ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

20 ANALYSIS QUALS
Problem 5: Construct a Borel subset E of the real line R such that for all
intervals [a, b] we have
0 < m(E ∩ [a, b]) < |b − a|
where m denotes Lebesgue measure.
Proof. We will use the Baire Category Theorem on the complete metric
space
X = {χE ∈ L 1 : E Borel .
First, to see that X is complete, it suffices to show that X is closed in L1 .
Suppose fn → f in L1 , where fn ∈ X. Then there is a subsequence fnk
that converges to f pointwise a.e. (cf. another common qual problem).
But as characteristic functions converge pointwise a.e. to f, we must have
that f is almost everywhere a characteristic function of some set F . As
every set differs by a measure zero set from a Borel set, we can assume that
F is Borel. Thus, f ∈ X, so X is complete.
Now for a non-trivial interval I = [a, b], we define the set UI = {χE ∈ X :
0 < m(E ∩ I) < m(I)}. We claim that UI is open and dense in X. First,
UI is dense in X: if χE ∈ UI, then either m(E ∩I) = 0 or m(E ∩I) = m(I).
In the former case, we can add a small amount of measure to E, and in the
latter case, we can remove a small amount of measure from E to obtain E ′
with E − E ′ L1 < ε, and E ′ ∈ X. This gives density. To see that UI is
open in X, we note that χE ↦→ m(E ∩ I) is a continuous function on L1 restricted to X. Thus, if 0 < m(E ∩ I) < m(I), then 0 < m(E ′ ∩ I) < m(I)
for all E ′ such that χE ′ is L1-close to χE, so UI is open.
Thus, by the Baire Category Theorem,
UI is dense in X,
I=[a,b],a,b∈Q
so in particular there exists a Borel F such that χF is in the intersection.
Since every non-trivial interval contains an interval with rational endpoints,
this F does the job, which completes the proof.
Problem 7: a) Define unitary operator on a complex Hilbert space.
b) Let S be a unitary operator on a complex Hilbert space. Using your
definition, prove that for every complex number |λ| < 1 the operator
S − λI is invertible. Here I denotes the identity operator.
c) For a fixed vector v in the Hilbert space and all {λ ∈ C : |λ < 1}, we
define
h(λ) = (S + λI)(S − λI) −1 v, v .
Show that Re h is a positive harmonic function. [You may not invoke
the spectral theorem - this is part of a proof of that theorem.]
Solution. a) An operator U : H → H is a unitary operator if U is
bounded, linear, and U ∗ U = UU ∗ = I.