1. Sequences and Series of Functions
Chapter 9
Sequences and Series of Functions

2. Section 9.1
Pointwise and
Uniform Convergence

3. for each x  S, and we say that ( fn) converges to f pointwise on S.
Definition 9.1.1
Let ( fn) be a sequence of functions defined on a subset S of Then ( fn) converges pointwise on S if for each x  S the sequence of numbers ( fn(x)) converges. If ( fn) converges pointwise on S, then we define f : S  by
for each x  S, and we say that ( fn) converges to f pointwise on S.
Question: What properties are preserved in taking the limit?
If each fn is continuous on a set S, is f continuous on S?
If each fn is integrable, is f integrable?
No,
not in
general
And is  f = lim  fn ?
If each fn is differentiable, is f differentiable?
And is f  = lim ?

4. To say that the functions fn are continuous
at x means that
Suppose S is an interval with x  S.
for all n.
If ( fn) converges to f pointwise on S, then
If the limit function f is continuous at x, then
equal??
Thus the limit of a sequence of continuous functions will be continuous at x
when these limits are equal.
That is, we wish to know when the order of the limit processes may be reversed.
Since derivatives and integrals are also defined in terms of limits, a similar
question applies: Can the order of the limits be reversed?

5. Example 9.1.2
For x  [0, 1] and n  , define fn(x) = x n.
When n = 1 we have f1(x) = x.
When n = 2 we have f2(x) = x2.
When n = 3 we have f3(x) = x3. x y 1
etc.
f 1
f 2
f 3

6. Example 9.1.2
For x  [0, 1] and n  , define fn(x) = x n.
Then for each x  [0,1) we have
lim n   fn(x) = 0
and lim n   fn(1) = 1.
Thus ( fn) is pointwise convergent on S = [0, 1], and the limit function f is given by x y 1
f 1
f 2
f 3
(1, f (1))
Each  fn is continuous
(and differentiable) on [0, 1],
but the limit function f is
neither continuous nor
differentiable at x = 1. ) f

7. In terms of the limit operations, we have
Go down at t < 1, then across toward x = 1.
First, go across each fn to where x = 1. x y
f 1 f
f 2
f 3
(1, f (1)) 1 )

8. Example 9.1.3
For n  2 and x  [0, 1], construct a sequence of functions fn as follows:
f2 graphs as a peak of height 2 and width 1.
f3 graphs as a peak of height 3 and width 2/3.
f4 graphs as a peak of height 4 and width 2/4, etc.
The limit function f is the constant function f (x) = 0 for all x. 1 2 4
For each n we have
f 4
f 3
But
f 2
So the integral of the limit is
not the limit of the integrals. f 

9. To reverse the limit operations, we need a stronger kind of convergence.
Definition 9.1.6
Let ( fn ) be a sequence of functions defined on a subset S of
Then ( fn ) converges uniformly on S to a function f if for each  > 0 there
exists a natural number N such that for all x  S, n  N implies that
| fn(x) – f (x)| <  .
The difference between pointwise and uniform convergence:
If ( fn ) converges pointwise to f on S, then for every  > 0 and each x  S,
there exists N  (depending on both  and x) such that | fn(x) – f (x)| < 
whenever n > N.
If ( fn ) converges uniformly to f on S, then it is possible for each  > 0
to find one number N that will work for all x  S.
This is the same type of quantifier reversal that produced uniform continuity
from continuity.

10. Geometrically, uniform convergence means that the entire graph of fn
must lie within the strip bounded by the graphs of f –  and f + . x y S
y = fn (x)
y = f (x) + 
y = f (x)
y = f (x) – 

11. Back to the first example:
And the second example: x y
f 1 f
f 2
(1, f (1)) 1 ) 1 2 4
f 2
f 3
f 4 )
f + 
f + 
f – 
f – 
In neither case can we get fn within  of f uniformly.

12. As we might expect, there is a Cauchy criterion for sequences of functions.
Theorem
Let ( fn ) be a sequence of functions defined on a subset S of There exists
a function f such that ( fn ) converges to f uniformly on S iff
for each  > 0 there exists N  such that | fn(x) – fm(x)| < 
for all x  S and all m, n  N.
Series of functions are treated similar to series of constants and power series.
If ( fn ) is a sequence of functions defined on a set S, the series is said
to converge pointwise (resp. uniformly) on S iff the sequence of partial sums
converges pointwise (resp. uniformly) on S.

13. There is a very useful test for establishing the uniform convergence
of a series of functions:
Theorem
(Weierstrass M-test)
Suppose that ( fn ) is a sequence of functions defined on S and (Mn) is a
sequence of nonnegative numbers such that
| fn(x)|  Mn for all x  S and all n  .
If  Mn converges, then  fn converges uniformly on S.
In the next section we will see that uniform convergence does what we want
it to do—namely, we can interchange the order of the limit operations.

14. Example 9.1.12 | sn(x)  sm(x)|  1 = .
Consider the series of functions where fn(x) = x n/n! for all x  .
In Example 8.3.7(a) we saw that this series is (pointwise) convergent on
We now show that the convergence is not uniform on , but given any t  ,
the convergence is uniform on the closed interval [ t, t].
To show that the convergence of the series   fn is not uniform on , we show that the sequence (sn) of partial sums does not satisfy the Cauchy criterion of Theorem
For example, let  = 1. Given any N  , we must find m, n  N and x  such that
| sn(x)  sm(x)|  1 = .
To this end, let m = N, n = N + 1, and x = n.
It follows that
Hence the series is not uniformly convergent on .

15. Example
Consider the series of functions where fn(x) = x n/n! for all x  .
In Example 8.3.7(a) we saw that this series is (pointwise) convergent on
We now show that the convergence is not uniform on , but given any t  ,
the convergence is uniform on the closed interval [ t, t].
On the other hand, if t  , then let Mn = t n/n!. For any x  [t, t] we have
Since  Mn is convergent (by the ratio test), it follows from the Weierstrass M-test that  x n/n! is uniformly convergent on [ t, t].