Real & Complex Analysis Qualifying
Exam Solution, Spring 2011
Shiu-Tang Li
January 1, 2013
A-1
P
(a)Let A be a collection of sequences: {(a1 , a2 , · · · )|ai ∈ Q, ∞
i=1 |ai | <
1
∞)}. For each (b1 , b2 , · · · ) ∈ l , we can find (a1 , a2 , · · · ) ∈ A s.t. |ai − bi | < 2i
when arbitrary is given.
(b)The techniques in (a) may be applied to this problem, too.
A-2
(a)Let g(x) : x 7→ |x|, R → R. Since g is a continuous function, g −1
maps every B ∈ B(R) to B(R). Therefore, |f |−1 = f −1 ◦ g −1 maps every
B ∈ B(R) to a measurable set in X.
(b)Let A be a set in X which is not measurable. It’s obvious that f (x) =
1A (x) − 1{X\A} (x) is non-measurable while |f | ≡ 1 is a measurable function.
1
A-3
Assume that at θ = 0, every continuous function f satisfies
f (1) = lim
N →∞
N
X
cn
n=−N
Z 2π
N
X
1
= lim
f (eiθ )e−inθ dθ
N →∞
2π
0
n=−N
Z 2π
−iN θ
1
(1 − ei(2N +1)θ )
iθ e
= lim
f (e )
dθ
N →∞ 2π 0
1 − eiθ
Z 2π
1
e−i(N +1/2)θ − ei(N +1/2)θ
= lim
f (eiθ )
dθ
N →∞ 2π 0
e−iθ/2 − eiθ/2
Z 2π
sin((N + 1/2)θ)
1
dθ
f (eiθ )
= lim
N →∞ 2π 0
sin(θ/2)
R 2π
Define wn := C[eiθ : 0 ≤ θ < 2π] → R by wn (f (eiθ )) = 0 f (eiθ ) sin((n+1/2)θ)
dθ.
sin(θ/2)
iθ
For each g ∈ C[e : 0 ≤ θ < 2π], |wn (g)| ≤ Mg for some Mg > 0 (Since
wn (g) → 2πg(1)). Therefore, kwn k ≤ M by uniform boundedness principle,
for C[eiθ : 0 ≤ θ < 2π] endowed with sup-norm is a Banach space. By defiR 2π
nition, 0 | sin((n+1/2)θ)
| dθ ≤ kwn k ≤ M , for every n ∈ N.
sin(θ/2)
However, this is wrong. Note that
Z π
Z 2π
sin((n + 1/2)θ)
sin((n + 1/2)θ)
| dθ ≥
|
| dθ
|
sin(θ/2)
sin(θ/2)
0
0
Z π
sin((n + 1/2)θ)
≥
|
| dθ
θ/2
0
Z (n+1/2)π
sin(z)
| dz
=2
|
z
0
n
X
1
≥2
> M when n is large enough,
jπ
j=1
which is a contradiction.
A-4
Assume that lim|x|→∞ f (x) 6= 0. Therefore, there is a sequence {xn } s.t.
|xn | → ∞ and |f (xn )| > a > 0 for all n large enough. Since f is uniformly
2
continuous, there exists a δ so that for |x − y| < δ, |f (x) − f (y)| < a/2. As
a result, for all n large enough, |f (y)| > a/2 for y ∈ Bδ (xn ). We may pick a
further subsequence
{xnP
} of {xn } so that these balls are mutually disjoint.
k
R
a
2
As a result, |f | dx ≥ ∞
n=1 2 · δ π = ∞, which contradicts the fact that
f ∈ L1 (R).
A-5
Apply Fatou’s lemma to f + fn − |f − fn |, we have
Z
Z
Z
Z
f dµ − lim sup
|f − fn | dµ ≥
lim inf(f + fn − |f − fn |) dµ = 2
f dµ.
2
X
X
Therefore, lim sup
R
X
X
X
|f − fn | dµ ≤ 0.
B-6
For each z0 on U, there is some r > 0 so that f (z) =
for all |z − z0 | < r. Taking conjugates, we have
f (z) =
∞
X
an (z − z0 )n
P∞
n=0
an (z − z0 )n
for all |z − z0 | < r
n=0
⇒g(z) = f (z) =
∞
X
an (z − z 0 )n
for all |z − z 0 | = |z − z0 | < r.
n=0
That is, g(z) has power point representation at z 0 . This shows g(z) is holomorphic on U .
B-7
We first claim that f has only finitely many poles on {z : |z| ≤ R}, for
otherwise there would be a cluster point of these poles, and it would be a
contradiction to the definition of meromorphic functions.
(1/z)
1
Consider g(z) = f(1/z)
Since the only
k on B0 (1/R) := {z : |z| < R }.
isolated singularity is 0, and |g(z)| ≤ C on B0 (1/R), 0 is a removable
singularity of g. Redefine g and we may write f (1/z) = (1/z)k g(z) on
a deleted neighborhood
of 0, where g is holomorphic on B0 (1/R). ThereP
n
fore, f (1/z) = ∞
b
n=−k n z on a deleted neighborhood of 0, and we define
3
h(z) =
Pk
n=1 b−n z
k
for z ∈ C.
Now we let all the poles ofP
f be {z1 , · · · , zm }. For each zj , we have Laun
rent series expansion f (z) = ∞
n=−N (j) aj,n (z − zj ) on a deleted neighborP
n
hood of zj , and we define fj (z) = −1
n=−N (j) aj,n (z −zj ) for every z ∈ C\{0}.
P
We claim that F (z) = f (z) − m
j=1 fj (z) − h(z) is a bounded entire function. To see F is entire, we note
P that F is holomorphic on each zj , since
both f (z) − fj (z) and fj (z) − m
j=1 fj (z) − h(z) are holomorphic on zj , and
this is true for every 1 ≤ j ≤ m. To see F is bounded, we only need to show
that f (z) − h(z) is bounded for all |z| large, which is equivalent to show that
f (1/z) − h(1/z) is P
bounded for all |z| small. By the definition of h, we have
n
f (1/z)−h(1/z) = ∞
n=0 bn z , which is obviously bounded on a neighborhood
of 0.
As a result, by
PLouville’s theorem, F (z) is a constant a on C. This shows
that f (z) = a + m
j=1 fj (z) + h(z), where each fj is a rational function and
h(z) is a polynomial, and therefore f (z) is also a rational function.
B-8
(i) ⇒ (ii): By definition, we have f (z) = (z − a)n g(z) on Ba (), where
n ≥ k and g(z) is holomorphic on Ba (), for some > 0. Take C =
supz∈Ba () |(z − a)n−k g(z)| = supz∈∂Ba () n−k |g(z)|.
f (z)
(ii) ⇒ (i): Let g(z) = (z−a)
k , we have |g(z)| ≤ C on Ba (), and g(z) is
holomorphic on Ba () \ {a}. Riemann’s theorem on removable singularities
imply that 0 is a removable singularity of g, and we may redefine g to be
holomorphic on Ba (). We may write f (z) = g(z)(z − a)k ; that is, 0 is a zero
of f of order ≥ k.
B-9
It is obvious that f (z) is differentiable on C \ {0}, and f (0) is not defined.
Thus 0 is the only singularity.
P∞
P∞
Notice that if both
n=1 an and
m=1 bm converge absolutely, then by
P
Tonelli’s theorem n,m |an bm | < ∞. Therefore, we may sum the double array {an bm }n,m in any order.
4
3
5
sin(z) = z − z3! + z5! − · · · for all z ∈ C, and sin( z1 ) = z1 − z313! + z515! − · · ·
for all z ∈ C \ {0}. Since for every z ∈ C \ {0}, both series are absolutely
convergent, we have
∞
∞
2n−1
X
X
1
1
n+1 z
sin(z) sin( ) =
(−1)
(−1)m+1 2m−1
z
(2n − 1)! m=1
z
(2m − 1)!
n=1
=
+
∞ X
∞
X
k=0 n=1
∞ X
∞
X
(−1)n+k+1
(−1)
k=0 n=1
n+1
z 2n+2k−1
1
(−1)n+1 2n−1
(2n − 1)!
z
(2n − 1)!
1
z 2n−1
(−1)n+k+1 2n+2k−1
(2n − 1)!
z
(2n + 2k − 1)!
and we find that the coefficient of 1/z in the above Laurent series is 0, so the
residue of z = 0 is 0.
B-10
RR
cos(x)
dx+
Let CR = Reiθ , θ goes from 0 to π. By the residue theorem, −R x2x+2x+10
R R x sin(x)
R
(−1+3i)e−3−i
zeiz
i −R x2 +2x+10 dx + CR z2 +2z+10 dz = 2πiRes(f ; −1 + 3i) = 2πi ·
=
6i
e−3 π
(− cos(1) + 3 sin(1) + 3i cos(1) + i sin(1)). Since
3
Z
|
CR
Z π
zeiz
Reiθ · eiR cos(θ)−R sin(θ)
dz|
=
|
iReiθ dθ|
2 ei2θ + 2Reiθ + 10
z 2 + 2z + 10
R
0
Z π
R2
≤
e−R sin(θ) 2
dθ
R − 2R − 10
0
Z π/2
R2
=2
e−R sin(θ) 2
dθ
R − 2R − 10
0
Z π/2
2
R2
≤
e−R· π θ 2
dθ
R − 2R − 10
0
π
R2
=
(1 − e−R ) 2
→ 0,
2R
R − 2R − 10
as R → ∞. Therefore, as R → ∞, we have
Z ∞
x sin(x)
e−3 π
dx
=
(3 cos(1) + sin(1)).
2
3
−∞ x + 2x + 10
5