Real & Complex Analysis Qualifying
Exam Solution, Spring 2009
Shiu-Tang Li
December 21, 2012
A-1
Refer to the solution of A-1(b) in Spring 2010 qualifying exam.
A-2
R
Since XR |f |p dµ ≤ kf kp∞ µ(X), kf kp ≤ kf k∞ . On the other hand, for every > 0, ( X |f |p dµ)1/p ≥ (kf k∞ −)µ(|f | > kf k∞ −)1/p ⇒ lim inf p→∞ kf kp ≥
kf k∞ − . Arbitrariness of and the fact that kf kp ≤ kf k∞ for every p > 0
shows that limp→∞ kf kp exists and equals kf k∞ .
A-3
Since T maps {x : kxkX = 1, x ∈ X} to a precompact set in Y , it follows
that T is a bounded linear operator (and is hence continuous). The fact that
T is onto and the open mapping theorem tells us T maps every open set in
X to some open set in Y .
Now for each y ∈ Y , y = T (x) for some x ∈ X. It follows that T (B1 (x))
is a open set that contains y, and it is percompact since B1 (x) is a bounded
set. As a result, T (B1 (x)) is a compact neighborhood of y (Since it contains
T (B1 (x))).
The above arguments show that Y is locally compact. Since Y is also a
topological vector space, it follows that Y is finite dimensional. It is not an
easy job to prove this; alternatively, we may use the fact that Y is a locally
1
compact normed linear space, of which the proof relies only Riesz’s lemma
and is easier.
A-4
See the solution of A-3 in Fall 2010 qualifying exam.
A-5
ˆ
Since f ∈ L1 (R),
convergence theoR∞
R ∞ f (x) ∈ C(R)R 1by use of dominated
rem. Therefore, −∞ |fˆ(x)| dx ≤ −1 |fˆ(x)| dx + −∞ |xfˆ(x)| dx < ∞, and
this shows fˆ(x) ∈ L1 (R),
R ∞and now Fourier inversion formula is applicable
and we have f (x) = √12π −∞ fˆ(y)eixy dy a.e. R.
R∞
√1
fˆ(y)eixy dy is
2π −∞
R∞
ihy
√1
fˆ(y)eixy e h−1 dy.
2π −∞
The final step is to show g(x) =
differentiable any-
where. Consider g(x+h)−g(h)
=
As h → 0, mean
h
value theorem along with dominated convergence theorem
R ∞ with dominating
√i
=
y fˆ(y)eixy dy.
function |2y fˆ(y)| tells us that limh→0 g(x+h)−g(h)
h
2π −∞
B-6
First we observe that
sin(z)
z
=
P
n z 2n
n=0 (−1) (2n+1)!
= 1+z 2
n z 2(n−2)
n=1 (−1) (2n+1)! .
P
= 1, and we have f (z)z → 1 as z → 0, which implies 0 is
Thus limz→0 sin(z)
z
a simple pole for f , and the residue at 0 is exactly limz→0 f (z)z = 1.
The residue formula implies that
R
γ
f (z) dz = 2πiRes(f ; 0) = 2πi.
B-7
See B-7 of Spring 2011. But this is much easier - we only need to subtract
principal part of each pole from f , to form a bounded entire function.
B-8
on Ω. Even if the line integral
R We MUST assume that f is continuous
1
f (z) dz = 0 for every piecewise-C closed curve, we still need to assume
γ
2
continuity to assure that Morera’s theorem works.
For each
z0 ∈ Ω, there is some > 0 so that Bz0 () ⊂ Ω. Define
R
F (z) = γ(z) f (w) dw on Bz0 (), where γ(z) is the line segment(s) starting
from z0 = a + bi to c + bi, then c + bi to c + di = z.
(z)
=
For h 6= 0 small enough, For every z ∈ Bz0 () we have F (z+h)−F
h
R
1
f
(w)
dw,
where
γ(z,
h)
is
the
line
segment(s)
starting
from
z
=
a
+
bi
h γ(z,h)
to c + bi, then c + bi to c + di = z + h, by the assumption of this problem.
Therefore,
Z
1
F (z + h) − F (z)
1
− f (z)| = |
|
f (w) dw − hf (z)|
h
h γ(z,h)
h
Z
1
f (w) − f (z) dw|
=|
h γ(z,h)
Z
1
≤
|f (w) − f (z)| dw
|h| γ(z0 ,h)
1
≤
|h| sup |f (w) − f (z)| → 0
as h → 0,
|h| w∈γ(z,h)
since supw∈γ(z,h) |f (w) − f (z)| → 0 due to continuity of f at z. This implies
F 0 (z) = f (z) on Bz0 ().
We can define F (z) on every small ball B ∈ Ω in this way. Consider F1
defined on B1 , F2 defined on B2 , B1 ∩ B2 6= ∅, but F1 6= F2 on B1 ∩ B2 , then
F10 (z) − F20 (z) = f (z) − f (z) = 0, so that F1 (z) − F2 (z) = c on B1 ∩ B2 . We
may redefine F2 by adding a constant to make F1 = F2 on B1 ∩ B2 .
S
Now we write
Ω
=
z∈Ω Bz (rz ), and by Lindelöf’s Theorem, we have
S∞
further Ω = n=1 Bn , where each Bn is of form Bz (rz ), and there is some
Fn (z) defined on Bn so that Fn0 = f on Bn . We may apply the technique
in the previous paragraph to redefine Fn (z) for every n ∈ N, and define
F (z) := Fn (z) on Bn . The proof is complete.
B-9
It seems that this problem is misstated.
3
B-10
Z
0
2π
1
dθ =
a + sin θ
Z
Z|z|=1
=
|z|=1
1
1
·
dz
iz a + (z − z −1 )/2i
2
dz
i2az + z 2 − 1
√
Solving √
i2az + z 2 − 1 = 0, we have z = −i2a ± 4a2 − 1i. Since a > 1, only
−i2a + 4a2 − 1i = √4a2−i
lies in the unit disk. By the residue theorem,
−1+2a
√
R
2
2
dz
=
2πiRes(
;
−i2a
+
4a2 − 1i) = 2πi √4a12 −1i =
2
2
i2az+z −1
|z|=1 i2az+z −1
√ 2π
.
4a2 −1
4