The Real And Complex Number Systems

(b) Prove that {f n } converges pointwise but not uniformly on [0, π/2] .
Proof: Note that each f n (x) is continuous on [0, π/2] , and the limit
function
{ 0 if x ∈ (0, π/2]
f =
.
1 if x = 0
Hence, by Theorem9.2, we know that {f n } converges pointwise but not
uniformly on [0, π/2] .
9.29 Let f n (x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let f n (x) = n if
1/n < x < 2/n. Prove that {f n } converges pointwise to 0 on [0, 1] but that
l.i.m. n→∞ f n ≠ 0 on [0, 1] .
Proof: It is clear that {f n } converges pointwise to 0 on [0, 1] . In order
to show that l.i.m. n→∞ f n ≠ 0 on [0, 1] , it suffices to note that
∫ 1
Hence, l.i.m. n→∞ f n ≠ 0 on [0, 1] .
Power series
0
f n (x) dx = 1 for all n.
9.30 If r is the radius of convergence if ∑ a n (z − z 0 ) n , where each a n ≠ 0,
show that
∣ ∣ ∣ ∣∣∣
lim inf a n ∣∣∣ ≤ r ≤ lim sup
a n ∣∣∣
n→∞ a n+1 n→∞ ∣ .
a n+1
Proof: By Exercise 8.4, we have
1
∣ lim n→∞ sup ∣ a n+1 ∣∣
≤ r =
a n
1
lim n→∞ sup |a n | 1 n
≤
1
lim n→∞ inf
∣ ∣ a n+1 ∣∣
.
a n
Since
and
1
∣
lim n→∞ sup
1
lim n→∞ inf
∣
∣ a n+1
a n
∣ a n+1
a n
∣ ∣∣
= lim
n→∞
inf
∣ ∣∣
= lim
n→∞
sup
∣ a n ∣∣∣
∣a n+1
∣ a n ∣∣∣
∣ ,
a n+1
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