The Real And Complex Number Systems

In addition, as c ≥ 1/2, if f n → 0 uniformly on [0, 1] , then given ε > 0, there
exists a positive integer N such that as n ≥ N, we have
which implies that as n ≥ N,
which contradicts to
|f n (x)| < ε for all x ∈ [0, 1]
|f n (x n )| < ε
{
lim f √ 1
n (x n ) = 2e
if c = 1/2
n→∞ ∞ if c > 1/2 . (**)
From (*) and (**), we conclude that only as c < 1/2, the seqences of
functions converges uniformly on [0, 1] .
In order to determine those c for which term-by-term integration on [0, 1] ,
we consider ∫ 1
n c
f n (x) dx =
2 (n + 1)
and ∫ 1
f (x) dx =
0
0
∫ 1
0
0dx = 0.
Hence, only as c < 1, we can integrate it term-by-term.
9.11 Prove that ∑ x n (1 − x) converges pointwise but not uniformly on
[0, 1] , whereas ∑ (−1) n x n (1 − x) converges uniformly on [0, 1] . This illustrates
that uniform convergence of ∑ f n (x) along with pointwise convergence
of ∑ |f n (x)| does not necessarily imply uniform convergence
of ∑ |f n (x)| .
Proof: Let s n (x) = ∑ n
k=0 xk (1 − x) = 1 − x n+1 , then
{ 1 if x ∈ [0, 1)
s n (x) →
.
0 if x = 1
Hence, ∑ x n (1 − x) converges pointwise but not uniformly on [0, 1] by Theorem
9.2 since each s n is continuous on [0, 1] .
Let g n (x) = x n (1 − x) , then it is clear that g n (x) ≥ g n+1 (x) for all x ∈
[0, 1] , and g n (x) → 0 uniformly on [0, 1] by Exercise 9.6. Hence, by Dirichlet’s
Test for uniform convergence, we have proved that ∑ (−1) n x n (1 − x)
converges uniformly on [0, 1] .
11