Sequences and Series Del- CSIR-NET Mathematical Science | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


1. Review
Remark 1.1. From now on, unless otherwise specied, R
n
refers to Euclidean space R
n
withn 1 a positive integer, and where we use the metric d
`
2
onR
n
. In particular,R refers
to the metric spaceR equipped with the metric d(x;y) =jxyj.
Proposition 1.2. Let (X;d) be a metric space. Let (x
(j)
)
1
j=k
be a sequence of elements ofX.
Let x;x
0
be elements of X. Assume that the sequence (x
(j)
)
1
j=k
converges to x with respect to
d. Assume also that the sequence (x
(j)
)
1
j=k
converges to x
0
with respect to d. Then x =x
0
.
Proposition 1.3. Let a < b be real numbers, and let f : [a;b]! R be a function which
is both continuous and strictly monotone increasing. Then f is a bijection from [a;b] to
[f(a);f(b)], and the inverse function f
1
: [f(a);f(b)]! [a;b] is also continuous and strictly
monotone increasing.
Theorem 1.4 (Inverse Function Theorem). LetX;Y be subsets ofR. Letf : X!Y be
bijection, so thatf
1
: Y!X is a function. Letx
0
2X andy
0
2Y such thatf(x
0
) =y
0
. If
f is dierentiable atx
0
, iff
1
is continuous aty
0
, and iff
0
(x
0
)6= 0, thenf
1
is dierentiable
at y
0
with
(f
1
)
0
(y
0
) =
1
f
0
(x
0
)
:
Date: February 14, 2015.
1
Page 2


1. Review
Remark 1.1. From now on, unless otherwise specied, R
n
refers to Euclidean space R
n
withn 1 a positive integer, and where we use the metric d
`
2
onR
n
. In particular,R refers
to the metric spaceR equipped with the metric d(x;y) =jxyj.
Proposition 1.2. Let (X;d) be a metric space. Let (x
(j)
)
1
j=k
be a sequence of elements ofX.
Let x;x
0
be elements of X. Assume that the sequence (x
(j)
)
1
j=k
converges to x with respect to
d. Assume also that the sequence (x
(j)
)
1
j=k
converges to x
0
with respect to d. Then x =x
0
.
Proposition 1.3. Let a < b be real numbers, and let f : [a;b]! R be a function which
is both continuous and strictly monotone increasing. Then f is a bijection from [a;b] to
[f(a);f(b)], and the inverse function f
1
: [f(a);f(b)]! [a;b] is also continuous and strictly
monotone increasing.
Theorem 1.4 (Inverse Function Theorem). LetX;Y be subsets ofR. Letf : X!Y be
bijection, so thatf
1
: Y!X is a function. Letx
0
2X andy
0
2Y such thatf(x
0
) =y
0
. If
f is dierentiable atx
0
, iff
1
is continuous aty
0
, and iff
0
(x
0
)6= 0, thenf
1
is dierentiable
at y
0
with
(f
1
)
0
(y
0
) =
1
f
0
(x
0
)
:
Date: February 14, 2015.
1
2. Sequences of Functions
As we have seen in analysis, it is often desirable to discuss sequences of points that
converge. Below, we will see that it is similarly desirable to discuss sequences of functions
that converge in various senses. There are many distinct ways of discussing the convergence of
sequences of functions. We will only discuss two such modes of convergence, namely pointwise
and uniform convergence. Before beginning this discussion, we discuss the limiting values
of functions between metric spaces, which should generalize our notion of limiting values of
functions on the real line.
2.1. Limiting Values of Functions.
Denition 2.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X!Y be a function, letx
0
2X be an adherent point ofE, and letL2Y . We say that
f(x) converges toL inY asx converges tox
0
inE, and we write lim
x!x
0
;x2E
f(x) =L,
if and only if, for every " > 0, there exists  = (") > 0 such that, if x2 E satises
d
X
(x;x
0
)<, then d
Y
(f(x);L)<".
Remark 2.2. So, f is continuous at x
0
if and only if
lim
x!x
0
;x2X
f(x) =f(x
0
): ()
And f is continuous on X if and only if, for all x
0
2X, () holds.
Remark 2.3. When the domain of x of the limit lim
x!x
0
;x2X
f(x) is clear, we will often
instead write lim
x!x
0
f(x).
The following equivalence is generalized from its analogue on the real line.
Proposition 2.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X! Y be a function, let x
0
2 X be an adherent point of E, and let L2 Y . Then the
following statements are equivalent.
 lim
x!x
0
;x2E
f(x) =L.
 For any sequence (x
(j)
)
1
j=1
in E which converges to x
0
with respect to the metric d
X
,
the sequence (f(x
(j)
))
1
j=1
converges to L with respect to the metric d
Y
.
Exercise 2.5. Prove Proposition 2.4.
Remark 2.6. From Propositions 2.4 and 1.2, the function f can converge to at most one
limit L as x converges to x
0
.
Remark 2.7. The notation lim
x!x
0
;x2E
f(x) implicitly refers to a convergence of the function
values f(x) in the metric space (Y;d
Y
). Strictly speaking, it would be better to write d
Y
somewhere next to the notation lim
x!x
0
;x2E
f(x). However, this omission of notation should
not cause confusion.
2.2. Pointwise Convergence and Uniform Convergence.
Denition 2.8 (Pointwise Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges pointwise to f on X if and only if, for every x2X, we have
lim
j!1
f
j
(x) =f(x):
Page 3


1. Review
Remark 1.1. From now on, unless otherwise specied, R
n
refers to Euclidean space R
n
withn 1 a positive integer, and where we use the metric d
`
2
onR
n
. In particular,R refers
to the metric spaceR equipped with the metric d(x;y) =jxyj.
Proposition 1.2. Let (X;d) be a metric space. Let (x
(j)
)
1
j=k
be a sequence of elements ofX.
Let x;x
0
be elements of X. Assume that the sequence (x
(j)
)
1
j=k
converges to x with respect to
d. Assume also that the sequence (x
(j)
)
1
j=k
converges to x
0
with respect to d. Then x =x
0
.
Proposition 1.3. Let a < b be real numbers, and let f : [a;b]! R be a function which
is both continuous and strictly monotone increasing. Then f is a bijection from [a;b] to
[f(a);f(b)], and the inverse function f
1
: [f(a);f(b)]! [a;b] is also continuous and strictly
monotone increasing.
Theorem 1.4 (Inverse Function Theorem). LetX;Y be subsets ofR. Letf : X!Y be
bijection, so thatf
1
: Y!X is a function. Letx
0
2X andy
0
2Y such thatf(x
0
) =y
0
. If
f is dierentiable atx
0
, iff
1
is continuous aty
0
, and iff
0
(x
0
)6= 0, thenf
1
is dierentiable
at y
0
with
(f
1
)
0
(y
0
) =
1
f
0
(x
0
)
:
Date: February 14, 2015.
1
2. Sequences of Functions
As we have seen in analysis, it is often desirable to discuss sequences of points that
converge. Below, we will see that it is similarly desirable to discuss sequences of functions
that converge in various senses. There are many distinct ways of discussing the convergence of
sequences of functions. We will only discuss two such modes of convergence, namely pointwise
and uniform convergence. Before beginning this discussion, we discuss the limiting values
of functions between metric spaces, which should generalize our notion of limiting values of
functions on the real line.
2.1. Limiting Values of Functions.
Denition 2.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X!Y be a function, letx
0
2X be an adherent point ofE, and letL2Y . We say that
f(x) converges toL inY asx converges tox
0
inE, and we write lim
x!x
0
;x2E
f(x) =L,
if and only if, for every " > 0, there exists  = (") > 0 such that, if x2 E satises
d
X
(x;x
0
)<, then d
Y
(f(x);L)<".
Remark 2.2. So, f is continuous at x
0
if and only if
lim
x!x
0
;x2X
f(x) =f(x
0
): ()
And f is continuous on X if and only if, for all x
0
2X, () holds.
Remark 2.3. When the domain of x of the limit lim
x!x
0
;x2X
f(x) is clear, we will often
instead write lim
x!x
0
f(x).
The following equivalence is generalized from its analogue on the real line.
Proposition 2.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X! Y be a function, let x
0
2 X be an adherent point of E, and let L2 Y . Then the
following statements are equivalent.
 lim
x!x
0
;x2E
f(x) =L.
 For any sequence (x
(j)
)
1
j=1
in E which converges to x
0
with respect to the metric d
X
,
the sequence (f(x
(j)
))
1
j=1
converges to L with respect to the metric d
Y
.
Exercise 2.5. Prove Proposition 2.4.
Remark 2.6. From Propositions 2.4 and 1.2, the function f can converge to at most one
limit L as x converges to x
0
.
Remark 2.7. The notation lim
x!x
0
;x2E
f(x) implicitly refers to a convergence of the function
values f(x) in the metric space (Y;d
Y
). Strictly speaking, it would be better to write d
Y
somewhere next to the notation lim
x!x
0
;x2E
f(x). However, this omission of notation should
not cause confusion.
2.2. Pointwise Convergence and Uniform Convergence.
Denition 2.8 (Pointwise Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges pointwise to f on X if and only if, for every x2X, we have
lim
j!1
f
j
(x) =f(x):
That is, for all x2X, we have
lim
j!1
d
Y
(f
j
(x);f(x)) = 0:
That is, for every x2X and for every "> 0, there exists J > 0 such that, for all j >J, we
have d
Y
(f
j
(x);f(x))<".
Remark 2.9. Note that, if we change the point x, then the limiting behavior of f
j
(x)
can change quite a bit. For example, let j be a positive integer, and consider the functions
f
j
: [0; 1]!R wheref
j
(x) =j for allx2 (0; 1=j), andf
j
(x) = 0 otherwise. Letf : [0; 1]!R
be the zero function. Then f
j
converges pointwise to zero, since for any x2 (0; 1], we have
f
j
(x) = 0 for all j > 1=x. (And f
j
(0) = 0 for all positive integers j.) However, given any
xed positive integer j, there exists an x such that f
j
(x) = j. Moreover,
R
1
0
f
j
= 1 for all
positive integers j, but
R
1
0
f = 0. So, we see that pointwise convergence does not preserve
the integral of a function.
Remark 2.10. Pointwise convergence also does not preserve continuity. For example, con-
sider f
j
: [0; 1]!R dened by f
j
(x) =x
j
, where j2N and x2 [0; 1]. Dene f : [0; 1]!R
so that f(1) = 1 and so that f(x) = 0 for x2 [0; 1). Then f
j
converges pointwise to f as
j!1, and each f
j
is continuous, but f is not continuous.
In summary, pointwise convergence doesn't really preserve any useful analytic quantities.
The above remarks show that some points are changing at much dierent rates than other
points as j !1. A stronger notion of convergence will then x these issues, where all
points in the domain are controlled simultaneously.
Denition 2.11 (Uniform Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges uniformly tof onX if and only if, for every"> 0, there exists
J > 0 such that, for all j >J and for all x2X we have d
Y
(f
j
(x);f(x))<".
Remark 2.12. Note that the dierence between uniform and pointwise convergence is that
we simply moved the quantier \for all x2 X" within the statement. This change means
that the integer J does not depend on x in the case of uniform convergence.
Remark 2.13. The sequences of functions from Remarks 2.9 and 2.10 do not converge
uniformly. So, pointwise convergence does not imply uniform convergence. However, uniform
convergence does imply pointwise convergence.
3. Uniform Convergence and Continuity
We saw that pointwise convergence does not preserve continuity. However, uniform con-
vergence does preserve continuity.
Theorem 3.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Let x
0
2 X. Suppose f
j
converges uniformly to f on X. Suppose that, for each j 1, we know that f
j
is continuous
at x
0
. Then f is also continuous at x
0
.
Page 4


1. Review
Remark 1.1. From now on, unless otherwise specied, R
n
refers to Euclidean space R
n
withn 1 a positive integer, and where we use the metric d
`
2
onR
n
. In particular,R refers
to the metric spaceR equipped with the metric d(x;y) =jxyj.
Proposition 1.2. Let (X;d) be a metric space. Let (x
(j)
)
1
j=k
be a sequence of elements ofX.
Let x;x
0
be elements of X. Assume that the sequence (x
(j)
)
1
j=k
converges to x with respect to
d. Assume also that the sequence (x
(j)
)
1
j=k
converges to x
0
with respect to d. Then x =x
0
.
Proposition 1.3. Let a < b be real numbers, and let f : [a;b]! R be a function which
is both continuous and strictly monotone increasing. Then f is a bijection from [a;b] to
[f(a);f(b)], and the inverse function f
1
: [f(a);f(b)]! [a;b] is also continuous and strictly
monotone increasing.
Theorem 1.4 (Inverse Function Theorem). LetX;Y be subsets ofR. Letf : X!Y be
bijection, so thatf
1
: Y!X is a function. Letx
0
2X andy
0
2Y such thatf(x
0
) =y
0
. If
f is dierentiable atx
0
, iff
1
is continuous aty
0
, and iff
0
(x
0
)6= 0, thenf
1
is dierentiable
at y
0
with
(f
1
)
0
(y
0
) =
1
f
0
(x
0
)
:
Date: February 14, 2015.
1
2. Sequences of Functions
As we have seen in analysis, it is often desirable to discuss sequences of points that
converge. Below, we will see that it is similarly desirable to discuss sequences of functions
that converge in various senses. There are many distinct ways of discussing the convergence of
sequences of functions. We will only discuss two such modes of convergence, namely pointwise
and uniform convergence. Before beginning this discussion, we discuss the limiting values
of functions between metric spaces, which should generalize our notion of limiting values of
functions on the real line.
2.1. Limiting Values of Functions.
Denition 2.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X!Y be a function, letx
0
2X be an adherent point ofE, and letL2Y . We say that
f(x) converges toL inY asx converges tox
0
inE, and we write lim
x!x
0
;x2E
f(x) =L,
if and only if, for every " > 0, there exists  = (") > 0 such that, if x2 E satises
d
X
(x;x
0
)<, then d
Y
(f(x);L)<".
Remark 2.2. So, f is continuous at x
0
if and only if
lim
x!x
0
;x2X
f(x) =f(x
0
): ()
And f is continuous on X if and only if, for all x
0
2X, () holds.
Remark 2.3. When the domain of x of the limit lim
x!x
0
;x2X
f(x) is clear, we will often
instead write lim
x!x
0
f(x).
The following equivalence is generalized from its analogue on the real line.
Proposition 2.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X! Y be a function, let x
0
2 X be an adherent point of E, and let L2 Y . Then the
following statements are equivalent.
 lim
x!x
0
;x2E
f(x) =L.
 For any sequence (x
(j)
)
1
j=1
in E which converges to x
0
with respect to the metric d
X
,
the sequence (f(x
(j)
))
1
j=1
converges to L with respect to the metric d
Y
.
Exercise 2.5. Prove Proposition 2.4.
Remark 2.6. From Propositions 2.4 and 1.2, the function f can converge to at most one
limit L as x converges to x
0
.
Remark 2.7. The notation lim
x!x
0
;x2E
f(x) implicitly refers to a convergence of the function
values f(x) in the metric space (Y;d
Y
). Strictly speaking, it would be better to write d
Y
somewhere next to the notation lim
x!x
0
;x2E
f(x). However, this omission of notation should
not cause confusion.
2.2. Pointwise Convergence and Uniform Convergence.
Denition 2.8 (Pointwise Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges pointwise to f on X if and only if, for every x2X, we have
lim
j!1
f
j
(x) =f(x):
That is, for all x2X, we have
lim
j!1
d
Y
(f
j
(x);f(x)) = 0:
That is, for every x2X and for every "> 0, there exists J > 0 such that, for all j >J, we
have d
Y
(f
j
(x);f(x))<".
Remark 2.9. Note that, if we change the point x, then the limiting behavior of f
j
(x)
can change quite a bit. For example, let j be a positive integer, and consider the functions
f
j
: [0; 1]!R wheref
j
(x) =j for allx2 (0; 1=j), andf
j
(x) = 0 otherwise. Letf : [0; 1]!R
be the zero function. Then f
j
converges pointwise to zero, since for any x2 (0; 1], we have
f
j
(x) = 0 for all j > 1=x. (And f
j
(0) = 0 for all positive integers j.) However, given any
xed positive integer j, there exists an x such that f
j
(x) = j. Moreover,
R
1
0
f
j
= 1 for all
positive integers j, but
R
1
0
f = 0. So, we see that pointwise convergence does not preserve
the integral of a function.
Remark 2.10. Pointwise convergence also does not preserve continuity. For example, con-
sider f
j
: [0; 1]!R dened by f
j
(x) =x
j
, where j2N and x2 [0; 1]. Dene f : [0; 1]!R
so that f(1) = 1 and so that f(x) = 0 for x2 [0; 1). Then f
j
converges pointwise to f as
j!1, and each f
j
is continuous, but f is not continuous.
In summary, pointwise convergence doesn't really preserve any useful analytic quantities.
The above remarks show that some points are changing at much dierent rates than other
points as j !1. A stronger notion of convergence will then x these issues, where all
points in the domain are controlled simultaneously.
Denition 2.11 (Uniform Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges uniformly tof onX if and only if, for every"> 0, there exists
J > 0 such that, for all j >J and for all x2X we have d
Y
(f
j
(x);f(x))<".
Remark 2.12. Note that the dierence between uniform and pointwise convergence is that
we simply moved the quantier \for all x2 X" within the statement. This change means
that the integer J does not depend on x in the case of uniform convergence.
Remark 2.13. The sequences of functions from Remarks 2.9 and 2.10 do not converge
uniformly. So, pointwise convergence does not imply uniform convergence. However, uniform
convergence does imply pointwise convergence.
3. Uniform Convergence and Continuity
We saw that pointwise convergence does not preserve continuity. However, uniform con-
vergence does preserve continuity.
Theorem 3.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Let x
0
2 X. Suppose f
j
converges uniformly to f on X. Suppose that, for each j 1, we know that f
j
is continuous
at x
0
. Then f is also continuous at x
0
.
Exercise 3.2. Prove Theorem 3.1. Hint: it is probably easiest to use the " denition of
continuity. Once you do this, you may require the triangle inequality in the form
d
Y
(f(x);f(x
0
))d
Y
(f(x);f
j
(x)) +d
Y
(f
j
(x);f
j
(x
0
)) +d
Y
(f
j
(x
0
);f(x
0
)):
Corollary 3.3. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Suppose (f
j
)
1
j=1
converges
uniformly to f on X. Suppose that, for each j 1, we know that f
j
is continuous on X.
Then f is also continuous on X.
Uniform limits of bounded functions are also bounded. Recall that a function f : X!Y
between metric spaces (X;d
X
) and (Y;d
Y
) is bounded if and only if there exists a radius
R> 0 and a point y
0
2Y such that f(x)2B
(Y;d
Y
)
(y
0
;R) for all x2X.
Proposition 3.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Suppose (f
j
)
1
j=1
converges
uniformly to f on X. Suppose also that, for each j 1, we know that f
j
is bounded. Then
f is also bounded.
Exercise 3.5. Prove Proposition 3.4.
3.1. The Metric of Uniform Convergence. We will now see one advantage to our ab-
stract approach to analysis on metric spaces. We can in fact talk about uniform convergence
in terms of a metric on a space of functions, as follows.
Denition 3.6. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let B(X;Y ) denote the set
of functions f : X ! Y that are bounded. Let f;g 2 B(X;Y ). We dene the metric
d
1
: B(X;Y )B(X;Y )! [0;1) by
d
1
(f;g) := sup
x2X
d
Y
(f(x);g(x)):
This metric is known as the sup norm metric or the L
1
metric. We also use d
B(X;Y)
as
a synonym for d
1
. Note that d
1
(f;g)<1 since f;g are assumed to be bounded.
Exercise 3.7. Show that the space (B(X;Y );d
1
) is a metric space.
Example 3.8. LetX = [0; 1] and letY =R. Consider the functionsf(x) =x andg(x) = 2x
where x2 [0; 1]. Then f;g are bounded, and
d
1
(f;g) = sup
x2[0;1]
jx 2xj = sup
x2[0;1]
jxj = 1:
Here is our promised characterization of uniform convergence in terms of the metric d
1
.
Proposition 3.9. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions inB(X;Y ). Letf2B(X;Y ). Then (f
j
)
1
j=1
converges uniformly to f onX if and
only if (f
j
)
1
j=1
converges to f in the metric d
B(X;Y)
.
Exercise 3.10. Prove Proposition 3.9.
Denition 3.11. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Dene the set of bounded
continuous functions from X to Y as
C(X;Y ) :=ff2B(X;Y ): f is continuousg:
Page 5


1. Review
Remark 1.1. From now on, unless otherwise specied, R
n
refers to Euclidean space R
n
withn 1 a positive integer, and where we use the metric d
`
2
onR
n
. In particular,R refers
to the metric spaceR equipped with the metric d(x;y) =jxyj.
Proposition 1.2. Let (X;d) be a metric space. Let (x
(j)
)
1
j=k
be a sequence of elements ofX.
Let x;x
0
be elements of X. Assume that the sequence (x
(j)
)
1
j=k
converges to x with respect to
d. Assume also that the sequence (x
(j)
)
1
j=k
converges to x
0
with respect to d. Then x =x
0
.
Proposition 1.3. Let a < b be real numbers, and let f : [a;b]! R be a function which
is both continuous and strictly monotone increasing. Then f is a bijection from [a;b] to
[f(a);f(b)], and the inverse function f
1
: [f(a);f(b)]! [a;b] is also continuous and strictly
monotone increasing.
Theorem 1.4 (Inverse Function Theorem). LetX;Y be subsets ofR. Letf : X!Y be
bijection, so thatf
1
: Y!X is a function. Letx
0
2X andy
0
2Y such thatf(x
0
) =y
0
. If
f is dierentiable atx
0
, iff
1
is continuous aty
0
, and iff
0
(x
0
)6= 0, thenf
1
is dierentiable
at y
0
with
(f
1
)
0
(y
0
) =
1
f
0
(x
0
)
:
Date: February 14, 2015.
1
2. Sequences of Functions
As we have seen in analysis, it is often desirable to discuss sequences of points that
converge. Below, we will see that it is similarly desirable to discuss sequences of functions
that converge in various senses. There are many distinct ways of discussing the convergence of
sequences of functions. We will only discuss two such modes of convergence, namely pointwise
and uniform convergence. Before beginning this discussion, we discuss the limiting values
of functions between metric spaces, which should generalize our notion of limiting values of
functions on the real line.
2.1. Limiting Values of Functions.
Denition 2.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X!Y be a function, letx
0
2X be an adherent point ofE, and letL2Y . We say that
f(x) converges toL inY asx converges tox
0
inE, and we write lim
x!x
0
;x2E
f(x) =L,
if and only if, for every " > 0, there exists  = (") > 0 such that, if x2 E satises
d
X
(x;x
0
)<, then d
Y
(f(x);L)<".
Remark 2.2. So, f is continuous at x
0
if and only if
lim
x!x
0
;x2X
f(x) =f(x
0
): ()
And f is continuous on X if and only if, for all x
0
2X, () holds.
Remark 2.3. When the domain of x of the limit lim
x!x
0
;x2X
f(x) is clear, we will often
instead write lim
x!x
0
f(x).
The following equivalence is generalized from its analogue on the real line.
Proposition 2.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces, let E be a subset of X, let
f : X! Y be a function, let x
0
2 X be an adherent point of E, and let L2 Y . Then the
following statements are equivalent.
 lim
x!x
0
;x2E
f(x) =L.
 For any sequence (x
(j)
)
1
j=1
in E which converges to x
0
with respect to the metric d
X
,
the sequence (f(x
(j)
))
1
j=1
converges to L with respect to the metric d
Y
.
Exercise 2.5. Prove Proposition 2.4.
Remark 2.6. From Propositions 2.4 and 1.2, the function f can converge to at most one
limit L as x converges to x
0
.
Remark 2.7. The notation lim
x!x
0
;x2E
f(x) implicitly refers to a convergence of the function
values f(x) in the metric space (Y;d
Y
). Strictly speaking, it would be better to write d
Y
somewhere next to the notation lim
x!x
0
;x2E
f(x). However, this omission of notation should
not cause confusion.
2.2. Pointwise Convergence and Uniform Convergence.
Denition 2.8 (Pointwise Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges pointwise to f on X if and only if, for every x2X, we have
lim
j!1
f
j
(x) =f(x):
That is, for all x2X, we have
lim
j!1
d
Y
(f
j
(x);f(x)) = 0:
That is, for every x2X and for every "> 0, there exists J > 0 such that, for all j >J, we
have d
Y
(f
j
(x);f(x))<".
Remark 2.9. Note that, if we change the point x, then the limiting behavior of f
j
(x)
can change quite a bit. For example, let j be a positive integer, and consider the functions
f
j
: [0; 1]!R wheref
j
(x) =j for allx2 (0; 1=j), andf
j
(x) = 0 otherwise. Letf : [0; 1]!R
be the zero function. Then f
j
converges pointwise to zero, since for any x2 (0; 1], we have
f
j
(x) = 0 for all j > 1=x. (And f
j
(0) = 0 for all positive integers j.) However, given any
xed positive integer j, there exists an x such that f
j
(x) = j. Moreover,
R
1
0
f
j
= 1 for all
positive integers j, but
R
1
0
f = 0. So, we see that pointwise convergence does not preserve
the integral of a function.
Remark 2.10. Pointwise convergence also does not preserve continuity. For example, con-
sider f
j
: [0; 1]!R dened by f
j
(x) =x
j
, where j2N and x2 [0; 1]. Dene f : [0; 1]!R
so that f(1) = 1 and so that f(x) = 0 for x2 [0; 1). Then f
j
converges pointwise to f as
j!1, and each f
j
is continuous, but f is not continuous.
In summary, pointwise convergence doesn't really preserve any useful analytic quantities.
The above remarks show that some points are changing at much dierent rates than other
points as j !1. A stronger notion of convergence will then x these issues, where all
points in the domain are controlled simultaneously.
Denition 2.11 (Uniform Convergence). Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let
(f
j
)
1
j=1
be a sequence of functions from X to Y . Let f : X! Y be another function. We
say that (f
j
)
1
j=1
converges uniformly tof onX if and only if, for every"> 0, there exists
J > 0 such that, for all j >J and for all x2X we have d
Y
(f
j
(x);f(x))<".
Remark 2.12. Note that the dierence between uniform and pointwise convergence is that
we simply moved the quantier \for all x2 X" within the statement. This change means
that the integer J does not depend on x in the case of uniform convergence.
Remark 2.13. The sequences of functions from Remarks 2.9 and 2.10 do not converge
uniformly. So, pointwise convergence does not imply uniform convergence. However, uniform
convergence does imply pointwise convergence.
3. Uniform Convergence and Continuity
We saw that pointwise convergence does not preserve continuity. However, uniform con-
vergence does preserve continuity.
Theorem 3.1. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Let x
0
2 X. Suppose f
j
converges uniformly to f on X. Suppose that, for each j 1, we know that f
j
is continuous
at x
0
. Then f is also continuous at x
0
.
Exercise 3.2. Prove Theorem 3.1. Hint: it is probably easiest to use the " denition of
continuity. Once you do this, you may require the triangle inequality in the form
d
Y
(f(x);f(x
0
))d
Y
(f(x);f
j
(x)) +d
Y
(f
j
(x);f
j
(x
0
)) +d
Y
(f
j
(x
0
);f(x
0
)):
Corollary 3.3. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Suppose (f
j
)
1
j=1
converges
uniformly to f on X. Suppose that, for each j 1, we know that f
j
is continuous on X.
Then f is also continuous on X.
Uniform limits of bounded functions are also bounded. Recall that a function f : X!Y
between metric spaces (X;d
X
) and (Y;d
Y
) is bounded if and only if there exists a radius
R> 0 and a point y
0
2Y such that f(x)2B
(Y;d
Y
)
(y
0
;R) for all x2X.
Proposition 3.4. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions from X to Y . Let f : X ! Y be another function. Suppose (f
j
)
1
j=1
converges
uniformly to f on X. Suppose also that, for each j 1, we know that f
j
is bounded. Then
f is also bounded.
Exercise 3.5. Prove Proposition 3.4.
3.1. The Metric of Uniform Convergence. We will now see one advantage to our ab-
stract approach to analysis on metric spaces. We can in fact talk about uniform convergence
in terms of a metric on a space of functions, as follows.
Denition 3.6. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let B(X;Y ) denote the set
of functions f : X ! Y that are bounded. Let f;g 2 B(X;Y ). We dene the metric
d
1
: B(X;Y )B(X;Y )! [0;1) by
d
1
(f;g) := sup
x2X
d
Y
(f(x);g(x)):
This metric is known as the sup norm metric or the L
1
metric. We also use d
B(X;Y)
as
a synonym for d
1
. Note that d
1
(f;g)<1 since f;g are assumed to be bounded.
Exercise 3.7. Show that the space (B(X;Y );d
1
) is a metric space.
Example 3.8. LetX = [0; 1] and letY =R. Consider the functionsf(x) =x andg(x) = 2x
where x2 [0; 1]. Then f;g are bounded, and
d
1
(f;g) = sup
x2[0;1]
jx 2xj = sup
x2[0;1]
jxj = 1:
Here is our promised characterization of uniform convergence in terms of the metric d
1
.
Proposition 3.9. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Let (f
j
)
1
j=1
be a sequence of
functions inB(X;Y ). Letf2B(X;Y ). Then (f
j
)
1
j=1
converges uniformly to f onX if and
only if (f
j
)
1
j=1
converges to f in the metric d
B(X;Y)
.
Exercise 3.10. Prove Proposition 3.9.
Denition 3.11. Let (X;d
X
) and (Y;d
Y
) be metric spaces. Dene the set of bounded
continuous functions from X to Y as
C(X;Y ) :=ff2B(X;Y ): f is continuousg:
Note that C(X;Y ) B(X;Y ) by the denition of C(X;Y ). Also, by Corollary 3.3,
C(X;Y ) is closed in B(X;Y ) with respect to the metric d
1
. In fact, more is true.
Theorem 3.12. Let (X;d
X
) be a metric space, and let (Y;d
Y
) be a complete metric space.
Then the space (C(X;Y );d
B(X;Y)
j
C(X;Y)C(X;Y)
) is a complete subspace of B(X;Y ). That
is, every Cauchy sequence of functions in C(X;Y ) converges to a function in C(X;Y ).
Exercise 3.13. Prove Theorem 3.12
4. Series of Functions and the Weierstrass M-test
For each positive integer j, let f
j
: X!R be a function. We will now consider innite
series of the form
P
1
j=1
f
j
. The most natural thing to do now is to determine in what sense
the series
P
1
j=1
f
j
is a function, and if it is a function, determine if it is continuous. Note
that we have restricted the range to beR since it does not make sense to add elements in a
general metric space. Power series and Fourier series perhaps give the most studied examples
of series of functions. If x2 [0; 1] and if a
j
are real numbers for all j 1, we want to make
sense of the series
P
1
j=1
a
j
cos(2jx). We want to know in what sense this innite series is
a function, and if it is a function, do the partial sums converge in any reasonable manner?
We will return to these issues later on.
Denition 4.1. Let (X;d
X
) be a metric space. For each positive integer j, let f
j
: X!R
be a function, and let f : X!R be another function. If the partial sums
P
J
j=1
f
j
converge
pointwise tof asJ!1, then we say that the innite series
P
1
j=1
f
j
converge pointwise
to f, and we write f =
P
1
j=1
f
j
. If the partial sums
P
J
j=1
f
j
converge uniformly to f as
J !1, then we say that the innite series
P
1
j=1
f
j
converge uniformly to f, and we
write f =
P
1
j=1
f
j
. (In particular, the notation f =
P
1
j=1
f
j
is ambiguous, since the nature
of the convergence of the series is not specied.)
Remark 4.2. If a series converges uniformly then it converges pointwise. However, the
converse is false in general.
Exercise 4.3. Let x2 (1; 1). For each integer j 1, dene f
j
(x) := x
j
. Show that the
series
P
1
j=1
f
j
converges pointwise, but not uniformly, on (1; 1) to the function f(x) =
x=(1x). Also, for any 0 < t < 1, show that the series
P
1
j=1
f
j
converges uniformly to f
on [t;t].
Denition 4.4. Letf : X!R be a bounded real-valued function. We dene the sup-norm
kfk
1
of f to be the real number
kfk
1
:= sup
x2X
jf(x)j:
Exercise 4.5. Let X be a set. Show thatk
k
1
is a norm on the space B(X;R).
Theorem 4.6 (Weierstrass M-test). Let (X;d) be a metric space and let (f
j
)
1
j=1
be a
sequence of bounded real-valued continuous functions on X such that the series (of real num-
bers)
P
1
j=1
kf
j
k
1
is absolutely convergent. Then the series
P
1
j=1
f
j
converges uniformly to
some continuous function f : X!R.