ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

16 ANALYSIS QUALS
(2) As W has positive measure, then by the definition of outer measure,
there exists a collection of closed intervals In such that W ⊂ In
and
∞
|In| ≤ (21/20)m(W ).
n=1
There exists n such that m(W ∩ I) ≥ (9/10)m(In), for if not, then we
have
m(W ) ≤ m(W ∩ In) < 9
10 m(In) ≤ 9
2120m(W ) < m(W ),
10
which is a contradiction.
Thus, let In = I = [a, b], so m(W ∩I) ≥ (9/10)m(I). Let W0 = W ∩I.
Then it suffices to show that W0 −W0 contains an open neighborhood.
Suppose not. Then for all n, choose an ∈ (−1/n, 1/n) \ (W0 − W0).
Consider W0 ∩ (W0 + an). If x ∈ W0 ∩ (W0 + an), then x = y + an for
y ∈ W0, so an ∈ W0 − W0. Hence, W0 ∩ (W0 + an) = ∅ for all n. But
W0 ∪ (W0 + an) ⊂ I ∪ (I + an) ⊂ [a − |an|, b + |an|],
and thus
m(W0 ∪ (W0 + an)) ≤ b − a + 2|an| → m(I)
as n → ∞. However, since an ∈ W0 − W0, we have that
m(W0 ∪ (W0 + an)) = 2m(W0) ≥ 18
10 m(I)
so we must have m(I) = 0, which is a contradiction. This completes
the proof.
Problem 7: Let H be a Hilbert space and let E be a closed convex subset of
H. Prove that there exists a unique element x ∈ E such that
x = inf y .
y∈E
Solution. Let d = infy∈E y. Choose {yn} ∈ E such that yn → d. Then
yn − ym 2 = 2 yn 2 + 2 ym 2 2
yn
− 4
+ ym
2 .
As E is convex, we have that yn+ym
2 ∈ E, so yn+ym
2
2 ≥ d2 . Thus,
yn − ym 2 ≤ 2 yn 2 + 2 ym 2 − 4d 2 → 0 as n, m → ∞.
Thus, {yn} is Cauchy, and since H is complete, the sequence converges to
a point x. Then x ∈ E since E is closed, and x = d. This proves existence.
If y also satisfies y = d, then
2
x + y
2 = 4d 2
− 4
x + y
2
0 ≤ x − y 2 = 2 x 2 + 2 y 2 − 4
2
≤ 4d 2 − 4d 2 = 0,
so x = y. This proves uniqueness.