Real & Complex Analysis Qualifying
Exam Solution, Fall 2010
Shiu-Tang Li
January 1, 2013
A-1
Note that {x : f (x) > a} = ∩b∈Q,b<a {x : f (x) > b} is measurable,
{x : f (x) < a} = ∩b∈Q,b>a {x : f (x) < b} is measurable, and {x : f (x) ∈ O}
is measurable for every open set O in R since O may be decomposed into a
countable union of open intervals.
Also, F := {A : f −1 (A) is measurable } is a sigma-algebra containing
all open sets in R. It means that B(R) ⊆ F . Thus, for every B ∈ B(R),
f −1 (B) is measurable and we’re done.
A-2
(1)
kfn gn − f gkL1 ≤ kfn gn − fn gkL1 + kfn g − f gkL1
≤ kfn kLp kgn − gkLq + kgkLq kfn − f kLp
≤ M1 kgn − gkLq + M2 kfn − f kLp
which goes to 0 as n → ∞.
(2) Since gn → g in L∞ , gn is bounded by M in L∞ -norm, and when
arbitrary > 0 is chosen there is some N > 0 so that kgn − gkL∞ < ,
1
kfn gn − f gkL1 ≤ kfn gn − fn gkL1 + kfn g − f gkL1
≤ kfn (gn − g)kL1 + k(fn − f )gkL1
≤ kfn kL1 + kM (fn − f )kL1
≤ M 0 + M k(fn − f )kL1
and the result follows from arbitrariness of .
A-3
Let x0n : y 7→ hxn , yi , H → C be a sequence of elements in H ∗ . Since
supn |x0n (y)| = supn | hxn , yi | ≤ My < ∞ for each y ∈ H, by uniform
boundedness principle we have Dkx0n k is Ebounded by M in n, Therefore,
M ≥ sup{y∈H,kyk=1} | hxn , yi | ≥ | xn , kxxnn k | = kxn k for every n.
A-4
If x ∈ LP ∩ Lr , then xχx>1 ∈ Lp ⊆ Lq and xχx≤1 ∈ Lr ⊆ Lq . By
Minkowski’s inequlity, kxkLq ≤ kxχx>1 kLq + kxχx≤1 kLq , giving us the result.
A-5
Define d := inf{|x − z|, z ∈ M }. Pick a sequence {yn } ⊆ M so that
|x − yn | < d + n1 . Apply parallelogram’s law to 12 (x − yn ) and 12 (x − ym ), we
m 2
k + 14 kyn − ym k2 .
have 12 (kx − yn k2 + kx − ym k2 ) = kx − yn +y
2
Thus, 41 kyn −ym k2 ≤ 12 (2d2 + 2d
+ 2d
+ n12 + m12 )−d2 , meaning {yn } is Cauchy
n
m
and hence yn → y ∈ M . Therefore, d ≤ kx − yk ≤ kx − yn k + ky − yn k → d
as n → ∞, which shows that y is closest to x than any other element in M is.
If both y, y 0 minimize the distance to x, then by parallelogram’s law
applied to x − y, x − y 0 , we have 4d2 = k2x − y − y 02 k2 + ky − y 0 k2 =
02
4kx − y+y
k2 + ky − y 0 k2 ≥ 4d2 + ky − y 0 k2 , forcing y = y 0 .
2
2
B-6
To solve
1
1 − cos(a + ib) = 1 − (ei(a+ib) + ei(a−ib) )
2
1
= 1 − (e−b (cos(a) + i sin(a)) + eb (cos(a) + i sin(a))) = 0,
2
first we observe that the imaginary part of both sides are zero, so sin(a) =
0. If cos(a) = −1, then we have 1 + 12 (e−b + eb ) = 0, which is impossible.
If cos(a) = 1, then we have 2 = eb + e−b , so b = 0. Therefore the zeros for
1 − cos(z) are {2kπ : k ∈ Z}.
B-7
(i) The only isolated singularity is −1. Let z = −1 + bi, we have
| sin(
z
−1 + bi
i
)| = | sin(
)| = | sin 1 + |
z+1
bi
b
i
i
1
= | (ei(1+ b ) − e−i(1+ b ) )|
2i
−1
1
1
= | (ei+ b − e−i+ b )| → ∞
2i
as b → 0 + .
However, if we let z = −1 + a,
| sin(
z
)| = | sin(1 − 1/a)| ≤ 1 for every a 6= 1.
z+1
Thus 0 is neither a pole or a removable singularity ⇒ it must be an essential
singularity.
(ii)sin(z) =
z ∈ C. Let z =
P∞
n=0
w
w+1
=
f (n) (1)
(z − 1)n =
n!
1
1 − 1+w
, we have
∞
(n)
n f (1)
(1
n=0 (−1)
n!
P∞
− z)n for every
−∞
X
X
w
f (n) (1)
f (−n) (1)
sin(
)=
(−1)n
(w + 1)−n =
(−1)n
(w + 1)n .
w+1
n!
(−n)!
n=0
n=0
(iii) The coefficient of
1
1+w
is −f 0 (1) = − cos( 21 ) ·
3
z+1−z
|
(1+z)2 z=1
= − 14 cos( 21 ).
B-8
See B-10 of Spring 2011.
B-9
See B-6 of Fall 2007.
B-10
Rouche’s theorem states that if f and g are holomorphic on a closed disc
D with circle C as its boundary, and |f (z)| > |g(z)| on C, then f and f + g
has the same number of zeros inside C.
Take f (z) = 8, g(z) = 2z 5 − z 3 + 3z 2 − z, we find that |f | > |g| on
{z : |z| = 1}, thus f + g = 2z 5 − z 3 + 3z 2 − z + 8 has the same zeros as f on
{z : |z| < 1}, namely no zeros. Since |2z 5 − z 3 + 3z 2 − z| =
6 8, f (z) + g(z) 6= 0
when |z| = 1. By the fundamental theorem of algebra, f (z) + g(z) has 5
zeros on C, thus it also has 5 zeros on {z : |z| > 1}.
4