Chapter 4

Continuous Maps

ed
ct
4.1 Continuity

te
Definition 4.1. Let (M, d) and (N, ρ) be two metric spaces, A Ď M and f : A Ñ N be a
ro
map. For a given x0 P A1 , we say that b P N is the limit of f at x0 , written
P
lim f (x) = b or f (x) Ñ b as x Ñ x0 ,
xÑx0
ht

␣ (8
if for every sequence txk u8
k=1 Ď Aztx0 u converging to x0 , the sequence f (xk ) k=1 converges
to b.
ig

Proposition 4.2. Let (M, d) and (N, ρ) be two metric spaces, A Ď M and f : A Ñ N be a
r
py

map. Then lim f (x) = b if and only if

xÑx0

@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), b) ă ε whenever 0 ă d(x, x0 ) ă δ and x P A .

Co

Proof. “ñ” Assume the contrary that D ε ą 0 such that for all δ ą 0, there exists xδ P A
with
0 ă d(xδ , x0 ) ă δ and ρ(f (xδ ), b) ě ε .
1
In particular, letting δ = , we can find txk u8
k=1 Ď Aztx0 u such that
k

1
0 ă d(xk , x0 ) ă and ρ(f (xk ), b) ě ε .
k

Then xk Ñ x0 as k Ñ 8 but f (xk ) Ñ̂ b as k Ñ 8, a contradiction.

104
§4.1 Continuity 105

“ð” Let txk u8

k=1 Ď Aztx0 u be such that xk Ñ x0 as k Ñ 8, and ε ą 0 be given. By
assumption,

D δ = δ(x0 , ε) ą 0 Q ρ(f (x), b) ă ε whenever 0 ă d(x, x0 ) ă δ and x P A .

Since xk Ñ x0 as k Ñ 8, D N ą 0 Q d(xk , x0 ) ă δ if k ě N . Therefore,

ρ(f (xk ), b) ă ε @k ě N

which suggests that lim f (xk ) = b. ˝

kÑ8

ed
Remark 4.3. Let (M, d) = (N, ρ) = (R, | ¨ |), A = (a, b), and f : A Ñ N . We write
lim f (x) and lim´ f (x) for the limit lim f (x) and lim f (x), respectively, if the later exist.

ct
xÑa+ xÑb xÑa xÑb
Following this notation, we have

te
lim f (x) = L ô @ ε ą 0, D δ ą 0 Q |f (x) ´ L| ă ε if 0 ă x ´ a ă δ and x P (a, b) ,
xÑa+

xÑb´
P ro
lim f (x) = L ô @ ε ą 0, D δ ą 0 Q |f (x) ´ L| ă ε if 0 ă b ´ x ă δ and x P (a, b) .

Definition 4.4. Let (M, d) and (N, ρ) be two metric spaces, A Ď M , and f : A Ñ N
be a map. For a given x0 P A, f is said to be continuous at x0 if either x0 P AzA1 or
ht

lim f (x) = f (x0 ).

xÑx0

Rn Ñ Rn
ig

Example 4.5. The identity map f : is continuous at each point of Rn .

x ÞÑ x
r

1
py

Example 4.6. The function f : (0, 8) Ñ R defined by f (x) = is continuous at each

x
point of (0, 8).
Co

Proposition 4.7. Let (M, d) and (N, ρ) be two metric spaces, A Ď M , and f : A Ñ N be
a map. Then f is continuous at x0 P A if and only if

@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X A .

Proof. Case 1: If x0 P A1 , then f is continuous at x0 if and only if

@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X Aztx0 u .

Since ρ(f (x0 ), f (x0 )) = 0 ă ε, we find that the statement above is equivalent to that

@ ε ą 0, D δ = δ(x0 , ε) ą 0 Q ρ(f (x), f (x0 )) ă ε whenever x P D(x0 , δ) X A .

106 CHAPTER 4. Continuous Maps

Case 2: Let x0 P AzA1 .

“ñ” then D δ ą 0 such that D(x0 , δ) X A = tx0 u. Therefore, for this particular δ, we
must have

ρ(f (x), f (x0 )) = 0 ă ε whenever x P D(x0 , δ) X A .

“ð” We note that if x0 P AzA1 , f is defined to be continuous at x0 . In other

words,
f is continuous at each isolated point. ˝

d
te
Remark 4.8. We remark here that Proposition 4.7 suggests that f is continuous at x0 P A
if and only if

ec
@ ε ą 0, D δ ą 0 Q f (D(x0 , δ) X A) Ď D(f (x0 ), ε) .

ot
Pr
ht
r ig

Remark 4.9. In general the number δ in Proposition 4.7 also depends on the function f .
py

For a function f : A Ñ R which is continuous at x0 P A, let δ(f, x0 , ε) denote the largest

( )
δ ą 0 such that if x P D(x0 , δ) X A, then ρ f (x), f (x0 ) ă ε. In other words,
Co

␣ ˇ ( ) (
δ(f, x0 , ε) = sup δ ą 0 ˇ ρ f (x), f (x0 ) ă ε if x P D(x0 , δ) X A .

This number provides another way for the understanding of the uniform continuity (in
Section 4.5) and the equi-continuity (in Section 5.5). See Remark 4.51 and Remark 5.51 for
further details.

Definition 4.10. Let (M, d) and (N, ρ) be metric spaces, and A Ď M . A map f : A Ñ N
is said to be continuous on the set B Ď A if f is continuous at each point of B.

Theorem 4.11. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
Then the following assertions are equivalent:
§4.1 Continuity 107

1. f is continuous on A.

2. For each open set V Ď N , f ´1 (V) Ď A is open relative to A; that is, f ´1 (V) = U X A
for some U open in M .

3. For each closed set E Ď N , f ´1 (E) Ď A is closed relative to A; that is, f ´1 (E) = F XA
for some F closed in M .

Proof. It should be clear that 2 ô 3 (left as an exercise); thus we show that 1 ô 2. Before
proceeding, we recall that B Ď f ´1 (f (B)) for all B Ď A and f (f ´1 (B)) Ď B for all B Ď N .

ed
“1 ñ 2” Let a P f ´1 (V). Then f (a) P V. Since V is open in (N, ρ), D εf (a) ą 0 such that
D(f (a), εf (a) ) Ď V . By continuity of f (and Remark 4.8), there exists δa ą 0 such

ct
that
( )

te
f (D(a, δa ) X A) Ď D f (a), εf (a) .

)
ro
Therefore, by Proposition 0.16, for each a P f ´1 (V), D δa ą 0 such that
( ( ( ))
D(a, δa ) X A Ď f ´1 f (D(a, δa ) X A) Ď f ´1 D f (a), εf (a) Ď f ´1 (V) . (4.1.1)
P
Let U = D(a, δa ). Then U is open (since it is the union of arbitrarily many
Ť
ht

aPf ´1 (V)
open balls), and
ig

(a) U Ě f ´1 (V) since U contains every center of balls whose union forms U ;
r

(b) U X A Ď f ´1 (V) by (4.1.1).

py

Therefore, U X A = f ´1 (V).
Co

“2 ñ 1” Let a P A and ε ą 0 be given. Define V = D(f (a), ε). By assumption there exists
U open in (M, d) such that f ´1 (V) = U X A . Since a P f ´1 (V), a P U ; thus by the
openness of U , D δ ą 0 such that D(a, δ) Ď U . Therefore, by Proposition 0.16 we have

f (D(a, δ) X A) Ď f (U X A) = f (f ´1 (V)) Ď V = D(f (a), ε)

which suggests that f is continuous at a for all a P A; thus f is continuous on A. ˝

Example 4.12. Let f : Rn Ñ Rm be continuous. Then x P Rn ˇ }f (x)}2 ă 1 is open since

␣ ˇ (

( )
x P Rn ˇ }f (x)}2 ă 1 = f ´1 D(0, 1) .
␣ ˇ (
108 CHAPTER 4. Continuous Maps

Remark 4.13. For a function f of two variable or more, it is important to distinguish the
continuity of f and the continuity in each variable (by holding all other variables fixed). For
example, let f : R2 Ñ R be defined by
1 if either x = 0 or y = 0,
"
f (x, y) =
0 if x ‰ 0 and y ‰ 0.
Observe that f (0, 0) = 1, but f is not continuous at (0, 0). In fact, for any δ ą 0, f (x, y) = 0
for infinitely many values of (x, y) P D((0, 0), δ); that is, |f (x, y) ´ f (0, 0)| = 1 for such
values. However if we consider the function g(x) = f (x, 0) = 1 or the function h(y) =

d
f (0, y) = 1, then g, h are continuous. Note also that lim f (x, y) does not exists but

te
(x,y)Ñ(0,0)
lim (lim f (x, y)) = lim 0 = 0.
xÑ0 yÑ0 xÑ0

ec
4.2 Operations on Continuous Maps

ot
Definition 4.14. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
Pr
and f, g : A Ñ V be maps, h : A Ñ R be a function. The maps f + g, f ´ g and hf ,
mapping from A to V, are defined by
ht

(f + g)(x) = f (x) + g(x) @x P A,

(f ´ g)(x) = f (x) ´ g(x) @x P A,
ig

(hf )(x) = h(x)f (x) @x P A.

r

f
The map : Aztx P A | h(x) = 0u Ñ V is defined by
py

h
(f ) f (x)
(x) = @ x P Aztx P A | h(x) = 0u .
Co

h h(x)
Proposition 4.15. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be maps, h : A Ñ R be a function. Suppose that x0 P A1 , and lim f (x) = a,
xÑx0
lim g(x) = b, lim h(x) = c. Then
xÑx0 xÑx0

lim (f + g)(x) = a + b ,
xÑx0

lim (f ´ g)(x) = a ´ b ,
xÑx0

lim (hf )(x) = ca ,

xÑx0
(f ) a
lim = if c ‰ 0 .
xÑx0 h c
§4.2 Operations on Continuous Maps 109

Corollary 4.16. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be maps, h : A Ñ R be a function. Suppose that f, g, h are continuous at
f
x0 P A. Then the maps f + g, f ´ g and hf are continuous at x0 , and is continuous at
h
x0 if h(x0 ) ‰ 0.
Corollary 4.17. Let (M, d) be a metric space, (V, } ¨ }) be a (real) normed space, A Ď M ,
and f, g : A Ñ V be continuous maps, h : A Ñ R be a continuous function. Then the maps
f
f + g, f ´ g and hf are continuous on A, and is continuous on Aztx P A | h(x) = 0u.
h
Definition 4.18. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and

d
f : A Ñ N , g : B Ñ P be maps such that f (A) Ď B. The composite function g ˝ f : A Ñ P

te
is the map defined by
( )

ec
(g ˝ f )(x) = g f (x) @x P A.

ot
Pr
ht
ig

Figure 4.1: The composition of functions

r
py

Theorem 4.19. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and
f : A Ñ N , g : B Ñ P be maps such that f (A) Ď B. Suppose that f is continuous at x0 ,
Co

and g is continuous at f (x0 ). Then the composite function g ˝ f : A Ñ P is continuous at

x0 .
Proof. Let ε ą 0 be given. Since g is continuous at f (x0 ), D r ą 0 such that
( )
g(D(f (x0 ), r) X B) Ď D (g ˝ f )(x0 ), ε .

Since f is continuous at x0 , D δ ą 0 such that

( )
f (D(x0 , δ) X A) Ď D f (x0 ), r .
( )
Since f (A) Ď B, f (D(x0 , δ) X A) Ď D f (x0 ), r X B; thus
( )
(g ˝ f )(D(x0 , δ) X A) Ď g(D(f (x0 ), r) X B) Ď D (g ˝ f )(x0 ), ε . ˝
110 CHAPTER 4. Continuous Maps

Corollary 4.20. Let (M, d), (N, ρ) and (P, r) be metric space, A Ď M , B Ď N , and
f : A Ñ N , g : B Ñ P be continuous maps such that f (A) Ď B. Then the composite
function g ˝ f : A Ñ P is continuous on A.

Alternative Proof of Corollary 4.20. Let W be an open set in (P, r). By Theorem 4.11,
there exists V open in (N, ρ) such that g ´1 (W) = V X B. Since V is open in (N, ρ), by
Theorem 4.11 again there exists U open in (M, d) such that f ´1 (V) = U X A. Then
( )
(g ˝ f )´1 (W) = f ´1 g ´1 (W) = f ´1 (V X B) = f ´1 (V) X f ´1 (B) = U X A X f ´1 (B) ,

ed
while the fact that f (A) Ď B further suggests that

ct
(g ˝ f )´1 (W) = U X A .

te
Therefore, by Theorem 4.11 we find that (g ˝ f ) is continuous on A. ˝

4.3 Images of Compact Sets under Continuous Maps

P ro
Theorem 4.21. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a
continuous map.
ht

1. If K Ď A is compact, then f (K) is compact in (N, ρ).

ig

2. Moreover, if (N, ρ) = (R, | ¨ |), then there exist x0 , x1 P K such that

r
py

␣ ˇ ( ␣ ˇ (
f (x0 ) = inf f (K) = inf f (x) ˇ x P K and f (x1 ) = sup f (K) = sup f (x) ˇ x P K .
Co

Proof. 1. Let tVα uαPI be an open cover of f (K). Since Vα is open, by Theorem 4.11 there
exists Uα open in (M, d) such that f ´1 (Vα ) = Uα X A. Since f (K) Ď Vα ,
Ť
αPI
ď ď
K Ď f ´1 (f (K)) Ď f ´1 (Vα ) = A X Uα
αPI αPI

which implies that tUα uαPI is an open cover of K. Therefore,

ď ď
D J Ď I, #J ă 8 Q K Ď A X Uα = f ´1 (Vα ) ;
αPJ αPJ

thus f (K) Ď f (f ´1 (Vα )) Ď Vα .

Ť Ť
αPJ αPJ
§4.3 Images of Compact Sets under Continuous Maps 111

2. By 1, f (K) is compact; thus sequentially compact. Corollary 3.5 then implies that
inf f (K) P f (K) and sup f (K) P f (K). ˝

Alternative Proof of Part 1. Let tyn u8

n=1 be a sequence in f (K). Then there exists txn un=1 Ď
8

K such that yn = f (xn ). Since K is sequentially compact, there exists a convergent subse-
quence txnk u8
k=1 with limit x P K. Let y = f (x) P f (K). By the continuity of f ,

( ) ( )
lim ρ ynk , y = lim ρ f (xnk ), f (x) = 0
kÑ8 kÑ8
␣ (8
which implies that the sequence ynk k=1 converges to y P f (K). Therefore, f (K) is sequen-

ed
tially compact. ˝

ct
Corollary 4.22 (The Extreme Value Theorem（極值定理）). Let f : [a, b] Ñ R be contin-
uous. Then f attains its maximum and minimum in [a, b]; that is, there are x0 P [a, b] and

te
x1 P [a, b] such that
␣ ˇ
f (x0 ) = inf f (x) ˇ x P [a, b]
( ro
␣ ˇ (
and f (x1 ) = sup f (x) ˇ x P [a, b] .
P
(4.3.1)

Proof. The Heine-Borel Theorem suggests that [a, b] is a compact set in R; thus Theorem
4.21 implies that f ([a, b]) must be compact in R. By the Heine-Borel Theorem again f ([a, b])
ht

is closed and bounded, so

ig

inf f ([a, b]) P f ([a, b]) and sup f ([a, b]) P f ([a, b])
r

which further imply (4.3.1). ˝

py

␣ ˇ
Remark 4.23. If f attains its maximum (or minimum) on a set B, we use max f (x) ˇ x P
(( ((
Co

␣ ˇ () ␣ ˇ ␣ ˇ ()
B or min f (x) ˇ x P B to denote sup f (x) ˇ x P B or inf f (x) ˇ x P B . Therefore,
(4.3.1) can be rewritten as
␣ ˇ ( ␣ ˇ (
f (x0 ) = min f (x) ˇ x P [a, b] and f (x1 ) = max f (x) ˇ x P [a, b] .

Example 4.24. Two norms } ¨ } and ~ ¨ ~ on a real vector space V are called equivalent if
there are positive constants C1 and C2 such that

C1 }x} ď ~x~ ď C2 }x} @x P V .

We note that equivalent norms on a vector space V induce the same topology; that is, if } ¨ }
and ~ ¨ ~ are equivalent norms on V, then U is open in the normed space (V, } ¨ }) if and
112 CHAPTER 4. Continuous Maps

only if U is open in the normed space (V, ~ ¨ ~). In fact, let U be an open set in (V, } ¨ }).
Then for any x P U , there exists r ą 0 such that

D}¨} (x, r) ” y P V ˇ }x ´ y} ă r Ď U .
␣ ˇ (

Let δ = C1 r. Then if z P D~¨~ (x, δ) ” y P V ˇ ~x ´ y~ ă δ ,

␣ ˇ (

1 1
}x ´ z} ď ~x ´ z~ ă ¨ C1 r = r
C1 C1

which implies that D~¨~ (x, δ) Ď D}¨} (x, r) Ď U . Therefore, U is open in (V, ~ ¨ ~). Similarly,

ed
if U is open in (V, ~ ¨ ~), then the inequality ~x~ ď C2 }x} suggests that U is open in (V, }¨}).
Claim: Any two norms on Rn are equivalent.

ct
Proof of claim: It suffices to show that any norm } ¨ } on Rn is equivalent to the two-norm

te
} ¨ }2 (check). Let tek unk=1 be the standard basis of Rn ; that is,

ek = ( 0,
P ro
¨ ¨ ¨ , 0 , 1, 0, ¨ ¨ ¨ , 0) .
looomooon
(k ´ 1) zeros

n
c n
Every x P Rn can be written as x = xi ei , and }x}2 = |xi |2 . By the definition of
ř ř
ht

i=1 i=1
norms and the Cauchy-Schwarz inequality,
ig

d
n
ÿ n
ÿ
}ei }2 ; (4.3.2)
r

}x} ď |xi |}ei } ď }x}2

py

i=1 i=1

c n
thus letting C2 = }ei }2 we have }x} ď C2 }x}2 .
ř
Co

i=1
On the other hand, define f : Rn Ñ R by
›ÿn ›
f (x) = }x} = › xi ei › .
› ›
i=1

Because of (4.3.2), f is continuous on Rn . In fact, for x, y P Rn ,

ˇ ˇ ˇ ˇ
ˇf (x) ´ f (y)ˇ = ˇ}x} ´ }y}ˇ ď }x ´ y} ď C2 }x ´ y}2

which guarantees the continuity of f on Rn . Let Sn´1 = x P Rn ˇ }x}2 = 1 . Then Sn´1 is a

␣ ˇ (

compact set in (Rn , } ¨ }2 ) (since it is closed and bounded); thus by Theorem 4.21 f attains
§4.3 Images of Compact Sets under Continuous Maps 113

its minimum on Sn´1 at some point a = (a1 , ¨ ¨ ¨ , an ). Moreover, f (a) ą 0 (since if f (a) = 0,
x
a = 0 R Sn´1 ). Then for all x P Rn , P Sn´1 ; thus
}x}2
( x )
f ě f (a) .
}x}2

The inequality above further implies that f (a)}x}2 ď f (x) = }x}; thus letting C1 = f (a) we
have C1 }x}2 ď }x}.

Remark 4.25.

d
1. Let f : R Ñ R be defined by f (x) = 0. Then f is continuous. Note that t0u Ď R is

te
compact (7 closed and bounded), but f ´1 (t0u) = R is not compact.

ec
2. Let f : R Ñ R be defined by f (x) = x2 . Then f is continuous. Note that C = t1u is
connected, but f ´1 (C) = t1, ´1u is not connected.

Remark 4.26.
ot
Pr
1. If K is not compact, then Theorem 4.21 is not true. Consider the following counter
1
example: K = (0, 1), f : K Ñ R defined by f (x) = . Then f (K) is unbounded.
x
ht

2. If f is not continuous, then Theorem 4.21 is not true either.

ig

(a) Counter example 1: f : K = [0, 1] Ñ R defined by

r

$
1
& if x ‰ 0,
py

f (x, y) = x
% 0 if x = 0.
Co

Then f (K) is unbounded ñ E x1 P K Q f (x1 ) = sup f (K).

(b) Counter example 2: f : [0, 1] Ñ R by
"
x if x ‰ 1,
f (x, y) =
0 if x = 1.

Then there is no x1 P [0, 1] such that f (x1 ) = sup f (x) = 1.

xP[0,1]

Example 4.27 (An example show that x0 , x1 in Theorem 4.21 are not unique). Let f :
[´2, 2] Ñ R be defined by f (x) = (x2 ´ 1)2 .

1. Critical point: f 1 (x) = 2(x2 ´ 1) ¨ 2x = 0 ô x = 0, ˘1.

114 CHAPTER 4. Continuous Maps

2. Comparison: f (0) = 1, f (1) = f (´1) = 0, f (2) = f (´2) = 9. Then

f (2) = f (´2) = sup f (x) and f (1) = f (´1) = inf f (x) .

xP[´2,2] xP[´2,2]

Corollary 4.28. Let (M, d) be a metric space, K Ď M be a compact set, and f : K Ñ R

be continuous. Then the set
␣ ˇ (
x P K ˇ f (x) is the maximum of f on K

ed
is a non-empty compact set.

Proof. Let M = sup f (K). Then the set defined above is f ´1 (tMu), and

ct
1. f ´1 (tMu) is non-empty by Theorem 4.21;

te
2. f ´1 (tMu) is closed since tMu is a closed set in (R, | ¨ |) and f is continuous on K.
ro
Lemma 3.11 suggests that f ´1 (tMu) is compact. ˝
P
4.4 Images of Connected and Path Connected Sets un-
ht

der Continuous Maps

ig

Definition 4.29. Let (M, d) be a metric space. A subset A Ď M is said to be path

r

connected if for every x, y P A, there exists a continuous map φ : [0, 1] Ñ A such that
py

φ(0) = x and φ(1) = y.

Co

y
A

Figure 4.2: Path connected sets

Example 4.30. A set A in a vector space V is called convex if for all x, y P A, the line
segment joining x and y, denoted by xy, lies in A. Then a convex set in a normed space is
path connected. In fact, for x, y P A, define φ(t) = ty + (1 ´ t)x. Then
§4.4 Images of Connected and Path Connected Sets under Continuous Maps 115

1. φ : [0, 1] Ñ xy Ď A, φ(0) = x, φ(1) = y;

2. φ : [0, 1] Ñ A is continuous.

xy = φ([0, 1])
1´t •
t y
• •
x
A

Figure 4.3: Convex sets

ed
Example 4.31. A set S in a vector space V is called star-shaped if there exists p P S

ct
such that for any q P S, the line segment joining p and q lies in S. A star-shaped set in a
normed space is path connected. In fact, for x, y P S, define

te
$ [ 1]
& 2tp + (1 ´ 2t)x ro if t P 0, ,
φ(t) = [ 2]
% (2t ´ 1)y + (2 ´ 2t)p if t P 1 , 1 .
2
P
Then
ht

1. φ : [0, 1] Ñ xp Y py Ď S, φ(0) = x, φ(1) = y;

ig

2. φ : [0, 1] Ñ A is continuous.
r

Theorem 4.32. Let (M, d) be a metric space, and A Ď M . If A is path connected, then A
py

is connected.

Proof. Assume the contrary that there are two open sets V1 and V2 such that
Co

1. A X V1 X V2 = H; 2. A X V1 ‰ H; 3. A X V2 ‰ H; 4. A Ď V1 Y V2 .

Since A is path connected, for x P A X V1 and y P A X V2 , there exists φ : [0, 1] Ñ A such

that φ(0) = x and φ(1) = y. By Theorem 4.11, there exist U1 and U2 open in (R, | ¨ |) such
that φ´1 (V1 ) = U1 X [0, 1] and φ´1 (V2 ) = U2 X [0, 1]. Therefore,

[0, 1] = φ´1 (A) Ď φ´1 (V1 ) Y φ´1 (V2 ) Ď U1 Y U2 .

Since 0 P U1 , 1 P U2 , and [0, 1] X U1 X U2 = φ´1 (A X V1 X V2 ) = H, we conclude that [0, 1]

is disconnected, a contradiction. ˝
116 CHAPTER 4. Continuous Maps

␣( 1 ) ˇˇ (
Example 4.33. Let A = x, sin x P (0, 1] Y (t0u ˆ [´1, 1]) . Then A is connected in
x
(R2 , } ¨ }2 ), but A is not path connected.
To see this, we assume the contrary that A is path connected !( such that there is )a
1 ) ˇˇ
continuous function φ : [0, 1] Ñ A such that φ(0) = (x0 , y0 ) P x, sin x P (0, 1)
ˇ x
␣ (
and φ(1) = (0, 0) P t0u ˆ [´1, 1]. Let t0 = inf t P [0, 1] ˇ φ(t) P t0u ˆ [´1, 1] . In other
words, at t = t0 the path touches 0 ˆ [´1, 1] for the “first time”. By the continuity of φ,
!(
1)ˇ
ˇ )
φ(t0 ) P t0u ˆ [´1, 1]. Since φ(0) R t0u ˆ [´1, 1], φ([0, t0 )) Ď x, sin ˇ x P (0, 1) .
x
( 1 )
Suppose that φ(t0 ) = (0, ȳ) for some ȳ P [´1, 1], and φ(t) = x(t), sin for 0 ď t ă t0 .

ed
x(t)
By the continuity of φ, there exists δ ą 0 such that if |t ´ t0 | ă δ, |φ(t) ´ φ(t0 )| ă 1. In

ct
particular,
( 1 )2
x(t)2 + sin ´ ȳ ă 1 @ t P (t0 ´ δ, t) .

te
x(t)
On the other hand, since φ is continuous, x(t) is continuous on [0, t0 ); thus by the fact that
ro
[0, t0 ) is connected, x([0, t0 )) is connected. Therefore, x([0, t0 )) = (0, x̄] for ˇsome x̄ ą 0. ˇSince
1
lim x(t) = 0, there exists ttn u8 n=1 P [0, t0 ) such that tn Ñ t0 as n Ñ 8 and sin ´ ȳ ˇ ě 1.
P
ˇ
tÑt0 x(tn )
For n " 1, tn P (t0 ´ δ, t0 ) but
ht

( 1 )2
x(tn )2 + sin ´ ȳ ě 1 ,
x(tn )
ig

a contradiction. !(
1)ˇ
)(
r

ˇ
On the other hand, A is the closure of the connected set B =x, sin ˇ x P (0, 1) the
py

x
( 1)
connectedness of B follows from the fact that the function ψ(x) = x, sin is continuous
) x
Co

on the connected set (0, 1) . Therefore, by Problem 9 of Exercise 8, A = B s is connected.

Theorem 4.34. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a

continuous map.

1. If C Ď A is connected, then f (C) is connected in (N, ρ).

2. If C Ď A is path connected, then f (C) is path connected in (N, ρ).

Proof. 1. Suppose that there are two open sets V1 and V2 in (N, ρ) such that

(a) f (C) X V1 X V2 = H; (b) f (C) X V1 ‰ H; (c) f (C) X V2 ‰ H; (d) f (C) Ď V1 Y V2 .

§4.4 Images of Connected and Path Connected Sets under Continuous Maps 117

By Theorem 4.11, there are U1 and U2 open in (M, d) such that f ´1 (V1 ) = U1 X A and
f ´1 (V2 ) = U2 X A. By (d),

C Ď f ´1 (f (C)) Ď f ´1 (V1 ) Y f ´1 (V2 ) = (U1 Y U2 ) X A Ď U1 Y U2 .

Moreover, by (a) we find that

C X U1 X U2 = C X (U1 X A) X (U2 X A) = C X f ´1 (V1 ) X f ´1 (V2 )

Ď f ´1 (f (C) X V1 X V2 ) = H

ed
which implies C X U1 X U2 = H. Finally, (b) implies that for some x P C, f (x) P V1 .
Therefore, x P f ´1 (V1 ) = U1 X A which suggests that x P U1 ; thus C X U1 ‰ H.

ct
Similarly, C X U2 ‰ H. Therefore, C is disconnected which is a contradiction.

te
2. Let y1 , y2 P f (C). Then D x1 , x2 P C such that f (x1 ) = y1 and f (x2 ) = y2 . Since C is
ro
path connected, D r : [0, 1] Ñ C such that r is continuous on [0, 1] and r(0) = x1 and
r(1) = x2 . Let φ : [0, 1] Ñ f (C) be defined by φ = f ˝ r. By Corollary 4.20 φ is
P
continuous on [0, 1], and φ(0) = y1 and φ(1) = y2 . ˝
ht

Corollary 4.35 (The Intermediate Value Theorem（中間值定理）). Let f : [a, b] Ñ R be

continuous. If f (a) ‰ f (b), then for all d in between f (a) and f (b), there exists c P (a, b)
ig

such that f (c) = d.

r

Proof. The closed interval [a, b] is connected by Theorem 3.38, so Theorem 4.34 implies that
py

f ([a, b]) must be connected in R. By Theorem 3.38 again, if d is in between f (a) and f (b),
then d belongs to f ([a, b]). Therefore, for some c P (a, b) we have f (c) = d. ˝
Co

Example 4.36. Let f : [0, 1] Ñ [0, 1] be continuous. Then D x0 P [0, 1] Q f (x0 ) = x0 .

Proof. Let g(x) = x ´ f (x). Then

1. g(0) = 0 or g(1) = 0 ñ x0 = 0 or 1.

2. g(0) ‰ 0 or g(1) ‰ 0 ñ g(0) ă 0 and g(1) ą 0. Since g : [0, 1] Ñ R is continuous,

D x0 P [0, 1] Q g(x0 ) = 0 ñ D x0 P (0, 1) Q f (x0 ) = x0 . ˝

Remark 4.37. Such an x0 in Example 4.36 is called a fixed-point of f .

118 CHAPTER 4. Continuous Maps

Example 4.38. Let f : [1, 2] Ñ [0, 3] be continuous, and f (1) = 0 and f (2) = 3. Then
D x0 P [1, 2] Q f (x0 ) = x0 .

Proof. Let g(x) = x ´ f (x). Then g : [1, 2] Ñ R is continuous. Moreover,

g(1) = 1 ´ f (1) = 1, g(2) = 2 ´ f (2) = ´1 ;

thus D x0 P (1, 2) Q g(x0 ) = 0. ˝

Example 4.39. Let p be a cubic polynomial; that is, p(x) = a3 x3 + a2 x2 + a1 x + a0 for some

ed
a0 , a1 , a2 P R and a3 ‰ 0. Then p has a real root x0 (that is, D x0 P R such that p(x0 ) = 0).

ct
Proof. Note that p is obviously continuous and R is connected. Write

te
( a a a )
p(x) = a3 x3 1 + 2 + 1 2 + 0 3 .
a3 x ro a3 x a3 x
α
Now lim = 0 if n ą 0 and β ‰ 0, so
xÑ˘8 βxn
P
( a2 a a )
lim 1+ + 12 + 03 = 1 .
ht

xÑ˘8 a3 x a3 x a3 x

Moreover,
ig

"
3 8 if a ą 0,
lim ax =
´8 if a ă 0.
r

xÑ8
py

Suppose that a ą 0. Then lim ax3 = 8 and lim ax3 = ´8 ñ D x, y P R Q p(x) ă 0 ă

xÑ8 xÑ´8
p(y). By Corollary 4.35 D r P R Q p(r) = 0. The case that a ă 0 is similar. ˝
Co

4.5 Uniform Continuity（均勻連續）

Definition 4.40. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be
a map. For a set B Ď A, f is said to be uniformly continuous on B if for any
two sequences txn u8 n=1 , tyn un=1 Ď B with the property that lim d(xn , yn ) = 0, one has
8
( ) nÑ8
lim ρ f (xn ), f (yn ) = 0.
nÑ8

Proposition 4.41. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a

map. If f is uniformly continuous on A, then f is continuous on A.
§4.5 Uniform Continuity 119

Proof. Let x0 P A X A1 , and txk u8

k=1 Ď A be a sequence such that xk Ñ x0 as k Ñ 8.

k=1 be a constant sequence with value x0 ; that is, yk = x0 for all k P N. Then
Let tyk u8
k=1 Ď A and d(xk , yk ) Ñ 0 as k Ñ 8. By the uniform continuity of f on A,
tyk u8
( ) ( )
lim ρ f (xk ), f (x0 ) = lim ρ f (xk ), f (yk ) = 0
kÑ8 kÑ8

which implies that f is continuous on x0 . ˝

Example 4.42. Let f : [0, 1] Ñ R be the Dirichlet function; that is,

0 if x P Q ,
"

ed
f (x) =
1 if x P QA .

and B = Q X [0, 1]. Then f is continuous nowhere in [0, 1], but f is uniformly continuous

ct
on B. However, the proposition above guarantees that if f is uniformly continuous on A,

te
then f must be continuous on A (Check why the proof of Proposition 4.41 does not go
through if B is a proper subset of A). ro
Example 4.43. The function f (x) = |x| is uniformly continuous on R.
P
Proof. By the triangle inequality,
ht

ˇ ˇ ˇ ˇ
ˇf (x) ´ f (y)ˇ = ˇ|x| ´ |y|ˇ ď |x ´ y| ;
ig

n=1 and tyn un=1 are sequences in R and lim |xn ´ yn | = 0, by the Sandwich
thus if txn u8 8
ˇ ˇ nÑ8
lemma we must have lim ˇf (xn ) ´ f (yn )ˇ = 0. ˝
r

nÑ8
py

1
Example 4.44. The function f : (0, 8) Ñ R defined by f (x) = is uniformly continuous
x
on [a, 8) for all a ą 0. However, it is not uniformly continuous on (0, 8).
Co

n=1 and tyn un=1 be sequences in [a, 8) such that lim |xn ´ yn | = 0. Then
Proof. Let txn u8 8
nÑ8

ˇ1 1ˇ |xn ´ yn | |xn ´ yn |
|f (xn ) ´ f (yn )| = ˇ ´ ˇ = ď Ñ0 as nÑ8
xn yn |xn yn | a2
which implies that f is uniformly continuous on [a, 8) if a ą 0. However, by choosing
1 1
xn = and yn = , we find that
n 2n
1
|xn ´ yn | = but |f (xn ) ´ f (yn )| = n ě 1 ;
2n
thus f cannot be uniformly continuous on (0, 8). ˝
120 CHAPTER 4. Continuous Maps

Remark 4.45. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : B Ď A Ñ N be a

map. Then the following four statements are equivalent:

(1) f is not uniformly continuous on B.

( )
(2) D txn u8
n=1 , tyn un=1 Ď B Q lim d(xn , yn ) = 0 and lim sup ρ f (xn ), f (yn ) ą 0.
8
nÑ8 nÑ8
( )
(3) D txn u8
n=1 , tyn un=1 Ď B Q lim d(xn , yn ) = 0 and lim ρ f (xn ), f (yn ) ą 0.
8
nÑ8 nÑ8

1 ( )
(4) D ε ą 0 Q @ n ą 0, D xn , yn P B and d(xn , yn ) ă Q ρ f (xn ), f (yn ) ě ε.

ed
n
Example 4.46. Let f : R Ñ R defined by f (x) = x2 . Then f is continuous in R but not

ct
1
uniformly continuous on R. Let ε = 1, xn = n, and yn = n + ,
2n

te
ˇf (xn ) ´ f (yn )ˇ = ˇn2 ´ (n + 1 )2 ˇ = ˇn2 ´ n2 ´ 1 ´ 1 ˇ ą 1
ˇ ˇ ˇ ˇ ˇ ˇ
@n ą 0.
2n ro 4n2
Example 4.47. The function f (x) = sin(x2 ) is not uniform continuous on R.
P
? ?
? π ? π
Proof. Let ε = 1, xn = 2n π + and yn = 2n π ´ . Then
8n 8n
ht

( ) ( 2 π )ˇˇ
ˇ sin(x2n ) ´ sin(yn2 )ˇ = ˇˇ sin 4n2 π + π + π π π
ˇ ˇ ˇ
2
´ sin 4n π ´ + 2
ˇ = 2 cos ;
2 64n 2 64n 64n2
ig

ˇ ˇ
thus if n is large enough, ˇ sin(x2n ) ´ sin(yn2 )ˇ ě 1. ˝
r

1
py

Example 4.48. The function f : (0, 1) Ñ R defined by f (x) = sin is not uniformly
x
continuous.
Co

( π )´1 ( π )´1
Proof. Let ε = 1, xn = 2nπ + and yn = 2nπ ´ . Then
2 2

ˇ sin 1 ´ sin 1 ˇ = 2 ,
ˇ ˇ
xn yn

π 1 1
while |xn ´ yn | = π2
= 1 ď for all n P N. ˝
4n2 π 2 ´ 4
(4n2 ´ 4 )π n

Theorem 4.49. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
For a set B Ď A, f is uniformly continuous on B if and only if
( )
@ ε ą 0, D δ ą 0 Q ρ f (x), f (y) ă ε whenever d(x, y) ă δ and x, y P B .
§4.5 Uniform Continuity 121

Proof. “ð” Suppose the contrary that f is not uniformly continuous on B. Then there are
two sequences txn u8
n=1 , tyn un=1 in B such that
8

( )
lim d(xn , yn ) = 0 but lim sup ρ f (xn ), f (yn ) ą 0 .
kÑ8 nÑ8

1 ( )
Let ε = lim sup ρ f (xn ), f (yn ) . Then by the definition of the limit and the limit
2 nÑ8
superior (or Proposition 1.121) we conclude that there exist subsequences txnk u8
k=1
and tynk u8
k=1 such that
( ) ( )
ρ f (xnk ), f (ynk ) ě lim sup ρ f (xn ), f (yn ) ´ ε = ε ą 0

ed
nÑ8

while lim d(xnk , ynk ) = 0, a contradiction.

ct
kÑ8

1
“ñ” Suppose the contrary that there exists ε ą 0 such that for all δ = ą 0, there exist

te
n
two points xn and yn P B such that

d(xn , yn ) ă
1 (
ro )
but ρ f (xn ), f (yn ) ě ε .
n
P
These points form two sequences txn u8 n=1 , tyn un=1 in B such that lim d(xn , yn ) = 0,
8
( ) nÑ8
while the limit of ρ f (xn ), f (yn ) , if exists, does not converges to zero as n Ñ 8. As
ht

a consequence, f is not uniformly continuous on B, a contradiction. ˝

ig

Remark 4.50. The theorem above provides another way (the blue color part) of defining
the uniform continuity of a function over a subset of its domain. Moreover, according to
r
py

this alternative definition, if f : A Ñ N is uniformly continuous on B Ď A, then

( ( δ) ) ( ε)
@ ε ą 0, D δ ą 0 Q @ b P M, f D b, X B Ď D c, for some c P N ;
Co

2 2
that is, the diameter of the image, under f , of subsets of B whose diameter is not greater
than δ is not greater than ε（在 B 中直徑不超過 δ 的子集合被函數 f 映過去之後，在對
應域中的直徑不會超過 ϵ）.

Remark 4.51. In terms of the number δ(f, x, ε) defined in Remark 4.9, the uniform conti-
nuity of a function f : A Ñ N is equivalent to that

δf (ε) ” inf δ(f, x, ε) ą 0 @ε ą 0.

xPA

The function δf (¨) is the inverse of the modulus of continuity of (a uniform continuous)
function f .
122 CHAPTER 4. Continuous Maps

Theorem 4.52. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be a map.
If K Ď A is compact and f is continuous on K, then f is uniformly continuous on K.

Proof. Let ε ą 0 be given. Since f is continuous on K,

( ) ε
@ a P K, D δ = δ(a) ą 0 Q ρ f (x), f (a) ă whenever x P D(a, δ) X A .
2
! (
δ(a) )
)
Then D a, is an open cover of K; thus
2 aPK

N
( δ)

ed
ď
D ta1 , ¨ ¨ ¨ , aN u Ď K Q K Ď D ai , i ,
2
i=1

ct
1
where δi = δ(ai ). Let δ = mintδ1 , ¨ ¨ ¨ , δN u. Then δ ą 0, and if x1 , x2 P K and d(x1 , x2 ) ă
2

te
( δ )
δ, there must be j = 1, ¨ ¨ ¨ , N such that x1 , x2 P B(aj , δj ). In fact, since x1 P D aj , j for
ro 2
some j = 1, ¨ ¨ ¨ , N , then

δj
P
d(x2 , aj ) ď d(x1 , x2 ) + d(x1 , aj ) ă δ + ă δj .
2
Therefore, x1 , x2 P D(aj , δj ) X A for some j = 1, ¨ ¨ ¨ , N ; thus
ht

( ) ( ) ( ) ε ε
ρ f (x1 ), f (x2 ) ď ρ f (x1 ), f (aj ) + ρ f (x2 ), f (aj ) ă + = ε . ˝
ig

2 2
Alternative proof. Assume the contrary that f is not uniformly continuous on K. Then ((3)
r
py

n=1 and tyn un=1 in K such that

of Remark 4.45 implies that) there are sequences txn u8 8

( )
lim d(xn , yn ) = 0 but lim ρ f (xn ), f (yn ) ą 0 .
Co

nÑ8 nÑ8

Since K is (sequentially) compact, there exist convergent subsequences txnk u8

k=1 and tynk uk=1
8

with limits x, y P K. On the other hand, lim d(xn , yn ) = 0, we must have x = y; thus by
nÑ8
the continuity of f (on K),
( ) ( ) ( )
0 = ρ f (x), f (x) = lim ρ f (xnk ), f (ynk ) = lim ρ f (xn ), f (yn ) ą 0 ,
kÑ8 nÑ8

a contradiction. ˝

Lemma 4.53. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be uniformly
␣ (8
continuous. If txk u8
k=1 Ď A is a Cauchy sequence, so is f (x k ) k=1
.
§4.5 Uniform Continuity 123

Proof. Let txk u8

k=1 be a Cauchy sequence in (M, d), and ε ą 0 be given. Since f : A Ñ N
is uniformly continuous,
( )
D δ ą 0 Q ρ f (x), f (y) ă ε whenever d(x, y) ă δ and x, y P A .

For this particular δ, D N ą 0 Q d(xk , xℓ ) ă δ if k, ℓ ě N . Therefore,

)
ρ(f (xk ), f (xℓ ) ă ε if k, ℓ ě N . ˝

Corollary 4.54. Let (M, d) and (N, ρ) be metric spaces, A Ď M , and f : A Ñ N be

ed
uniformly continuous. If N is complete, then f has a unique extension to a continuous
function on A;
s that is, D g : A
s Ñ N such that

ct
(1) g is uniformly continuous on A;
s

te
(2) g(x) = f (x) for all x P A;

(3) if h : A
ro
s Ñ N is a continuous map satisfying (1) and (2), then h = g.
P
Proof. Let x P AzA.
s Then D txk u8 Ď A such that xk Ñ x as k Ñ 8. Since txk u8
␣ (k=1
8
k=1
is Cauchy, by Lemma 4.53 f (xk ) k=1 is a Cauchy sequence in (N, ρ); thus is convergent.
ht

Moreover, if tzk u8
k=1 Ď A is another sequence converging to x, we must have d(xk , zk ) Ñ 0
␣ (8 ␣ (8
as k Ñ 8; thus ρ(f (xk ), f (zk )) Ñ 0 as k Ñ 8, so the limit of f (xk ) k=1 and f (zk ) k=1
ig

must be the same.

r

Define g : A
s Ñ N by
py

if x P A ,
#
f (x)
Co

g(x) =
lim f (xk ) if x P AzA,
s and txk u8
k=1 Ď A converging to x as k Ñ 8 .
kÑ8

Then the argument above shows that g is well-defined, and (2), (3) hold.
Let ε ą 0 be given. Since f : A Ñ N is uniformly continuous,
( ) ε
D δ ą 0 Q ρ f (x), f (y) ă whenever d(x, y) ă 2δ and x, y P A .
3
Suppose that x, y P A
s such that d(x, y) ă δ. Let txk u8 , tyk u8 Ď A be sequences
k=1 k=1
converging to x and y, respectively. Then D N ą 0 such that
δ δ ( ) ε ( ) ε
d(xk , x) ă , d(yk , y) ă and ρ f (xk ), g(x) ă , ρ f (yk ), g(y) ă @k ě N .
2 2 3 3
124 CHAPTER 4. Continuous Maps

In particular, due to the triangle inequality,

δ δ
d(xN , yN ) ď d(xN , x) + d(x, y) + d(y, yN ) ă + δ + = 2δ ;
2 2
( ) ε
thus ρ f (xN ), f (yN ) ă . As a consequence,
3
( ) ( ) ( ) ( ) ε ε ε
ρ g(x), g(y) ď ρ g(x), f (xN ) + ρ f (xN ), f (yN ) + ρ f (yN ), f (y) ă + + = ε . ˝
3 3 3

4.6 Differentiation of Functions of One Variable

d
te
Definition 4.55. A function f : (a, b) Ñ R is said to be differentiable at x0 if there exists
a number m such that

ec
f (x) ´ f (x0 ) ´ m(x ´ x0 )
lim = 0.
xÑx0 x ´ x0

ot
The (unique) number m is usually denoted by f 1 (x0 ), and is called the derivative of f at
x0 .
Pr
Remark 4.56. The derivative of f at x0 can be computed by

f (x) ´ f (x0 )
ht

f 1 (x0 ) = lim .
xÑx0 x ´ x0
Remark 4.57. By the definition of the limit of functions, f : (a, b) Ñ R is differentiable at
ig

x0 P (a, b) if and only if there exists m P R, denoted by f 1 (x0 ), such that

r
py

ˇ ˇ
@ ε ą 0, D δ ą 0 Q ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ ď ε|x ´ x0 | if |x ´ x0 | ă δ .

Definition 4.58. A function f : (a, b) Ñ R is said to be differentiable (on (a, b)) if f is

Co

differentiable at each x0 P (a, b).

Proposition 4.59. Suppose that a function f : (a, b) Ñ R is differentiable at x0 . Then f

is continuous at x0 .
f (x) ´ f (x0 )
Proof. For x ‰ x0 , f (x) ´ f (x0 ) = ¨ (x ´ x0 ); thus Proposition 4.15 implies that
x ´ x0
( ) f (x) ´ f (x0 )
lim f (x) ´ f (x0 ) = lim ¨ lim (x ´ x0 ) = f 1 (x0 ) ¨ 0 = 0 . ˝
xÑx0 xÑx0 x ´ x0 xÑx0

Theorem 4.60. Suppose that functions f, g : (a, b) Ñ R are differentiable at x0 , and k P R

is a constant. Then
§4.6 Differentiation of Functions of One Variable 125

1. (kf )1 (x0 ) = kf 1 (x0 ).

2. (f ˘ g)1 (x0 ) = f 1 (x0 ) ˘ g 1 (x0 ).

3. (f g)1 (x0 ) = f 1 (x0 )g(x0 ) + f (x0 )g 1 (x0 ).

( f )1 f 1 (x0 )g(x0 ) ´ f (x0 )g 1 (x0 )
4. (x0 ) = if g(x0 ) ‰ 0.
g g(x0 )2

Theorem 4.61 (Chain Rule). Suppose that a function f : (a, b) Ñ R is differentiable at x0 ,

and g : (c, d) Ñ R is differentiable at y0 = f (x0 ) P (c, d). Then g ˝ f is differentiable at x0 ,

ed
and
(g ˝ f )1 (x0 ) = g 1 (f (x0 ))f 1 (x0 ) .

ct
Proof. Let ε ą 0 be given. Since f : (a, b) Ñ R is differentiable at x0 and g : (c, d) Ñ R is

te
differentiable at y0 = f (x0 ),
ˇ ˇ !
D δ1 ą 0 Q ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ ď min 1,
ro ε
)
|x ´ x0 | if |x ´ x0 | ă δ1
2(1 + |g 1 (y0 )|)
P
and
ˇ ˇ ε|y ´ y0 |
ht

D δ2 ą 0 Q ˇg(y) ´ g(y0 ) ´ g 1 (y0 )(y ´ y0 )ˇ ď if |y ´ y0 | ă δ2 .

2(1 + |f 1 (x0 )|)
Moreover, by Proposition 4.59 f is continuous at x0 ; thus
ig

ˇ ˇ
r

D δ3 ą 0 Q ˇf (x) ´ f (x0 )ˇ ă δ2 if |x ´ x0 | ă δ3 and x P (a, b) .

py

Let δ = mintδ1 , δ3 u, and denote f (x) by y. Then if |x ´ x0 | ă δ, we have |y ´ y0 | ă δ2 and

Co

ˇ ˇ ˇ ˇ
ˇ(g ˝ f )(x) ´ (g ˝ f )(x0 ) ´ g 1 (y0 )f 1 (x0 )(x ´ x0 )ˇ = ˇg(y) ´ g(y0 ) ´ g 1 (y0 )f 1 (x0 )(x ´ x0 )ˇ
ˇ ˇ
= ˇg(y) ´ g(y0 ) ´ g 1 (y0 )(y ´ y0 ) + g 1 (y0 )(f (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ
ˇ ˇ
εˇf (x) ´ f (x0 )ˇ ˇˇ 1 ˇ ε|x ´ x0 |
ď + g (y 0 ) ˇ
2(1 + |f 1 (x0 )|) 2(1 + |g 1 (y0 )|)
ε ( 1
) ε
ď |x ´ x 0 | + |f (x 0 )||x ´ x 0 | + |x ´ x0 | = ε|x ´ x0 | .
2(1 + |f 1 (x0 )|) 2

By Remark 4.57, g ˝ f is differentiable at x0 with derivative g 1 (f (x0 ))f 1 (x0 ). ˝

Proposition 4.62. If f : (a, b) Ñ R is differentiable at x0 P (a, b) and f attains a local

minimum or maximum at x0 , then f 1 (x0 ) = 0.
126 CHAPTER 4. Continuous Maps

Proof. W.L.O.G. we assume that f attains its local minimum at x0 . Then f (x) ´ f (x0 ) ě 0
for all x P I, where I is an open interval containing x0 . Therefore,

f (x) ´ f (x0 ) f (x) ´ f (x0 )

f 1 (x0 ) = lim = lim´ ď0
xÑx0 x ´ x0 xÑx0 x ´ x0

and
f (x) ´ f (x0 ) f (x) ´ f (x0 )
f 1 (x0 ) = lim = lim+ ě 0.
xÑx0 x ´ x0 xÑx0 x ´ x0
As a consequence, f 1 (x0 ) = 0. ˝

ed
Theorem 4.63 (Rolle). Suppose that a function f : [a, b] Ñ R is continuous, and is
differentiable on (a, b). If f (a) = f (b), then D c P (a, b) such that f 1 (c) = 0.

ct
Proof. By the Extreme Value Theorem, there exists x0 and x1 in [a, b] such that

te
f (x0 ) = min f ([a, b]) and f (x1 ) = max f ([a, b]) .
ro
Case 1. f (x0 ) = f (x1 ), then f is constant on [a, b]; thus f 1 (x) = 0 for all x P (a, b).
P
Case 2. One of f (x0 ) and f (x1 ) is different from f (a). W.L.O.G. we may assume that
ht

f (x0 ) ‰ f (a). Then x0 P (a, b), and f attains its global minimum at x0 . By Proposi-
tion
ig

4.62, f 1 (x0 ) = 0. ˝
r
py

Theorem 4.64 (Cauchy’s Mean Value Theorem). Suppose that functions f, g : [a, b] Ñ R
are continuous, and f, g : (a, b) Ñ R are differentiable. If g(a) ‰ g(b) and g 1 (x) ‰ 0 for all
Co

x P (a, b), then there exists c P (a, b) such that

f 1 (c) f (b) ´ f (a)

1
= .
g (c) g(b) ´ g(a)

Proof. Consider the function

( )( ) ( )( )
h(x) ” f (x) ´ f (a) g(b) ´ g(a) ´ f (b) ´ f (a) g(x) ´ g(a) .

Then h : [a, b] Ñ R is continuous, and is differentiable on (a, b). Moreover, h(b) = h(a) = 0.
By Rolle’s theorem, there exists c P (a, b) such that
( ) ( )
h1 (c) = f 1 (c) g(b) ´ g(a) ´ f (b) ´ f (a) g 1 (c) = 0 . ˝
§4.6 Differentiation of Functions of One Variable 127

Corollary 4.65 (Mean Value Theorem). Suppose that a function f : [a, b] Ñ R is continu-
ous, and f : (a, b) Ñ R is differentiable. Then there exists c P (a, b) such that
f (b) ´ f (a)
f 1 (c) = .
b´a
Proof. Apply the Cauchy Mean Value Theorem for the case that g(x) = x. ˝
Corollary 4.66. Suppose that a function f : [a, b] Ñ R is continuous and f 1 (x) = 0 for all
x P (a, b). Then f is constant.
Proof. Let x P (a, b) be given. By Mean Value Theorem, there exists c P (a, x) such that

ed
f (x) ´ f (a) = f 1 (c)(x ´ a) = 0 .
Therefore, f (x) = f (a); thus for all x P (a, b), f (x) = f (a). Now by continuity, f (b) =

ct
lim f (x) = f (a). ˝
xÑb´

te
Corollary 4.67 (L’Hôspital’s rule). Let f, g : (a, b) Ñ R be differentiable functions. Suppose
ro f 1 (x)
that for some x0 P (a, b), f (x0 ) = g(x0 ) = 0, g 1 (x) ‰ 0 for all x ‰ x0 , and the limit lim
xÑx0 g 1 (x)
f (x)
P
exists. Then the limit lim also exists, and
xÑx0 g(x)
f (x) f 1 (x)
lim = lim 1 .
ht

xÑx0 g(x) xÑx0 g (x)

Proof. We first note that g(x) ‰ g(x0 ) for all x ‰ x0 since if not, the Mean Value Theorem
ig

implies that the existence of c in between x and x0 such that g 1 (c) = 0 which contradicts
to the condition that g 1 (x) ‰ 0 for all x ‰ x0 . By Cauchy’s Mean Value Theorem, for all
r
py

x P (a, b) and x ‰ x0 , there exists ξ = ξ(x) in between x and x0 such that

f (x) f (x) ´ f (x0 ) f 1 (ξ)
= = 1
Co

g(x) g(x) ´ g(x0 ) g (ξ)

Since ξ Ñ x0 as x Ñ x0 , we have
f (x) f 1 (ξ) f 1 (x)
lim = lim 1 = lim 1 . ˝
xÑx0 g(x) ξÑx0 g (ξ) xÑx0 g (x)

Theorem 4.68 (Taylor). Suppose that for some k P N, f : (a, b) Ñ R be (k + 1)-times

differentiable and c P (a, b). Then for all x P (a, b), there exists d in between c and x such
that
k
ÿ f (j) (c) f (k+1) (d)
f (x) = (x ´ c)j + (x ´ c)(k+1) ,
j=0
j! (k + 1)!
where f (j) denotes the j-th derivative of f .
128 CHAPTER 4. Continuous Maps

k f (j) (c)
Proof. Let g(x) = f (x) ´ (x ´ c)j , and h(x) = (x ´ c)k+1 . Then for 1 ď j ď k,
ř
j=0 j!

g (j) (c) = h(j) (c) = 0 ;

thus by the Cauchy mean value theorem (Theorem 4.64), there exists ξ1 in between x and
c, ξ2 in between ξ1 and c, ¨ ¨ ¨ , ξk+1 in between ξk and c such that

g(x) g(x) ´ g(c) g 1 (ξ1 ) g 1 (ξ1 ) ´ g 1 (c) g 2 (ξ2 )

= = 1 = 1 = = ¨¨¨
h(x) h(x) ´ h(c) h (ξ1 ) h (ξ1 ) ´ h1 (c) h2 (ξ2 )

d
g (k) (ξk ) g (k) (ξk ) ´ g (k) (c) g (k+1) (ξk+1 ) f (k+1) (ξk+1 )
= (k) = (k) = = .

te
h (ξk ) h (ξk ) ´ h(k) (c) h(k+1) (ξk ) (k + 1)!

Letting d = ξk+1 we conclude the theorem.

ec
˝

Example 4.69. A function f : [a, b] Ñ R is said to be Lipschitz continuous if D M ą 0

such that
ot
Pr
|f (x1 ) ´ f (x2 )| ď M |x1 ´ x2 | @ x1 , x2 P [a, b] .

If the derivative of a differentiable function f : (a, b) Ñ R is bounded; that is, D M ą 0

Q |f 1 (x)| ď M for all x P (a, b), then the Mean Value Theorem implies that f is Lipschitz
ht

continuous. A Lipschitz continuous function must be uniformly continuous.

ig

increasing
decreasing
Definition 4.70. A function f : (a, b) Ñ R is said to be (on (a, b))
r

strictly increasing
py

strictly decreasing
ď
ě
Co

if f (x1 ) f (x2 ) if a ă x1 ă x2 ă b. f is said to be monotone if f is either increasing

ă
ą
or decreasing on (a, b), and strictly monotone if f is either strictly increasing or strictly
decreasing.

Theorem 4.71. Suppose that f : (a, b) Ñ R is differentiable.

1. f is increasing on (a, b) if and only if f 1 (x) ě 0 for all x P (a, b).

2. f is decreasing on (a, b) if and only if f 1 (x) ď 0 for all x P (a, b).

3. If f 1 (x) ą 0 for all x P (a, b), then f is strictly increasing.

§4.7 Integration of Functions of One Variable 129

4. If f 1 (x) ă 0 for all x P (a, b), then f is strictly decreasing.

Theorem 4.72 (Inverse Function Theorem). Let f : (a, b) Ñ R be differentiable, and f 1

be sign-definite; that is, f 1 (x) ą 0 for all x P (a, b) or f 1 (x) ă 0 for all x P (a, b). Then
f : (a, b) Ñ f ((a, b)) is a bijection, and f ´1 , the inverse function of f , is differentiable on
f ((a, b)), and
1
(f ´1 )1 (f (x)) = 1 @ x P (a, b) . (4.6.1)
f (x)
Proof. W.L.O.G. we assume that f 1 (x) ą 0 for all x P (a, b). By Theorem 4.71 f is strictly
increasing; thus f ´1 exists.

ed
Claim: f ´1 : f ((a, b)) Ñ (a, b) is continuous.
Proof of claim: Let y0 = f (x0 ) P f ((a, b)), and ε ą 0 be given. Then f ((x0 ´ ε, x0 + ε)) =

ct
( )
f (x0 ´ ε), f (x0 + ε) since f is continuous on (a, b) and (x0 ´ ε, x0 + ε) is connected. Let

te
(
δ = mintf (x0 ) ´ f (x0 ´ ε), f (x0 + ε) ´ f (x0 ) . Then δ ą 0, and
( ro )
(y0 ´ δ, y0 + δ) = f (x0 ) ´ δ, f (x0 ) + δ Ď f ((x0 ´ ε, x0 + ε)) ;

thus by the injectivity of f ,

P
f ´1 ((y0 ´ δ, y0 + δ)) Ď f ´1 (f ((x0 ´ ε, x0 + ε))) = (x0 ´ ε, x0 + ε) = (f ´1 (y0 ) ´ ε, f ´1 (y0 ) + ε) .
ht

The inclusion above implies that f ´1 is continuous at y0 .

Writing y = f (x) and x = f ´1 (y). Then if y0 = f (x0 ) P f ((a, b)),
ig

f ´1 (y) ´ f ´1 (y0 ) x ´ x0
r

= .
y ´ y0 f (x) ´ f (x0 )
py

Since f ´1 is continuous on f ((a, b)), x Ñ x0 as y Ñ y0 ; thus

Co

f ´1 (y) ´ f ´1 (y0 ) x ´ x0 1
lim = lim = 1
yÑy0 y ´ y0 xÑx 0 f (x) ´ f (x0 ) f (x0 )
which implies that f ´1 is differentiable at y0 . ˝

4.7 Integration of Functions of One Variable

Definition 4.73. Let A Ď R be a bounded subset. A collection P of finitely many points
tx0 , x1 , ¨ ¨ ¨ , xn u is called a partition of A if inf A = x0 ă x1 ă ¨ ¨ ¨ ă xn´1 ă xn = sup A.
The mesh size of the partition P, denoted by }P}, is defined by
␣ ˇ (
}P} = max xk ´ xk´1 ˇ k = 1, ¨ ¨ ¨ , n .
130 CHAPTER 4. Continuous Maps

Definition 4.74. Let A Ď R be a bounded subset, and f : A Ñ R be a bounded function.

For any partition P = tx0 , x1 , ¨ ¨ ¨ , xn u of A, the upper sum and the lower sum of f
with respect to the partition P, denoted by U (f, P) and L(f, P) respectively, are numbers
defined by

n
ÿ n´1
ÿ
U (f, P) = sup fs(x)(xk ´ xk´1 ) = sup fs(x)(xk+1 ´ xk ) ,
k=1 xP[xk´1 ,xk ] k=0 xP[xk ,xk+1 ]
n
ÿ n´1
ÿ
L(f, P) = inf fs(x)(xk ´ xk´1 ) = inf fs(x)(xk+1 ´ xk ) ,
xP[xk´1 ,xk ] xP[xk ,xk+1 ]

ed
k=1 k=0

where fs is an extension of f given by

ct
"
f (x) x P A ,
fs(x) = (4.7.1)

te
0 x R A.

The two numbers

P ro
ż
f (x)dx ” inf U (f, P) ˇ P is a partition of A ,
␣ ˇ (
A
ht

and ż
f (x)dx ” sup L(f, P) ˇ P is a partition of A
␣ ˇ (
ig

A
r

are called the upper integral and lower integral of f over A, respective. The function
py

ż ż
f is said to be Riemann (Darboux) integrable (over A) if f (x)dx = f (x)dx, and
ż A A
Co

in this case, we express the upper and lower integral as f (x)dx, called the integral of f
A
over A. The upper integral, the lower integral, and the integral of f over [a, b] sometimes
żb żb żb
are also denoted by f (x)dx, f (x)dx, and f (x)dx.
a a a

żb żb
Example 4.75. f (x)dx and f (x)dx are not always the same. For example, define
a a
f : [0, 1] Ñ R by
1 if x P [0, 1]zQ,
"
f (x) =
0 if x P [0, 1] X Q.

Let P = t0 = x0 ă x1 ă ¨ ¨ ¨ ă xn = 1u be any partition on [0, 1]. Then for any k =

§4.7 Integration of Functions of One Variable 131

0, 1, ¨ ¨ ¨ , n ´ 1, sup f (x) = 1 and inf f (x) = 0; thus

xP[xk ,xk+1 ] xP[xk ,xk+1 ]

n´1
ÿ n
ÿ
U (f, P) = sup f (x)(xk ´ xk´1 ) = (xk ´ xk´1 )
k=0 xP[xk ,xk+1 ] k=0
= (x1 ´ x0 ) + (x2 ´ x1 ) + ¨ ¨ ¨ + (xn ´ xn´1 ) = xn ´ x0 = 1 ´ 0 = 1

and n
ÿ
L(f, P) = 0(xi ´ xi´1 ) = 0 .
i=1
As a consequence,

ed
ż1
f (x)dx = inf U (f, P) ˇ P is a partition on [0, 1] = 1 ,
␣ ˇ (

ct
ż01
f (x)dx = sup L(f, P) ˇ P is a partition on [0, 1] = 0 ;
␣ ˇ (

te
0

hence f is not Riemann integrable over [0, 1].

ro żb
Example 4.76. Suppose f : [a, b] Ñ R is integrable and f ě 0 on [a, b], then
P
f (x)dx ě 0.
a
Reason: Since f ě 0 on [a, b] ñ sup f (x) ě 0 for k = 0, 1, . . . , n ´ 1. Therefore,
xP[xk ,xk+1 ]
ht

U (f, P) ě 0 for all partition P on [a, b], so

żb żb
ig

f (x)dx = inftU (f, P) ˇ P is a partition on [a, b] ě 0 .

ˇ (
f (x)dx =
a a
r

Definition 4.77. A partition P 1 of a bounded set A Ď R is said to be a refinement of

py

another partition P if P Ď P 1 .
Co

Proposition 4.78. Let A Ď R be a bounded subset, and f : A Ñ R be a bounded function.

If P and P 1 are partitions of A and P 1 is a refinement of P, then

L(f, P) ď L(f, P 1 ) ď U (f, P 1 ) ď U (f, P) .

Proof. Let fs be the extension of f given by (4.7.1). Suppose that P = tx0 , x1 , ¨ ¨ ¨ , xn u, P 1 =

ty0 , y1 , ¨ ¨ ¨ , ym u, and P Ď P 1 . For any fixed k = 0, 1, ¨ ¨ ¨ , n ´ 1, either P 1 X (xk , xk+1 ) = H
or P 1 X (xk , xk+1 ) ‰ H.

1. If P 1 X (xk , xk+1 ) = H, then xk = yℓ and xk+1 = yℓ+1 for some ℓ. Therefore,

sup fs(x)(xk+1 ´ xk ) = sup fs(x)(yℓ+1 ´ yℓ ).

xP[xk ,xk+1 ] xP[yℓ ,yℓ+1 ]
132 CHAPTER 4. Continuous Maps

2. If P 1 X (xk , xk+1 ) = tyℓ+1 , yℓ+2 , ¨ ¨ ¨ , yℓ+p u, then xk = yℓ and xk+1 = yℓ+p+1 . Therefore,
p+1
ÿ
sup fs(x)(yℓ+i ´ yℓ+i´1 ) = sup fs(x)(yℓ+1 ´ yℓ )
i=1 xP[yℓ+i´1 ,yℓ+i ] xP[yℓ ,yℓ+1 ]

+ sup fs(x)(yℓ+2 ´ yℓ+1 ) + ¨ ¨ ¨ + sup fs(x)(yℓ+p+1 ´ yℓ+p )

xP[yℓ+1 ,yℓ+2 ] xP[yℓ+p ,yℓ+p+1 ]

ď sup fs(x)(yℓ+1 ´ yℓ ) + sup fs(x)(yℓ+2 ´ yℓ+1 ) + ¨ ¨ ¨

xP[xk ,xk+1 ] xP[xk ,xk+1 ]

+ sup fs(x)(yℓ+p+1 ´ yℓ+p ) = sup fs(x)(xk+1 ´ xk ) .

xP[xk ,xk+1 ] xP[xk ,xk+1 ]

d
In either case,

te
ÿ
sup fs(x)(yℓ ´ yℓ´1 ) ď sup fs(x)(xk+1 ´ xk ) .

ec
[yℓ´1 ,yℓ ]Ď[xk ,xk+1 ] xP[yℓ´1 ,yℓ ] xP[xk ,xk+1 ]

As a consequence,

U (f, P ) =
1
m´1
ÿ
sup ot n´1
ÿ ÿ
Pr
fs(x)(yℓ+1 ´ yℓ ) = fs(x)(yℓ ´ yℓ´1 )
ℓ=0 xP[yℓ ,yℓ+1 ] k=0 [yℓ´1 ,yℓ ]Ď[xk ,xk+1 ]
n´1
ÿ
ď sup fs(x)(xk+1 ´ xk ) = U (f, P) .
ht

k=0 xP[xk ,xk+1 ]

Similarly, L(f, P) ď L(f, P 1 ); thus the fact that L(f, P 1 ) ď U (f, P 1 ) concludes the proposi-
ig

tion. ˝
r

Corollary 4.79. Let f : [a, b] Ñ R be a function bounded by M ; that is, |f (x)| ď M for all
py

a ď x ď b. Then for all partitions P1 and P2 of [a, b],

żb żb
´M (b ´ a) ď L(f, P1 ) ď f (x)dx ď U (f, P2 ) ď M (b ´ a) .
Co

f (x)dx ď
a a
żb żb
Proof. It suffices to show that f (x)dx ď f (x)dx. By the definition of infimum and
a a
supremum, for any given ε ą 0, D partitions P
s and Pr such that
żb żb żb żb
ε ε
f (x)dx ´ ă L(f, P) ď
s f (x)dx and f (x)dx ď U (f, P) ă
r f (x)dx + .
a 2 a a a 2

Let P = Ps Y P.
r Then P is a refinement of both P
s and P;
r thus
żb żb
ε ε
f (x)dx ´ ă L(f, P) ď L(f, P) ď U (f, P) ď U (f, P) ă
s r f (x)dx + .
a 2 a 2
§4.7 Integration of Functions of One Variable 133

żb żb
Since ε ą 0 is given arbitrarily, we must have f (x)dx ď f (x)dx. ˝
a a

Proposition 4.80 (Riemann’s condition). Let A Ď R be a bounded set, and f : A Ñ R be

a bounded function. Then f is Riemann integrable over A if and only if

@ ε ą 0, D a partition P of A Q U (f, P) ´ L(f, P) ă ε .

Proof. “ñ” Let ε ą 0 be given. Since f is integrable over A,

ż
inf U (f, P) = sup L(f, P) =

d
f (x)dx ;
P: Partition of A P: Partition of A A

te
thus there exist P1 and P2 , partitions of A, such that

ec
ż ż ż
ε ε
f (x)dx ´ ă L(f, P1 ) ď f (x)dx ď U (f, P2 ) ă f (x)dx + .
A 2 A A 2

ot
Let P = P1 Y P2 . Then P is a refinement of P1 and P2 ; thus
Pr
ż ż
ε
f (x)dx ´ ă L(f, P1 ) ď L(f, P) ď f (x)dx
A 2 A ż
ε
ď U (f, P) ď U (f, P2 ) ă f (x)dx +
ht

A 2
which implies that U (f, P) ´ L(f, P) ă ε.
ig

“ð” We note that for any partition P of A,

r
py

ż ż
L(f, P) ď f (x)dx ď f (x)dx ď U (f, P) ;
A A
Co

so we have that for all partition P of A,

ż ż
f (x)dx ´ f (x)dx ă U (f, P) ´ L(f, P) .
A A

Let ε ą 0 be given. By choosing P so that U (f, P) ´ L(f, P) ă ε, we conclude that

ż ż
f (x)dx ´ f (x)dx ă ε .
A A

ż ż
Since ε ą 0 is given arbitrarily, f (x)dx = f (x)dx; thus f is Riemann integrable
A A
over A. ˝
134 CHAPTER 4. Continuous Maps

Proposition 4.81. Suppose that f, g : [a, b] Ñ R are Riemann integrable, and k P R. Then
żb żb
1. kf is Riemann integrable, and (kf )(x)dx = k f (x)dx.
a a
żb żb żb
2. f ˘ g are Riemann integrable, and (f ˘ g)(x)dx = f (x)dx ˘ g(x)dx.
a a a
żb żb
3. If f ď g for all x P [a, b], then f (x)dx ď g(x)dx.
a a

4. If f is also Riemann integrable over [b, c], then f is Riemann integrable over [a, c],

ed
and żc żb żc
f (x)dx = f (x)dx + f (x)dx . (4.7.2)

ct
a a b

ˇż b ˇ żb

te
5. The function |f | is also Riemann integrable, and ˇ f (x)dxˇ ď |f (x)|dx.
ˇ ˇ
a a

Proof. 1. Case 1. k ě 0. We note that

P ro
inf (kf )(x) = k inf f (x) and sup (kf )(x) = k sup f (x) .
xP[xi´1 ,xi ] xP[xi´1 ,xi ] xP[xi´1 ,xi ] xP[xi´1 ,xi ]
ht

Then
n
ig

ÿ
L(kf, P) = inf (kf )(x)(xi ´ xi´1 )
xP[xi´1 ,xi ]
i=1
r

n
py

ÿ
= k inf f (x)(xi ´ xi´1 ) = kL(f, P) .
xP[xi´1 ,xi ]
i=1
Co

Similarly, U (kf, P) = kU (f, P) for every partition P. So

żb
(kf )(x)dx = sup L(kf, P) = k sup L(f, P)
a P: Partition of [a, b] P: Partition of [a, b]
żb żb
=k f (x)dx = k f (x)dx .
a a

żb żb
Similarly, (kf )(x)dx = k f (x)dx. Hence kf is integrable and
a a

żb żb żb żb
(kf )(x)dx = (kf )(x)dx = k f (x)dx = k f (x)dx .
a a a a
§4.7 Integration of Functions of One Variable 135

Case 2. k ă 0. We have

inf (kf )(x) = k sup f (x) and sup (kf )(x) = k inf f (x) .
xP[xi´1 ,xi ] xP[xi´1 ,xi ] xP[xi´1 ,xi ] xP[xi´1 ,xi ]

Then L(kf, P) = kU (f, P) and U (kf, P) = kL(f, P); thus

żb
(kf )(x)dx = sup L(kf, P) = sup kU (f, P)
a P: Partition of [a, b] P: Partition of [a, b]
żb żb
=k inf U (f, P) = k f (x)dx = k f (x)dx .
P: Partition of [a, b]

ed
a a
żb żb
Similarly, (kf )(x)dx = k f (x)dx. Hence kf is Riemann integrable over [a, b] and

ct
a a
żb żb żb żb

te
(kf )(x)dx = (kf )(x)dx = k f (x)dx = k f (x)dx .
a a a a

2. We prove the case of summation. For ant partition P, we have

P ro
n
ÿ
L(f + g, P) = inf (f + g)(x)(xi ´ xi´1 )
xP[xi´1 ,xi ]
ht

i=1
ÿn n
ÿ
ě inf f (x)(xi ´ xi´1 ) + inf g(x)(xi ´ xi´1 )
ig

xP[xi´1 ,xi ] xP[xi´1 ,xi ]

i=1 i=1
= L(f, P) + L(g, P) .
r
py

Similarly, U (f + g, P) ď U (f, P) + U (g, P). Therefore,

L(f, P) + L(g, P) ď L(f + g, P) ď U (f + g, P) ď U (f, P) + U (g, P) .

Co

(4.7.3)

Let ε ą 0 be given. By Proposition 4.80, D P1 , P2 partitions of [a, b] such that

ε ε
U (f, P1 ) ´ L(f, P1 ) ă and U (g, P2 ) ´ L(g, P2 ) ă .
2 2
Let P = P1 Y P2 . By (4.7.3),

U (f + g, P) ´ L(f + g, P) ď (U (f, P) + U (g, P)) ´ (L(f, P) + L(g, P))

= (U (f, P) ´ L(f, P)) + (U (g, P) ´ L(g, P))
ε ε
ď (U (f, P1 ) ´ L(f, P1 )) + (U (g, P2 ) ´ L(g, P2 )) ă + = ε.
2 2
136 CHAPTER 4. Continuous Maps

By Proposition 4.80, f + g is Riemann integrable over [a, b].

żb żb żb
To see (f + g)(x)dx = f (x)dx + g(x)dx, we note that by Proposition 4.78,
a a a

ε
U (f, P) ď L(f, P) + U (f, P1 ) ´ L(f, P1 ) ă L(f, P) +
2
żb żb
ε ε
ď f (x)dx + = f (x)dx +
a 2 a 2
żb
ε
and similarly, U (g, P) ă g(x)dx + . Therefore, by (4.7.3),
a 2

ed
żb żb
(f + g)(x)dx = (f + g)(x)dx ď U (f + g, P)

ct
a a
żb żb
ď U (f, P) + U (g, P) ă f (x)dx + g(x)dx + ε . (4.7.4)

te
a a

On the other hand, ro żb

ε ε
L(f, P) ą U (f, P) ´
P
ě f (x)dx ´
2 a 2

and
ht

żb
ε ε
L(g, P) ą U (g, P) ´ ě g(x)dx ´ ;
2 a 2
ig

hence by (4.7.3),
r

żb żb
(f + g)(x)dx ě L(f + g, P) ě L(f, P) + L(g, P)
py

(f + g)(x)dx =
a a
żb żb
(4.7.5)
Co

ą f (x)dx + g(x)dx ´ ε .
a a

By (4.7.4) and (4.7.5),

żb żb żb żb żb
f (x)dx + g(x)dx ´ ε ă (f + g)(x)dx ă f (x)dx + g(x)dx + ε .
a a a a a
żb żb żb
Since ε ą 0 is arbitrary, (f + g)(x)dx = f (x)dx + g(x)dx.
a a a

3. Let P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn = bu be a partition of [a, b]. Define

mi (f ) = inf f (x) and mi (g) = inf g(x) .

xP[xi´1 ,xi ] xP[xi´1 ,xi ]
§4.7 Integration of Functions of One Variable 137

Since f (x) ď g(x) on [a, b], mi (f ) ď mi (g). As a consequence, for any partition P,
n
ÿ n
ÿ
L(f, P) = mi (f )(xi ´ xi´1 ) ď mi (g)(xi ´ xi´1 ) = L(g, P) ;
i=1 i=1

thus taking the infimum over all partition P,

żb żb żb żb
f (x)dx = f (x)dx = sup L(f, P) ď sup L(g, P) = g(x)dx = g(x)dx .
a a P P a a

4. Let ε ą 0 be given. Since f is Riemann integrable of [a, b] and [b, c], there exist a
partition P1 over [a, b] and a partition P2 of [b, c] such that

ed
ε ε
U (f, P1 ) ´ L(f, P1 ) ă and U (f, P2 ) ´ L(f, P2 ) ă .
2 2

ct
Let P = P1 Y P2 . Then P is a partition of [a, c] such that

U (f, P) ´ L(f, P) = U (f, P1 ) + U (f, P2 ) ´ L(f, P1 ) ´ L(f, P2 ) ă ε .

te
Therefore, Proposition 4.80 implies that f is Riemann integrable over [a, c].
ro
żc żb żc
Now we show that f (x)dx = f (x)dx + f (x)dx. To simplify the notation,
P
a a b
we let żc żb żc
A= f (x)dx, B= f (x)dx, C= f (x)dx .
ht

a a b
Let ε ą 0 be given. Then D partition P = tx0 , x1 , ¨ ¨ ¨ , xn u of [a, c] such that
ig

A ď U (f, P) ă A + ε .
r

Let P 1 = P Y tbu. Then P 1 is a refinement of P. Moreover,

py

U (f, P 1 ) = U (f, P1 ) + U (f, P2 ) ,

Co

where P1 = P 1 X [a, b] and P2 = P 1 X [b, c] are partitions of [a, b] and [b, c] whose union
is P. Therefore,

B + C ď U (f, P1 ) + U (f, P2 ) = U (f, P 1 ) ď U (f, P) ă A + ε .

On the other hand, D partition P1 of [a, b] and partition P2 of [b, c] such that
ε ε
B ď U (f, P1 ) ă B + and C ď U (f, P2 ) ă C + .
2 2
Let P = P1 Y P2 . Then P is a partition of [a, c]. Therefore,

A ď U (f, P) = U (f, P1 ) + U (f, P2 ) ă B + C + ε .

Therefore, @ ε ą 0, B + C ă A + ε and A ă B + C + ε; thus A = B + C.

138 CHAPTER 4. Continuous Maps

5. Note that for any interval [α, β],

sup |f (x)| ´ inf |f (x)| ď sup f (x) ´ inf f (x) ; (Check!)

xP[α,β] xP[α,β] xP[α,β] xP[α,β]

thus for any partition P of [a, b],

U (|f |, P) ´ L(|f |, P) ď U (f, P) ´ L(f, P) .

Therefore, Proposition 4.80 implies that |f | is Riemann integrable over [a, b]. More-
over, since ´|f (x)| ď f (x) ď |f (x)| for all x P [a, b], by 3 we have

ed
żb żb żb
´ |f (x)|dx ď f (x)dx ď |f (x)|dx . ˝

ct
a a a

Remark 4.82. The proof of 4 in Proposition 4.81 in fact also shows that if a ă b ă c, then

te
żc żb żc
f (x)dx = f (x)dx + f (x)dx .
ro
a a b

Similar proof also implies that

P
żc żb żc
f (x)dx = f (x)dx + f (x)dx .
ht

a a b

ża żb
ig

Remark 4.83. If a ă b, we let the number f (x)dx denote the number ´ f (x)dx. Then
b a
(4.7.2) holds for all a, b, c P R.
r
py

Example 4.84. Let f : [0, 1] Ñ R be defined by

1 if x P Q,
Co

"
f (x) =
´1 if x P QA .

Then f (x) is not Riemann integrable over [0, 1] since U (f, P ) = 1 and L(f, P ) = ´1.
However |f (x)| ” 1, thus |f | is Riemann integrable. In other words, if |f | is integrable, we
cannot know whether f is integrable or not.

Theorem 4.85. If f : [a, b] Ñ R is continuous, then f is Riemann integrable.

Proof. Let ε ą 0 be given. Theorem 4.52 implies that

ε
D δ ą 0 Q |f (x) ´ f (y)| ă whenever |x ´ y| ă δ and x, y P [a, b] .
2(b ´ a)
§4.7 Integration of Functions of One Variable 139

Let P be a partition with mesh size less than δ. Then

n
ÿ ( )
U (f, P) ´ L(f, P) = sup f (x) ´ inf f (x) (xk ´ xk´1 )
xP[xk´1 ,xk ] xP[xk´1 ,xk ]
k=1
n
ε ÿ ε
ď (xk ´ xk´1 ) = (xn ´ x0 ) ă ε ;
2(b ´ a) k=1 2(b ´ a)

thus by Proposition 4.80 f is Riemann integrable over [a, b]. ˝

Corollary 4.86. If f : (a, b) Ñ R is continuous and f is bounded on [a, b], then f is

ed
Riemann integrable over [a, b].
[ ε ε ]

ct
Proof. Let |f (x)| ď M for all x P [a, b], and ε ą 0 be given. Since f : a + , b´ ÑR
8M 8M
is continuous, by Theorem 4.85 f is Riemann integrable; thus

te
[ ε ε ] ε
D P 1 : partition of a + ,b ´ Q U (f, P 1 ) ´ L(f, P 1 ) ă .
ro
8M 8M 2
Let P = P 1 Y ta, bu. Then
P
U (f, P) ´ L(f, P)
( ) ε ε ( ) ε
ht

ă sup f (x) ´ inf f (x) + + sup f (x) ´ infε f (x)

ε
xP[a,a+ 8M ]
ε
xP[a,a+ 8M ] 8M 2 ε
xP[b´ 8M ,b] xP[b´ 8M ,b] 8M
ig

ε ε ε
ď 2M ¨ + + 2M ¨ = ε;
8M 2 8M
r
py

thus Proposition 4.80 implies that f is Riemann integrable over [a, b]. ˝

Corollary 4.87. If f : [a, b] Ñ R is bounded and is continuous at all but finitely many
Co

points of [a, b], then f is Riemann integrable.

Proof. Let tc1 , ¨ ¨ ¨ , cN u be the collection of all discontinuities of f in (a, b) such that c1 ă
c2 ă ¨ ¨ ¨ ă cN . Let a = c0 and b = cN +1 . Then for all k = 0, 1, ¨ ¨ ¨ , N , f : (ck , ck+1 ) is
continuous and f : [ck , ck+1 ] is bounded; thus f is Riemann integrable by Corollary 4.87.
Finally, 4 of Proposition 4.81 implies that f is Riemann integrable over [a, b]. ˝

Theorem 4.88. Any increasing or decreasing function on [a, b] is Riemann integrable.

Proof. Let f : [a, b] Ñ R be a monotone function, and ε ą 0 be given. W.L.O.G. we may

assume that f (b) ‰ f (a). Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of [a, b] with mesh size
140 CHAPTER 4. Continuous Maps

ε
less than . Then
|f (b) ´ f (a)|
n
ÿ ( )
U (f, P) ´ L(f, P) = sup f (x) ´ inf f (x) (xk ´ xk´1 )
xP[xk´1 ,xk ] xP[xk´1 ,xk ]
k=1
n
ÿ ˇ ˇ
ˇf (xk ) ´ f (xk´1 )ˇ ε ˇ ˇ ε
ă = ˇf (b) ´ f (a)ˇ = ε;
k=1
|f (b) ´ f (a)| |f (b) ´ f (a)|

thus Proposition 4.80 implies that f is Riemann integrable over [a, b]. ˝

Definition 4.89. A continuous function F : [a.b] Ñ R is called an anti-derivative（反導函數）of

ed
f : [a, b] Ñ R if F is differentiable on (a, b) and F 1 (x) = f (x) for all x P (a, b).

ct
Theorem 4.90 (Fundamental Theorem of Calculus（微積分基本定理）). Let f : [a, b] Ñ R
be continuous. Then f has an anti-derivative F , and

te
żb
f (x)dx = F (b) ´ F (a) .
a
P ro żb
Moreover, if G is any other anti-derivative of f , we also have f (x)dx = G(b) ´ G(a).
a
żx
ht

Proof. Define F (x) = f (y)dy, where the integral of f over [a, x] is well-defined because
a
of continuity of f on [a, x]. We first show that F is differentiable on (a, b).
ig

Let x0 P (a, b) and ε ą 0 be given. Since [a, b] is compact,

r

ˇ ˇ ε
D δ1 ą 0 Q ˇf (x) ´ f (y)ˇ ă whenever |x ´ y| ă δ1 and x, y P [a, b] .
py

2
Let δ = mintδ1 , x0 ´ a, b ´ x0 u. By 4 of Proposition 4.81, if x, x0 P (a, b),
Co

żx żx ż x0
f (y)dy = f (y)dy ´ f (y)dy = F (x) ´ F (x0 ) ;
x0 a a

thus if 0 ă |x ´ x0 | ă δ,
ˇ F (x) ´ F (x ) ˇ ˇ 1 żx ˇ ˇ 1 żx( ) ˇˇ
0
´ f (x0 )ˇ = ˇ f (y)dx ´ f (x0 )ˇ = ˇ f (y) ´ f (x0 ) dy ˇ
ˇ ˇ ˇ ˇ ˇ
x ´ x0 x ´ x 0 x0 x ´ x 0 x0
ˇ
ż maxtx0 ,xu ż maxtx0 ,xu
1 ˇ ˇ 1 ε
ď ˇf (y) ´ f (x0 ) dy ď
ˇ dy ă ε .
|x ´ x0 | mintx0 ,xu |x ´ x0 | mintx0 ,xu 2

F (x) ´ F (x0 )
Therefore, lim = f (x0 ) for all x0 P (a, b), so F 1 (x) = f (x) for all x P (a, b).
xÑx0 x ´ x0
§4.7 Integration of Functions of One Variable 141

Next we show that F is continuous at x = a and x = b. This is simply because of the

boundedness of f on [a, b] which implies that
ˇ ˇ ˇż x ˇ żx
lim sup ˇF (x) ´ F (a)ˇ = lim sup ˇ f (t)dtˇ ď max |f (x)| ¨ lim sup 1dt = 0
ˇ ˇ
xÑa+ xÑa+ a xP[a,b] xÑa+ a

and
ˇ ˇ ˇż b ˇ żb
lim sup F (x) ´ F (b) = lim sup ˇ f (t)dtˇ ď max |f (x)| ¨ lim sup 1dt = 0 .
ˇ ˇ ˇ ˇ
xÑb´ xÑb´ x xP[a,b] xÑb´ x

Therefore, F is an anti-derivative of f .

ed
Now suppose that G is another anti-derivative of f . Then (G ´ F )1 (x) = 0 for all
x P (a, b). By Corollary 4.66, (G ´ F )(x) = (G ´ F )(a) for all x P [a, b]; thus G(b) ´ G(a) =

ct
F (b) ´ F (a). ˝

te
Example 4.91. If f is only integrable but not continuous, then the function
żx
P
F (x) =
a
ro f (t)dt

is not necessarily differentiable. For example, consider

0 if 0 ď x ď 1,
"
f (x) =
ht

1 if 1 ă x ď 2.
Then
ig

"
0 if 0 ď x ď 1,
F (x) =
x ´ 1 if 1 ď x ď 2.
r
py

so F is continuous on [0, 2] but not differentiable at x = 1.

Theorem 4.92. Let f : [a, b] Ñ R be differentiable. If f 1 is Riemann integrable over [a, b],
Co

żb
then f 1 (x)dx = f (b) ´ f (a).
a

Proof. Let P = tx0 , x1 , ¨ ¨ ¨ xn u be a partition of [a, b]. Since f : [a, b] Ñ R is differentiable,

by the Mean Value Theorem there exists tξ1 , ¨ ¨ ¨ , ξn u with the property that xk ă ξk+1 ă
xk+1 for all k = 0, 1 ¨ ¨ ¨ , n ´ 1 such that

f 1 (ξk+1 )(xk+1 ´ xk ) = f (xk+1 ) ´ f (xk ) @ k = 0, 1, ¨ ¨ ¨ , n ´ 1 .

Therefore,
n´1
ÿ n´1
ÿ n´1
ÿ
1 1
inf f (x)(xk+1 ´ xk ) ď f (ξk+1 )(xk+1 ´ xk ) ď sup f 1 (x)(xk+1 ´ xk ) .
xP[xk ,xk+1 ]
k=0 k=0 k=0 xP[xk ,xk+1 ]
142 CHAPTER 4. Continuous Maps

n´1 ř(
n´1 )
Since f 1 (ξk+1 )(xk+1 ´ xk ) = f (xk+1 ) ´ f (xk ) = f (b) ´ f (a), the inequality above
ř
k=0 k=0
implies that

L(f 1 , P) ď f (b) ´ f (a) ď U (f 1 , P) for all partitions P of [a, b] ;

thus by the definition of the upper and the lower integrals,

żb żb
1
f (x)dx ď f (b) ´ f (a) ď f 1 (x)dx .
a a

ed
We then conclude the theorem by the identity
żb żb żb
1 1
f (x)dx = f (x)dx = f 1 (x)dx

ct
a a a

te
since f 1 is Riemann integrable. ˝

Definition 4.93. Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of a bounded set A Ď R. A

ro
collection of points tξ1 , ¨ ¨ ¨ , ξn u is called a sample set for the partition P if ξk P [xk´1 , xk ]
P
for all k = 1, ¨ ¨ ¨ , n.
Let f : A Ñ R be a bounded function with extension fs given by (4.7.1). A Riemann
ht

sum of f for the partition P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn = bu of A is a sum which takes the

form
ig

n´1
ÿ
fs(ξk )(xk+1 ´ xk ) ,
r

k=0
py

where the set Ξ = tξ0 , ξ1 , ¨ ¨ ¨ , ξn´1 u is a sample set for P.

Theorem 4.94 (Darboux). Let f : A Ñ R be a bounded function with extension fs given by

Co

(4.7.1). Then f is Riemann integrable over A if and only if there exists I P R such that for
every given ε ą 0, there exists δ ą 0 such that if P is a partition of A satisfying }P} ă δ,
then any Riemann sum of f for the partition P lies in the interval (I ´ ε, I + ε). In other
words, f is Riemann integrable over A if and only if for every given ε ą 0, there exists δ ą 0
such that there exists I P R such that
ˇ n´1 ˇ
ˇÿ s
f (ξ )(x ´ x ) ´ I ˇăε (4.7.6)
ˇ
ˇ k+1 k+1 k
k=0

whenever P = tx0 , x1 , ¨ ¨ ¨ , xn u is a partition of A satisfying }P} ă δ and tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is

a sample set for P.
§4.7 Integration of Functions of One Variable 143

Proof. “ð” Suppose the right-hand side statement is true. Let ε ą 0 be given. Then there
exists δ ą 0 such that if P is a partition of A satisfying }P} ă δ, then for all sets of
sample points tξ1 , ¨ ¨ ¨ , ξn u with respect to P, we must have
ˇ n´1 ˇ ε
ˇÿ s
f (ξk+1 )(xk+1 ´ xk ) ´ I ˇ ă .
ˇ
4
ˇ
k=0

Let P = tx0 , x1 , ¨ ¨ ¨ , xn u be a partition of A with }P} ă δ. Choose sets of sample

points tξ1 , ¨ ¨ ¨ , ξn u and tη1 , ¨ ¨ ¨ , ηn u with respect to P such that
ε

ed
(a) sup fs(x) ´ ă fs(ξk+1 ) ď sup fs(x);
xP[xk ,xk+1 ] 4(xn ´ x0 ) xP[xk ,xk+1 ]
ε
(b) inf fs(x) + inf

ct
ą fs(ηk+1 ) ě fs(x).
xP[xk ,xk+1 ] 4(xn ´ x0 ) xP[xk ,xk+1 ]

te
Then
n´1
ÿ ro n´1
ÿ [ ε ]
U (f, P) = sup fs(x)(xk+1 ´ xk ) ă fs(ξk+1 ) + (xk+1 ´ xk )
4(xn ´ x0 )
k=0 xP[xk ,xk+1 ] k=0
P
n´1 n´1
ÿ ε ÿ ε ε ε
= fs(ξk+1 )(xk+1 ´ xk ) + (xk+1 ´ xk ) ă I + + = I +
4(xn ´ x0 ) 4 4 2
k=0 k=0
ht

and
ig

n´1
ÿ n´1
ÿ [ ε ]
L(f, P) = inf fs(x)(xk+1 ´ xk ) ą fs(ηk+1 ) ´ (xk+1 ´ xk )
xP[xk ,xk+1 ] 4(xn ´ x0 )
r

k=0 k=0
py

n´1 n´1
ÿ ε ÿ ε ε ε
= fs(ηk+1 )(xk+1 ´ xk ) ´ (xk+1 ´ xk ) ą I ´ ´ = I ´ .
4(xn ´ x0 ) 4 4 2
k=0 k=0
Co

ε ε
As a consequence, I ´ ă L(f, P) ď U (f, P) ă I + ; thus U (f, P) ´ L(f, P) ă ε.
2 2
ż
“ñ” Let ε ą 0 be given, and I = fs(x)dx. Since f is Riemann integrable over A, there
A
ε
exists a partition P1 = ty0 , y1 , ¨ ¨ ¨ , ym u of A such that U (f, P1 ) ´ L(f, P1 ) ă . Define
2
ε
! )
δ = min |y1 ´ y0 |, |y2 ´ y1 |, ¨ ¨ ¨ , |ym ´ ym´1 |, ( ) .
4m sup f (A) ´ inf f (A) + 1

If P = tx0 , x1 , ¨ ¨ ¨ , xn u is a partition of A with }P} ă δ, then at most 2m intervals

of the form [xk , xk+1 ] contains one of these yj ’s, and each such interval [xk , xk+1 ] can
only contain one of these yj ’s. Let P 1 = P Y P1 .
144 CHAPTER 4. Continuous Maps

ε
Claim: U (f, P) ´ U (f, P 1 ) ă .
2
Proof of claim: We note that
n´1
ÿ
U (f, P) = sup fs(x)(xk+1 ´ xk )
k=0 xP[xk ,xk+1 ]
ÿ ÿ
= sup fs(x)(xk+1 ´ xk ) + sup fs(x)(xk+1 ´ xk )
0ďkďn´1 with xP[xk ,xk+1 ] 0ďkďn´1 with xP[xk ,xk+1 ]
P1 X[xk ,xk+1 ]=H P1 X[xk ,xk+1 ]‰H

and

d
ÿ
U (f, P 1 ) = sup fs(x)(xk+1 ´ xk )

te
0ďkďn´1 with xP[xk ,xk+1 ]
P1 X[xk ,xk+1 ]=H
[ ]

ec
ÿ
+ sup fs(x)(yj ´ xk ) + sup fs(x)(xk+1 ´ yj ) .
0ďkďn´1 with xP[xk ,yj ] xP[yj ,xk+1 ]
P1 X[xk ,xk+1 ]=yj

Therefore,
ot
Pr
( ) ÿ
U (f, P) ´ U (f, P 1 ) ď sup f (A) ´ inf f (A) (xk+1 ´ xk )
0ďkďn´1 with
P1 X[xk ,xk+1 ]‰H
( ) ε
ă 2m sup f (A) ´ inf f (A) δ ď .
ht

2
ε
On the other hand, the inequality U (f, P1 ) ´ L(f, P1 ) ă implies that
ig

2
ε
U (f, P1 ) ´ I ă
r

.
2
py

As a consequence,
Co

U (f, P) ´ I ď U (f, P) ´ I + U (f, P1 ) ´ U (f, P 1 ) ă ε .

Therefore, for any sample set tξ1 , ¨ ¨ ¨ , ξn u with respect to P,

n´1
ÿ
fs(ξk+1 )(xk+1 ´ xk ) ď U (f, P) ă I + ε .
k=0

Similar argument can be used to show that

n´1
ÿ
fs(ξk+1 )(xk+1 ´ xk ) ě L(f, P) ą I ´ ε ;
k=0

thus (4.7.6) is established. ˝

§4.8 Exercises 145

Theorem 4.95 (Change of Variable Formula). Let g : [a, b] Ñ R be a one-to-one continu-

ously differentiable function, and f : g([a, b]) Ñ R be Riemann integrable. Then (f ˝ g)g 1 is
also Riemann integrable, and
ż żb
f (y) dy = f (g(x))|g 1 (x)| dx .
g([a,b]) a

Proof. We only prove the case that f is continuous on g([a, b]), and the general case is
covered by Theorem 8.65 (which will be proved in detail).
W.L.O.G. we can assume that g 1 (x) ě 0 for all x P [a, b] so that g([a, b]) = [g(a), g(b)].

ed
Let F be an anti-derivative of f . Then F is differentiable, and the chain rule implies that
d
(F ˝ g)(x) = (F 1 ˝ g)(x)g 1 (x) = (f ˝ g)(x)g 1 (x) .

ct
dx
Therefore, the fundamental theorem of Calculus implies that

te
ż ż g(b) żb
d
f (y)dy = f (y)dy = F (g(b)) ´ F (g(a)) =
ro (F ˝ g)(x)dx
g([a,b]) g(a) a dx
żb
P
= (f ˝ g)(x)g 1 (x)dx . ˝
a
ht

4.8 Exercises
ig

§4.1 Continuity
Started from this section, for all n P N Rn always denotes the normed space (Rn , } ¨ }2 ).
r
py

Problem 4.1. Use whatever methods you know to find the following limits:
1 (? ? )
1. lim+ (1 + sin 2x) x ; 2. lim 1 + x + x2 ´ 1 ´ x + x2 ;
Co

xÑ0 xÑ´8

πx (π x )
3. lim (2 ´ x)sec 2 ; 4. lim x ´ sin´1 ? ;
xÑ1 xÑ8 2 x2 + 1
( ( x )x ) ( ax ´ 1 ) x1
5. lim x e´1 ´ ; 6. lim , where a ą 0 and a ‰ 1.
xÑ8 x+1 xÑ8 x(a ´ 1)
Problem 4.2. Complete the following.

1. Find a function f : R2 Ñ R such that

lim lim f (x, y) and lim lim f (x, y)

xÑ0 yÑ0 yÑ0 xÑ0

exist but are not equal.

146 CHAPTER 4. Continuous Maps

2. Find a function f : R2 Ñ R such that the two limits above exist and are equal but f
is not continuous.

3. Find a function f : R2 Ñ R that is continuous on every line through the origin but is
not continuous.

Problem 4.3. Complete the following.

R2 Ñ R
1. Show that the projection map f : is continuous.
(x, y) ÞÑ x

ed
2. Show that if U Ď R is open, then A = (x, y) P R2 ˇ x P U is open.
␣ ˇ (

ct
3. Give an example of a continuous function f : R Ñ R and an open set U Ď R such
that f (U ) is not open.

te
Problem 4.4. Show that f : A Ñ Rm , where A Ď Rn , is continuous if and only if for every
ro
B Ď A,
P
f (cl(B) X A) Ď cl(f (B)) .

Problem 4.5. Let } ¨ } be a norm on Rn , and f : Rn Ñ R be defined by f (x) = }x}. Show

ht

that f is continuous on (Rn , } ¨ }2 ).

Hint: Show that |f (x) ´ f (y)| ď C}x ´ y}2 for some fixed constant C ą 0.
r ig

Problem 4.6. Let T : Rn Ñ Rm satisfy T (x + y) = T (x) + T (y) for all x, y P Rn .

py

1. Show that T (rx) = rT (x) for all r P Q and x P Rn .

Co

2. Suppose that T is continuous on Rn . Show that T is linear; that is, T (cx + y) =

cT (x) + T (y) for all c P R, x, y P Rn .

3. Suppose that T is continuous at some point x0 in Rn . Show that T is continuous on

Rn .

4. Suppose that T is bounded on some open subset of Rn . Show that T is continuous on

Rn .

5. Suppose that T is bounded from above (or below) on some open subset of Rn . Show
that T is continuous on Rn .
§4.8 Exercises 147

6. Construct a T : R Ñ R which is discontinuous at every point of R, but T (x + y) =

T (x) + T (y) for all x, y P R.

Problem 4.7. Let (M, d) be a metric space, A Ď M , and f : A Ñ R. For a P A1 , define

␣ ˇ (
lim inf f (x) = lim+ inf f (x) ˇ x P D(a, r) X Aztau ,
xÑa rÑ0
␣ ˇ (
lim sup f (x) = lim+ sup f (x) ˇ x P D(a, r) X Aztau .
xÑa rÑ0

Complete the following.

ed
1. Show that both lim inf f (x) and lim sup f (x) exist (which may be ˘8), and
xÑa xÑa

ct
lim inf f (x) ď lim sup f (x) .
xÑa xÑa

te
Furthermore, there exist sequences txn u8
n=1 , tyn un=1 Ď Aztau such that txn un=1 and
8 8

8
tyn un=1 both converge to a, and ro
lim f (xn ) = lim inf f (x) and lim f (yn ) = lim sup f (x) .
P
nÑ8 xÑa nÑ8 xÑa

n=1 Ď Aztau be a convergent sequence with limit a. Show that

2. Let txn u8
ht

lim inf f (x) ď lim inf f (xn ) ď lim sup f (yn ) ď lim sup f (x) .
ig

xÑa nÑ8 nÑ8 xÑa

3. Show that lim f (x) = ℓ if and only if

r

xÑa
py

lim inf f (x) = lim sup f (x) = ℓ .

xÑa xÑa
Co

4. Show that lim inf f (x) = ℓ P R if and only if the following two conditions hold:
xÑa

(a) for all ε ą 0, there exists δ ą 0 such that ℓ ´ ε ă f (x) for all x P D(a, δ) X Aztau;
(b) for all ε ą 0 and δ ą 0, there exists x P D(a, δ) X Aztau such that f (x) ă ℓ + ε.

Formulate a similar criterion for limsup and for the case that ℓ = ˘8.

5. Compute the liminf and limsup of the following functions at any point of R.
$
& 0 if x P QA ,
(a) f (x) = 1 q
% if x = with (p, q) = 1, q ą 0, p ‰ 0 .
p p
148 CHAPTER 4. Continuous Maps

x if x P Q ,
"
(b) f (x) =
´x if x P QA .
Problem 4.8. Let (M, d) be a metric space, and A Ď M . A function f : A Ñ R is called
lim inf f (x) ě f (a) ,
lower semi-continuous xÑa
at a P A if and is called lower/upper
upper semi-continuous lim sup f (x) ď f (a) ,
xÑa
semi-continuous on A if f is lower/uppser semi-continuous at a for all a P A.

1. Show that if f : A Ñ R is lower semi-continuous on A, then f ´1 ((´8, r]) is closed

relative to A. Also show that if f : A Ñ R is upper semi-continuous on A, then

ed
f ´1 ([r, 8)) is closed relative to A.

ct
2. Show that f is lower semi-continuous at a if and only if for all convergent sequences
n=1 Ď A and trn un=1 Ď R satisfying f (xn ) ď rn for all n P N, we have
txn u8 8

te
( )
f lim xn ď lim rn .
ro
nÑ8 nÑ8

3. Let tfα uαPI be a family of lower semi-continuous functions on A. Prove that f (x) =
P
sup fα (x) is lower semi-continuous on A.
αPI
ht

4. Let f : A Ñ R be given. Define

ig

f ˚ (x) = lim sup f (y) and f˚ (x) = lim inf f (y) .

yÑx yÑx
r

Show that f ˚ is upper semi-continuous and f˚ is lower semi-continuous, and f˚ (x) ď

py

f (x) ď f ˚ (x) for all x P A. Moreover, if g is a lowe semi-continuous function on A

such that g(x) ď f (x) for all x P A, then g ď f˚ .
Co

§4.2 Operations on Continuous Maps

Problem 4.9.

Problem 4.10.

§4.3 Images of Compact Sets under Continuous Maps

Problem 4.11. Complete the following.

1. Show that if f : Rn Ñ Rm is continuous, and B Ď Rn is bounded, then f (B) is

bounded.
§4.8 Exercises 149

2. If f : R Ñ R is continuous and K Ď R is compact, is f ´1 (K) necessarily compact?

3. If f : R Ñ R is continuous and C Ď R is connected, is f ´1 (C) necessarily connected?

Problem 4.12. Consider a compact set K Ď Rn and let f : K Ñ Rm be continuous and

one-to-one. Show that the inverse function f ´1 : f (K) Ñ K is continuous. How about if K
is not compact but connected?

Problem 4.13. Let (M, d) be a metric space, K Ď M be compact, and f : K Ñ R be lower

semi-continuous (see Problem 4.8 for the definition). Show that f attains its minimum on

ed
K.

ct
§4.4 Images of Connected and Path Connected Sets under Continuous Maps

te
Problem 4.14. Let D Ď Rn be an open connected set, where n ą 1. If a, b and c are any
three points in D, show that there is a path in G which connects a and b without passing
ro
through c. In particular, this shows that D is path connected and D is not homeomorphic
to any subset of R.
P
Problem 4.15.
ht

§4.5 Uniform Continuity

ig

Problem 4.16. Check if the following functions on uniformly continuous.

r
py

1. f : (0, 8) Ñ R defined by f (x) = sin log x.

1
Co

2. f : (0, 1) Ñ R defined by f (x) = x sin .

x
?
3. f : (0, 8) Ñ R defined by f (x) = x.

4. f : R Ñ R defined by f (x) = cos(x2 ).

5. f : R Ñ R defined by f (x) = cos3 x.

6. f : R Ñ R defined by f (x) = x sin x.

sin(xa )
Problem 4.17. Find all positive numbers a and b such that the function f (x) = is
1 + xb
uniformly continuous on [0, 8).
150 CHAPTER 4. Continuous Maps

Problem 4.18. Find all positive numbers a and b such that the function f (x, y) = |x|a |y|b
is uniformly continuous on R2 .

Problem 4.19. Let f : Rn Ñ Rm be continuous, and lim f (x) = b exists for some b P Rm .
|x|Ñ8
Show that f is uniformly continuous on Rn .

Problem 4.20. Suppose that f : Rn Ñ Rm is uniformly continuous. Show that there exists
a ą 0 and b ą 0 such that }f (x)}Rm ď a}x}Rn + b.
q(x)
Problem 4.21. Let f (x) = be a rational function define on R, where p and q are two

ed
p(x)
polynomials. Show that f is uniformly continuous on R if and only if the degree of q is not

ct
more than the degree of p plus 1.

Problem 4.22. Suppose that f : R Ñ R is a continuous periodic function; that is, D p ą 0

te
such that f (x + p) = f (x) for all x P R (and f is continuous). Show that f is uniformly
ro
continuous on R.
P
Problem 4.23. Let (a, b) Ď R be an open interval, and f : (a, b) Ñ Rm be a function.
Show that the following three statements are equivalent.
ht

1. f is uniformly continuous on (a, b).

ig

2. f is continuous on (a, b), and both limits lim+ f (x) and lim´ f (x) exist.
r

xÑa xÑb
py

ˇ f (x) ´ f (y) ˇ
ˇ ˇ ˇ ˇ
3. For all ε ą 0, there exists N ą 0 such that f (x) ´ f (y) ă ε whenever ˇ
ˇ ˇ ˇą
x´y
N.
Co

Problem 4.24. Suppose that f : [a, b] Ñ R is Hölder continuous with exponent α;

that is, there exist M ą 0 and α P (0, 1] such that

|f (x1 ) ´ f (x2 )| ď M |x1 ´ x2 |α @ x1 , x2 P [a, b] .

Show that f is uniformly continuous on [a, b]. Show that f : [0, 8) Ñ R defined by
? 1
f (x) = x is Hölder continuous with exponent .
2

Problem 4.25. A function f : A ˆ B Ñ Rm , where A Ď R and B Ď Rp , is said to be

separately continuous if for each x0 P A, the map g(y) = f (x0 , y) is continuous and for
§4.8 Exercises 151

y0 P B, h(x) = f (x, y0 ) is continuous. f is said to be continuous on A uniformly with

respect to B if
› ›
@ ε ą 0, D δ ą 0 Q ›f (x, y) ´ f (x0 , y)›2 ă ε whenever }x ´ x0 }2 ă δ and y P B .

Show that if f is separately continuous and is continuous on A uniformly with respect to

B, then f is continuous on A ˆ B.

Problem 4.26. Let (M, d) be a metric space, A Ď M , and f, g : A Ñ R be uniformly

continuous on A. Show that if f and g are bounded, then f g is uniformly continuous on A.

ed
Does the conclusion still hold if f or g is not bounded?

ct
§4.6 Differentiation of Functions of One Variable

te
Problem 4.27. Show that f : (a, b) Ñ R is differentiable at x0 P (a, b) if and only if there
exists m P R, denoted by f 1 (x0 ), such that ro
ˇ ˇ
@ ε ą 0, D δ ą 0 Q ˇf (x) ´ f (x0 ) ´ f 1 (x0 )(x ´ x0 )ˇ ď ε|x ´ x0 | whenever |x ´ x0 | ă δ .
P
d
Problem 4.28. Suppose that f, g : R Ñ R are differentiable, and f ě 0. Find f (x)g(x) .
dx
ht

Problem 4.29. Suppose α and β are real numbers, β ą 0 and f : [´1, 1] Ñ R is defined
ig

by
xα sin(x´β ) if x ‰ 0 ,
"
f (x) =
r

0 if x = 0 .
py

Prove the following statements.

Co

1. f is continuous if and only if α ą 0.

2. f 1 (0) exists if and only if α ą 1.

3. f 1 is bounded if and only if α ě 1 + β.

4. f 1 is continuous if and only if α ą 1 + β.

5. f 2 (0) exists if and only if α ą 2 + β.

6. f 2 is bounded if and only if α ě 2 + 2β.

7. f 2 is continuous if and only if α ą 2 + 2β.

152 CHAPTER 4. Continuous Maps

Problem 4.30 (The inverse statement of the chain rule). Let f : (a, b) Ñ R be continuous
and g : (c, d) Ñ R be differentiable at y0 = f (x0 ) P (c, d). Show that if (g˝f ) is differentiable
at x0 and g 1 (y0 ) ‰ 0, then f is differentiable at x0 .

Problem 4.31. Let f : R Ñ R be a polynomial, and f has a double root at a and b. Show
that f 1 (x) has at least three roots in [a, b].

Problem 4.32. Let f : R Ñ R be differentiable. Assume that for all x P R, 0 ď f 1 (x) ď

f (x). Show that g(x) = e´x f (x) is decreasing. If f vanishes at some point, conclude that f

ed
is zero.

Problem 4.33. Let f : R Ñ R be twice differentiable. Suppose that f (x + h) ´ f (x) =

ct
hf 1 (x + θh) for all x, h P R, where θ is independent of h. Show that f is a quadratic

te
polynomial.
ro
Problem 4.34. Let f be a differentiable function defined on some interval I of R. Prove
that f 1 maps connected subsets of I into connected set; that is, f 1 has the intermediate
P
value property.
ht

Problem 4.35. Let f : R Ñ R be a polynomial, and f has a double root at a and b. Show
that f 1 (x) has at least three roots in [a, b].
ig

Problem 4.36. Let f : [´1, 1] Ñ R be a function such that x2 + f (x)2 = 1 for all |x| ď 1.
r

␣ ˇ (
Define C = x ˇ |x| ď 1, f is continuous at x . Show that C contains at least 2 points and
py

C X (´1, 1) is an open set. Hence if f is continuous at more than 2 points, it is continuous

at uncountably many points.
Co

Problem 4.37. Let f, g : R Ñ R be differentiable functions. Suppose that lim f (x) =

xÑ8
f 1 (x)
lim g(x) = 0, g 1 (x) ‰ 0 for all x P R, and the limit lim 1 exists. Show that the limit
xÑ8 xÑ8 g (x)
f (x)
lim also exists, and
xÑ8 g(x)
f (x) f 1 (x)
lim = lim 1 .
xÑ8 g(x) xÑ8 g (x)

Problem 4.38. Let f, g : (a, b) Ñ R be differentiable functions. Show that if lim+ f (x) =
xÑa
f 1 (x) f (x)
lim+ g(x) = 8, g 1 (x) ‰ 0 for all x P (a, b), and the limit lim+ 1 exists, then lim+
xÑa xÑa g (x) xÑa g(x)
§4.8 Exercises 153

exists and
f (x) f 1 (x)
lim+ = lim+ 1 . (‹)
xÑa g(x) xÑa g (x)
f 1 (x)
Hint: Let L = lim+ and ϵ ą 0 be given. Choose c P (a, b) such that
xÑa g 1 (x)
ˇ f 1 (x) ˇ ϵ
´ Lˇ ă @a ă x ă c.
ˇ ˇ
ˇ 1
g (x) 2

Then for a ă x ă c, the Cauchy mean value theorem implies that for some ξ P (x, c) such
that

ed
f (x) ´ f (c) f 1 (ξ)
= 1 .
g(x) ´ g(c) g (ξ)

ct
Show that there exists δ ą 0 such that a + δ ă c and

te
ˇ f (x) ´ f (c) f (x) ˇ ϵ
´ ˇă @ x P (a, a + δ)
ˇ ˇ
g(x) ´ g(c) g(x) 2
ˇ

and then conclude (‹).

P ro
Problem 4.39. Let f : (a, b) Ñ R be k-times differentiable, and c P (a, b). Let hk : (a, b) Ñ
R be given by
ht

k
ÿ f (j) (c)
hk (x) = f (x) ´ (x ´ c)j .
j!
ig

j=0

hk (x)
Show that lim = 0.
r

xÑc (x ´ c)k
py

Problem 4.40. Two metric spaces (M, d) and (N, ρ) are called homeomorphics if there
exists a continous map f : M Ñ N , called a homeomorphism between M and N , such
Co

that f is one-to-one and onto, and its inverse f ´1 is also continuous. Homeomorphic metric
spaces have the same topological properties. In the following problems, (M, d) and (N, ρ)
are two metric spaces.

1. Suppose that M is compact, and f : M Ñ N is one-to-one and onto. Show that f is

a homeomorphism between M and N .

2. Suppose that f is a homeomorphism between M and N . Show that the restriction of

f to any subset A Ď M establishes a homeomorphism between A and f (A).

3. Determine which of the following pairs of metric spaces is homeomorphic.

154 CHAPTER 4. Continuous Maps

(a) M = (a, b] Ď R and N = R.

(b) M is an open ball in Rn and N = Rn .
(c) M = R and N = Rn .
(d) M = [0, 1] ˆ [0, 1] Ď R2 and N = [0, 1] Ď R.
(e) M = (x, y) P R2 ˇ x2 + y 2 = 1 and N = [0, 1] Ď R.
␣ ˇ (

(f) M = (x, y) P R2 ˇ x2 + y 2 = 1 and N = (x, y) P R2 ˇ x2 + xy + y 2 = 1 .

␣ ˇ ( ␣ ˇ (

(g) M = R2 and N = R3 .

ed
4. Let I Ď R be an interval and f : I Ñ R be a one-to-one continuous function. Show

ct
that f must be strictly monotonic in I and f is a homeomorphism between I and
f (I).

te
If I Ď Rn for n ą 1 and f : I Ñ Rn is continuous and one-to-one, can we still assert
that f is homeomorphism between I and f (I)?
P ro
§4.7 Integration of Functions of One Variable

Problem 4.41. Let f : [a, b] Ñ R be a bounded function, and Pn denote the division of
ht

[a, b] into 2n equal sub-intervals. Show that f is Riemann integrable over [a, b] if and only if
ig

lim U (f, Pn ) = lim L(f, Pn ) .

r

nÑ8 nÑ8
py

Problem 4.42. Let f, g : [a, b] Ñ R be functions, where g is continuous, and f be non-

negative, bounded, Riemann integrable over [a, b]. Show that
Co

1. f g is Riemann integrable.

2. D x0 P (a, b) such that

żb żb
f (x)g(x)dx = g(x0 ) f (x)dx .
a a

Problem 4.43. Let f : [a, b] Ñ R be differentiable and assume that f 1 is Riemann inte-
żb
grable. Prove that f 1 (x) dx = f (b) ´ f (a).
a
Hint: Use the Mean Value Theorem.
§4.8 Exercises 155

Problem 4.44. Suppose that f : [a, b] Ñ R is Riemann integrable, m ď f (x) ď M for all
x P [a, b], and φ : [m, M ] Ñ R is continuous. Show that φ ˝ f is Riemann integrable on [a, b].

Problem 4.45 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.

1. Let f : R2 zt(0, 0)u Ñ R satisfy lim f (x, axn ) = 0 for all a P R, n P N and lim f (0, y) =
xÑ0 yÑ0
0. Then lim f (x, y) = 0.
(x,y)Ñ(0,0)

2. There exists a function f : R Ñ R which is continuous only at three points of R.

ed
3. Let f : R Ñ R. Then f is continuous on R if and only if its graph (x, f (x)) ˇ x P R
␣ ˇ (

ct
is closed in R2 .

te
4. Let I1 and I2 be open intervals in R. Then f : I1 Ñ I2 is a diffeomorphism if and only
if f is differentiable and f 1 (x) ‰ 0 for all x P I1 .
ro
5. Let f : [a, b] Ñ R be a function. If f 2 is Riemann integrable, then f is Riemann
P
integrable.
?
6. Let f : [a, b] Ñ R be a function. If f is Riemann integrable, then
ht

3
f is Riemann
integrable.
ig

1
7. Let f (x) = sin be defined on (0, 1]. Then no matter how we define f (0), f is always
x
r

Riemann integrable on [0, 1].

py
Co