Problems and Solutions in - Mathematics - University of Hawaii

1.2 1994 November 16 1 REAL ANALYSIS
1.2 1994 November 16
Masters students: Do any 5 problems.
Ph.D. students: Do any 6 problems.
1. Let E be a normed linear space. Show that E is complete if and only if, whenever ∞ 1 xn < ∞, then ∞ converges to an s ∈ E.
Solution: Suppose E is complete. Let {xn} ⊂ E be absolutely convergent; i.e., xn < ∞. We must
∞
xn := lim
n=1
Let SN = N
n=1 xn. Then, for any j ∈ N,
SN+j − SN =
N→∞
n=1
N+j
n=N+1
1 xn
N
xn = s ∈ E. (6)
xn
≤
N+j
n=N+1
xn → 0
as N → ∞, since xn < ∞. Therefore, {SN } is a Cauchy sequence. Since E is complete, there is an s ∈ E
such that ∞ n=1 xn = lim
N→∞ SN = s.
Conversely, suppose whenever ∞ 1 xn < ∞, then ∞ 1 xn converges to an s ∈ E. Let {yn} ⊂ E be a Cauchy
sequence. That is, yn − ym → 0 as n, m → ∞. Let n1 < n2 < · · · be a subsequence such that
Next observe, for k > 1,
and
ynk
n, m ≥ nj ⇒ yn − ym < 2 −j .
k−1
= yn1 + (yn2 − yn1 ) + (yn3 − yn2 ) + · · · + (ynk − ynk−1 ) = yn1 +
By hypothesis, this implies that
ynk
∞
j=1
ynj+1
− yn1 =
− ynj <
k−1
j=1
(ynj+1
∞
2 −j = 1
j=1
− ynj ) → s ∈ E,
j=1
(ynj+1
− ynj ),
as k → ∞. We have thus found a subsequence {ynk } ⊆ {yn} having a limit in E. Finally, since {yn} is Cauchy,
it is quite easy to verify that {yn} must converge to the same limit. This proves that every Cauchy sequence in E
converges to a point in E. ⊓⊔
2. Let fn be a sequence of real continuous functions on a compact Hausdorff space X. Show that if f1 ≥ f2 ≥ f3 ≥
· · · , and fn(x) → 0 for all x ∈ X, then fn → 0 uniformly.
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