The Real And Complex Number Systems

which implies that
∣ ∣ |g n (x) − g| =
x
∣∣∣ ∣1 + nx∣ = 1 ∣∣∣
1
+ n < 1 n < ε.
x
So, g n → 0 uniformly on (0, 1) .
9.6 Let f n (x) = x n . The sequence {f n (x)} converges pointwise but not
uniformly on [0, 1] . Let g be continuous on [0, 1] with g (1) = 0. Prove that
the sequence {g (x) x n } converges uniformly on [0, 1] .
Proof: It is clear that f n (x) = x n converges NOT uniformly on [0, 1]
since each term of {f n (x)} is continuous on [0, 1] and its limit function
{ 0 if x ∈ [0, 1)
f =
1 if x = 1.
is not a continuous function on [0, 1] by Theorem 9.2.
In order to show {g (x) x n } converges uniformly on [0, 1] , it suffices to
shows that {g (x) x n } converges uniformly on [0, 1). Note that
lim g (x)
n→∞ xn = 0 for all x ∈ [0, 1).
We partition the interval [0, 1) into two subintervals: [0, 1 − δ] and (1 − δ, 1) .
As x ∈ [0, 1 − δ] : Let M = max x∈[0,1] |g (x)| , then given ε > 0, there is a
positive integer N such that as n ≥ N, we have
M (1 − δ) n < ε
which implies that for all x ∈ [0, 1 − δ] ,
|g (x) x n − 0| ≤ M |x n | ≤ M (1 − δ) n < ε.
Hence, {g (x) x n } converges uniformly on [0, 1 − δ] .
As x ∈ (1 − δ, 1) : Since g is continuous at 1, given ε > 0, there exists a
δ > 0 such that as |x − 1| < δ, where x ∈ [0, 1] , we have
|g (x) − g (1)| = |g (x) − 0| = |g (x)| < ε
which implies that for all x ∈ (1 − δ, 1) ,
|g (x) x n − 0| ≤ |g (x)| < ε.
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