ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 3
Proof. (a) First, we note that P(Z), the power set of Z, contains continuum
many elements. For α ∈ P(Z), α = ∅, let xα = χα, the
characteristic function of α. Then |xα L ∞ = 1. But for α = β, there
exists n ∈ α∆β. Then |xα(n) − xβ(n)| = 1, so xα − xβ L ∞ ≥ 1.
(b) Assume for sake of contradiction that l ∞ (Z) ∗ has a countable dense
subset {fn}. That is, l ∞ (Z) ∗ is separable.
For each n, choose yn ∈ l ∞ (Z) such that |fn(yn)| ≥ 1
2 fn. Let S be
the set of linear combinations of {yn} with rational coefficients. Then
S is countable. We claim S = l ∞ (Z). If not, S is a closed proper
subspace of l ∞ (Z). By the Hahn-Banach theorem, we may choose
f ∈ l ∞ (Z) ∗ such that f(y) = 0 for all y ∈ S, but f = 1.
Choose {fnj } for our countable dense set such that fnj − f → 0 as
j → ∞. But
fnj − f ≥ fnj (ynj ) − f(ynj ) = fnj (ynj ) ≥ 1
fnj .
2
Thus, we must have
fnj → 0, so f = 0, which contradictions the
assumption that S was a proper subspace. Hence, l∞ (Z) is separable.
Thus, we let {yn} be a countable dense set in l∞ (Z). We have that
for every ε > 0 and for every non-empty α ∈ P(Z), there exists n
such that yn − xαL∞ < ε. But there are uncountably many xα with
xα − xβ ≥ 1 if α = β, so by the Pigeonhole Principle (and Axiom of
Choice), the xα’s cannot all be approximated by the yn’s, so we have
a contradiction to the assumption that l∞ (Z) ∗ has a countable dense
subset.
(c) By definition, l1 (Z) reflexive means that l1 (Z) ∼ = (l1 (Z)) ∗∗ . But l1 (Z)
is separable, and (l1 (Z)) ∗∗ ∼ = (l∞ (Z)) ∗ is not separable by part (b).
Hence, l1 (Z) is not reflexive.
Note: I lost a point on the qual with this solution for not elaborating
on the fact that l1 (Z) is separable. I don’t know whether they wanted
me to prove this, or just mention a countable dense set. For the record,
the set of linear combinations with rational coefficients of the sequences
ek consisting of all zeroes with 1 in the kth spot is a countable dense
set.
Problem 6: Suppose µ and ν are finite positive (regular) Borel measures on
R n . Prove the existence and uniqueness of the Lebesgue decomposition:
There are a unique pair of positive Borel measures µa and µs so that
µ = µa + µs, µa