Real & Complex Analysis Qualifying
Exam Solution, Fall 2008
Shiu-Tang Li
January 1, 2013
A-1
P
(a) It’s obvious that Ta is linear. (Ta (x))n ≤ ni=1 |ai xi | ≤ supn |xn |·kak1 .
Therefore, supn |(Ta (x))n | ≤ supn |xn | · kak1 , so Ta is bounded.
∞
(b) We have shown kTa k ≤ kakP
s.t. yi = sgn(ai ),
1 in (a). Choose y ∈ l
n
we have kTa k ≥ kTa (y)k∞ = supn i=1 |ai | = kak1 .
A-2
(a) By Holder’s inequality, kf kL2 (X) k1kL2 (X) ≥ kf kL1 (X) . for every f ∈
L2 (X) ⊂ L1 (X). Thus, kik ≤ k1kL2 (X) = µ(X)1/2 . On the other hand,
kik ≥ ki(µ(X)−1/2 )kL1 (X) = µ(X)1/2 .
(b) L1 (X) = L2 (X) implies that L1 (X) ⊂ L2 (X). See the solution of A-2
in Spring 2010 qualifying exam.
A-3
(⇒)Let f = cg. For every h ∈ H s.t (g, h) = 0, we have (f, h) = c(g, h) =
0, so there exists no h that satisfies both conditions.
(⇐) Define G = span{g}, which is closed linear subspace space in H, we
may write H = G ⊕ G⊥ and G⊥ is closed. Let f 6= cg for every c ∈ C, we
may write f = g0 + g1 , where g0 ∈ G and g1 ∈ G⊥ , g1 6= 0. Define h := kgg11k2
and we may find that h satisfies (g, h) = 0 for all g ∈ G and (f, h) = 1.
1
A-4
Note that on (x, y) ∈ [0, 1] × [0, 1], |f (x)χx>y | ≤ |f (x)|, and ∞ >
R
R1R1
|f (x)| dx dy = [0,1]×[0,1] |f (x)| d(x × y) by Tonelli’s theorem, which
0 0
shows Rf (x) ∈ L1 ([0, 1]×[0,
by Fubini’s theR 1])
R and so is f (x)χx>y . RTherefore,
R
orem, [0,1] xf (x) dx = [0,1] [0,1] f (x)χx>y dy dx = [0,1] [0,1] f (x)χx>y dx dy =
R
R
f (x) dx dy.
[0,1] [y,1]
A-5
Assume that Φ(f ) = k 6= 0 for some f ∈ Cc (R) (Since Cc (R) is dense
in
C
Pn 0 (R) in sup-norm.) Assume that supp(f ) ⊂ (−M, M ). Consider fn :=
j=−n f (x + 2jM ), and it is easy to find that kf kC0 (R) = kfn kC0 (R) for all n.
However, Φ(fn ) = (2n + 1)k, and this shows Φ is not bounded.
B-6
i2z
is holomorphic
Let CR = Reiθ , θ goes from 0 to π. Since f (z) = e −1−i2z
z2
on C (except for z = 0, but it is a removable singularity so we can redefine
RR
R R sin(2x)−2x
R ei2z −1−i2z
f ). Therefore, −R cos(2x)−1
dx+i
dx+
dz = 0...(*).
2
2
x
x
z2
−R
CR
R
R
R
i2z
Since | CR e z2−1 dz| ≤ πR · R22 , and CR −i2z
dz = −2i CR z1 dz = 2π, we
z2
let R → ∞ and take real parts in (*) to obtain
Z ∞
Z ∞
sin2 (x)
cos(2x) − 1
dx
+
2π
=
0
=
−2
dx + 2π.
x2
x2
−∞
−∞
It follows that
R∞
−∞
sin2 (x)
x2
dx = π ⇒
R∞
0
sin2 (x)
x2
dx = π/2.
B-7
Assume not. Let D \ {a} be a deleted neighborhood of a. Then there
is some r > 0, w ∈ C, so that |f (z) − w| > r for all z ∈ D \ {a}. Define
1
g(z) = f (z)−w
, we have that |g(z)| < 1r for z ∈ D \ {a}. As a result,
a is a removable singularity of g. If g(a) = 0, then a is a pole for f , a
contradiction. If g(a) 6= 0, then a is a removable singularity for f , which is
again a contradiction.
2
B-8
Approach 1 Apply Rouche’s theorem on the circle |z − R|2 = R2 for
arbitrary R > 0.
Approach 2 Let z = a + bi. We may rewrite the equation as α − a −
e cos(b) − ib + i sin(b) = 0. If π/2 > b > 0, then by MVT, | sin(b)/b| =
| cos(b0 )| < 1, where b0 lies between 0 and b. If b ≥ π/2, then we also have
| sin(b)/b| < 1. Similar for b < 0. As a result, 0 is the only real root for
sin(b) = b, and the equation becomes
−a
α − a − e−a = 0.
Let f (x) = x + e−x . Since f (0) = 1 < α and f (α) > α, by intermediate
value theorem we can find x0 so that f (x0 ) = α, which proves the existence of
this equation. To prove uniqueness, we consider x0 > y 0 > 0 be two different
solutions, and we have
0
0
0
x0 − y 0 = e−y − e−x = e−z (x0 − y 0 ) < x0 − y 0
for some y 0 < z 0 < x0 , by MVT, and this is a contradiction. This proves
the uniqueness.
B-9
Since φ is a bijective (and also holomorphic) map, the inverse φ−1 exists,
and it is continuous by open mapping theorem. Therefore, w = φ(z) → w0 =
φ(z 0 ) implies z → z 0 , and we have
lim 0
w→w
−1
w − w0
φ−1 (w) − φ−1 (w0 )
=
lim
w→w0 φ−1 (w) − φ−1 (w 0 )
w − w0
φ(z) − φ(z 0 ) −1
= lim 0
w→w
z − z0
−1
,
= (φ0 (z 0 ))−1 = φ0 (φ−1 (w0 ))
which is well-defined because φ0 (z) 6= 0 for all z ∈ Ω.
Let f : Ω → D be an arbitrary holomorphic map. f ◦ φ−1 is a holomorphic map from D → D, so by Schwarz’s lemma we have 1 ≥ |(f ◦
3
−1
0
−1
φ ) (0)| = |f (φ (0))(φ ) (0)| = |f (φ (0)) φ (φ (0))
|. Therefore,
−1 0
0
−1
−1 0
0
−1
|φ0 (z0 )| ≥ |f 0 (φ−1 (0))| = |f 0 (z0 )|.
B-10
We would prove that g(z) = f (1/z) = U (z) + iV (z) is holomorphic on
|z| > 1. To see this, we let z = x + iy and f (z) = u(x, y) + iv(x, y), we have
y
y
x
x
g(z) = u( x2 +y
2 , x2 +y 2 ) − iv( x2 +y 2 , x2 +y 2 ), and for |z| > 1,
∂ x
y
x
y
x2 + y 2 − 2x2
u( 2
,
)
=
u
(
,
)
·
x
∂x
x + y 2 x2 + y 2
x2 + y 2 x2 + y 2
(x2 + y 2 )2
x
y
−2xy
+ uy ( 2
, 2
)· 2
2
2
x +y x +y
(x + y 2 )2
y
x2 + y 2 − 2x2
x
y
2xy
x
,
)
·
+ vx ( 2
, 2
)· 2
= vy ( 2
2
2
2
2
2
2
2
2
x +y x +y
(x + y )
x +y x +y
(x + y 2 )2
∂
x
y
=
− v( 2
, 2
) ,
2
∂y
x + y x + y2
and
∂ x
y
x
y
−2xy
u( 2
, 2
) = ux ( 2
, 2
)· 2
2
2
2
2
∂y
x +y x +y
x +y x +y
(x + y 2 )2
x
y
x2 + y 2 − 2y 2
+ uy ( 2
,
)
·
x + y 2 x2 + y 2
(x2 + y 2 )2
x
y
−2xy
x
y
x2 + y 2 − 2y 2
= vy ( 2
,
)
·
−
v
(
,
)
·
x
x + y 2 x2 + y 2 (x2 + y 2 )2
x2 + y 2 x2 + y 2
(x2 + y 2 )2
∂ x
y
=
v( 2
,
)
.
∂x
x + y 2 x2 + y 2
Therefore, we have Ux (z), Uy (z), Vx (z), Vy (z) all exist and are continuous on
|z| > 1, and for each |z| > 1, Ux (z) = Vy (z) and Uy (z) = −Vx (z). This proves
−1
g(z) = f (1/z) is holomorphic on |z| > 1, and hence h(z) := f (1/z)
is
holomorphic on |z| > 1 except for those z such that g(z) = f (1/z) = 0.
However, there are only finitely such z’s, for otherwise { z1 : g(z) = 0} would
have a cluster point in the unit disk, and f ≡ 0 by uniqueness theorem. Let
these z’s be {z1 , · · · , zn }, and they are poles of h.
Define F (z) = f (z) when |z| ≤ 1, and F (z) = h(z) when |z| > 1, z ∈
/
{z1 , · · · , zn }. Since F (z) is holomorphic on C \ {z : |z| = 1} ∪ {z1 , · · · , zn },
4
and F (z) is continuous on |z| = 1 by the fact that |f (z)| = 1 on |z| = 1,
Morera’s theorem shows us F (z) is holomorphic on C \ {z1 , · · · , zn }, and the
proof is complete.
5