Real & Complex Analysis Qualifying
Exam Solution, Spring 2008
Shiu-Tang Li
Finished: January 1, 2013
Last updated: November 25, 2013
A-1
R
Pick M > 0 so that {f >M } f dµ < /2 by monotone convergence theorem.
and everything is fine now.
Now let δ = 2M
A-2
Z 1
Z 1
Z 1
2
2
1/2
f (y) dy)2 dx)1/2
kKk∞ (
kT f k2 = ( (T f (x)) dx) ≤ (
0
0
Z 1
Z0 1
kKk2∞ (
f (y)2 dy) dx)1/2
≤(
0
Z 1
Z0 1
2
1/2
≤(
kKk∞ dx) (
f (y)2 dy)1/2
0
0
= kKk∞ kf k2 .
A-3
We first claim that self-adjoint operators are bounded (Hellinger-Toeplitz).
To see this we consider un ∈ H for every n ∈ N, and un → u, T un → v.
Since (T un , w) → (v, w) and (T un , w) = (un , T w) → (u, T w) = (T u, w) for
every w ∈ H, it follows that T u = v, and closed graph theorem shows us
that T is bounded.
1
Let B be a bounded set in H. Our goal is to show that T (B) is precompact. Pick a sequence {T (xn )} from T (B), we may write each T (xn ) =
an1 y1 + · · · + anm ym , where {y1 , · · · , ym } is an orthonormal set that spans
T (H). Since T is bounded, |T (xn )| is bounded in n, and thus supi,j |aij | < ∞.
by Bolzano-Weierstrass theorem we can find a subsequence
Pm {T (xnj )}j so that
anj ,k → bk ∈ R for 1 ≤ k ≤ m. Therefore, T (xnj ) → k=1 bk yk . Therefore
T (B) is precompact, or relative compact.
The final task is to show V = K ⊥ . For each u ∈ H, v ∈ K, (T u, v) =
(u, T v) = (u, 0) = 0. Thus V ⊂ K ⊥ .
On the other hand, since V and K are both closed linear spaces, we have
H = V ⊕ V ⊥ = K ⊕ K ⊥ . Besides, since every element w in V ⊥ satisfies
(w, T y) = 0 for every y ∈ H, we have (T w, y) = 0 for every y ∈ H and thus
T w = 0. This shows V ⊥ ⊂ K. Now for every x ∈
/ V , we have x = v + v 0 ,
where v ∈ V , v 0 ∈ V ⊥ , and v 0 6= 0. Since V ⊂ K ⊥ and V ⊥ ⊂ K, it follows
that x ∈
/ K ⊥ . Therefore, K ⊥ ⊂ V . The proof of V = K ⊥ is complete.
A-4
(The proof may also be found in Theorem 5.14 of [Rudin], but I adopt a
slightly different approach here.)
Given > 0, for every f ∈ L1 [−π, π], we may find M > 0 large so that
kf − fM kL1 [−π,π] < , where fM := (f ∧ M ) ∨ (−M ).
We let µ be the Lebesgue measure on [−π, π]. By Lusin’s theorem (Theorem 9. above), since µ(x : fM (x) 6= 0) < ∞, there is a function g ∈ Cc [−π, π]
such that µ(fM 6= g) < /M , and supx∈[−π,π] |g(x)| ≤ supx∈[−π,π] |fM (x)| ≤
M . It follows that kg − fM kL1 [−π,π] < 2, which shows C[−π, π] is dense in
L1 [−π, π].
Since C[−π, π] contains all uniformly continuous
functions on [−π, π], for
Pm
every g ∈ C[−π, π] we may find h(x) = i=1 ai χ[bi ,bi+1 ) , −π = b1 < b2 <
· · · < bm+1 = π so that h is close to g in L1 -norm.
2
Consider
Z
π
m
X
ai χ[bi ,bi+1 ) e
inx
dx =
−π i=1
=
m
X
i=1
m
X
Z
bi+1
ai
einx dx
bi
ai
i=1
1 inbi+1
(e
− einbi )
in
which → 0 as n → ∞. Therefore, for every f ∈ L1 [−π, π] we may pick h of
the form above and kf − hkL1 [−π,π] < δ, and thus
Z
π
|
inx
f (x)e
−π
Z
π
dx| ≤
Z
π
|f (x) − h(x)| dx + |
−π
h(x)einx dx|
−π
≤ 2δ
for all n large enough. This completes the proof.
A-5
For every x ∈ l∞ , kT xkl1∗ = supkykl1 =1 |
kxkl∞ . This shows that kT k ≤ 1.
P
i
xi yi | ≤ supkykl1 =1 kxkl∞
P
i
|yi | =
For x, y ∈ l∞ , x 6= y, ∃i so that xi 6= yi . Define ei so that (ei )k = δik . We
have T x(z) = xi 6= yi = T y(z).
P
To prove that T is onto, for every x ∈ l1∗ , decompose x = i zi (ei )∗ ,
where (ei )∗ ∈ l1∗ and (ei )∗ (ej ) = δij . We find that supi |zi | < ∞; otherwise,
|x(en )|
we may pick a subsquence {znj }j so that |znj | → ∞, and ken kj 1 = |znj |
j l
is unbounded in j, contradicts the fact that x is bounded. Now we have
z = (z1 , z2 , · · · ) ∈ l∞ and T z = x. The proof is complete.
B-6
Hard.
B-7
P
n zn
Since cos(z) = 12 (eiz +e−iz ) is an entire function, cos(z) = ∞
n=0 (−1) (2n)!
P∞
n z n−2
for every z ∈ C. Therefore, when z 6= 0, cos(z)
=
n=0 (−1) (2n)! =
z2
P∞
zn
n
n=−2 (−1) (2(n+2))! , which is the Laurent series for F .
3
B-8
For every piecewise C 1 closed curve γ 0 ⊂ 4,
Z
Z Z
Z bZ d
f (z) dz =
φ(w)g(w − z) dw dz =
φ(w(s))g(w(s) − z(t))w0 (s)z 0 (t) ds dt
γ0
γ0 γ
dZ b
Z
=
c
Z
φ(w(s))g(w(s) − z(t))w0 (s)z 0 (t) dt ds
Z
φ(w(s))g(w(s) − z))w0 (s) dz ds
γ0
c
=
c
a
d
=
Z
a
d
0 · w0 (s) ds = 0,
c
since as a function of z, φ(w(s))g(w(s) − z)) is analytic in the region enclosed
by γ 0 . As a result, by Morera’s theorem, f is analytic on 4. (Actually it
suffices to show the result above for every triangular closed curve.)
B-9
Assume that f (z) 6= 0 for every z ∈ C, and lim inf |z|→∞ |f (z)| > 0. Since
for each pole zk , limz→zk |f (z)| = ∞, it follows that inf z∈C |f (z)| > 0 ⇒
1
1
supz∈C |f (z)|
< ∞. Since now f (z)
is a bounded entire function, f (z) is a
constant function by Louville’s theorem, which contradicts the assumption
of the problem.
B-10
Define g(w) = 1r (f (Rw + z) − f (z)). Note that g maps the unit disk to
itself, g(0) = 0, and g is holomorphic. By Schwarz lemma, we have
(i)|g(z)| ≤ |z|, and if the equality holds for some z0 6= 0 in the unit disk,
then g is a rotation.
(ii)|g 0 (0)| ≤ 1, and if the equality holds, then g is a rotation.
By (ii), |g 0 (0)| = | 1r f 0 (Rw + z)R|w=0 | ≤ 1 ⇒ |f 0 (0)| ≤
4
r
.
R