The Real And Complex Number Systems

(b) Prove that h n (x) does not converges uniformly on any bounded interval.
Proof: Write
{
h n (x) =
a + a n
x
n
(
1 +
1
( n)
if x = 0 or x is irrational
1 +
1
+ ) 1
b bn if x is rational, say x =
a
b
Then
{ 0 if x = 0 or x is irrational
lim h n (x) = h (x) =
n→∞ a if x is rational, say x = a .
b
Hence, if h n (x) converges uniformly on any bounded interval I, then h n (x)
converges uniformly on [c, d] ⊆ I. So, given ε = max (|c| , |d|) > 0, there is a
positive integer N such that as n ≥ N, we have
max (|c| , |d|) > |h n (x) − h (x)|
{ ∣ ( )∣ x
= n 1 +
1 ∣
n =
|x| ∣
n 1 +
1 ∣ ( n if x ∈ Q c ∩ [c, d] or x = 0
∣ a
n 1 +
1
+ )∣ 1 ∣
b bn if x ∈ Q ∩ [c, d] , x =
a
b
which implies that (x ∈ [c, d] ∩ Q c or x = 0)
max (|c| , |d|) > |x|
n ∣ 1 + 1 n∣ ≥ |x|
n
≥
max (|c| , |d|)
n
which is absurb. So, h n (x) does not converges uniformly on any bounded
interval.
9.3 Assume that f n → f uniformly on S, g n → f uniformly on S.
(a) Prove that f n + g n → f + g uniformly on S.
Proof: Since f n → f uniformly on S, and g n → f uniformly on S, then
given ε > 0, there is a positive integer N such that as n ≥ N, we have
.
and
|f n (x) − f (x)| < ε 2
|g n (x) − g (x)| < ε 2
for all x ∈ S
for all x ∈ S.
Hence, for this ε, we have as n ≥ N,
|f n (x) + g n (x) − f (x) − g (x)| ≤ |f n (x) − f (x)| + |g n (x) − g (x)|
< ε for all x ∈ S.
3