition IAn I < IAn+ll for all n. Since there are no finite accumlll1ation points, we see that IAnI ~ 00 as n ~ 00. Suppose that the conditions of Theorem 9.4.1 are satisfied, and that Am and An are distinct eigenvalues with corresponding eigenfunctions Ym, Yn, respectively. Then, £Ym = -AmP(X)Ym, £Yn

= -r-AnP(x)Yn,

and therefore Ym£Yn - Yn£Ym = (Am - An)P(X)YmYn.

The above equation indicates th~t

1

(Ym(X)£Yn(X) - Yn(X)£Ym(X)) dx

[a,b]

= (Ym, £Yn) - (Yn, £Ym)

= (Am -

An)

1

P(X)Ym(X)Yn(X) dx.

[alb]

Now, the eigenfunctions are real, so that (Ym, £Yn) since £ is self-adjoint, (Am - An)

1

P(X)Ym(X)Yn(x}dx

[a,b]

i i

= (£Yn, Ym) -

= (£Yn,Ym), and (Yn, £Ym)

= O.

192

9. Hilbert Spaces and L 2

By hypothesis, the eigenvalues are distinct (Am =f. An), and the above calculation shows that [

ira,b]

P(X)Ym (X)Yn (X) dx =

o.

(9.10)

If P = I, the above equation im(plies that Ym ..L Yn for m =f. n, and thus the eigenfunctions are orthgonal. By hypothesis we have that p(x) > 0, so that in any event the set of functions {.JPYn} is orthogonal. As the eigenfunctions are by definition nontrivial solutions, we know that IIPYn 112 =f. 0, so that this set of functions can always be normalized to form an orthonormal set in L 2 [a, b]. The Sturm-Liouville problem thus produces eigenfundtions from which orthonormal sets can be derived. Given a continuous function p positive on the interval [a, b], it is always possible to define anothet norm for L 2 [a, b], viz., pllYII2 = { [

ira,b]

P(X)y2(X)dx}1/2

(9.11)

Since P is positive and continuous on [a, b], there exist numbers Pm and PM such that 0 < Pm < p(x) tS PM for all x E [a, b]. This implies that PmllyII2 < pllYII2 < PMIIYII2,

so that the p 1\·112 norm is equivalent to the 1\·112 norm. Consequently, the vector space L 2 [a, b] equippe