ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 39
where the inequality follows since ˆ f(0) = 0 by assumption, and the equality
also comes from Parseval.
If equality holds, then ˆ f(n) = 0 for all n, so f is identically zero, as seen
below in problem 8 of this exam.
Problem 2: Is there a closed uncountable subset of R which contains no
rational numbers? Prove your answer.
Solution. Let {qn} be an enumeration of the rational numbers. Then we
have that
∞
Q = B(qn, 2 −n )
n=1
is a countable union of open sets, and hence is open. Thus, R\Q is closed
and contains no rational numbers. Furthermore,
∞
µ(Q) ≤ 2 1−n < ∞,
n=1
so R\Q has strictly positive Lebesgue measure, and hence is uncountable.
Problem 3: (a) Prove that a complete normed vector space (a Banach space)
is either finite-dimensional or has uncountable dimension in the vector
space sense, i.e., it is not generated as finite linear combinations of
elements of some countable subset.
(b) Use part (a) to give an example of a vector space that cannot be given
the structure of a Banach space, that is, it is not complete in any norm.
Solution. (a) Assume X is a an infinite-dimensional Banach space, and
assume for sake of contradiction that (xi)is a countable Hamel basis
for X. Let Vn = span{x1, . . . , xn}. Then X = ∞
i=1 Vn. Note that
Vn is finite dimensional, so it is isomorphic as a Banach space to R n ,
and thus is a closed subspace. However, Vn is nowhere dense, as Vn
contains no open balls in V . Thus, by the Baire Category Theorem,
Vn = X, a contradiction.
(b) Let V be the vector space consisting of all infinite sequences with
finitely many non-zero terms. Then V has a countable Hamel basis:
let xi be the sequence of all zeros with a 1 in the ith coordinate. Every
element of V can be written uniquely as a finite linear combination of
the xi. Thus, by part (a), V cannot be a Banach space.
Problem 4: Prove that non every subset of [0, 1] is Lebesgue measurable.
Solution. Consider the relation x ∼ y is x − y ∈ Q. This is an equivalence
relation. For α ∈ [0, 1], let Eα be the equivalence class of α. Then [0, 1] ⊂
α Eα. By the axiom of choice, we may choose exactly one element xα ∈
[0, 1] from each equivalence class, and let N be the set of these elements.
We claim that N is not Lebesgue measurable.
Assume for sake of contradiction that N is measurable. Let {rk} be an