The Real And Complex Number Systems

which implies that
∣ lim sup 1 k∑
( ) j
f − lim inf 1 k∑
( ) ∣ j ∣∣∣
f
k→∞ k k k→∞ k k
j=1
j=1
≤
∣ lim sup 1 k∑
( )
∣
j ∣∣∣∣ f − S (f n )
k→∞ k k ∣ + lim inf 1
k→∞ k
j=1
k∑
( )
j f − S (f n )
k ∣
≤ 2ε by (3) and (4)
3
< ε. (5)
Note that (3)-(5) imply that the existence of S (f) . Also, (3) or (4) implies
that S (f) = S. So, we complete the proof.
9.18 Let f n (x) = 1/ (1 + n 2 x 2 ) if 0 ≤ x ≤ 1, n = 1, 2, ... Prove that {f n }
converges pointwise but not uniformly on [0, 1] . Is term-by term integration
permissible
Proof: It is clear that
lim f n (x) = 0
n→∞
for all x ∈ [0, 1] . If {f n } converges uniformly on [0, 1] , then given ε = 1/3,
there exists a positive integer N such that as n ≥ N, we have
which implies that
|f n (x)| < 1/3 for all x ∈ [0, 1]
∣ ∣∣∣
f N
( 1
N
)∣ ∣∣∣
= 1 2 < 1 3
j=1
which is impossible. So, {f n } converges pointwise but not uniformly on [0, 1] .
Since {f n (x)} is clearly uniformly bounded on [0, 1] , i.e., |f n (x)| ≤ 1
for all x ∈ [0, 1] and n. Hence, by Arzela’s Theorem, we know that the
sequence of functions can be integrated term by term.
9.19 Prove that ∑ ∞
n=1 x/nα (1 + nx 2 ) converges uniformly on every finite
interval in R if α > 1/2. Is the convergence uniform on R
Proof: By A.P. ≥ G.P., we have
x
∣n α (1 + nx 2 ) ∣ ≤ 1
17
2n α+ 1 2
for all x.