Problems and Solutions in - Mathematics - University of Hawaii

1.3 1998 April 3 1 REAL ANALYSIS
The converse of this lemma is also true. 18
Lemma 1.3 If f : R → R is increasing and f ′ ∈ L 1 ([a, b]), then x
a f ′ (t)dt ≤ f(x) − f(a).
To finish the ⇐ direction of the proof, by lemma 1.2, it suffices to show that
R f ′ dm = 1 implies b
a f ′ (t)dt =
f(b) − f(a) holds for all −∞ < a < b < ∞. By lemma 1.3, we have b
a f ′ (t)dt ≤ f(b) − f(a), so we need only
show that strict inequality cannot hold. Suppose, by way of contradiction, that b
a f ′ (t)dt < f(b) − f(a) holds for
some −∞ < a < b < ∞. Then,
1 =
R
f ′ dm =
a
−∞
f ′ dm +
b
a
f ′ dm +
∞
b
f ′ dm
< [f(a) − f(−∞)] + [f(b) − f(a)] + [f(∞) − f(b)]
= f(∞) − f(−∞) = 1.
This contradiction proves that
R f ′ dm = 1 implies b
a f ′ (t)dt = f(b) − f(a) holds for all −∞ < a < b < ∞, as
desired.
(⇒) Now assume f ∈ AC[a, b] for all −∞ < a < b < ∞. We must show
f(∞) − f(−∞) = 1, so this is equivalent to showing
x
lim f
x→∞
−x
′ (t)dm(t) = lim [f(x) − f(−x)].
x→∞
R f ′ dm = 1. By assumption
Let x ∈ R, x > 0, and f ∈ AC[−x, x]. Then we claim f(x) − f(−x) = x
−x f ′ dm. Assuming the claim is true
(see Royden [6], p. 110 for the proof), we have
x
1 = lim [f(x) − f(−x)] = lim
x→∞ x→∞
f ′
(t)dm(t) = f ′ dm.
9. Let F be a continuous linear functional on the space L 1 [−1, 1], with the property that F (f) = 0 for all odd
functions f in L 1 [−1, 1]. Show that there exists an even function φ such that
F (f) =
1
−1
−x
f(x)φ(x) dx, for all f ∈ L 1 [−1, 1].
[Hint: One possible approach is to use the fact that any function in L p [−1, 1] is the sum of an odd function and an
even function.]
Solution: Since F ∈ L 1 [−1, 1] ∗ , then by the Riesz representation theorem 19 there is a unique h ∈ L ∞ [−1, 1]
such that
F (f) =
1
−1
f(x)h(x) dx (∀f ∈ L 1 [−1, 1])
18 See Folland [4] for a nice, concise treatment of the fundamental theorem of calculus for Lebesgue integration.
19 See problem 3 of section 1.5.
16
R
⊓⊔