The Real And Complex Number Systems

y A.P. ≥ G.P. for all real x. Hence,
∣ x ∣∣∣
∣ ≤ 1
1 + nx 2 2 √ n
which implies that f n → f uniformly on R.
(c) In what subintervals of R does f ′ n → g uniformly
Proof: Since each f n ′ =
1−nx2 is continuous on R, and the limit function
(1+nx 2 ) 2
g is continuous on R − {0} , then by Theorem 9.2, the interval I that we
consider does not contains 0. Claim that f n ′ → g uniformly on such interval
I = [a, b] which does not contain 0 as follows.
Consider ∣ ∣ ∣∣∣ 1 − nx 2 ∣∣∣ 1
(1 + nx 2 ) 2 ≤
1 + nx ≤ 1
2 na , 2
so we know that f ′ n → g uniformly on such interval I = [a, b] which does not
contain 0.
9.15 Let f n (x) = (1/n) e −n2 x 2 if x ∈ R, n = 1, 2, ... Prove that f n → 0
uniformly on R, that f ′ n → 0 pointwise on R, but that the convergence of
{f ′ n} is not uniform on any interval containing the origin.
Proof: It is clear that f n → 0 uniformly on R, that f n ′ → 0 pointwise
on R. Assume that f n ′ → 0 uniformly on [a, b] that contains 0. We will prove
that it is impossible as follows.
We may assume that 0 ∈ (a, b) since other cases are similar. Given ε = 1,
e
then there exists a positive integer N ′ such that as n ≥ max ( )
N ′ , 1 b := N
(⇒ 1 ≤ b), we have N
|f n ′ (x) − 0| < 1 for all x ∈ [a, b]
e
which implies that
∣ ∣∣∣
2 Nx
e (Nx)2 ∣ ∣∣∣
< 1 e
for all x ∈ [a, b]
which implies that, let x = 1 N , 2
e < 1 e
which is absurb. So, the convergence of {f ′ n} is not uniform on any interval
containing the origin.
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