Haddad W.M. Nonlinear Dynamical Systems and Control: A Lyapunov-Based Approach

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 55

Proof. Since f : D → R

is continuous on th e compact set D

follows that α = max

x∈D

f(x) exists. Now, suppose, ad absurdum, that f

is not uniformly Lipschitz continuous on D

. Th en , for every L > 0, there

exists x, y ∈ D

such th at kf(y) − f(x)k > Lky − xk. In particular, there

exist sequences {x

}

∞

n=1

and {y

}

∞

n=1

⊆ D

such that

kf(y

) − f(x

)k > nky

− x

k, n ∈ Z

. (2.82)

Now, since D

is a compact set, it follows from the Bolzano-Lebesgue

theorem (Theorem 2.4) that there exist subsequences {x

}

∞

k=1

and {y

}

∞

k=1

such that x

→ x

∗

∈ D

and y

→ y

∗

∈ D

as k → ∞ . Since n

∈ Z

, it

follows that

∗

− x

∗

k = lim

k→∞

−x

k ≤

kf(y

) −f (x

)k ≤

2α

, (2.83)

and hence, x

∗

= y

∗

. Now, since f is Lipschitz continuous on D, there exists

a neighborhood N ⊂ D of x

∗

and a Lipschitz constant L = L(x

∗

) > 0 such

that

kf(y) − f (x)k ≤ Lky − xk, x, y ∈ N. (2.84)

Since x

→ x

∗

and y

→ y

∗

as k → ∞ it follows that f or suﬃciently large

n, x

and y

∈ N, and hence,

kf(y

) −f(x

)k ≤ Lky

− x

k. (2.85)

For n ≥ L, (2.85) contradicts (2.82). Hence, f is uniformly Lipschitz

continuous on D.

The following proposition is a dir ect consequence of Proposition 2.22.

Proposition 2.24. Let D ⊆ R

and let f : D → R

be continuous on

D. If f

′

exists and is continuous on D, then f is Lipschitz continuous on D.

Proof. It follows from Theorem 2.15 that the existence and continuity

of f

′

on D implies that f is continuously diﬀerentiable on D. Now, the r esult

is a direct consequence of Proposition 2.22.

Proposition 2.25. Let D ⊆ R

, C ⊂ D be a convex set, and let f :

D → R

. Suppose f

′

exists and is continuous on D. If there exists L > 0

such that kf

′

(x)k ≤ L for all x ∈ C, then f is uniformly Lipschitz continuous

on C.

Proof. It follows from Theorem 2.15 that the existence and continuity

of f

′

on D implies that f is continuously diﬀerentiable on D. Now, it follows

as in the proof of Proposition 2.22 with B

) replaced by C that

kf(x) −f (y)k ≤ Lkx −yk, x, y ∈ C,

which implies the result since L is independent of x and y.

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56 CHAPTER 2

Finally, we provide necessary and suﬃcient conditions for a function

f to be globally Lipschitz continuous.

Proposition 2.26. Let f : R

→ R

be continuously diﬀerentiable on

. Then f is globally Lipschitz continuous if and only if there exists L > 0

such that kf

′

(x)k ≤ L for all x ∈ R

Proof. Suﬃciency is a d irect consequence of Proposition 2.25 with

C = R

. To show n ecessity suppose f is globally Lipschitz continuous with

the Lipschitz constant L. Now, since f (·) is continuously diﬀerentiable on

, it follows from Deﬁnition 2.34 and Theorem 2.15 that for every x ∈ R

and ε > 0, there exists δ > 0 s uch that



f(y) − f (x) −

∂f

∂x

(x)(y − x)



≤ εky − xk, y ∈ B

(x). (2.86)

Next, note that



∂f

∂x

(x)(y − x)



≤ kf(y) − f(x)k +



f(y) − f (x) −

∂f

∂x

(x)(y − x)



≤ (L + ε)ky − xk, y ∈ B

(x), (2.87)

and h en ce,



∂f

∂x

(x)z



≤ (L + ε)kzk, z ∈ B

(0). (2.88)

Thus ,



∂f

∂x

(x)



= max

z∈B

(0)

∂f

∂x

(x)zk

kzk

≤ (L + ε), (2.89)

where k · k : R

n×n

→ R is the equi-induced matrix norm induced by the

vector norm k · k : R

→ R. Finally, since ε is arbitrary it follows th at

∂f

∂x

(x)k ≤ L, w hich proves the result.

Note that kf

′

(x)k ≤ L, x ∈ R

, is equivalent to saying that f

′

uniformly bounded, that is, the bound is independent of x in R

2.5 Vector Spaces and Banach Spaces

In this section, we introduce concepts involving linear vector sp aces, normed

linear spaces, and Banach spaces. First, however, we introduce the concepts

of binary operations, semigroups, groups, rings, and ﬁelds. Let V be a set

which may be ﬁnite or inﬁnite. A binary operation ◦ takes any two elements

x, y ∈ V and transforms them to z = x ◦ y, not necessarily in V. A binary

operation ◦ is said to be closed if, for every x, y ∈ V, x ◦ y ∈ V. Hence, a

closed binary operation is a mapping φ : V×V → V deﬁned by φ(x, y) = x◦y.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 57

An algebraic system (V, ◦) is a nonempty set V with the binary operation

◦ : V × V → V.

A semigroup (S, ◦) is a n on empty set S with associativity of the binary

operation ◦ such that S is closed. That is, x ◦ y ∈ S for all x, y ∈ S, and

x ◦(y ◦z) = (x ◦y) ◦z for all x, y, z ∈ S. A group (G, ◦) is a nonempty set G

with a binary operation ◦ s uch that i) G is closed under ◦, ii) ◦ is associative

in G, iii) there exists an identity element 1 ∈ G such that x ◦1 = x = 1 ◦ x

for all x ∈ G, and iv) for each x ∈ G, there exists a unique inverse x

−1

∈ G

such that x ◦ x

−1

= 1 = x

−1

◦ x. An Abelian group is a group (G, ◦) with

commutativity of the binary operation, that is, x ◦y = y ◦x for all x, y ∈ G.

A ring (R, +, ·) is a nonempty set R with the two binary operations

of addition (+) and multiplication (·) connected by distributive laws. That

is, x ·(y + z) = x · y + x · z and (y + z) · x = y · x + y · z for all x, y, z ∈ R.

Here we assume that (R, +) is an Abelian group with the identity element

denoted by 0 ∈ R and (R, ·) is a semigroup with respect to multiplication

with the identity element 1 ∈ R. Furthermore, 0 is referred to as the zero

element and the elements x ∈ R, x 6= 0, are referred to as nonzero elements.

Finally, a ﬁeld F is a commutative ring with 1 ∈ F and with the nonzero

elements in F forming a group under the binary operation of multiplication.

For the following d eﬁ nition we let F denote a ﬁeld. For example, F can

denote the ﬁeld of real numbers, the ﬁeld of complex numbers, the binary

ﬁeld, the ﬁeld of rational fun ctions, etc.

Deﬁnition 2.36. A linear vector space (V, +, ·) over a ﬁeld F is a set

V with the (addition and multiplication) operations + : V × V → V and

· : F ×V → V such th at the following axioms hold:

i) (Commutativity of addition): x + y = y + x for all x, y ∈ V.

ii) (Associativity of addition): x + (y + z) = (x + y) + z for all x, y, z ∈ V.

iii) (Existence of additive identity): For every x ∈ V, there exists a unique

element 0 ∈ V such that 0 + x = x + 0 = x.

iv) (Existence of additive inverse): For every x ∈ V, there exists −x ∈ V

such that x + (−x) = 0.

v) α(βx) = (αβ)x for all α, β ∈ F and x ∈ V.

vi) α(x + y) = αx + αy for all α ∈ F and x, y ∈ V.

vii) (α + β)x = αx + βx for all α, β ∈ F and x ∈ V.

viii) For every x ∈ V and the identity element 1 ∈ F, 1 · x = x.

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58 CHAPTER 2

Henceforth, unless otherwise stated, we choose F = R. There are many

examples of linear vector spaces. As a ﬁrst example note that the set R

constitutes a linear vector space with addition and scalar multiplication

deﬁned in the usual way. In particular, for x = [x

, . . . , x

]

, y =

, . . . , y

]

, and α ∈ R, with addition and scalar multiplication deﬁned

x + y

△

= [x

+ y

, . . . , x

+ y

]

, (2.90)

αx

△

= [αx

, . . . , αx

]

, (2.91)

x, y satisfy Axioms i)–viii) of Deﬁnition 2.36, and hence, R

is a real linear

vector space. As another example consider vector-valued functions f(·)

deﬁned on [a, b], that is, f : [a, b] → R

. Then the set of such functions

constitutes a linear vector space since αf (·) can be deﬁned by

(αf)(t)

△

= αf (t), t ∈ [a, b], (2.92)

and (f + g)(·) can be deﬁned by

(f + g)(t)

△

= f (t) + g(t), t ∈ [a, b], (2.93)

where add ition and scalar multiplication in the right-hand side of (2.92) and

(2.93) are deﬁned as in (2.90) and (2.91). Many diﬀerent linear vector spaces

can be created by specifying which functions are allowed. For examples,

we may admit continuous functions, diﬀerentiable functions, or bounded

functions. Each of these cases gives a d iﬀerent vector space.

Deﬁnition 2.37. Let V be a linear vector space and let S ⊂ V. Then

S is a subspace if αx + βy ∈ S for all x, y ∈ S and α, β ∈ F.

Example 2.24. Consider the subset of R

given by H = {(x

, x

) ∈

: x

≥ 0}. Note that H is not a subspace since H is not an additive

group, that is, H is not closed on taking negatives. However, the boundary

of H given by ∂H = {(x, 0) : x ∈ R} ⊂ R

is a subspace of R

. △

Let {x

, . . . , x

} ⊂ R

. Then the set of all vectors of the form α

···+ α

, wh ere α

∈ R, i = 1, . . . , m, is a subs pace of R

. This subspace

is called the span of {x

, . . . , x

} and is denoted by span{x

, . . . , x

Proposition 2.27. Let M and N be subspaces of a linear vector space

V. Then M ∩N is a subspace of V.

Proof. Note that 0 ∈ M ∩ N since 0 ∈ M and 0 ∈ N. Hence,

M∩N 6= Ø. Now, if x, y ∈ M∩N, then x, y ∈ M and x, y ∈ N. Next, for

every α, β ∈ F the vector αx + βy ∈ M and αx + βy ∈ N since M and N

are subspaces. Hence, αx + βy ∈ M∩ N.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 59

Deﬁnition 2.38. Let V be a linear vector space and let M ⊂ V and

N ⊂ V. Then M + N = {x + y ∈ V : x ∈ M, y ∈ N} is called the sum of

M and N.

Proposition 2.28. Let M and N be subspaces of a linear vector space

V. Then M + N is a subspace.

Proof. First, note that 0 ∈ M + N since 0 ∈ M and 0 ∈ N. Next,

let x, y ∈ M + N. I n this case, there exist vectors u, v ∈ M and w, z ∈ N

such that x = u + w and y = v + z. Now, for every α, β ∈ F, αx + βy =

(αu + βv) + (αw + βz), and hence, αx + βy ∈ M+ N for all x, y ∈ M+ N

and α, β ∈ F.

Next, we introduce the notion of distance for a linear vector space.

As seen in Section 2.2, since norms provide a natural measure of distance,

endowing a linear vector space with a norm gives a normed linear space.

Deﬁnition 2.39. A normed linear space over a ﬁ eld F is a linear vector

space X with a norm ||| · ||| : X → R such that the following axioms hold:

i) |||x||| ≥ 0, x ∈ X.

ii) |||x||| = 0 if and only if x = 0.

iii) |||αx||| = |α||||x|||, x ∈ X, α ∈ F.

iv) |||x + y||| ≤ |||x||| + |||y|||, x, y ∈ X.

In Deﬁnition 2.39 the notation ||| · ||| is used to denote a vector or

matrix operator (function) norm as opposed to k·k, which is used to denote

a vector or matrix norm . In the case where X = R

or X = C

, |||·||| should

be interpreted as k ·k. (The reason for this change in notation will become

clear later.)

We can create a normed function space by assigning a norm to a linear

vector space. For example, for f : [0, ∞) → R deﬁne

|||f ||| = sup

t∈[0,∞)

|f(t)|, (2.94)

where we allow only continuous functions f (·). Then it can be easily veriﬁed

that the conditions in Deﬁnition 2.39 with the norm ||| ·||| are satisﬁed. It is

important to note that in general many diﬀerent norms can be assigned to

the same linear space.

Example 2.25. Consider the linear vector s pace R

with norm k·k

∞

→ R deﬁned by kxk

∞

= max

i=1,...,n

(i)

|. Note that k · k

∞

satisﬁes

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60 CHAPTER 2

Axioms i)–iii) of Deﬁnition 2.39 tr ivially. To show iv) let x = [x

, . . . , x

]

and y = [y

, . . . , y

]

. Now,

kx + yk

∞

= max

i=1,...,n

(i)

+ y

(i)

| ≤ max

i=1,...,n

(i)

|+ max

i=1,...,n

(i)

| = kxk

∞

+ kyk

∞

(2.95)

and hence, iv) holds. Hence, the linear vector space R

endowed with the

inﬁnity norm is a normed linear space. I n a similar fashion it can be shown

that R

endowed with the absolute sum norm and the Euclidean norm are

normed linear spaces. △

In dyn amical sy s tem th eory, and in particular controlled dynamical

systems, the system input vector u(t) ∈ R

, t ≥ 0, is often constrained to

lie in some compact set U ⊂ R

. In addition, in certain app lications, some

or all of th e components of u(t) may be discontinuous with respect to t. In

this case, there might exist a ﬁnite or inﬁnite number of times t where the

vector ﬁeld F (·, ·), and hence, the derivative of the state vector ˙x in (2.1),

is discontinuous. To formally address such cases it is useful to quantify the

size of a set.

Let X be a normed vector space over R

and let S ⊂ X. In addition,

let {A

: i ∈ I} be a (possibly inﬁnite) set of subsets such that S ⊆

i∈I

If, for each i ∈ I, A

is an open set, then {A

: i ∈ I} is called a cover

(or an open cover) of S. Furthermore, if {A

: i ∈ I} = {A

, . . . , A

} is a

ﬁnite s et, then {A

, . . . , A

} is called a ﬁnite cover of S.

For a collection

S of bounded subsets of R

, the measure µ : S → [0, ∞) assigns to each set

in S a nonnegative number. (See [288] and P roblem 2.71 for the properties

of µ(S).) A subset S ⊂ R

is a set of measure zero if and only if for all

ε > 0, there exists a cover {A

, A

, . . .} of S, where A

is an open (or

closed) n-dimensional symmetric polytope (or cube), such that th e sum of

their volumes vol(A

) is less than ε, that is,

∞

i=1

vol(A

) < ε. Any ﬁnite or

countably inﬁnite set of points in R

has measure zero. A statement about

S ⊆ R

holds almost everywhere (a.e.) in S if and only if the statement

holds for every element in S except for a subset of S with zero measure.

A measure on R

is a natural generalization of th e notion of length of

a line segment for n = 1, the area of a rectangle for n = 2, and th e volume of

a parallelepiped for n = 3. In particular, for a closed interval I = [a, b], the

length ℓ(I) of the set I is given by ℓ(I) = b − a. Analogous ly, the volume

of an n-dimensional interval I = [a

, b

] × ··· × [a

, b

] ⊂ R

is given by

vol(I) = Π

i=1

− a

). For n = 1, ℓ(I) = vol(I).

If S ⊂ X, where X is a normed space, then S is compact if and only if every open cover of

S contains a ﬁnite subcollection of open sets that also covers S. In the case where S ⊂ R

, this

deﬁnition is equivalent to the compactness deﬁnition given in Deﬁnition 2.14.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 61

Deﬁnition 2.40. Let S ⊆ R

. The outer measure of S is deﬁned by

∗

(S) = inf

∈I

vol(I

), (2.96)

where the inﬁmum in (2.96) is taken over all countable collections I =

}

∞

j=1

of closed intervals such that S ⊂

∞

j=1

Note that µ

∗

) = ∞ w hile µ

∗

(Ø) = 0.

Deﬁnition 2.41. Let S ⊆ R

. S is Lebesgu e measurable, or simply

measurable, if, for each ε > 0, there exists an open set Q ⊆ R

such that

S ⊂ Q and

∗

(Q\S) < ε. (2.97)

If S is measurable, then the Lebesgue measure (or measure) of S, denoted

by µ(S), is given by µ(S)

△

= µ

∗

(S).

It follows from Deﬁnition 2.41 that a measurable set can be approx-

imated arbitrarily closely, in terms of its outer measure, by open subsets

of R

. Hence, every open set in R

is measurable. More precisely, the

measurable sets of R

are deﬁned as the members of the smallest family of

sets of R

containing all open, closed, and measure-zero sets of R

, as well as

every diﬀerence, countable un ion, and countable intersection of its members.

Sets that are not measurable are exceptional and seldom encountered in

applications.

As noted in Section 2.4, if f : D → R is continuous on D ⊆ R

, then

{x ∈ D : f(x) > α} is open for every α ∈ R. Using the notions of measurable

sets, one can deﬁne measurable functions as a generalization of continuous

functions.

Deﬁnition 2. 42. Let D ⊆ R

be measurable. The function f : D → R

is measurable on D if, for every α ∈ R, {x ∈ D : f (x) > α} is measurable.

It can be shown that a measurable function is continuous almost

everywhere in its d omain of deﬁnition. On an interval I ⊂ R, a real-valued

function f : I → R

is a measurable function on I if and only if it is the

pointwise limit, except for a set of measure zero, of a sequence of piecewise

constant functions on I. That is, if there exists a sequence of s tep functions

}

∞

n=0

on I such that lim

n→∞

(x) = f(x) almost everywhere on I, then

f is a measurable function on I.

In the theory of Riemann integration, a piecewise continuous function

over the compact interval I = [a, b] ⊂ R, a continuous function over

the interval I, or a bounded function over the interval I with at most

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62 CHAPTER 2

countable many discontinuities in every ﬁnite subinterval of I is integrable.

Speciﬁcally, we can ignore sets of measur e zero points at which the function

is discontinuous since the value of the fu nction can be altered arbitrarily

at these points without altering the value of the integral. That is, if

f : I ⊂ R → R

and

f(t)dt =

f(t)dt (2.98)

exists and is ﬁnite, then

f(t)dt =

I\S

f(t)dt, (2.99)

where S ⊂ I = [a, b] has measure zero. If f : I → R

and g : I → R

are

equal almost everywhere in I, that is, f(t) = g(t) for all t ∈ I except on a

set of measure zero, then

f(t)dt =

g(t)dt. (2.100)

In light of the above discussion, the following th eorem, due to

Lebesgue, presents necessary and suﬃcient conditions for Riemann inte-

grability of a bounded function.

Theorem 2.19. Let I ⊂ R be a compact interval and assume f : I →

is bounded. Then f is Riemann integrable on I if and only if th e set of

points at w hich f is discontinuous has zero measure.

It follows from Theorem 2.19 that a bounded function f on a compact

interval I is Riemann integrable on I if and only if f is continuous almost

everywhere on I.

Example 2.26. The functions f : R → {0, 1} and g : R → {0, 1}

deﬁned by

f(x) =



1, x ∈ Z,

0, otherwise,

and

g(x) =



1, x ∈ Q,

0, otherwise,

where Q

△

= {x : x = p/q, p, q ∈ Z, q 6= 0}, are zero almost everywhere in

R. In addition, f (x) = g(x) almost everywhere in R. Since a set consisting

of a single point has measure zero (see Problem 2.72), it follows that every

countable su bset of R has measure zero. In particular, the set of integers

Z and the set of rational numbers Q have measure zero. However, even

though Q is countable, it is not the set of discontinuities of g, whereas Z

is the set of discontinuities of f. Since Q is dense in R it follows that g is

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 63

discontinuous at every point in R. Hence, it follows from Theorem 2.19 that

the Riemann integral of g does not exist on every compact interval of R,

whereas the Riemann integral of f does exist. △

As shown by Theorem 2.19, the concept of the Riemann integral

applies to piecewise continuous, continuous, and bounded functions with

at most countable many discontinuities over an interval I ⊂ R. However,

the Riemann integral of a general measurable f unction f, wherein f is

discontinuous everywhere or when f is deﬁned on an abstr act measurable

space (that is, an arbitrary set X equipped with a measure), cannot be

formed. For such functions, one has to use the more general notion of

Lebesgue integration [288]. The Lebesgue integral allows the integration of

more general (i.e., m easurable) functions. In particular, for a function f

deﬁned on a closed interval I ⊂ R, the Riemann integral of f is formed by

dividing I into subintervals, thereby grouping together neighboring points

of R. In contrast, the Lebesgue integral of f is formed by grouping together

points of R at which the function f takes neighboring values; that is, the

range of the function f rather than its domain is partitioned. Hence, the

Lebesgue integral is an extension of the Riemann integral in th e sense that

whenever the R iemann integral exists its value is equal to the Lebesgue

integral.

For a given fun ction f : D → R, where D ⊆ R

, the Lebesgue integral

of f is d en oted by

f(x)dx. (2.101)

If f : D → R is measurable, where D ⊆ R

is measurable, and if (2.101)

exists and is ﬁnite, then we say that f is Lebesgue integrable on D. In this

case, the set of integrable functions on D is denoted by

L(D)

△



f :

f(x)dx < ∞



. (2.102)

The following theorem gives necessary and suﬃcient conditions for Lebesgue

integrability.

Theorem 2.20. Let D be measurable and assume f : D → R is

measurable on D. Then f is Lebesgue integrable over D if and only if

|f| is Lebesgue integrable over D.

If (2.101) exists where D ⊆ R

is measurable and f : D → R is

measurable on D, then



f(x)dx



≤

|f(x)|dx. (2.103)

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64 CHAPTER 2

To see this, ﬁ rst n ote that, by Theorem 2.20, |f(x)|, x ∈ D, is Lebesgue

integrable over D. Next, let y =

f(x)dx and note that αy = |y|, where

|α| = 1. Now, since |αf| = |f |,



f(x)dx



= α

f(x)dx =

αf(x)dx ≤

|αf(x)|dx =

|f(x)|dx,

which proves (2.103). Inequality (2.103) is k nown as the absolute value

theorem for integrals. In light of (2.103), the set of Lebesgue integrable

functions on D deﬁned by (2.102) becomes

L(D) =



f : f is measurable and

|f(x)|dx < ∞



. (2.104)

Note that the measur ability of f in (2.104) is included since it is possible

for |f | to be integrable but not measurable.

Finally, if D = I, where I is a measurable s ubset of R, f : I → R is

measurable on I, and f ∈ L(I), th en

f(t)dt =

I\S

f(t)dt, (2.105)

where S ⊂ I has measur e zero. Hence, the value of the integral over I is

not aﬀected by removing the set S of measure zero from I. In addition, if

f(t) = g(t) almost everywhere in I, where g : I → R is measurable on I

and g ∈ L (I), then

f(t)dt =

g(t)dt. (2.106)

Analogous results hold for functions f and g deﬁned on D ⊆ R

Example 2.27. Consider the Dirichlet function g : [0, 1] → {0, 1}

deﬁned in Example 2.26. Since the set of rational numbers Q is countable

and therefore has measure zero, it follows that g(x) = 0 almost everywhere

on [0, 1]. Since the function 0 is Riemann integrable on [0, 1], it is also

Lebesgue integrable on [0, 1]. Hence, by (2.106),

[0,1]

g(x)dx =

0dx = 0,

and therefore g is Lebesgue integrable on [0, 1]. △

Next, we consider some special normed vector spaces. Speciﬁcally, we

consider the normed function space of all functions f : [0, ∞) → R

and

with norm deﬁned by

|||f |||

p,q

△



∞

kf(t)k



1/p

, 1 ≤ p < ∞, (2.107)