ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

ANALYSIS QUALS 7
Thus, as |h ′ (0)| ≤ 1, we have that
f ′ ( i
2 )
≤ (1 − i f( 2 ) √ 2 ) i + 1 2
(1 − |g(0)| )
1 + i
,
and so
f ′ ( i
2 )
≤ (1 − i f( 2 ) √ 2 √ 2 )( i + 1 − i − 1 )
i 1 +
.
We have that equality holds if and only if h is a rotation. To maximize
the right-hand-side of the above, we may take f( i
2 ) = 0 (which of course
eliminates the need for f2). Then we conclude that
sup
f ′ ( i
2 )
=
√ i + 1 2 − √ i − 1 2
1 + i
.
This completes the proof.
Fall 2010
Problem 1: For this problem, consider just Lebesgue measurable functions
f : [0, 1] → R together with the Lebesgue measure.
(a) State Fatou’s lemma (no proof required).
(b) State and prove the Dominated Convergence Theorem.
(c) Give an example where fn(x) → 0 a.e., but fn(x)dx → 1.
Solution.
(a) Fatou: Let {fn} be a sequence of nonegative functions such that f =
lim fn exists a.e. Then f(x)dx ≤ lim inf fn(x)dx.
(b) DCT: Let {fn} be a sequence of functions such that fn → f pointwise
a.e., and suppose there exists Lebesgue integrable g with fn} ≤ g a.e.
for every n. Then fn(x)dx → f(x)dx.
2
2
Proof: g + fn → g + f a.e., and g + fn is nonnegative a.e., so by
Fatou, g + f ≤ lim inf (g + fn). Then
g + f ≤ g + lim inf
so we have that
f ≤ lim inf
Similarly, g − fn → g − f a.e. and g − fn is nonnegative a.e., so by
Fatou, g − f ≤ lim inf (g − fn). Then
g − f ≤ g − lim sup
so we have that
lim sup
fn ≤
Hence, f = lim fn.
(c) Consider the pointwise-linear bump functions fn defined to be 0 at 0,
n at 1/n, 0 at 2/n, and linear between.
2
fn
f
fn
fn