Problems and Solutions in - Mathematics - University of Hawaii

1.7 2007 November 16 1 REAL ANALYSIS
Therefore, µ(B0) = 0, since µ ≪ ν, which proves that f(x) > 0, [µ]-a.e.
If B1 = {x ∈ X : f(x) ≥ 1}, then
ν(B1) =
B1
f d(µ + ν) ≥ (µ + ν)(B1) = µ(B1) + ν(B1).
Since ν is finite by assumption, we can subtract ν(B0) from both sides to obtain µ(B1) = 0. This proves f(x) < 1,
[µ]-a.e. ⊓⊔
6. 34 Suppose that 1 < p < ∞ and that q = p/(p − 1).
(a) Let a1, a2, . . . be a sequence of real numbers for which the series anbn converges for all real sequences {bn}
satisfying the condition |bn| q < ∞. Prove that |an| p < ∞.
(b) Discuss the cases of p = 1 and p = ∞. Prove your assertions.
Solution: (a) For each k ∈ N, define Tk : ℓq(N) → R by Tk(b) = k
n=1 anbn, for b ∈ ℓq(N). Then {Tk} is a
family of pointwise bounded linear functionals. That is, each Tk is a linear functional, and, for each b ∈ ℓq(N),
there is an Mb ≥ 0 such that |Tk(b)| ≤ Mb holds for all k ∈ N. To see this, simply note that a convergent sequence
of real numbers is bounded, and, in the present case, we have
Sk
k
∞
anbn → anbn = x ∈ R.
n=1
Thus, {Tk(b)} = {Sk} is a convergent sequence of reals, so, if N ∈ N is such that k ≥ N implies |Sk − x| < 1,
and if M ′ b is defined to be max{|Sk| : 1 ≤ k ≤ N}, then, for any k ∈ N,
n=1
|Tk(b)| ≤ Mb max{M ′ b, x + 1}.
Next note that ℓq is a Banach space, so the (Banach-Steinhauss) principle of uniform boundedness implies that
there is a single M > 0 such that Tk ≤ M for all k ∈ N. In other words,
(∃ M > 0) (∀b ∈ ℓq) (∀k ∈ N) |Tk(b)| ≤ Mb.
Define let T (b) ∞
n=1 anbn = limk→∞ Tk(b), which exists by assumption. Since |·| is continuous, we conclude
that limk→∞ |Tk(b)| = |T (b)|. Finally, since |Tk(b)| ≤ Mb for all k ∈ N, we have |T (b)| ≤ Mb. That is T is
a bounded linear functional on ℓq(N).
Now, by the Riesz representation theorem, if 1 ≤ q < ∞, then any bounded linear functional T ∈ ℓ ∗ q is uniquely
representable by some α = (α1, α2, . . . ) ∈ ℓp as
T (b) =
∞
αnbn. (29)
n=1
On the other hand, by definition, T (b) = ∞
n=1 anbn, for all b ∈ ℓq. Since the representation in (29) is unique,
a = α ∈ ℓp. That is,
n |an| p < ∞. ⊓⊔
34 On the original exam this question asked only about the special case p = q = 2.
36