Spaces of Continuous Functions as examples - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


1 The space of continuous functions
Whileyouhavehadratherabstractde?nitionsofsuchconceptsasmetricspacesand
normed vector spaces, most of 1530, and also 1540, are about the spaces R
n
. This
is what is sometimes called ?classical analysis?, about ?nite dimensional spaces,
and provides the essential background to graduate analysis courses. More and
more, however, students are being introduced to some in?nite dimensional spaces
earlier. This section is about one of the most important of these spaces, the space
of continuous functions from some subset A of a metric space M to some normed
vector space N: The text gives a careful de?nition, calling the space C(A;N).
The simplest case is when M =R (=R
1
). Let A be a subset of R. Then let
C(A;R) =ff :A!R j f is continuousg:
In the most common applications A is a compact interval. Thus C([0;1];R) is the
space of all continuous f : [0;1]!R:
When A  R and N = R; C(A;R) is often shortened to C(A): However any
carefulwritermakesitclearwhatismeantbythisnotation. Moregenerally,A could
be a subset of any metric space M; andf could mapM toR or toR
m
:
De?nition 1 If AM, then
C(A;R
m
) =ff :A!R
m
j f is continuousg:
It is easy to turn C(A;R
m
) into a vector space, by de?ning addition and scalar
multiplication in the usual way: (f +g)(x) := f (x) +g(x), (cf)(x) = c(f (x)):
You probably ran into this even in linear algebra. (I use := to denote an equation
which de?nes the quantity on the left.)
Page 2


1 The space of continuous functions
Whileyouhavehadratherabstractde?nitionsofsuchconceptsasmetricspacesand
normed vector spaces, most of 1530, and also 1540, are about the spaces R
n
. This
is what is sometimes called ?classical analysis?, about ?nite dimensional spaces,
and provides the essential background to graduate analysis courses. More and
more, however, students are being introduced to some in?nite dimensional spaces
earlier. This section is about one of the most important of these spaces, the space
of continuous functions from some subset A of a metric space M to some normed
vector space N: The text gives a careful de?nition, calling the space C(A;N).
The simplest case is when M =R (=R
1
). Let A be a subset of R. Then let
C(A;R) =ff :A!R j f is continuousg:
In the most common applications A is a compact interval. Thus C([0;1];R) is the
space of all continuous f : [0;1]!R:
When A  R and N = R; C(A;R) is often shortened to C(A): However any
carefulwritermakesitclearwhatismeantbythisnotation. Moregenerally,A could
be a subset of any metric space M; andf could mapM toR or toR
m
:
De?nition 1 If AM, then
C(A;R
m
) =ff :A!R
m
j f is continuousg:
It is easy to turn C(A;R
m
) into a vector space, by de?ning addition and scalar
multiplication in the usual way: (f +g)(x) := f (x) +g(x), (cf)(x) = c(f (x)):
You probably ran into this even in linear algebra. (I use := to denote an equation
which de?nes the quantity on the left.)
However it is also very useful to de?ne a norm, so that C(A;R
m
) is a normed
vector space. This is a little trickier. We shall only do so when A is a compact
set. In that case, Theorem 4.2.2 (in the text) implies that each f 2 C(A;R
m
) is
bounded. (It is a good thought exercise to look at Theorem 4.2.2, observe that the
word ?bounded?does not appear, and explain to yourself why it implies that in the
case we are considering, f is bounded.) This allows us to de?ne a norm. With an
eye toward the application we will give shortly, we take m = 1:
De?nition 2 If AR is compact, and f 2C(A;R); let
jjfjj = max
x2A
jf (x)j:
Note that in this de?nition I am assuming that the ?maximum?of f is de?ned.
This is justi?ed by the Maximum-Minimum theorem, Theorem 4.4.1. If I say that
the maximum of a function f is c; this means that there is a particular x in the
domain of f such that f (x) =c; and furthermore, there is no x with f (x)>c: Be
sure you are straight on the di¤erence between a maximum and supremum.
The text gives the de?nition required for more generalA andN: As an example,
consider C(I;R) where I is the interval (1;1). In that case f : I ! R can be
continuous andyet not have amaximum. Youshouldbeabletothinkof anexample
which is a bounded function. More signi?cantly, f might not be bounded.
In that case,C(I;R) is not turned into a normed space in a natural way. (There
may be some peculiar way to do it. ) Instead, for general A; the text considers
C
b
(A;R) :=ff 2C(A;R) j f is boundedg:
Note that this is a vector space. Then for f 2C
b
(A;R) we let
jjfjj = sup
x2A
jf (x)j:
(What theorem tells us this exists?) Some homework at the end of the section is
concerned with this.
Theorem 5.5.1 implies that if A is compact then C(A;R) is a normed linear
space, as is C
b
(A;R) for general A: We will mainly be concerned with part (ii) of
this theorem, since I think that in all our examplesN will be a normed vector space.
Once we de?ne a norm, we can de?ne convergence of sequences.
De?nition 3 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if
lim
k!1
jjf
k
fjj = 0:
Page 3


1 The space of continuous functions
Whileyouhavehadratherabstractde?nitionsofsuchconceptsasmetricspacesand
normed vector spaces, most of 1530, and also 1540, are about the spaces R
n
. This
is what is sometimes called ?classical analysis?, about ?nite dimensional spaces,
and provides the essential background to graduate analysis courses. More and
more, however, students are being introduced to some in?nite dimensional spaces
earlier. This section is about one of the most important of these spaces, the space
of continuous functions from some subset A of a metric space M to some normed
vector space N: The text gives a careful de?nition, calling the space C(A;N).
The simplest case is when M =R (=R
1
). Let A be a subset of R. Then let
C(A;R) =ff :A!R j f is continuousg:
In the most common applications A is a compact interval. Thus C([0;1];R) is the
space of all continuous f : [0;1]!R:
When A  R and N = R; C(A;R) is often shortened to C(A): However any
carefulwritermakesitclearwhatismeantbythisnotation. Moregenerally,A could
be a subset of any metric space M; andf could mapM toR or toR
m
:
De?nition 1 If AM, then
C(A;R
m
) =ff :A!R
m
j f is continuousg:
It is easy to turn C(A;R
m
) into a vector space, by de?ning addition and scalar
multiplication in the usual way: (f +g)(x) := f (x) +g(x), (cf)(x) = c(f (x)):
You probably ran into this even in linear algebra. (I use := to denote an equation
which de?nes the quantity on the left.)
However it is also very useful to de?ne a norm, so that C(A;R
m
) is a normed
vector space. This is a little trickier. We shall only do so when A is a compact
set. In that case, Theorem 4.2.2 (in the text) implies that each f 2 C(A;R
m
) is
bounded. (It is a good thought exercise to look at Theorem 4.2.2, observe that the
word ?bounded?does not appear, and explain to yourself why it implies that in the
case we are considering, f is bounded.) This allows us to de?ne a norm. With an
eye toward the application we will give shortly, we take m = 1:
De?nition 2 If AR is compact, and f 2C(A;R); let
jjfjj = max
x2A
jf (x)j:
Note that in this de?nition I am assuming that the ?maximum?of f is de?ned.
This is justi?ed by the Maximum-Minimum theorem, Theorem 4.4.1. If I say that
the maximum of a function f is c; this means that there is a particular x in the
domain of f such that f (x) =c; and furthermore, there is no x with f (x)>c: Be
sure you are straight on the di¤erence between a maximum and supremum.
The text gives the de?nition required for more generalA andN: As an example,
consider C(I;R) where I is the interval (1;1). In that case f : I ! R can be
continuous andyet not have amaximum. Youshouldbeabletothinkof anexample
which is a bounded function. More signi?cantly, f might not be bounded.
In that case,C(I;R) is not turned into a normed space in a natural way. (There
may be some peculiar way to do it. ) Instead, for general A; the text considers
C
b
(A;R) :=ff 2C(A;R) j f is boundedg:
Note that this is a vector space. Then for f 2C
b
(A;R) we let
jjfjj = sup
x2A
jf (x)j:
(What theorem tells us this exists?) Some homework at the end of the section is
concerned with this.
Theorem 5.5.1 implies that if A is compact then C(A;R) is a normed linear
space, as is C
b
(A;R) for general A: We will mainly be concerned with part (ii) of
this theorem, since I think that in all our examplesN will be a normed vector space.
Once we de?ne a norm, we can de?ne convergence of sequences.
De?nition 3 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if
lim
k!1
jjf
k
fjj = 0:
Asthetextpointsout,thefollowingassertionisthenobviousfromthede?nitions
ofconvergenceofasequenceinC
b
(A;R)anduniformconvergenceasitwasdiscussed
in section 5.1.
Proposition 4 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if and only
if
lim
k!1
f
k
(x) =f (x)
uniformly on A:
Why? and does it make any di¤erence if A is compact? If A is compact, what
is the relation between C(A;R) and C
b
(A;R)? What can you say about these two
spaces if you don? t know that A is compact (it might be, or not).
2 Examples
1. n = 1;A = [0;1]. The spaceC([0;1];R) is often denoted byC([0;1]); the image
spaceR being understood. Clearly, f (x) =x
2
+cosx+e
x
is inC(A;R); but
g(x) =

1
x
if 0<x 1
0 if x = 0
is not.
2. n = 1; A = (0;1) . Then g(x) =
1
x
for 0 < x < 1 is in C(A;R): Is g in
C
b
(A;R)?
3. n = 2; A = [0;1][0;1). Is x
y
inC
b
(A;R)? What about y
x
?
3 The Ascoli-Arzela theorem
This is one of the most important theorems in analysis. As one application, it tells
us what subsets of C(A;R) are compact.
First let? s recall how to decide if a subset S of R
n
is compact. Since R
n
is a
metric space, the Bolzano-Weierstrass theorem tells us thatS is compact if and only
if every sequence of points inS has a convergent subsequence. Then the Heine-Borel
theorem says that a set R
n
is compact if and only if it is closed and bounded.
We saw that C([0;1]) is a normed linear space. Let
S =ff 2C([0;1]) j jjfjj 1g:
Page 4


1 The space of continuous functions
Whileyouhavehadratherabstractde?nitionsofsuchconceptsasmetricspacesand
normed vector spaces, most of 1530, and also 1540, are about the spaces R
n
. This
is what is sometimes called ?classical analysis?, about ?nite dimensional spaces,
and provides the essential background to graduate analysis courses. More and
more, however, students are being introduced to some in?nite dimensional spaces
earlier. This section is about one of the most important of these spaces, the space
of continuous functions from some subset A of a metric space M to some normed
vector space N: The text gives a careful de?nition, calling the space C(A;N).
The simplest case is when M =R (=R
1
). Let A be a subset of R. Then let
C(A;R) =ff :A!R j f is continuousg:
In the most common applications A is a compact interval. Thus C([0;1];R) is the
space of all continuous f : [0;1]!R:
When A  R and N = R; C(A;R) is often shortened to C(A): However any
carefulwritermakesitclearwhatismeantbythisnotation. Moregenerally,A could
be a subset of any metric space M; andf could mapM toR or toR
m
:
De?nition 1 If AM, then
C(A;R
m
) =ff :A!R
m
j f is continuousg:
It is easy to turn C(A;R
m
) into a vector space, by de?ning addition and scalar
multiplication in the usual way: (f +g)(x) := f (x) +g(x), (cf)(x) = c(f (x)):
You probably ran into this even in linear algebra. (I use := to denote an equation
which de?nes the quantity on the left.)
However it is also very useful to de?ne a norm, so that C(A;R
m
) is a normed
vector space. This is a little trickier. We shall only do so when A is a compact
set. In that case, Theorem 4.2.2 (in the text) implies that each f 2 C(A;R
m
) is
bounded. (It is a good thought exercise to look at Theorem 4.2.2, observe that the
word ?bounded?does not appear, and explain to yourself why it implies that in the
case we are considering, f is bounded.) This allows us to de?ne a norm. With an
eye toward the application we will give shortly, we take m = 1:
De?nition 2 If AR is compact, and f 2C(A;R); let
jjfjj = max
x2A
jf (x)j:
Note that in this de?nition I am assuming that the ?maximum?of f is de?ned.
This is justi?ed by the Maximum-Minimum theorem, Theorem 4.4.1. If I say that
the maximum of a function f is c; this means that there is a particular x in the
domain of f such that f (x) =c; and furthermore, there is no x with f (x)>c: Be
sure you are straight on the di¤erence between a maximum and supremum.
The text gives the de?nition required for more generalA andN: As an example,
consider C(I;R) where I is the interval (1;1). In that case f : I ! R can be
continuous andyet not have amaximum. Youshouldbeabletothinkof anexample
which is a bounded function. More signi?cantly, f might not be bounded.
In that case,C(I;R) is not turned into a normed space in a natural way. (There
may be some peculiar way to do it. ) Instead, for general A; the text considers
C
b
(A;R) :=ff 2C(A;R) j f is boundedg:
Note that this is a vector space. Then for f 2C
b
(A;R) we let
jjfjj = sup
x2A
jf (x)j:
(What theorem tells us this exists?) Some homework at the end of the section is
concerned with this.
Theorem 5.5.1 implies that if A is compact then C(A;R) is a normed linear
space, as is C
b
(A;R) for general A: We will mainly be concerned with part (ii) of
this theorem, since I think that in all our examplesN will be a normed vector space.
Once we de?ne a norm, we can de?ne convergence of sequences.
De?nition 3 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if
lim
k!1
jjf
k
fjj = 0:
Asthetextpointsout,thefollowingassertionisthenobviousfromthede?nitions
ofconvergenceofasequenceinC
b
(A;R)anduniformconvergenceasitwasdiscussed
in section 5.1.
Proposition 4 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if and only
if
lim
k!1
f
k
(x) =f (x)
uniformly on A:
Why? and does it make any di¤erence if A is compact? If A is compact, what
is the relation between C(A;R) and C
b
(A;R)? What can you say about these two
spaces if you don? t know that A is compact (it might be, or not).
2 Examples
1. n = 1;A = [0;1]. The spaceC([0;1];R) is often denoted byC([0;1]); the image
spaceR being understood. Clearly, f (x) =x
2
+cosx+e
x
is inC(A;R); but
g(x) =

1
x
if 0<x 1
0 if x = 0
is not.
2. n = 1; A = (0;1) . Then g(x) =
1
x
for 0 < x < 1 is in C(A;R): Is g in
C
b
(A;R)?
3. n = 2; A = [0;1][0;1). Is x
y
inC
b
(A;R)? What about y
x
?
3 The Ascoli-Arzela theorem
This is one of the most important theorems in analysis. As one application, it tells
us what subsets of C(A;R) are compact.
First let? s recall how to decide if a subset S of R
n
is compact. Since R
n
is a
metric space, the Bolzano-Weierstrass theorem tells us thatS is compact if and only
if every sequence of points inS has a convergent subsequence. Then the Heine-Borel
theorem says that a set R
n
is compact if and only if it is closed and bounded.
We saw that C([0;1]) is a normed linear space. Let
S =ff 2C([0;1]) j jjfjj 1g:
ObviouslyS is bounded. To see thatS is closed, suppose thatff
n
gS andf
n
!g
inC([0;1]): (This was de?ned in the previous section.) I claim thatjjgjj 1:
Suppose not, andjjgjj> 1: Let =jjgjj1; or
jjgjj = +1 (1)
Then there is anN such that fornN;jjf
n
gjj<

2
: In this case, by the triangle
inequality,
jjgjj =jjgf
n
+f
n
jjjjgf
n
jj+jjf
n
jj

2
+1;
which contradicts (1):
Hence,S isclosedandbounded. LetusseeifS iscompact. Considerthesequence
ff
n
g wheref
n
(x) =x
n
for 0x 1: Thenf
n
2S for eachn: Also.
lim
n!1
f
n
(x) =

0 if 0x< 1
1 if x = 1:
Hence,ff
n
g does not converge to a continuous function. Further, no subsequence of
ff
n
g converges to a continuous function. (Do you think this needs further proof? If
so, ?ll in the details!)
This shows thatS is not compact. The Heine-Borel theoremdoes not holdinthe
in?nite dimensional normed linear space C([0;1]). Some other testable condition is
needed to guarantee that a set S is compact.
De?nition 5 A sequenceff
n
g of functions inC([0;1]) is uniformly bounded if there
is an M > 0 such that jjf
n
jjM for n = 1;2;3;


:
De?nition 6 A sequence S of functions in C([0;1]) is equicontinuous if for each
"> 0 there is a  > 0 such that if x and y are in [0;1] with jxyj<, and f 2S;
then
jf (x)f (y)j<": (2)
Inotherwords,inthede?nitionofcontinuityofafunction,the doesnotdepend
onx;y orthefunctionf chosenfromS:(Comparethiswiththede?nitionofuniform
continuity of a function.)
One example of an equicontinuous set of functions is a set S of functions in
C([0;1]) such that every function f in S satis?es a Lipschitz condition in [0;1] and
the Lipschitz constantL can be chosen to be the same for everyf inS. In this case,
if"> 0 is given then choose =
"
L
and check that (2) holds: The does not depend
onf:
Page 5


1 The space of continuous functions
Whileyouhavehadratherabstractde?nitionsofsuchconceptsasmetricspacesand
normed vector spaces, most of 1530, and also 1540, are about the spaces R
n
. This
is what is sometimes called ?classical analysis?, about ?nite dimensional spaces,
and provides the essential background to graduate analysis courses. More and
more, however, students are being introduced to some in?nite dimensional spaces
earlier. This section is about one of the most important of these spaces, the space
of continuous functions from some subset A of a metric space M to some normed
vector space N: The text gives a careful de?nition, calling the space C(A;N).
The simplest case is when M =R (=R
1
). Let A be a subset of R. Then let
C(A;R) =ff :A!R j f is continuousg:
In the most common applications A is a compact interval. Thus C([0;1];R) is the
space of all continuous f : [0;1]!R:
When A  R and N = R; C(A;R) is often shortened to C(A): However any
carefulwritermakesitclearwhatismeantbythisnotation. Moregenerally,A could
be a subset of any metric space M; andf could mapM toR or toR
m
:
De?nition 1 If AM, then
C(A;R
m
) =ff :A!R
m
j f is continuousg:
It is easy to turn C(A;R
m
) into a vector space, by de?ning addition and scalar
multiplication in the usual way: (f +g)(x) := f (x) +g(x), (cf)(x) = c(f (x)):
You probably ran into this even in linear algebra. (I use := to denote an equation
which de?nes the quantity on the left.)
However it is also very useful to de?ne a norm, so that C(A;R
m
) is a normed
vector space. This is a little trickier. We shall only do so when A is a compact
set. In that case, Theorem 4.2.2 (in the text) implies that each f 2 C(A;R
m
) is
bounded. (It is a good thought exercise to look at Theorem 4.2.2, observe that the
word ?bounded?does not appear, and explain to yourself why it implies that in the
case we are considering, f is bounded.) This allows us to de?ne a norm. With an
eye toward the application we will give shortly, we take m = 1:
De?nition 2 If AR is compact, and f 2C(A;R); let
jjfjj = max
x2A
jf (x)j:
Note that in this de?nition I am assuming that the ?maximum?of f is de?ned.
This is justi?ed by the Maximum-Minimum theorem, Theorem 4.4.1. If I say that
the maximum of a function f is c; this means that there is a particular x in the
domain of f such that f (x) =c; and furthermore, there is no x with f (x)>c: Be
sure you are straight on the di¤erence between a maximum and supremum.
The text gives the de?nition required for more generalA andN: As an example,
consider C(I;R) where I is the interval (1;1). In that case f : I ! R can be
continuous andyet not have amaximum. Youshouldbeabletothinkof anexample
which is a bounded function. More signi?cantly, f might not be bounded.
In that case,C(I;R) is not turned into a normed space in a natural way. (There
may be some peculiar way to do it. ) Instead, for general A; the text considers
C
b
(A;R) :=ff 2C(A;R) j f is boundedg:
Note that this is a vector space. Then for f 2C
b
(A;R) we let
jjfjj = sup
x2A
jf (x)j:
(What theorem tells us this exists?) Some homework at the end of the section is
concerned with this.
Theorem 5.5.1 implies that if A is compact then C(A;R) is a normed linear
space, as is C
b
(A;R) for general A: We will mainly be concerned with part (ii) of
this theorem, since I think that in all our examplesN will be a normed vector space.
Once we de?ne a norm, we can de?ne convergence of sequences.
De?nition 3 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if
lim
k!1
jjf
k
fjj = 0:
Asthetextpointsout,thefollowingassertionisthenobviousfromthede?nitions
ofconvergenceofasequenceinC
b
(A;R)anduniformconvergenceasitwasdiscussed
in section 5.1.
Proposition 4 A sequence ff
k
gC
b
(A;R) converges to f 2C
b
(A;R) if and only
if
lim
k!1
f
k
(x) =f (x)
uniformly on A:
Why? and does it make any di¤erence if A is compact? If A is compact, what
is the relation between C(A;R) and C
b
(A;R)? What can you say about these two
spaces if you don? t know that A is compact (it might be, or not).
2 Examples
1. n = 1;A = [0;1]. The spaceC([0;1];R) is often denoted byC([0;1]); the image
spaceR being understood. Clearly, f (x) =x
2
+cosx+e
x
is inC(A;R); but
g(x) =

1
x
if 0<x 1
0 if x = 0
is not.
2. n = 1; A = (0;1) . Then g(x) =
1
x
for 0 < x < 1 is in C(A;R): Is g in
C
b
(A;R)?
3. n = 2; A = [0;1][0;1). Is x
y
inC
b
(A;R)? What about y
x
?
3 The Ascoli-Arzela theorem
This is one of the most important theorems in analysis. As one application, it tells
us what subsets of C(A;R) are compact.
First let? s recall how to decide if a subset S of R
n
is compact. Since R
n
is a
metric space, the Bolzano-Weierstrass theorem tells us thatS is compact if and only
if every sequence of points inS has a convergent subsequence. Then the Heine-Borel
theorem says that a set R
n
is compact if and only if it is closed and bounded.
We saw that C([0;1]) is a normed linear space. Let
S =ff 2C([0;1]) j jjfjj 1g:
ObviouslyS is bounded. To see thatS is closed, suppose thatff
n
gS andf
n
!g
inC([0;1]): (This was de?ned in the previous section.) I claim thatjjgjj 1:
Suppose not, andjjgjj> 1: Let =jjgjj1; or
jjgjj = +1 (1)
Then there is anN such that fornN;jjf
n
gjj<

2
: In this case, by the triangle
inequality,
jjgjj =jjgf
n
+f
n
jjjjgf
n
jj+jjf
n
jj

2
+1;
which contradicts (1):
Hence,S isclosedandbounded. LetusseeifS iscompact. Considerthesequence
ff
n
g wheref
n
(x) =x
n
for 0x 1: Thenf
n
2S for eachn: Also.
lim
n!1
f
n
(x) =

0 if 0x< 1
1 if x = 1:
Hence,ff
n
g does not converge to a continuous function. Further, no subsequence of
ff
n
g converges to a continuous function. (Do you think this needs further proof? If
so, ?ll in the details!)
This shows thatS is not compact. The Heine-Borel theoremdoes not holdinthe
in?nite dimensional normed linear space C([0;1]). Some other testable condition is
needed to guarantee that a set S is compact.
De?nition 5 A sequenceff
n
g of functions inC([0;1]) is uniformly bounded if there
is an M > 0 such that jjf
n
jjM for n = 1;2;3;


:
De?nition 6 A sequence S of functions in C([0;1]) is equicontinuous if for each
"> 0 there is a  > 0 such that if x and y are in [0;1] with jxyj<, and f 2S;
then
jf (x)f (y)j<": (2)
Inotherwords,inthede?nitionofcontinuityofafunction,the doesnotdepend
onx;y orthefunctionf chosenfromS:(Comparethiswiththede?nitionofuniform
continuity of a function.)
One example of an equicontinuous set of functions is a set S of functions in
C([0;1]) such that every function f in S satis?es a Lipschitz condition in [0;1] and
the Lipschitz constantL can be chosen to be the same for everyf inS. In this case,
if"> 0 is given then choose =
"
L
and check that (2) holds: The does not depend
onf:
Forinstance, wecanletS bethesequenceff
n
goffunctionsgivenbyf
n
(x) =
x
n
n
:
Then f
0
n
(x) = x
n1
; so jf
0
n
(x)j  1 for every n and every x 2 [0;1]: By the mean
value theorem we can choose L = 1:
On the other hand, we will show that fx
n
g is a uniformly bounded sequence
(on [0;1]), but not an equicontinuous one. Uniform boundedness is obvious. For
equicontinuity, suppose that " =
1
2
and that there is a  such that (2) holds for
every n and every x and y in [0;1] with jxyj < . Choose y = 1 and x = 1

2
:
Since lim
n!1

1

2


n
= 0; we can choose n so large that f
n
(1)f
n

1

2


=
1

1

2


n
>
1
2
; contradicting (2). Hence, S is not equicontinuous. As we saw, S
does not have a convergent subsequence.
Theorem 7 (Ascoli-Arzela) IfS =ff
n
g is a uniformly bounded and equicontinuous
sequence of functions in C([0;1]); then S has a subsequence which converges in
C([0;1]):
Remark 8 It is important that the limit function is also in C([0;1]): But this is
what we mean when we say that a subsequence ?converges in C([0;1]).?
The proof (for a more general case) is in the text, and will be given in class.
The practical application of this theorem is often easier than the example above
mightsuggest. Iff isdi¤erentiableon [0;1];andforsome;jf
0
(x)jLforeveryx2
[0;1];thenLisaLipschitzconstantforf: Thisfollowsfromthemeanvaluetheorem.
Hence, for a sequenceff
n
g, if eachf
n
is di¤erentiable on [0;1]; andjf
0
n
(x)jL for
every n and every x 2 [0;1]; then ff
n
g is equicontinuous. If in addition, there is
a K such that jjf
n
jj  K for every n; then ff
n
g is equicontinuous and uniformly
bounded, and some subsequence offf
n
g converges inC([0;1]).
Turning to compactness, if S C([0;1]); every f 2S is di¤erentiable on [0;1];
and there are numbersL andB such thatjjfjjB andjf
0
(x)jL for everyf 2S
and everyx2 [0;1]; thenS is a compact subset of C([0:1]).
As an example of a compact set we can consider
S =

g j g(x) =
Z
x
0
f (s)ds for some f 2C([0;1]) with jjfjj 1

:
You should use the Ascoli-Arzela theorem to show that S is compact.
4 Contraction mapping theorem
This is another result that can be used to show that sequences converge. It is easier
to prove than Ascoli-Arzela and gives a stronger result. However there are many