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

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 45

for all t ∈ I

except at a ﬁnite number of points at which the left-side limit

f(t

−

)

△

= lim

h→0,h>0

f(t+h) and the right-side limit f(t

)

△

= lim

h→0,h>0

f(t−

h) exist. f is left continuous at t if f (t

−

) = f(t), and right continuous

at t if f(t

) = f(t). The function is left (respectively, right) piecewise

continuous if f is left (respectively, right) continuous at every point in the

interior of I and if I has a left endpoint t

, then f (t

) = f(t

), while, if

I has a right endpoint t

, then f (t

−

) = f (t

). The function f : I → R

piecewise continuously diﬀerentiable on I if f is piecewise continuous and

its ﬁrst derivative exists and is piecewise continuous on each subinterval

= {t : t

k−1

< t < t

}, k = 1, . . . , n, of I, where t

= t

Note that a function f is piecewise continuous on [t

, t

] i) if it is

continuous on [t

, t

] at all but a ﬁnite number of points on (t

, t

); ii) if f

is discontinuous at t

∗

∈ (t

, t

), th en the left and right limits of f (t) exist as

t approaches t

∗

from the left and the right; an d iii) if f is discontinuous at

∗

, then f (t

∗

) is equal to either the left or the right limit of f(t).

Example 2.19. Consider the function f : R → {0, 1} given by

f(x) =



1, x 6= 0,

0, x = 0.

(2.52)

Clearly, f is not continuous at x = 0. The left-side limit f (0

−

) and right-

side limit of f exist and satisfy f (0

−

) = f (0

) = 1. Hence, f is piecewise

continuous on R. However, f is neither left piecewise continuous nor right

piecewise continuous on R. △

Proposition 2.18. Let I ⊂ R be a compact interval and let f : I → R

be piecewise continuous on I. Then f(I) is bounded.

Proof. Since f is piecewise continuous on a compact interval I ⊆ R, by

deﬁnition, f is continuous on a ﬁnite number of intervals [t

, t

), (t

, t

), . . . ,

n−1

, t

], where ∪

k=1

k−1

, t

] = I. Furthermore, for every k ∈ {1, . . . , n},

the left-side and the right-side limits of f (t) exist as t approaches t

from

the left and the right. Hence, η

△

= sup

t∈(t

k−1

)

|f(t)| < ∞, which implies

that

|f(t)| ≤ max



max

k∈{1,...,n}

, max

k∈{0,1,...,n}

f(t

)



proving th at f (I) is bounded.

The next deﬁnition introduces the concept of Lipschitz continuity.

Deﬁnition 2.33. Let D ⊆ R

and let f : D → R

. Then f is Lipschitz

continuous at x

∈ D if there exists a Lipschitz constant L = L(x

) > 0 and

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46 CHAPTER 2

a neighborhood N ⊂ D of x

such that

kf(x) − f(y)k ≤ Lkx − yk, x, y ∈ N. (2.53)

f is Lipschitz continuous on D if f is Lipschitz continuous at every point in

D. f is unif ormly Lipschitz continuous on D if there exists L > 0 such that

kf(x) − f(y)k ≤ Lkx − yk, x, y ∈ D. (2.54)

Finally, f is globally Lipschitz continuous if f is uniformly Lipschitz

continuous on D = R

Note that a function can be Lipschitz continuous on D but not

uniformly Lipschitz continuous on D since the L ipschitz condition might

not hold for the same Lipschitz constant on D. For example, f (x) = x

is Lipschitz continuous on R but not uniformly Lipschitz continuous on R.

However, if we restrict the domain to the closed interval [0, 1], then f(x) = x

is uniformly Lipschitz continuous over [0, 1] with Lipschitz constant L = 2.

To see this, note that |x

−y

| = |x + y| |x −y| ≤ 2|x −y| for all x, y ∈ [0, 1].

For f : R → R the Lips chitz condition gives

|f(x) −f (y)|

|x −y|

≤ L, (2.55)

which implies that any line segment joining two arb itrary points on the graph

of f cannot have a slope greater than ±L. Hence, any f unction having an

inﬁnite slope at a point x

∈ D is not Lipschitz continuous at x

. This of

course rules out all discontinuous functions from being Lipschitz continuous

at the point of discontinuity. Alternatively, if f is Lipschitz continuous at

a point x

∈ D, then f is clearly continuous at x

. The converse, however,

is not true. The same remarks hold for uniform Lipschitz continuity. In

particular, if f is uniformly Lipschitz continuous on D, then f is uniformly

continuous on D. Once again, the converse of this statement is not tru e.

Example 2.20. Consider the function f : R → R given by

f(x) =



sin(1/x), x 6= 0,

0, x = 0.

(2.56)

For x 6= 0 and y = 0, it follows that

|f(x) −f (y)| = |f (x)|

= |x

sin(1/x)|

≤ x

≤ ρ|x|

= ρ|x − y|, |x| ≤ ρ. (2.57)

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 47

Similarly, for x = 0 and y 6= 0, it follows that

|f(x) − f(y)| ≤ ρ|x − y|, |y| ≤ ρ. (2.58)

For x 6= 0 and y 6= 0, it follows that

|f(x) − f(y)| = |x

sin(1/x) − y

sin(1/y)|

= |(x

− y

) sin(1/x) + y

(sin(1/x) −sin(1/y))|

≤ |(x

− y

) sin(1/x)| + |y

(sin(1/x) − sin(1/y))|

≤ (|x + y| + |y/x|)|x − y|.

Hence, f is Lipschitz continuous on R. △

In light of the above discussion we have the following propositions.

Proposition 2.19. Let D ⊆ R

and f : D → R

. If f is Lipschitz

continuous at x

∈ D, then f is continuous at x

Proof. Sin ce f is Lipschitz continuous at x

there exists L = L(x

) > 0

and a neighborhood N ⊂ D of x

such that

kf(x) − f(y)k ≤ Lkx − yk, x, y ∈ N.

Now, for every ε > 0 let δ = δ(ε, x

) = ε/L. Hence, for all kx

− yk < δ,

it follows that kf (x

) − f(y)k < Lδ = ε. The result is now immediate from

the deﬁnition of continuity.

The following result shows that uniform Lips chitz continuity implies

uniform continuity.

Proposition 2.20. Let D ⊆ R

and let f : D → R

. If f is uniformly

Lipschitz continuous on D, then f is uniformly continuous on D.

Proof. Since f is uniformly Lipschitz continuous on D it follows th at

there exists L > 0 s uch that

kf(x) − f(y)k ≤ Lkx − yk, x, y ∈ D. (2.59)

Now, for a given ε > 0 let δ = ε/L. Hence, for all kx−yk < δ, it follows that

kf(x) −f (y)k < Lδ = ε. The result is now immediate from the deﬁn ition of

uniform continuity.

Lipschitz continuity is essentially a smoothness property almost

everywhere on D. Speciﬁcally, if f is continuously d iﬀerentiable at x

∈ D,

then f is Lipschitz continuous at x

. The converse, however, is not true. A

simple counterexample is f (x) = |x|, x ∈ R, which is Lip schitz continuous at

x = 0 but not continuously diﬀerentiable at x = 0. Alternatively, if the (left

or right) derivative exists at some point x

, then it does not automatically

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48 CHAPTER 2

imply that f is Lipschitz continuous at x

since the existence of the derivative

of f at x

does not automatically imp ly diﬀerentiability of f at x

. Of course,

if f

′

(x) is continuously diﬀerentiable and bounded on D, then f is uniformly

Lipschitz continuous on D.

Next, we extend the above observations to vector-valued functions.

However, ﬁrst we need the following deﬁnitions and r esults from mathemat-

ical analysis.

Deﬁnition 2.34. L et D ⊆ R

be an open set. The function f : D →

is diﬀerentiable at x

∈ D if there exists a linear transformation Df(x

) ∈

n×m

satisfying

lim

khk→0

kf(x

+ h) − f(x

) − Df(x

)hk

khk

= 0, (2.60)

where x

+ h ∈ D an d h 6= 0.

The linear transformation Df(x

) is called the derivative of f at x

We usually denote Df (x

) by

∂f

∂x

) or f

′

). A function f : R

→ R

is continuously diﬀerentiable at x

if Df(x

) exists and is continuous at x

A function f is continuously diﬀerentiable on D ⊆ R

if f is continuously

diﬀerentiable at all x ∈ D. A function f : R

→ R

is diﬀerentiable on

if f is diﬀerentiable at every point x ∈ R

. If f is diﬀerentiable on

, then f is continuous on R

. To see this, let x

∈ R

and assume f is

diﬀerentiable at x

. In this case, kf(x

+ h) −f (x

)k/khk tends to a deﬁnite

limit Df(x

) as khk → 0, and hence, kf (x

+ h) − f(x

)k → 0 as khk → 0,

which implies that f is continuous at x

. However, n ot every continuous

function has a derivative at every point in D ⊆ R

Example 2.21. Consider the function f : R → R given by f(x) = |x|.

Clearly, f is continuous at x = 0. However, for h > 0,

lim

h→0

f(h) − f(0)

h −0

= 1, (2.61)

whereas for h < 0,

lim

h→0

−

f(h) −f (0)

−h −0

= −1. (2.62)

Hence, f

′

−

) 6= f

′

), and hence, f is not diﬀerentiable at x = 0. △

Another important distinction is between diﬀerentiability and continu-

ous diﬀerentiability. In particular, if a function f is continuous on D and its

derivative exists at every point on D, then f is not necessarily continuously

diﬀerentiable on D. That is, it is not necessarily true that f

′

is continuous

on D.

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 49

Example 2.22. Consider the function f : R → R given in Example

2.20. The derivative of f is given by

′

(x) =



2x sin(1/x) −cos(1/x), x 6= 0,

0, x = 0,

(2.63)

and hence, f is diﬀerentiable everywhere on R. However, lim

x→0

′

(x) does

not exist since f

′

is unbounded in every neighborhood of the origin. Thus, f

′

is not continuous at x = 0, and hence, although f is diﬀerentiable everywhere

on R, it is not continuously diﬀerentiable at x = 0. △

Deﬁnition 2.35. Let X and Y be open subsets of R

and let f : X →

Y. Then f is a homeomorphism or a topological mapping of X onto Y if

f is a one-to-one continuous map with a continuous inverse. Furthermore,

X and Y are said to be homeomorphic or topologically equivalent. f is

a diﬀeomorphism of X onto Y if f is a homeomorphism and f and f

−1

are continuously diﬀerentiable. In this case, X and Y are said to be

diﬀeomorphic.

Example 2.23. Consider the function f : (−1, 1) → R deﬁned by

f(x) = tan(

πx

). Clearly, f is one-to-one and continuous on (−1, 1). The

inverse function f

−1

: R → (−1, 1) given by x =

tan

−1

f(x) is also

continuous, and hence, f is a homeomorphism and (−1, 1) and R are

homeomorphic. △

If khk in (2.60) is suﬃciently small, then x + h ∈ D, since D is open.

Hence, f(x + h) is deﬁned, f (x + h) ∈ R

, and, sin ce Df(x) : R

→ R

is a

linear operator, Df(x)h ∈ R

. The next proposition shows that the linear

transformation Df(x

) ∈ R

n×m

satisfying (2.60) is unique.

Proposition 2.21. Let D and f : D → R

be as in Deﬁnition 2.34 and

let x ∈ D. Then Df (x) : R

→ R

satisfying (2.60) is unique.

Proof. Suppose, ad absurdum, that (2.60) holds with Df (x) = X and

Df(x) = Y , where X 6= Y . Deﬁne Z

△

= X − Y and note that

kZhk = kf(x + h) − f (x) + Y h − f(x + h) + f(x) − Xhk

≤ kf(x + h) − f (x) −Xhk + k −f (x + h) + f(x) + Y hk

= kf(x + h) − f (x) −Xhk + kf (x + h) − f(x) −Y hk,

which implies that kZhk/khk → 0 as khk → 0. Now, for ﬁxed h 6= 0,

it follows that kεZhk/kεhk → 0 as ε → 0, and hence, Zh = 0 for every

h ∈ R

. Thus, Z = X − Y = 0, which leads to a contradiction.

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50 CHAPTER 2

Note that (2.60) can be rewritten as

f(x + h) − f (x) = f

′

(x)h + r(h), (2.64)

where the remainder r(h) satisﬁes

lim

khk→0

kr(h)k

khk

= 0. (2.65)

Hence, for a ﬁxed x ∈ D ⊆ R

and suﬃciently small h, the left-hand side

of (2.64) is approximately equal to f

′

(x)h. If f : D → R

and D is as in

Deﬁnition 2.34, and if f is diﬀerentiable on D, then for every x ∈ D, f

′

(x)

is a linear transformation on R

into R

. Note that if follows from (2.64)

that f is continuous at any point at which f is diﬀerentiable.

Next, we introd uce the notion of a partial derivative of the function

f = [f

, . . . , f

]

: D → R

. Let {e

, . . . , e

} and {v

, . . . , v

} be normalized

bases of R

and R

, respectively. I n this case,

f(x) =

i=1

(x)v

, x ∈ D, (2.66)

or, equivalently, f

(x) = f

(x)v

, i = 1, . . . , n. Now, for x ∈ D and i =

1, . . . , n, j = 1, . . . , m, deﬁne

(x)

△

= lim

ε→0

(x + εe

) − f

(x)

, (2.67)

whenever the limit on the right-hand side exists. Hence, noting that f

(x) =

, . . . , x

) it follows that D

(x) is the derivative of f

with r espect

to x

, keeping the other variables ﬁxed. We usually d en ote D

(x) by

∂f

∂x

and call

∂f

∂x

a partial derivative. The next theorem shows that if f

is diﬀerentiable at x ∈ D, then its partial derivatives exist at x, and they

completely determine f

′

(x).

Theorem 2.14. Let D ⊆ R

be open and f = [f

, . . . , f

]

: D → R

and suppose f is diﬀerentiable at x ∈ D. T hen D

(x), i = 1, . . . , n,

j = 1, . . . , m, exist, and

′

(x)e

i=1

(x)v

, j = 1, . . . , m. (2.68)

Proof. Let j ∈ {1, . . . , m} and note that since f is diﬀerentiable at x,

f(x + εe

) − f(x) = f

′

(x)(εe

) + r(εe

), (2.69)

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 51

where kr(εe

)k/ε → 0 as ε → 0. Since f

′

(x) is linear it follows that

lim

ε→0

f(x + εe

) − f(x)

= f

′

(x)e

, (2.70)

or, equivalently,

lim

ε→0

i=1

(x + εe

) − f

(x)

= f

′

(x)e

, (2.71)

where {e

, . . . , e

} and {v

, . . . , v

} are normalized bases of R

and R

respectively. Now, since each quotient in (2.71) has a limit as ε → 0, it

follows that D

(x), i = 1, . . . , n, j = 1, . . . , m, exist. Finally, (2.68) follows

from (2.71).

The next theorem gives necessary and suﬃcient conditions for a

function to be continuously diﬀerentiable on D ⊆ R

Theorem 2.15. Let D ⊆ R

be open and f = [f

, . . . , f

]

: D → R

Then f is continuously diﬀerentiable on D if and only if

∂f

∂x

, i = 1, . . . , n,

j = 1, . . . , m, exist and are continuous on D.

Proof. Assume f is continuously diﬀerentiable on D and note that

it follows from (2.68) that D

(x) = (f

′

(x)e

)

for all i = 1, . . . , n, j =

1, . . . , m, and all x ∈ D, where {e

, . . . , e

} and {v

, . . . , v

} are normalized

bases of R

and R

, respectively. Hence, for i = 1, . . . , n, j = 1, . . . , m, and

x, y ∈ D,

(y) − D

(x) = [(f

′

(y) −f

′

(x))e

]

. (2.72)

Now, usin g the Cauchy-Schwarz inequality and the fact that kv

k = ke

k = 1

it follows fr om (2.72) that for i = 1, . . . , n, j = 1, . . . , m, and x, y ∈ D,

(y) −D

(x)| = |[(f

′

(y) −f

′

(x))e

]

≤ k[f

′

(y) −f

′

(x)]e

≤ kf

′

(y) − f

′

(x)k, (2.73)

which shows that D

(x), i = 1, . . . , n, j = 1, . . . , m, is continuous on D.

To prove the converse, it su ﬃces to consider the case where n = 1. Let

x ∈ D and ε > 0, and n ote that since D is open, there exists r > 0 such that

(x) ⊂ D. Now, it follows from the continuity of D

f(x), j = 1, . . . , m,

that r can be chosen so that |D

f(y) − D

f(x)| < ε/m, y ∈ B

(x), and

j = 1, . . . , m.

Next, let h =

j=1

be such that khk < r and let u

= h

+ ··· +

, k ∈ {1, . . . , m}, with u

△

= 0. Now, writing the diﬀerence f(x + h) −

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52 CHAPTER 2

f(x) as a telescoping sum yields

f(x + h) − f (x) =

j=1

[f(x + u

) −f(x + u

j−1

)]. (2.74)

Since ku

k < r for k ∈ {1, . . . , m} and since B

(x) is convex, it follows th at

: z

= µ

(x + u

j−1

) + (1 − µ

)(x + u

), µ

∈ (0, 1)} ⊂ B

(x). Using

the mean value theorem from calculus and noting that u

= u

j−1

+ h

, it

follows that the jth summand of (2.74) is equal to h

f(x+u

j−1

+µ

Hence, for j = 1, . . . , m,

f(x + u

j−1

+ µ

) −h

f(x)| < |h

|ε/m.

Now, it follows from (2.74) that



f(x + h) −f(x) −

j=1

f(x)



≤

j=1

|ε ≤ khk

ε (2.75)

for all h s uch that khk

< r.

Equation (2.75) implies that f is diﬀerentiable at x and f

′

(x) is a linear

map assigning the number

j=1

f(x) to the vector h =

j=1

Now, since the gradient f

′

(x) = [D

f(x), . . . , D

f(x)] and D

f(x), j =

1, . . . , m, are continuous functions on D, it follows that f is continuously

diﬀerentiable on D.

In the case where f : R

→ R is continuously diﬀerentiable,

∂f

∂x

′

(x) = [

∂f

∂x

, . . . ,

∂f

∂x

] ∈ R

1×n

is called the gradient of f at x. Alternatively,

for a continuously d iﬀerentiable function f : R

→ R

∂f

∂x

= f

′

(x) ∈ R

n×m

is called the Jacobian of f at x and is given by

′

(x) =







∂f

∂x

(x)

∂f

∂x

(x) ···

∂f

∂x

(x)

∂f

∂x

(x)

∂f

∂x

(x)

∂f

∂x

(x) ··· ···

∂f

∂x

(x)







. (2.76)

Higher-order derivatives f

(r)

) of a function f : D → R

are deﬁned in a

similar way, and it can be shown that similar theorems to Theorem 2.15 hold

for a fu nction to be r-times continuously diﬀerentiable on D. For details the

interested reader is referred to [373, p. 235].

Next, we present the chain rule for vector-valued functions. In

particular, let D ⊆ R

, let f : D → R

be continuously d iﬀerentiable

at x

∈ D, and let g : f(D) → R

, where f (D) is open and g is

continuously diﬀerentiable at f(x

). Then the function h : D → R

given by

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DYNAMICAL SYSTEMS AND DIFFERENTIAL EQUATIONS 53

h(x) = g(f(x)) is continuously diﬀerentiable at x

, and

∂h

∂x

) is given by

∂h

∂x

) =

∂g

∂f

(f(x

))

∂f

∂x

). (2.77)

Finally, we present three important results from mathematical analysis. Th e

proofs of these theorems can be found in any analysis or advanced calculus

textbook, and hence, are n ot given here.

Theorem 2.16 (Mean Value Theorem). Let D ⊆ R

be open, let

f : D → R

be continuously diﬀerentiable in D, and assume there exist

x, y ∈ D s uch th at L

△

= {z : z = µx + (1 − µ)y, µ ∈ (0, 1)} ⊂ D. Then for

every v ∈ R

there exists z ∈ L such that

[f(y) − f (x)] = v

′

(z)(y − x)]. (2.78)

Theorem 2.17 (Inverse Function Theorem). Let D ⊆ R

be open,

let f : D → R

be continuously diﬀerentiable on D, and assume there exists

∈ D such that det f

′

) 6= 0. Then there exists a neighborhood N of

f(x

) and a unique function g : N → D such that f (g(y)) = y for all y ∈ N.

Furth ermore, the function g is continuously diﬀerentiable on N.

It follows from the inverse function theorem that if f : R

→ R

is a

bijection with det f

′

(x) = 0 for some x ∈ R

, then f

−1

is not diﬀerentiable

at f(x). To see this, assume, ad absurdum, that f

−1

is diﬀerentiable at f(x)

and note that (f

−1

◦ f)(x) = x. Now, it follows fr om the chain rule (2.77)

that

∂f

−1

∂f

(f(x))

∂f

∂x

(x) = Df

−1

(f(x)) ◦Df (x)

= D(f

−1

◦f )(x)

= I

, (2.79)

and hence, det Df

−1

(f(x)) det Df (x) = 1. This contradicts the fact that

det Df(x) = 0.

Theorem 2.18 (Implicit Function Theorem). Let D ⊆ R

and Q ⊆

be open, and let f : D × Q → R

. Suppose f(x

, y

) = 0 for (x

, y

) ∈

D×Q and s uppose the Jacobian m atrix

∂f

∂x

, y

) is non s ingular. Then th ere

exist neighborhoods U ⊂ D of x

and V ⊂ Q of y

such that f (x, y) = 0,

y ∈ V, has a unique solution x ∈ U. Moreover, there exists a continuously

diﬀerentiable function g : V → U at y = y

such that x = g(y).

The implicit function theorem can be used to guarantee the existence

of a feedback controller that provides a characterization of an equilibr ium

manifold. Speciﬁcally, consider the dynamical system (2.1) with F (t, x, u) =

F (x, u), where F : D × U → R

and D ⊆ R

and U ⊆ R

. If there exists

NonlinearBook10pt November 20, 2007

54 CHAPTER 2

∈ U such that F (x

, u

) = 0, then (x

, u

) is an equilibrium point of (2.1).

Since F is an implicit function, it follows that to ﬁnd a feedback control law

to solve the implicit equation F (x, u) = 0 we requ ire the existence of a

function φ : D → R

such th at the explicit equation u = φ(x) is satisﬁed.

If such a function exists, then F (x, φ(x)) = 0. Now, suppose there exists

, u

) ∈ D×U such that F (x

, u

) = 0. If det

∂F

∂u

, u

) 6= 0, then it follows

from the implicit function theorem that there exist open neighborhoods X

and U of x

and u

, respectively, such that, for each x ∈ X, there exists a

unique feedback control law φ : X → R

such th at i) φ(x

) = u

, ii) φ is

continuously diﬀerentiable in X, and iii) F (x, φ(x)) = 0 for all x ∈ X.

Next, using the above r esults we concretize our pr evious observations

on scalar Lipschitz continuous functions to vector-valued functions. Our ﬁrst

result shows that continuous diﬀerentiability implies Lipschitz continuity.

Proposition 2.22. Let D ⊆ R

be open and let f : D → R

. If f is

continuously diﬀerentiable on D, then f is Lipschitz continuous on D.

Proof. Since D is open , for a given x

∈ D th ere exists ε > 0 such that

) ⊂ D. Now, since f

′

(x) is continuous on D it follows from Theorem

2.13 that L = max

kx−x

k≤ε

′

(x)k

′

exists, where k · k

′

is the equi-induced

matrix norm generated by the vector norm k·k. Next, for some x, y ∈ B

set z = y − x. Since B

) is a convex set, it follows that x + µz ∈ B

µ ∈ [0, 1]. Now, deﬁne the function g : [0, 1] → R

by g(µ)

△

= f (x + µz).

Using the chain rule, it follows that g

′

(µ) = f

′

(x + µz)z, and hence,

f(y) − f (x) = g(1) − g(0) =

′

(µ)dµ =

′

(x + µz)zdµ. (2.80)

Thus ,

kf(y) − f (x)k ≤

′

(x + µz)zkdµ

≤

′

(x + µz)k

′

kzkdµ

≤ Lkzk

= Lky − xk, (2.81)

for all x, y ∈ B

The next result sh ows that if D is compact, then L ipschitz continuity

implies uniform Lipschitz continuity.

Proposition 2.23. Let D ⊆ R

, let D

⊂ D be a compact set, and let

f : D → R

. If f is Lipschitz continuous on D, then f is uniformly Lipschitz

continuous on D