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