ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 15
Then B has the same zeroes as f, and |B(z)| = 1 when |z| = r. Then
log f(z)
B(z) is harmonic, so by the mean value principle, we have that
log
f(0)
2π
1
B(0)
= log
2π
iθ
f(re ) dθ.
But
0
log
f(0)
B(0)
= log |f(0)| − log zj
,
r
and thus
log |f(0)| = 1
2π
log
2π 0
f(re iθ ) dθ +
log
j
But zj
r < 1, so
j log zj
r ≤ 0, so we have that
1
2π
2π
0
log |f(re iθ )|dθ ≥ log |f(0)|.
j
.
r
(b) If |{θ ∈ [0, 2π] : f(e iθ ) = 0}| > 0, then 2π
0 log f(re iθ ) dθ → −∞ as
r → 1, which contradicts the inequality in (a).
Problem 5: (a) For f ∈ L 2 (R) and {xn} converging to zero, define fn(x) =
f(x + xn). Show that {fn} converges to f in L 2 .
(b) Let W ⊂ R be Lebesgue measurable of positive measure. Show that
W − W = {x − y : x, y ∈ W } contains an open neighborhood of the
origin.
Solution. (a) We want to show that fn − f 2 → 0. Since Cc(R) is dense
in L 2 , choose g ∈ Cc(R) such that f − g 2 < ε. Then
fn − f 2 ≤ fn − gn 2 + gn − g 2 + g − f 2 < 2ε + gn − g 2
Chose δ so that |g(x) − g(y)| < ε when |x − y| < δ, and choose N so
that if n > N, |xn| < δ. Then for n > N, we have that
gn − g 2
2 =
zj
|g(x + xn) − g(x)|
R
2 dx ≤ ε 2 2supp(g).
Thus, fn − f2 → 0 as n → ∞.
(b) Here are two solutions, neither of which uses part (a).
(1) We use the fact that
f(x) = χW (y)χW (y − x)dy
R
is continuous, which is Q3 in Fall 2004. Then since W has positive
measure, f(0) > 0. As f is continuous, there exists δ > 0 such
that f(t) > 0 for t ∈ (−δ, δ). Then for each t in this interval, we
have that
R χW (y)χW (y − t)dy > 0, so there exists yt such that
χW (yt) = χW (yt − t) = 1. Thus yt ∈ W and yt − t ∈ W . Hence
t = yt − (yt − t) ∈ W − W . Hence (−δ, δ) ⊂ W − W .