Real & Complex Analysis Qualifying
Exam Solution, Spring 2010
Shiu-Tang Li
January 1, 2013
A-1
R1
(a) Given > 0, P (|fn − fm | > ) < 1p 0 |fn (x) − fm (x)|p dx. Therefore,
for each j ∈ N, there exists Nj such that P (|fn − fm | > 21j ) < 21j for every
n, m ≥ Nj . If we choose n = Nj , m = Nj+1 , and
define Aj := {|fNj −fNj+1 | >
S∞
1
}, we find that {fNj (x)} is Cauchy on ( m=n Am )c for every n; thus
2j
{fNj (x)} is Cauchy a.e. x, actually. As a result,gj (x) := fNj (x) → f (x)
for some finite a.e. f and the proof is complete.
(b)Let f1 = χ[0,1/2] , f2 = χ[1/2,1] , f3 = χ[0,1/4] , f4 = χ[1/4,2/4] , f5 = χ[2/4,3/4] ,
f6 = χ[3/4,1] , f7 = χ[0,1/8] , f8 = χ[1/8,2/8] , · · · , and so on.
A-2
(a)Approach 1. We would show that the inclusion map i is a closed
operator. If fn → f in Lp (X), and i(fn ) = fn → g in Lq (X), then by the
application of Chebeshev’s inequality we have fn → f in measure and fn → g
in measure. Thus, f = g a.e. X. By closed graph theorem, The inclusion i
is continuous linear operator and hence it is bounded.
Approach 2. The idea is basically due to Shiang Tang. Assume that
i is not bounded.
there exists kfn kp = 1 for all n but kfn kp ≥ n2n .
P Thus,
Consider f = n=1 21n |fn |. It is obvious that f ∈ Lp (X); however, for every n ∈ N, kf kq ≥ k 21n fn kq ≥ n, so f ∈
/ Lq (X), contradicts the fact that
p
q
L (X) ⊂ L (X).
(b) Assume that inf E 0 ∈M 0 µ(E 0 ) = 0, so we could pick a sequence of
An ∈ M 0 so that µ(An ) → 0. Let fn = µ(An )−1/p χAn , and it is easy to
1
check that kfn kp = 1 for every n ∈ N. However, kfn kq = µ(An )1/q−1/p → ∞
as n → ∞. (a) above is thus violated.
Remarks. To prove conversely, consider inf E 0 ∈M 0 µ(E 0 ) = m > 0. For
every Rf ∈ Lp , we claim
that
f is bounded and hence
f ∈ Lq . Otherwise,
R
P
P
∞
∞
∞ > PX |f |p dµ = n=0 X |f |p χ{n≤|f |<n+1} dµ ≥ n=0 np µ(n ≤ |f | < n +
1) ≥ ∞
of f will result in
n=1 µ(n ≤ |f | < n + 1) = ∞, since unboundednessP
infinitely many positive terms that are larger than m in ∞
n=1 µ(n ≤ |f | <
n + 1), and this is a contradiction.
A-3
(a) Decompose H = V ⊕ V ⊥ . For every h = v + w, h0 = v 0 + w0 , where
v, v 0 ∈ V and w, w0 ∈ V ⊥ ,
(i)πV (h) = πV2 (h) = v – idempotent
kvk
V (h)k
= khk
≤ 1 for all h ∈ H. – Norm
(ii) Since khk2 = kvk2 + kw2 k, kπkhk
less than 1.
(iii) (πV (h), h0 ) = (v, v 0 + w0 ) = (v + w, v 0 ) = (h, πV (h0 )). – Self adjoint
(b) Let N (P ) be the null space of P . we claim that N (P ) is closed, for
if an ∈ N (P ) and an → a ∈ N (P ), P (an ) = 0 → P (a) = 0 by continuity of P .
Now we decompose H = N (P ) ⊕ N (P )⊥ . For every a ∈ N (P ), b ∈
N (P )⊥ , 0 = (P (a), b) = (a, P (b)), hence P (b) ∈ N (P )⊥ for all b ∈ N (P )⊥ .
We claim that b = P (b) for all b ∈ N (P )⊥ . Consider P (b − P (b)) =
P (b) − P 2 (b) = P (b) − P (b) = 0, showing b − P (b) ∈ N (P ). However,
b − P (b) also belongs to N (P )⊥ due to the arguments above. The only possibility is that b − P (b) = 0.
Therefore, for every u = a + b ∈ H, a ∈ N (P ) and b ∈ N (P )⊥ , P (a + b) =
P (b) = b, and the proof is complete.
A-4
P
(a) Norm-preserving: For every ξ ∈ l∞ , kξkl∞ = 1, Λη (ξ) ≤ i |ηi | =
kηkl1 . Thus kηk(l∞ )∗ ≤ kηkl1 . Conversely, if we choose ξ so that ξi = sgn(ηi ),
2
then kηk(l∞ )∗ ≥ Λη (ξ) =
P
i
|ηi | = kηkl1 .
P
P
Injection: if i ηi ξi = i ηi0 ξi for every ξi ∈ l∞ , then we may choose ξ
so that ξj = 1 and ξk = 0 for k 6= j. Thus ηi = ηi0 for every i and hence η = η 0 .
Not onto: let c ∈ (l∞ )∗ be such that c(ξ) = lim supn ξn and assume that
c = Λη for some η ∈ l1 . If we choose ξ such that ξk vanishes in every subscript
k but some j, then we have ηj = 0. Arbitrariness of j shows that η = 0 and
c is a zero linear functional, which is a contradiction.
(b)µ : σ−finite positive measure; 1 ≤ q < ∞; p is chosen s.t.
For a proof, see Rudin page 127.
1
p
+
1
q
= 1.
A-5 (Poincare’s recurrence theorem)
Given E ∈ M , let the collection of sets satisfying this property be
A = AE , and note that A is measurable. We claim that {T −n (A)}n are
pairwise disjoint. If, x ∈ T −n (A) ∩ T −m (A), m > n, then T n (x) ∈ A and
T m−n T n (x) ∈ A, a contradiction to the definition of A.
P
P
S
−n
−n
(A)) = ∞
(A)) = ∞
Therefore, ∞ > µ(X) ≥ µ( ∞
n=1 µ(A),
n=1 µ(T
n=1 T
the only possibility is that µ(A) = 0.
B-6
(a) Res(f ; 1) = limz→1 (z − 1)f (z) = 6+2
= R1; Res(f ; −1) = limz→−1 (z +
2·4
−6+2
1)f (z) = −2·2 = 1; by the residue theorem, γ f (z) dz = 2πi(Res(f ; 1) +
Res(f ; −1)) = 4πi.
3
(b)
1
1
z
· (1 − 2 )−1 · (1 + )−1
2
3z
z
3
∞
∞
m
X
X
1
1
mz
= (6z + 2) · 2 ·
(−1) m
3z n=0 z 2n m=0
3
f (z) = (6z + 2) ·
=
=
∞
X
2 · (−1)m
n,m=0
∞
X
k=−∞
m−2 1 z m−1 2
m 1 z
(−1)
+
z 2n 3m
3
z 2n 3m
X
2(−1)m k
2(−1)m
z .
+
m
m+1
3
3
m,n≥0:m−2−2n=k
m,n≥0:m−1−2n=k
X
Where the interchange of summation is due to absolute and uniform converP∞ 2(−1)2n P∞ 2(−1)2n+1
−1
gence.
The
coefficient
of
z
is
+ n=0 32n+2 = 2. Therefore,
n=0
32n
R
R 2
f (z) dz = γ z dz = 4πi.
γ
B-7
Z
2π
2n
Z
1
1
( (z + z −1 ))2n dz
iz
|z|=1 2
Z
(z 2 + 1)2n
1
dz
= n
i4 |z|=1 z 2n+1
cos θ dθ =
0
1
(z 2 + 1)2n
2π
2n
= 2πi · n Res(
; 0) = n ·
.
2n+1
i4
z
4
n
B-8
(a) Let f 0 (z0 ) = 0. If f 0 (z) ≡ 0 on U , then f is a constant, and of course
it is not injective. So by uniqueness theorem we may assume that there is
some neighborhood Bz0 (R) of z0 such that f 0 (z) 6= 0 for z ∈ Bz0 (R) \ {0}.
Since f 0 (z0 ) = 0, we have f (z) = f (z0 ) + (z − z0 )k g(z) in a neighborhood
B := Bz0 (r) of z0 , where k ≥ 2, and g is holomorphic and non-vanishing on
B. And here we choose r < R. Let m := 12 supξ∈∂B |(z − z0 )k g(z)| > 0, we
have that (z − z0 )k g(z) + m and (z − z0 )k g(z) have both k zeros on B by
4
Rouche’s theorem. Therefore, f (z) − f (0) + m has k zeros on B.
Since (f (z)−f (0)+m)0 = f 0 (z) 6= 0 on B \{0}, These k zeros {z1 , · · · , zk }
of f (z) − f (0) + m on B are all distinct, and obviously none of them equals
0. Therefore, f (z1 ) = · · · = f (zk ), thus f is not injective.
(b) False. Let f (z) = z 2 defined on an open connected set which contains
{1, −1} but does not contain {0}.
B-9
2+z
Consider a(z) = 2−z
, which is analytic on HD2 := {z : |z| < 2, Re z > 0},
and it maps HD2 to Q := {z = a + bi : a > 0, b > 0}. To see this, notice that
Arg(2+z) = Arg(z−(−2)), and if we pick −2,2, and z ∈ HD2 as vertices of a
triangle, it is an obtuse triangle. Therefore, 0 < Arg(2 + z) − Arg(2 − z) < π2 ,
2+w
and this shows 2+z
∈ Q. Besides, for any z ∈ Q, z = 2−w
⇒ w = 2z−2
∈
2−z
z+1
2z−2
−1
−1
HD2 , we may define a (z) : Q → HD2 by a (z) = z+1 , which is the
inverse map of a. Therefore, a is a conformal map (analytic and bijective)
from HD2 to Q.
Let b(z) = z 2 : Q → H := {z : Im z > 0}. It is easy to check that b is
also a conformal map.
i−z
Finally, we define c(z) = i+z
, which is analytic on H, and for z = a + bi,
b > 0, |i − z| = | − a + (1 − b)i| < |a + (1 + b)i| = |i + z|, which shows
that c : H → D := {z : |z| < 1}. To see c(z) is onto and 1 − 1, for any
1−r2 −2i sin(θ)
1−reiθ
i−z
w = reiθ ∈ D, r < 1, w = i+z
⇒ z = i 1+re
iθ = i 1+r 2 +2r cos(θ) ∈ H. Thus,
we may define the inverse map of c by c−1 (z) = i 1−z
: D → H. These facts
1+z
show that c is a conformal map from H to D.
Therefore, F (z) := (c ◦ b ◦ a)(z) =
2+z 2
i−( 2−z
)
2+z 2
i+( 2−z
)
is a conformal map from HD2
to D. If φ(z) is another conformal map from HD2 to D, then φ ◦ F −1 is a
a−z
conformal map from D to D, which takes form eiθ 1−az
for some a ∈ D. As
a−F
(z)
a result, φ(z) = eiθ 1−aF (z) .
5
B-10
(w)
Define g(ζ) = f (ζ)−f
, g(w) = f 0 (w). Since g is continuous on the unit
ζ−w
disk D, and is holomorphic on D \ {0}, thus g is holomorphic on D by Morera’s theorem.
2M
As a result, when z 6= w, we have |g(w)| ≤ supζ∈∂B0 ( 1+r ) |g(ζ)| ≤ (1+r)/2
.
2
4M
That is, |f (z) − f (w)| ≤ 1+r |z − w|, and the constant is independent of our
choices of z, w.
6