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

2. Section 9.2
Applications of
Uniform Convergence

3. Proof: The argument is based on the inequality
Theorem 9.2.1
Let ( fn) be a sequence of continuous functions defined on a set S and suppose that ( fn) converges uniformly on S to a function f : S  Then f is continuous on S.
Proof: The argument is based on the inequality
The idea is to make the first and last terms small by using the uniform convergence of ( fn ) and choosing n sufficiently large.
Once an n is selected, the continuity of fn will enable us to make the middle term small by requiring x to be close to c.
To see this, let c  S and let  > 0.
Then there exists N  such that n  N implies that | fn(x) – f (x) | <  /3 for all x  S.
Since c  S, we also have | fn(c) – f (c) | <  /3 whenever n  N.
In particular, we have | fN (x) – f (x) | <  /3 and | fN (c) – f (c) | <  /3.
Since fN  is continuous at c, there exists a  > 0 such that
| fN (x) – fN (c) | <  /3 whenever | x – c | <  and x  S.
Thus for all x  S with | x – c | <  we have
| f (x)  f (c) |  | f (x)  fN (x) | + | fN (x)  fN (c) | + | fN (c)  f (c) |
Hence f is continuous at c, and since c is any point in S, we conclude that f is continuous on S. 

4. When we apply Theorem 1 to series of functions, we get the following corollary.
Let be a series of functions defined on a set S. Suppose that each fn is continuous on S and that the series converges uniformly to a function f on S. Then
is continuous on S.
Proof:
Since each partial sum is continuous, Theorem implies that f , the limit of the partial sums, is also continuous. 
One useful application of Theorem is in showing that a given sequence does not converge uniformly.

5. For example, let fn(x) = x n for x  [0, 1].
Then each fn is continuous on [0, 1],
but the limit function is not continuous at x = 1.
Thus the convergence cannot be uniform on [0, 1]. f
(1, f (1)) x y 1
f 1
f 2
f 3 )

6. “The limit of the integrals is the integral of the limit.”
Theorem 9.2.4
Let ( fn) be a sequence of continuous functions defined on an interval [a, b ] and suppose that ( fn) converges uniformly on [a, b ] to a function f. Then
“The limit of the integrals is the integral of the limit.”
Proof:
Theorem implies that f is continuous on [a, b], so the functions f, fn , and fn  f are all integrable on [a, b] by Theorem
Given any  > 0, since ( fn) converges uniformly to f on [a, b], there exists a natural number N such that
for all x  [a, b ] and all n  N.
(We may assume that a  b, for otherwise the integrals are all zero and the result is trivial.)
Thus whenever n  N, we have
It follows that 

7. “The integral of the sum is the sum of the integrals.”
Corollary 9.2.5
Let be a series of functions defined on an interval [a, b]. Suppose that each fn is continuous on [a, b] and that the series converges uniformly to a function f on [a, b ]. Then
“The integral of the sum is the sum of the integrals.”
Proof:
Let be the nth partial sum of the series so that (sn) converges uniformly to f
on [a, b].
Then we have
Theorem 9.2.4
Definition of sn
The integral of a finite sum is the sum of the integrals.
Definition of an infinite series

8. Example 9.2.6
Consider the geometric series and suppose that 0 < r < 1.
When t  [r, r], then the series converges to 1/(1 + t), so
In fact, since |  t n |  r n and  r n converges when 0 < r < 1, the Weierstrass M -test (9.1.11) shows that the convergence is uniform on any interval [r, r], where 0 < r < 1.
Thus, if x  ( 1, 1), we can integrate the series term by term from 0 to x:
But from calculus we know that
So for any x in ( –1, 1) we have
In Section 9.3 we extend this to include x = 1:
The next theorem deals with the derivatives of a convergent sequence of functions.

9. Theorem 9.2.7
Suppose that ( fn) converges to f on an interval [a, b]. Suppose also that each exists and is continuous on [a, b], and the sequence converges uniformly on [a, b]. Then for each x  [a, b].
Note that it is the sequence of derivatives that must converge uniformly, not the original functions.
Corollary 9.2.8
Let be a series of functions that converges to a function f on an interval [a, b]. Suppose that, for each n, exists and is continuous on [a, b] and that the series of derivatives is uniformly convergent on [a, b]. Then
f (x) = for all x  [a, b].
One of the surprising results in real analysis is that there exists a continuous function that does not have a derivative at any point. This is our next theorem.

10. Theorem 9.2.9
There exists a continuous function defined on that is nowhere differentiable.
Proof:
Define g (x) = | x | for x  [ 1, 1] and extend the definition of g to all of by requiring that g (x + 2) = g (x) for all x.
It is not difficult to show (in the text) that g is continuous on
For each integer n  0, let
Then, g0(x) = g(x),
In general, each gn + 1 oscillates four times as fast, but only three-fourths as high, as gn.
Now define f on by 1 1 3 2 2 4
g2(x)
g1(x)
g0(x) = g(x)

11. For all x we have 0  g (x)  1, so that 0  gn(x)  (3/4)n.
Thus the Weierstrass M -test implies that the series converges uniformly to f on .
Since each gn is continuous on , Corollary implies that f is continuous on .
The proof (in the text) that f is nowhere differentiable is tedious, but we can see from the first few partial sums what is happening.
As n  , we get “corners” at more and more points.
In the limit function, f , there is a corner at every point and it is nowhere differentiable. 1 1 3 2 2 4
g0(x) + g1(x) + g2(x)
g0(x) + g1(x)
g0(x)
g2(x)
g1(x)