ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

2 ANALYSIS QUALS
(b) WLOG, assume 0 ≤ y ≤ x ≤ 1. Then
x y
|Fn(x) − Fn(y)| =
fn(t)dt − fn(t)dt
0
0
x
=
fn(t)dt
y
1
=
fn(t)χ [y,x](t)dt
≤
≤
0
1
0
1
Hence, Fn ∈ C([0, 1]).
Similarly,
0
|fn(t)| χ [y,x](t)dt
|fn(t)| 2 dt
1/2 1
χ [y,x](t)
0
2 1/2
dt
= fn L 2 (x − y) 1/2 → 0 as x − y → 0.
|F (x) − F (y)| ≤ f L 2 (x − y) 1/2 ,
so F ∈ C([0, 1]) as well. It is left to show that Fn → F uniformly on
[0, 1].
Note that x ↦→ |Fn(x) − F (x)| is a continuous function on [0, 1], so
as [0, 1] is compact, it attains a maximum. Thus, there exists y ∈ [0, 1]
such that
|Fn(x) − F (x)| ≤ |Fn(y) − F (y)|
for every x ∈ [0, 1]. Let ε > 0. Note that χ [0,y] ∈ L 2 ([0, 1]). As fn → f
weakly, we may choose N such that
1
1
fn(t)χ [0,y](t)dt − f(t)χ [0,y](t)dt
< ε when n ≥ N.
0
0
Let x ∈ [0, 1] and suppose n ≥ N. Then we have that
1
1
|Fn(x) − F (x)| ≤ |Fn(y) − F (y)| =
fn(t)χ [0,y](t)dt − f(t)χ [0,y](t)dt
< ε.
Thus, Fn → F uniformly.
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Problem 5: (a) Show that l ∞ (Z) contains continuum many functions xα :
Z → R obeying
xα l ∞ = 1 and xα − xβ l ∞ ≥ 1 whenever α = β.
(b) Deduce (assuming the axiom of choice) that the Banach space dual of
l ∞ (Z) cannot contain a countable dense subset.
(c) Deduce that l 1 (Z) is not reflexive.
0