Post on 04-Mar-2018
Real & Complex Analysis QualifyingExam Solution, Spring 2008
Shiu-Tang Li
Finished: January 1, 2013Last updated: November 25, 2013
A-1
Pick M > 0 so that∫{f>M} f dµ < ε/2 by monotone convergence theorem.
Now let δ = ε2M
and everything is fine now.
A-2
‖Tf‖2 = (
∫ 1
0
(Tf(x))2 dx)1/2 ≤ (
∫ 1
0
‖K‖2∞(
∫ 1
0
f(y) dy)2 dx)1/2
≤ (
∫ 1
0
‖K‖2∞(
∫ 1
0
f(y)2 dy) dx)1/2
≤ (
∫ 1
0
‖K‖2∞ dx)1/2(
∫ 1
0
f(y)2 dy)1/2
= ‖K‖∞‖f‖2.
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, Tun → v.Since (Tun, w) → (v, w) and (Tun, w) = (un, Tw) → (u, Tw) = (Tu,w) forevery w ∈ H, it follows that Tu = v, and closed graph theorem shows usthat T is bounded.
1
Let B be a bounded set in H. Our goal is to show that T (B) is pre-compact. Pick a sequence {T (xn)} from T (B), we may write each T (xn) =an1y1 + · · · + anmym, where {y1, · · · , ym} is an orthonormal set that spansT (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 {T (xnj
)}j so thatanj ,k → bk ∈ R for 1 ≤ k ≤ m. Therefore, T (xnj
) →∑m
k=1 bkyk. ThereforeT (B) is precompact, or relative compact.
The final task is to show V = K⊥. For each u ∈ H, v ∈ K, (Tu, v) =(u, Tv) = (u, 0) = 0. Thus V ⊂ K⊥.
On the other hand, since V and K are both closed linear spaces, we haveH = V ⊕ V ⊥ = K ⊕ K⊥. Besides, since every element w in V ⊥ satisfies(w, Ty) = 0 for every y ∈ H, we have (Tw, y) = 0 for every y ∈ H and thusTw = 0. This shows V ⊥ ⊂ K. Now for every x /∈ V , we have x = v + v′,where v ∈ V , v′ ∈ V ⊥, and v′ 6= 0. Since V ⊂ K⊥ and V ⊥ ⊂ K, it followsthat 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 aslightly different approach here.)
Given ε > 0, for every f ∈ L1[−π, π], we may find M > 0 large so that‖f − fM‖L1[−π,π] < ε, where fM := (f ∧M) ∨ (−M).
We let µ be the Lebesgue measure on [−π, π]. By Lusin’s theorem (Theo-rem 9. above), since µ(x : fM(x) 6= 0) <∞, there is a function g ∈ Cc[−π, π]such that µ(fM 6= g) < ε/M , and supx∈[−π,π] |g(x)| ≤ supx∈[−π,π] |fM(x)| ≤M . It follows that ‖g − fM‖L1[−π,π] < 2ε, which shows C[−π, π] is dense inL1[−π, π].
Since C[−π, π] contains all uniformly continuous functions on [−π, π], forevery g ∈ C[−π, π] we may find h(x) =
∑mi=1 aiχ[bi,bi+1), −π = b1 < b2 <
· · · < bm+1 = π so that h is close to g in L1-norm.
2
Consider ∫ π
−π
m∑i=1
aiχ[bi,bi+1)einx dx =
m∑i=1
ai
∫ bi+1
bi
einx dx
=m∑i=1
ai1
in(einbi+1 − einbi)
which → 0 as n → ∞. Therefore, for every f ∈ L1[−π, π] we may pick h ofthe form above and ‖f − h‖L1[−π,π] < δ, and thus
|∫ π
−πf(x)einx dx| ≤
∫ π
−π|f(x)− h(x)| dx+ |
∫ π
−πh(x)einx dx|
≤ 2δ
for all n large enough. This completes the proof.
A-5
For every x ∈ l∞, ‖Tx‖l1∗ = sup‖y‖l1=1 |∑
i xiyi| ≤ sup‖y‖l1=1 ‖x‖l∞∑
i |yi| =‖x‖l∞ . This shows that ‖T‖ ≤ 1.
For x, y ∈ l∞, x 6= y, ∃i so that xi 6= yi. Define ei so that (ei)k = δik. Wehave Tx(z) = xi 6= yi = Ty(z).
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,
we may pick a subsquence {znj}j so that |znj
| → ∞, and|x(enj )|‖enj ‖l1
= |znj|
is unbounded in j, contradicts the fact that x is bounded. Now we havez = (z1, z2, · · · ) ∈ l∞ and Tz = x. The proof is complete.
B-6
Hard.
B-7
Since cos(z) = 12(eiz+e−iz) is an entire function, cos(z) =
∑∞n=0(−1)n zn
(2n)!
for every z ∈ C. Therefore, when z 6= 0, cos(z)z2
=∑∞
n=0(−1)n zn−2
(2n)!=∑∞
n=−2(−1)n zn
(2(n+2))!, which is the Laurent series for F .
3
B-8
For every piecewise C1 closed curve γ′ ⊂ 4,∫γ′f(z) dz =
∫γ′
∫γ
φ(w)g(w − z) dw dz =
∫ b
a
∫ d
c
φ(w(s))g(w(s)− z(t))w′(s)z′(t) ds dt
=
∫ d
c
∫ b
a
φ(w(s))g(w(s)− z(t))w′(s)z′(t) dt ds
=
∫ d
c
∫γ′φ(w(s))g(w(s)− z))w′(s) dz ds
=
∫ d
c
0 · w′(s) ds = 0,
since as a function of z, φ(w(s))g(w(s)−z)) is analytic in the region enclosedby γ′. As a result, by Morera’s theorem, f is analytic on 4. (Actually itsuffices 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. Sincefor each pole zk, limz→zk |f(z)| = ∞, it follows that infz∈C |f(z)| > 0 ⇒supz∈C
1|f(z)| < ∞. Since now 1
f(z)is a bounded entire function, f(z) is a
constant function by Louville’s theorem, which contradicts the assumptionof 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)| ≤ 1, and if the equality holds, then g is a rotation.
By (ii), |g′(0)| = |1rf ′(Rw + z)R|w=0| ≤ 1 ⇒ |f ′(0)| ≤ r
R.
4