Math 6321
Functions of a Real Variable
Dr. Vaughn Climenhaga, PGH 665
Spring 2016
HOMEWORK 1
Due in class Fri, Jan. 29.
This HW covers §5.1 and most of §6.1 in Folland, and §15.1–2 and §18.1 in Bass.
1. Bass, 15.4. Consider the measure space ([0, 1], B, m), where B is the Borel σ-algebra
and m is Lebesgue measure, and suppose f is a measurable function. Prove that
kf kp → kf k∞ as p → ∞.
2. Folland, 6.5. Suppose 0 < p < q < ∞. Then Lp 6⊂ Lq iff X contains sets of
arbitrarily small positive measure, and Lq 6⊂ Lp iff X contains sets of arbitrarily large
finite measure. (For the “if” implication: In the first case there is a disjoint sequence
{En } with 0 < µ(En ) < 2−n , and in the
P second case there is a disjoint sequence {En }
with 1 ≤ µ(En ) < ∞. Consider f = an 1En for suitable constants an .) What about
the case q = ∞?
3. Bass, 18.2. Show that Lp ([0, 1]) is separable, that is, there is a countable dense subset,
if 1 ≤ p < ∞. Show that L∞ ([0, 1]) is not separable.
4. Folland, 5.7. Let X be a Banach space.
(a) If T ∈ L(X, X) and kI − T kP< 1, where I is the identity operator, then T is
n
−1
.
invertible; in fact, the series ∞
0 (I − T ) converges in L(X, X) to T
(b) If T ∈ L(X, X) is invertible and kS − T k < kT −1 k−1 , then S is invertible. Thus
the set of invertible operators is open in L(X, X).
5. Folland, 5.8. Let (X, M) be a measurable space, and let M (X) be the space of
complex measures on (X, M). Then kµk = |µ|(X) is a norm on M (X) that makes
M (X) into a Banach space. (Use Theorem 5.1 in Folland, which is Theorem 1.4 in
our notes.)
6. Folland, 5.9. Let C k ([0, 1]) be the space of functions on [0, 1] possessing continuous
derivatives up to order k on [0, 1], including one-sided derivatives at the endpoints.
(a) If f ∈ C([0, 1]), then f ∈ C k ([0, 1]) iff f is k times continuously differentiable
on (0, 1) and limx&0 f (j) (x) and limx%1 f (j) (x) exist for j ≤ k. (The mean value
theorem is useful.)
P
(b) kf k = k0 kf (j) ku is a norm on C k ([0, 1]) that makes C k ([0, 1]) into a Banach
space. (Use induction on k. The essential point is that if {fn } ⊂ C 1 ([0, 1]),
fn → f uniformly, and fn0 → g uniformly, then f ∈ CR 1 ([0, 1]) and f 0 = g. The
x
easy way to prove this is to show that f (x) − f (0) = 0 g(t) dt.)