H

Prelim Problems in Analysis Hamza Ruzayqat Email: [email protected] Date: May 1, 2016

MATH 545 and MATH 546 Prelim Problems

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Chapter 1

Real Analysis 1.1

Review

Note 1.1.1. Some of the problems here are not prelim problems, but they are good candidates. Proposition 1.1.1. Suppose O ⊂ R is open. Then O can be written as the countable union of disjoint S open intervals {Ii }i = {(ai , bi )}i , i.e. O = ∞ i=1 Ii . Definition 1.1.1. . i . Gδ set is a countable intersection of open sets, i.e. for G ∈ Gδ , G = ∩∞ i=1 Oi where Oi is open. ii . Fσ set is a countable union of closed sets, i.e. for F ∈ Fσ , F = ∪∞ i=1 Fi where Fi is closed. Note that in both cases we can have finite intersection/union, for example in (i), by letting Oi = Si for i = 1, ..., n and opens sets {Si }i and set Oi = ∅ for i > n. Theorem 1.1.2. Let E ⊆ R be a given set. Then the following five statements are equivalent: (i) E is Lebesgue measurable. (ii) Given  > 0, there exists an open set O with E ⊆ O and m∗ (O \ E) < . (iii) Given  > 0, there exists a closed set F ⊆ E with m∗ (E \ F ) < . (iv) There is a G in Gδ with E ⊆ G and m∗ (G \ E) = 0. (v) There is an F in Fσ with F ⊆ E and m∗ (E \ F ) = 0. If m∗ (E) is finite, the above statements are equivalent to: (vi) Given  > 0, there is a finite union U of open intervals such that m∗ (U 4E) < . (Here for two sets A, B, we define their symmetric difference A4B = (A \ B) ∪ (B \ A)) Theorem 1.1.3. Let B be the Borel σ-algebra and M the Lebesgue σ-algebra. Then B ⊂ M. Thus if f is Borel-measurable, then it is Lebesgue measurable. Since continuous functions are Borel functions, then they are Lebesgue-measurable. 3

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Proposition 1.1.4. Suppose f ≥ 0 is a non-negative and measurable function. Then there exists a sequence of non-negative measurable simple functions sn increasing to f . Theorem 1.1.5. Egoroff ’s Theorem: Let (X, A, µ) be a finite measure space (i.e. µ(X) < ∞). If {fn } is a sequence of measurable functions that converge to a function f a.e. (for each x, fn (x) → f (x) point-wisely a.e. x) on a measurable set E, then given  > 0, there is a subset A ⊂ E with µ(A) <  such that {fn } converges to f uniformly on E \ A. Proof: For n and k in N, define the set En,k by the union  [  1 En,k = x ∈ E |fm (x) − f (x)| ≥ . k m≥n These sets get smaller as n increases, meaning that En+1,k is always a subset of En,k , because the first union involves fewer sets. A point x, for which the sequence {fm (x)} converges to f (x), cannot be in every En,k for a fixed k. Hence by the assumption of µ-almost everywhere point-wise convergence on E, µ

\

 En,k

=0

n∈N

for every k. Let  > 0, then since µ(E) < ∞ and µ(En,k ) → 0 as n → ∞, there exists, for each k, some natural number nk such that µ(Enk ,k ) <

ε . 2k

Define A=

[

Enk ,k

k∈N

Appealing to the sigma additivity of µ and using the geometric series, we get X X ε µ(A) ≤ µ(Enk ,k ) < = ε. k 2 k∈N k∈N If x ∈ / A (i.e. x ∈ E \ A), then x ∈ / Enk ,k for all k. Since Enk +1,k ⊂ Enk ,k , then for all n ≥ nk , we have |fn − f | < i.e. fn → f uniformly on E \ A. 

1 k 

Theorem 1.1.6. Lusin’s Theorem: Let (X, A, µ) be a measure space. Let f : [a, b] → R be a measurable function on [a, b]. Then given  > 0, there is a continuous function φ on [a, b] such that µ{x : f (x) 6= φ(x)} < . (every measurable function is a continuous function on nearly all its domain). P Definition 1.1.2. Let (X, A, µ) be a measure space. If s = ni=1 ai χEi is a non-negative measurable simple function, define the Lebesgue integral of s to be Z n X s dµ = ai µ(Ei ). i=1

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Definition 1.1.3. Let (X, A, µ) be a measure space. If f ≥ 0 is a measurable function, define Z  Z f dµ = sup s dµ : 0 ≤ s ≤ f, s simple . Let f be measurable and let f + = max(f, 0) and f − = max(−f, 0). Provided not both infinite, define Z Z Z + f dµ = f dµ − f − dµ.

R

f + dµ and

R

f − dµ are

Note 1.1.2. |f | = f + + f − . R Note 1.1.3. If X f dµ < ∞, i.e. f ∈ L1 , then f is finite µ-a.e.. Theorem 1.1.7. Monotone Convergence Theorem (MCT): Let (X, A, µ) be a measure space. Suppose fn is a sequence of non-negative increasing measurable functions, i.e. 0 ≤ f1 (x) ≤ f2 (x) ≤ ... for all x, and limn→∞ fn (x) = f (x) for all x (or a.e. x). Then Z Z lim fn dµ = f dµ. n→∞

Theorem 1.1.8. Fatou’s Lemma: Let (X, A, µ) be a measure space. Suppose the fn are non-negative and measurable. Then Z Z lim inf fn dµ ≤ lim inf fn dµ. n→∞

n→∞

If limn→∞ fn (x) = f (x) for all x (or a.e.), then Z Z f dµ ≤ lim inf fn dµ. n→∞

Theorem 1.1.9. Dominated Convergence Theorem (DCT): Suppose that fn are measurable realvalued functions and fn (x) → f (x) for each x(or a.e. x). Suppose there exists a nonnegative integrable function g such that |fn (x)| ≤ g(x) for all x. Then Z Z lim fn dµ = f dµ. n→∞

Note 1.1.4. Notice that in the previous three theorems we need point-wise convergence fn (x) → f (x) for all x or a.e. x. Definition 1.1.4. Sense of Vitali: Let J be a collection of intervals. Then we say that J covers a set E in the sense of Vitali, if for each  > 0 and any x ∈ E, there is an interval I ∈ J such that x ∈ I and `(I) < . Lemma 1.1.10. Vitali: Let E be a set of finite outer measure and J a collection of intervals that cover E in the sense of Vitali. Then, given  > 0, there is a finite disjoint collection {I1 , ..., IN } of intervals in J such that ! N [ µ∗ E \ Ii < . i=1

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Theorem 1.1.11. Let f be an increasing real-valued function on the interval [a, b]. Then f is differentiable a.e., the derivative f 0 is integrable, and Z b f 0 (x) dm(x) ≤ f (b) − f (a). a

Definition 1.1.5. Total Variation: Let f be a real-valued function defined on the interval [a, b]. Let the class P be a class of all possible partitions of [a, b], i.e. P = {P = {x0 , . . . , xn }|P is a partition of [a, b]}. Then the total variation of f on the interval [a, b] is Tab f

= sup P ∈P

n X

|f (xi ) − f (xi−1 )|.

i=1

Definition 1.1.6. Bounded Variation (BV): A real-valued function f is said to be of bounded variation (BV function) on [a, b] if its total variation is finite, i.e. f ∈ BV ([a, b]) ⇐⇒ Tba (f ) < ∞. Theorem 1.1.12. A real function f is of bounded variation in an interval [a, b], (f ∈ BV ([a, b])), if and only if it can be written as the difference f = f1 − f2 of two non-decreasing functions. (Hence all increasing functions are of BV and hence their derivatives are exists a.e.) Corollary 1.1.13. If f ∈ BV ([a, b]), then f 0 exists a.e. x ∈ [a, b], i.e., f 0 is in L1 (m, [a, b]). (Hence if f is an increasing function, then f 0 exists a.e. and it is integrable on [a, b]) Theorem 1.1.14. Fundamental Theorem of Calculus I (FTC I): Let f be an integrable function on [a, b], and suppose that Z x F (x) = F (a) + f (t) dm(t). a 0

Then F (x) = f (x) a.e. x in [a, b]. Definition 1.1.7. Absolute Continuity: A real-valued function f : [a, b] → R is said to be absolutely continuous on [a, b] (or f ∈ AC([a, b])) if, ∀ > 0

∃δ > 0 :

If

n X

|xi − xi−1 | < δ

=⇒

i=1

n X

|f (xi ) − f (xi−1 )| < ,

i=1

where {(xi−1 , xi )}ni=1 is a collection of disjoint open intervals. Lemma 1.1.15. If f ∈ AC([a, b]), then f ∈ BV ([a, b]) (and hence f 0 ∈ L1 (m). Theorem 1.1.16. Fundamental Theorem of Calculus II (FTC II): If f ∈ AC([a, b]) (i.e. f 0 exists a.e.), then Z x f 0 (t) dm(t) = f (x) − f (a). a

Definition 1.1.8. σ-Finite Measure: Let (X, A, µ) be a measure space. The measure µ is called σfinite if X is the countable union of measurable subsets of A with finite measure. (e.g R = ∪k∈N (−k, k))

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Definition 1.1.9. Complete Measure Space: A measure space (X, A, µ) is said to be complete if B ∈ A, µ(B) = 0, and A ⊂ B, imply that A ∈ A. (e.g Lebesgue measure is complete, while Lebesgue measure restricted to B the σ-algebra of Borel sets is not complete.) Definition 1.1.10. Semi-algebra S: A nonempty class S ⊂ P(X) is semi-algebra if i. If A and B in S, then A ∩ B ∈ S. ii. If A ∈ S, then Ac = a finite disjoint union of sets in S, i.e. Ac = ∪ni=1 Ai and Ai ’s are disjoint in S. Definition 1.1.11. Algebra C Generated by Semi-algebra S: C is an algebra generated by semialgebra S if: i. ∅ ∈ C. ii. If C ∈ C, then C = ∪ni=1 Ci , where {Ci }ni=1 is a finite collection of disjoint sets in S. iii. C is contained in any algebra containing S, i.e. if C 0 is an algebra and S ⊂ C 0 , then C ⊂ C 0 . I.e., C is the minimum algebra contains S. Definition 1.1.12. Rectangle: Let (X, A, µ) and (Y, B, ν) be two complete measure spaces, and consider the direct product X ×Y of X and Y . If A ⊂ X and B ⊂ Y , we call R = A×B a rectangle. If A ∈ A and B ∈ B, we call R a measurable rectangle. The collection R of measurable rectangles is a semi-algebra. Let R = A × B be a measurable rectangle and let λ = (µ × ν). Then λ(R) = (µ × ν)(A × B) = µ(A).ν(B). Definition 1.1.13. Product Outer Measure (µ×ν)∗ : Let (X, A, µ) and (Y, B, ν) be complete measure spaces. Let R0 be the algebra generated by the semi-algebra R as in Definition 1.1.11. Then λ = µ × ν can be extended on R0 with outer measure λ∗ = (µ × ν)∗ : P(X × Y ) → [0, ∞] defined by: (∞ ) ∞ X [ λ∗ (E) = inf λ(Ei ) : Ei ∈ R0 for i ∈ N, and E ⊆ Ei . i=1

i=1

And a set E is said to be (µ × ν)-measurable if for any set S ⊆ X × Y , we have: (µ × ν)∗ (S) = (µ × ν)∗ (S ∩ E) + (µ × ν)∗ (S ∩ E c ). Note 1.1.5. If X = Y = R and µ = ν = m the Lebesgue measure, then µ × ν is called two-dimensional Lebesgue measure for the plane. If E ∈ Rσδ , then E is (µ × ν)-measurable. Definition 1.1.14. Let Ri ∈ R (set of measurable rectangles) for each i ∈ N. Then, A ∈ Rσ implies ∞ A = ∪∞ i=1 Ri . Let Ai ∈ Rσ for each i ∈ N. Then, B ∈ Rσδ implies B = ∩i=1 Ai . Theorem 1.1.17. Suppose (X, A, µ) and (Y, B, ν) are complete measure spaces, (µ × ν)∗ is the outer measure on X ×Y , and (µ×ν) is the restriction of (µ×ν)∗ to the (µ×ν)∗ -measurable sets. Let E ⊆ X ×Y . Suppose either (µ × ν)∗ (E) < ∞ or (µ × ν)∗ (E) = ∞ and X × Y is σ-finite. The following are equivalent: (i) E is (µ × ν)-measurable.

MATH 545 and MATH 546 Prelim Problems

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(ii) For all  > 0, there exists S ∈ Rσ such that E ⊂ S and (µ × ν)∗ (S \ E) < . (iii) There exists T ∈ Rσδ such that E ⊂ T and (µ × ν)∗ (T \ E) = 0. Theorem 1.1.18. Fubini Theorem: Let (X, A, µ) and (Y, B, ν) be two complete measure spaces and f an integrable function on X × Y . Then i. For almost all x the function fx defined by fx (y) = f (x, y) is an integrable function on Y . For almost all y the function fy defined by fy (x) = f (x, y) is an integrable function on X. R R ii. Y f (x, y) dν(y) and X f (x, y) dµ(x) are integrable functions on X and Y respectively. iii.

Z Z



f d(µ × ν) =

f dν dµ = X

Y

Z Z

Z X×Y

 f dµ dν.

Y

X

Theorem 1.1.19. Tonelli Theorem: Let (X, A, µ) and (Y, B, ν) be two σ-finite measure spaces, and let f ≥ 0 be a measurable function on X × Y . Then i. For almost all x the function fx defined by fx (y) = f (x, y) is a measurable function on Y . For almost all y the function fy defined by fy (x) = f (x, y) is a measurable function on X. R R ii. Y f (x, y) dν(y) and X f (x, y) dµ(x) are measurable functions on X and Y respectively. iii.

Z Z



f d(µ × ν) =

f dν dµ = X

Y

Z Z

Z X×Y

 f dµ dν.

Y

X

Definition 1.1.15. Signed Measure: By a signed measure on the measurable space (X, A) we mean an extended real-valued set function ν defined for the sets of A and satisfying the following conditions: i. ν assumes at most one of the values +∞, −∞. ii. ν(∅) = 0. P∞ S iii. ν ( ∞ n=1 ν(En ) for any sequence En of disjoint measurable sets, the equality taken to n=1 En ) = S mean that the series on the right converges absolutely if ν ( ∞ n=1 En ) is finite and that it properly diverges otherwise. Definition 1.1.16. Positive and Negative Sets: We say that a set A is a positive set with respect to a signed measure ν if A is measurable and for every measurable subset E of A we have ν(E) ≥ 0. Similarly, a set B is called a negative set if it is measurable and every measurable subset E of it has nonpositive ν measure. A set that is both positive and negative with respect to ν is called a null set. A measurable set A is a null set if and only if every measurable subset E ⊂ A has ν(E) = 0. Note 1.1.6. Every measurable subset of a positive set is again positive, and the union of a countable collection of positive sets is also positive. Proof: 1) Let A be a positive set, then for every measurable subset E of A we have ν(E) ≥ 0. We need to prove that E is also positive. E is measurable as mentioned and if C ⊂ E is measurable, then ν(C) ≥ 0 since C ⊂ E ⊂ A. 2) Let A = ∪i∈N Ai , Ai is positive. Let E ⊂ A be measurable. Set Ei = E ∩ Ai ∩ Aci−1 ∩ Aci−2 ... ∩ Ac1 , then {Ei } are disjoint sets, Ei ⊂ Ai is measurable, and E = ∪i∈N Ei . Since Ei ⊂ Ai and Ai is positive, ν(Ei ) ≥ 0 and hence P ν(E) = i∈N ν(Ei ) ≥ 0.#

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Lemma 1.1.20. Let E be a measurable set such that 0 < ν(E) < ∞. Then there is a positive set A contained in E with ν(A) > 0. Theorem 1.1.21. Hahn Decomposition Theorem: Let ν be a signed measure on the measurable space (X, A). Then there is a positive set A and a negative set B such that X = A ∪ B and A ∩ B = ∅. Note 1.1.7. Hahn Decomposition is not unique. Definition 1.1.17. Singular Measure ν ⊥ µ: Two measures ν and µ defined on (X, A) are said to be mutually singular (in symbols ν ⊥ µ) if there are disjoint measurable sets A and B with X = A ∪ B such that ν(A) = µ(B) = 0. (e.g let A and B be as in Theorem 1.1.21 and ν + (E) = ν(E ∩ A) and ν − (E) = −ν(E ∩ B), then ν + and ν − are mutually singular.) We sometimes read ν ⊥ µ as: ν is singular with respect to µ. Theorem 1.1.22. Jordan Decomposition: Let ν be a signed measure on the measurable space (X, A). Then there are two mutually singular measures ν + and ν − on (X, A) such that ν = ν + − ν − . Moreover, there is only one such pair of mutually singular measures. Note 1.1.8. Since ν assumes at most one of the values +∞ and −∞, either ν + or ν − must be finite. Moreover, the measure |ν| defined by |ν|(E) = ν + (E) + ν − (A) is called the absolute value or total variation of ν. Definition 1.1.18. Absolute Continuity of Measure ν << µ: Let ν and µ defined on (X, A). The measure ν is said to be absolutely continuous with respect to the measure µ ( in symbols: ν << µ) if A ∈ A with µ(A) = 0 =⇒ ν(A) = 0. Note 1.1.9. In the case of signed measures µ and ν, we say ν << µ if |ν| << |µ|. Also in this case, ν ⊥ µ if |ν| ⊥ |µ|. Theorem 1.1.23. Radon-Nikodym Theorem (R-N): Let (X, A, µ) be a σ-finite measure space (so µ is a positive measure), and let ν be a measure defined on (X, A) with ν << µ. Then there is a nonnegative measurable function f : X → [0, ∞) such that for any measurable set E ⊂ A we have Z f dµ. ν(E) = E

The function f is unique in the sense that if g is any measurable function with this property then g = f dν µ-a.e.. f is usually written as . dµ Theorem 1.1.24. Lebesgue Decomposition: Let (X, A, µ) be a σ-finite measure space (so µ is a positive measure), and let ν be a σ-finite measure defined on (X, A). Then we can find a measure ν0 with ν0 ⊥ µ, and a measure ν1 with ν1 << µ, such that ν = ν0 + ν1 . The measures ν0 and ν1 are unique. Definition 1.1.19. A normed linear space is called complete if every Cauchy sequence in the space converges, that is, if for each Cauchy sequence {fn } in the space there is an element f in the space such that fn → f . A complete normed linear space is called a Banach space. Theorem 1.1.25. Riesz-Fischer Theorem: The Lp spaces are complete. Theorem 1.1.26. The set of continuous functions with compact support, i.e f (x) is defined on a compact set like [−n, n] and zero outside, is dense in Lp .

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Useful Inequalities: 1. Minkowski Inequality: If f and g are in Lp with 1 ≤ p ≤ ∞, then so is f + g and kf + gkp ≤ kf kp + kgkp . If 1 < p < ∞, then equality can hold only if there are nonnegative constants α and β such that βf = αg. 2. Minkowski Inequality for 0 < p < 1: If f and g are two nonnegative functions in Lp with 0 < p < 1, then kf + gkp ≥ kf kp + kgkp . 3. Holder Inequality: If p and q are nonnegative extended real numbers such that 1 1 + = 1, p q and if f ∈ Lp and g ∈ Lq , then f g ∈ L1 and kf gk1 ≤ kf kp kgkq . Equality holds if and only if there are some constants α and β, not both zero, such that α|f |p = β|g|q a.e.. 4. Jensen Inequality: Let φ be a convex function on (−∞, ∞) and f ∈ L1 [0, 1]. Then Z  Z φ f (x) dx ≤ φ(f (x)) dx.

1.2

Real Analysis Problems

1. Let r ∈ (0, ∞) and pk ∈ (0, ∞] for k ∈ (1, 2, ..., n) such that measurable functions {fk }nk=1 with fk ∈ Lpk we have

n

n

Y Y

kfk kpk .

fk ≤

k=1

r

Pn

k=1

r = 1. Show that for all pk

k=1

SOLUTION: We do this problem by induction. For n = 1, 2 it is obvious. Suppose it holds for n − 1 and let’s prove it holds for n. W.L.O.G let p1 ≤ p2 ≤ ... ≤ pn . Assume fk ∈ Lpk for k = 1, 2, ..., n. Then we have two cases: Case(1): pn = ∞.

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Pn−1 1 1 = . So k=1 pk r Z 1/r r r r r = |f1 | |f2 | ...|fn−1 | |fn | dx

Hence fn ∈ L∞ , i.e. kfn k∞ = ess sup |fn | < ∞, and

n

Y



fk

k=1

r

n−1

Y

≤ fk kfn k∞

k=1

r

≤kf1 kp1 kf2 kp2 ...kfn−1 kpn−1 kfn k∞ n Y ≤ kfk kpk .

By induction hyposthisis

k=1

Case(2): pn < ∞. Let

1 1 pn 1 Pn−1 r = k=1 . Then + = 1 and thus q = . So, q pk q pn /r pn − r

r Z n

Y

r

fk = |f1 f2 ...fn−1 | |fn |r dx

k=1

r

=k |f1 f2 ...fn−1 |r |fn |r k1 ≤k |f1 f2 ...fn−1 |r kq k |fn |r kpn /r (H¨older’s) Z 1/q Z r/pn qr pn = |f1 f2 ...fn−1 | dx |fn | dx Take the power 1/r on both sides, we get:

n

1/pn 1/qr Z Z

Y

qr pn |fn | dx |f1 f2 ...fn−1 | dx

fk ≤

k=1

r

= kf1 f2 ...fn−1 kqr kfn kpn Since

P 1 1 = n−1 , then by induction hypothesis k=1 qr pk kf1 f2 ...fn−1 kqr ≤

n−1 Y

kfk kpk

k=1

and therefore

n n

Y

Y

f ≤ kfk kpk .

k

k=1

r

k=1

 2. Find all constants K > 0 for which the following holds: If (X, M, µ) is any positive measure space and f : X → R is µ integrable satisfying

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R f dµ < K for all E ∈ M, then kf k1 < 1. E SOLUTION: Z

Z

Z

|f | dµ =

kf k1 = X

f dµ − {x∈X:f (x)≥0}

f dµ {x∈X:f (x)<0}

Z Z ≤ f dµ − f dµ {x∈X:f (x)≥0} {x∈X:f (x)<0} Z Z ≤ f dµ + f dµ {x∈X:f (x)≥0}

{x∈X:f (x)<0}

< 2K 1 So 0 < K ≤ . 2



P 3. (a) Let p ∈ [1, ∞) and suppose {fn }∞ n=1 ⊂ L (R) is a sequence that converges to 0 in the p norm. Prove that we may find a subsequence {fnk }∞ k=1 such that fnk → 0 a.e.

(b) Show that any sequence {fn }∞ n=1 of non-negative integrable functions on [0, 1] with R1 2 1 f dµ ≤ n3 must converge to zero a.e. x ∈ [0, 1]. 0 n (c) Is there a sequence {fn } of non-negative integrable functions on [0, 1] satisfying R1 2 f dµ → 0 which does not converge to zero a.e.? 0 n SOLUTION: (a) So we have kfn kp < ∞ and kfn kp → 0. Thus we can find n1 ∈ N such that for all n ≥ n1 , we have kfn kp < 1/2. Having found this n1 we may find n2 ∈ N such that n2 > n1 and for all n ≥ n2 , we have kfn kp < 1/22 . Continue in this manner, we inductively find nk ∈ N such that nk > nk−1 and for all n ≥ nk , we have kfn kp < 1/2k . So we got a subsequence {fnk }∞ k=1 such Pm p k that kfnk kp < 1/2 for all k ∈ N. Now let fm = k=1 |fnk | , then {fm } is a monotone increasing sequence, hence by MCT, Z X Z ∞ m X p |fnk | dµ = lim |fnk |p dµ R m→∞ k=1 Z X m

R k=1

= lim

m→∞

= lim

m→∞

=

∞ X k=1

P∞

R k=1 m XZ k=1

|fnk |p dµ |fnk |p dµ

R

kfnk kpp <

∞ X 1 1 = < ∞. pk p−1 2 2 k=1

P∞

So k=1 |fnk |p is integrable, hence k=1 |fnk |p is finite for a.e. x ∈ R, hence |fnk |p → 0 as k → 0 a.e. x ∈ R, and thus fnk → 0 as k → 0 a.e. x ∈ R. 

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(b) As in (a), Z

∞ 1X

0

Thus

P∞

k=1

fn2

dµ =

k=1

∞ Z X k=1

0

1

fn2

∞ X 1 dµ ≤ < ∞. 3 n k=1

fn2 < ∞ a.e. x ∈ [0, 1], hence fn2 → 0 a.e. x ∈ [0, 1], hence fn → 0 a.e. x ∈ [0, 1]. 

(c) Consider the sequence {En }∞ n=1 of subsets of [0, 1] defined by: E1 = [0, 21 ], E2 = [ 12 , 1], E3 = [0, 31 ], E4 = [ 13 , 32 ], E5 = [ 23 , 1], E6 = [0, 41 ], E7 = [ 14 , 42 ], E8 = [ 24 , 34 ], E9 = [ 34 , 1], ... Let fn (x) = χEn (x). Then, Z

1

fn2

Z

Z

2

0

1

χEn dµ = µ(En ) → 0 as n → ∞.

(χEn ) dµ =

dµ =

0

1

0

But for all x ∈ [0, 1], lim fn (x) does not exist because if for example x ∈ E6 , then already x n→∞

in E1 and E3 . Therefore, f1 (x) = f3 (x) = f6 (x) = 1 and fn (x) = 0 for n ∈ N, n 6= 1, 3, 6. So, if x ∈ [0, 1], then as n → ∞ x is already in infinitely many Ei ’s, and hence the limit does not exist.  4. For each δ > 0 and each f ∈ L1 (0, ∞), find (with proof ) Z 1 xδ f (tx) dx. lim t t→∞

0

SOLUTION: Let u = tx, then dx = du/t. Now we have: Z 1 Z t  δ u δ f (u) du. x f (tx) dx = lim lim t t→∞ 0 t→∞ t 0 Let ft (u) = (u/t)δ f (u)χ(0,t] , then limt→∞ ft = 0. Since 0 < u ≤ t,  u δ |f (u)|χ(0,t] ≤ |f (u)|χ(0,∞] ≡ g(u), |ft (u)| = t R R∞ and g(u) du = 0 |f (u)| du < ∞ so g is integrable. Therefore, by DCT we have Z 1 Z t  δ u δ lim t x f (tx) dx = lim f (u) du t→∞ t→∞ 0 t 0 Z ∞ Z ∞  u δ = lim f (u)χ(0,t] du = 0 du = 0. t→∞ t 0 0 

MATH 545 and MATH 546 Prelim Problems

14

5. Suppose (X, M, µ) is a measure space and that f is real-valued measurable on X. Prove that if Z f dµ ≥ 0 E

for every E ∈ M, then f ≥ 0 a.e. on X. SOLUTION: Let An = {x ∈ X : f (x) < −1/n} for each n ∈ N. Since f is µ-measurable, An is µ-measurable. Let sn = − n1 χAn for each n ∈ N. Then 0 ≤ −sn < −f χAn . Hence Z Z Z 1 0 ≤ µ(An ) = (−sn ) dµ < (−f χAn ) dµ = (−f ) dµ ≤ 0, n X X An R R using the definition of the integral and the assumption that X f dµ ≥ 0 and hence X (−f ) dµ ≤ 0. Hence µ(An ) = 0 for each n ∈ N. Note that {x ∈ X : f (x) < 0} =

∞ [

{x ∈ X : f (x) < −1/n} =

∞ [

An .

n=1

n=1

Since µ(An ) = 0 for each n ∈ N, µ({x ∈ X : f (x) < 0}) = 0 (a countable union of sets of measure 0 has measure 0). Hence, f (x) ≥ 0 a.e. x ∈ X.  6. Let f be an integrable function on a measure space (X, M, µ). Show that given  > 0, there is a δ > 0 such that for each µ−measurable set E with µ(E) < δ, we have Z Z |f | dµ < . f dµ ≤ E E SOLUTION:

Let  > 0. Define fn (x) = f (x) . By (MCT)

( f (x) if f (x) ≤ n , Then fn (x) ≤ n for all n ∈ N and fn (x) ↑ n otherwise Z X

f (x) dµ = lim

n→∞

Z

fn (x) dµ.

X

Hence, there exists n ∈ N such that Z Z  f (x) dµ − fn (x) dµ < 2 X X Z Z Z
fn (x) dµ <  ⇒ f − fn dµ ≤ f (x) dµ − 2 X X X

MATH 545 and MATH 546 Prelim Problems

Choose δ <

15

 . Then 2n Z Z Z f dµ = f dµ ≤ E E E Z ≤


f − fn dµ +

Z

fn dµ

E


f − fn dµ + n µ(E)

X

 < +n δ 2   < + = . 2 2  7. Suppose f : [0, 1] → R is increasing. f ∈ AC([0, 1]).

Suppose

R1 0

f 0 dm = f (1) − f (0).

Prove that

SOLUTION: Let x ∈ (0, 1] be arbitrary. Consider the interval [0, x]. Then since f is increasing on [0, 1], it is increasing on [0, x]. By Theorem 1.1.11, f 0 exists a.e. on [0, x], f 0 is Lebesgue integrable on [0, x] Rx Rx and 0 f 0 (t) dm(t) ≤ f (x) − f (0). However, 0 f 0 (t) dm(t) < f (x) − f (0) cannot happen. Proof: Suppose it happens, then because f 0 is inntegrable, by FTC(I) differentiating both sides of the strict inequality with respect to x gives f 0 (x) < f 0 (x) a.e. which is a contradiction. So Z x f 0 (t) dm(t) = f (x) − f (0). 0

Claim. If for all  > 0, there is δ > 0 such that for each m-measurable set E ⊂ [0, 1] with m(E) < δ we have Z |f 0 | dm < , E

then f ∈ AC([0, 1]). Proof: We will prove this claim for f 0 above. Let  > 0, then since f 0 is integrable, by the previous R problem such a δ exists. Let E ⊂ [0, 1] be such that m(E) < δ, then E |f 0 | dm < . Let {(ai , bi )}ni be a finite collection of disjoint open intervals in [0, 1] (ordered in a way such that bi < ai+1 for P i = 1, 2, ..., n − 1) such that E = ∪ni=1 (ai , bi ). Then one has ni=1 (bi − ai ) = m(E) < δ. Notice that Z bi Z ai Z bi Z 0 0 0 f (bi ) − f (ai ) = f (t) dm(t) − f (t) dm(t) = f (t) dm(t) = f 0 (t) dm(t). 0

0

ai

(ai ,bi )

And thus n X i=1

|f (bi ) − f (ai )| ≤

n Z X i=1

which proves that f ∈ AC([0, 1]). 

(ai ,bi )

0

Z

|f | dm =

0

Z

|f | dm = ∪n i=1 (ai ,bi )

E

|f 0 | dm < 

MATH 545 and MATH 546 Prelim Problems

16

 8. For x, y ∈ [0, 1], let f (x, y) =

∞ X √

n [sin(xy)]n .

n=1

(a) Show that the series defining f (x, y) converges for every (x, y) ∈ [0, 1] × [0, 1]. (b) Determine whether f is Lebesgue integrable on [0, 1] × [0, 1]. SOLUTION: (a) Note that for all (x, y) ∈ [0, 1] × [0, 1] we have √ √ √ √ n [sin(xy)]n = n [sin(xy)]n ≤ n [sin(1)]n = n [sin(1)]n , P √ and hence if we prove that ∞ n [sin(1)]n is absolutely convergent, then the series defining n=1 P √ n [sin(1)]n converges we use f (x, y) must converge by direct comparison test. To prove ∞ n=1 ratio test as follows: √ √ n + 1 [sin(1)]n+1 n + 1 [sin(1)]n+1 √ √ = lim lim n→∞ n→∞ | n [sin(1)]n | n [sin(1)]n r 1 = sin(1) lim 1 + = sin(1) < 1. n→∞ n P∞ √ Thus by ratio test n=1 n [sin(1)]n converges, thus it is absolutely convergent and hence by comparison test f (x, y) converges.  (b) We need to prove that f (x, y) is Lebesgue measurable on [0, 1] × [0, 1] and Z

∞ X √

n [sin(xy)]n dm(x, y) < ∞.

[0,1]×[0,1] n=1

√ Note that n [sin(xy)]n is continuous for any n ∈ N and (x, y) ∈ [0, 1]×[0, 1], thus it is Lebesgue P √ measurable. Let fm (x, y) = m n [sin(xy)]n , then fm is Lebesgue measurable because it is n=1 sum of Lebesgue measurable functions. Note that f (x, y) = limm→∞ fm which is a limit of Lebesgue measurable functions, hence it is Lebesgue measurable. Since ([0, 1], M, m) is finite, f (x, y) is Lebesgue measurable on [0, 1] × [0, 1] and f (x, y) ≥ 0, we can apply Tonelli’s Theorem to have Z

Z ∞ X √ n n [sin(xy)] dm(x, y) =

[0,1]×[0,1] n=1

0

1

Z 0

∞ 1X √

n [sin(xy)]n dm(x)dm(y).

n=1

Note that the sequence {fm (x, y)}∞ m=1 is nonegative increasing sequence of functions goes to

MATH 545 and MATH 546 Prelim Problems

17

f (x, y) as m goes to infinity. Thus we can apply MCT Z 1Z 1X Z ∞ ∞ X √ √ n n [sin(xy)] dm(x, y) = n [sin(xy)]n dm(x)dm(y) 0

[0,1]×[0,1] n=1

1

lim

≤ 0

Z =

n=1 0 ∞ Z 1 1X

n [sin(xy)]n dm(x)dm(y)

√

n [sin(1.y)]n dm(x)dm(y)

n=1 0 ∞ 1X √

n [sin(1.y)]n dm(y)

0

=

√

n [sin(xy)]n dm(x)dm(y) (MCT)

n [sin(xy)]n dm(x)dm(y)

0

≤

0 n=1 m XZ 1

0 m→∞ n=1 0 Z 1X ∞ Z 1 √

Z

=

m 1X √

Z

0 m→∞

Z

=

n=1

lim

= =

0 1

Z

n=1 1

∞ Z X

√

n=1 0 ∞ Z 1 X

√

n=1 ∞ X

n [sin(1.y)]n dm(y) (MCT again) n [sin(1)]n dm(y)

0

√

n [sin(1)]n < ∞ (by part a).

n=1

 9. Evaluate the following limit: Z lim

n→∞

n



1−

0

x x n  ln 2 + cos dm(x). n n

SOLUTION:   x n  x x n Let fn (x) = 1 − χ[0,n] . Then |fn | = fn ≤ e−x ln(3)χ[0,∞) ≡ g, since 1 − ln 2 + cos is n n −x n an increasing bounded by e which actually its limit. Also, since 1 ≤ 2 + cos(x/n) ≤ 3,  sequence x we have ln 2 + cos ≤ ln(3). Notice that limn→∞ fn = ln(3)e−x . Since g is integrable then by n DCT Z n Z  x n  x x n  x lim 1− ln 2 + cos dm(x) = lim 1 − ln 2 + cos χ[0,n] dm(x) n→∞ 0 n→∞ n n n n Z ∞ = ln(3) e−x dm(x) = ln(3). 0

 10. If q < p + 1 find (with proof ) Z lim

n→∞

0

1

nxp + xq dx. xp + nxq

MATH 545 and MATH 546 Prelim Problems

18

SOLUTION: nxp + xq χ[0,1] , then for all n ∈ N, fn ≥ 0 and xp + nxq nxp + xq nxp + xq 1 fn = p χ ≤ = xp−q + ≤ xp−q + 1 ≡ g. [0,1] q q x + nx nx n But g is integrable since Z 1 1 p − q + 2 xp−q+1 p−q (x + 1) dx = + x = < ∞, p−q+1 p−q+1 0 0 where this integral is valid because p − q > −1. Also, notice that the point-wise limit of fn is xq p x + p q nx + x lim p χ[0,1] = lim p n χ[0,1] = xp−q χ[0,1] . n→∞ x + nxq n→∞ x + xq n Hence by DCT, Z 1 p Z nx + xq nxp + xq lim dx = lim χ[0,1] dx n→∞ 0 xp + nxq n→∞ xp + nxq Z 1 1 xp−q dx = = . p−q+1 0 Again this is true because p − q > −1. Let fn =

 11. Suppose f is Lebesgue measurable on the product spce [0, ∞) × [0, ∞) and g ∈ L1 ([0, ∞)). Prove that if |xf (x, y)| ≤ g(x) for all y ∈ [0, ∞), then Z ∞ Z ∞   t sup f (x, y) cos dx dt < ∞. y y>0 0

t

SOLUTION: Notice that Z ∞ Z ∞ Z   t ≤ f (x, y) cos dx dt y 0

t

0

∞

Z t

∞

  Z ∞Z ∞ t |f (x, y)| cos dx dt ≤ |f (x, y)| dx dt y 0 t Z ∞Z ∞ = |f (x, y)|χ{x≥t} dx dt. 0

0

Since f is Lebesgue measurable and the absolute value function |.| is continuous, |f (x, y)| is Lebesgue measurable. Since t ≤ x could be a singleton when x = t or an interval otherwise which both are Lebesgue measurable, we have |f (x, y)|χ{x≥t} is Lebesgue measurable. Also, we know that [0, ∞) is σ-finite. Hence we can apply Tonelli’s Theorem on |f (x, y)|χ{x≥t} as follows Z ∞Z ∞ Z ∞Z ∞ |f (x, y)| χ{x≥t} dx dt = |f (x, y)| χ{x≥t} dt dx | {z } | {z } 0 0 0 0 x varies and t fixed Z ∞ Z ∞ t varies and x fixed = |f (x, y)| χ{x≥t} dt dx 0

0

MATH 545 and MATH 546 Prelim Problems

19 ∞

Z

x

Z |f (x, y)|

=

dt dx 0

Z0 ∞

x|f (x, y)| dx

= 0

Z

∞

≤

g(x) dx 0

Z

∞

|g(x)| dx < +∞.

= 0

Taking the supremum over y > 0 will not change the answer because it does not depend on y.  12. Suppose f is nonnegative and Lebesgue measurable on [0, 1]. If Z b f (x) dx = b − a a

for every [a, b] ⊂ [0, 1] prove that f = 1 a.e. on [0, 1]. SOLUTION: Since [a, b] ⊂ [0, 1], we can choose a = 0 and b = 1, and by the fact that f is nonnegative on [0, 1], we have Z 1 Z 1 |f (x)| dx = f (x) dx = 1 − 0 = 1 < ∞, 0

0

1

hence f ∈ L ([0, 1]). Let 0 < x ≤ 1, then [0, x] ⊂ [0, 1]. Then by the given assumption, we have Z x f (t) dt. x=0+ 0

Then by FTC(I) we have (x)0 = 1 = f (x) a.e. x in [0, 1].  13. Let f ∈ L1 ((0, ∞)) be real-valued function, show that   Z ∞ f (tx) lim t sin f (tx) dx t→∞ t 0 exists and find its value. SOLUTION: Let y = tx, then we have Z lim t t→∞

0

∞

 sin

f (tx) t



Z f (tx) dx = lim

t→∞

∞

 sin

0

f (y) t

 f (y) dy

MATH 545 and MATH 546 Prelim Problems

 Let ft (y) = sin

f (y) t



20

  f (y) f (y). Then, limt→∞ ft = 0, and |ft | = sin |f (y)| ≤ |f (y)| ≡ g(y). t

But Z

Z

∞

g(y) dy =

|f (y)| dy < ∞,

since f ∈ L1 (0, ∞).



Z

0

By DCT, Z

∞

lim t

t→∞

 sin

0

f (tx) t

∞



 f (y) f (tx) dx = lim sin f (y) dy t→∞ 0 t   Z ∞ f (y) lim sin = f (y)dy t→∞ t 0 Z ∞ = 0 dy = 0. 0

 14. Find with proof Z

n

lim

n→∞

−n

1 dx. (1 + x2 )n

SOLUTION: Notice that since 1/(1 + x2 )n is even for all n, the integral is equivalent to Z n 1 dx. 2 lim n→∞ 0 (1 + x2 )n Let fn = 1/(1 + x2 )n χ[0,n] , then the point-wise limit is limn→∞ fn = 0 everywhere except at x = 0 we have limn→∞ fn = 1, i.e. fn → 0 a.e.. By using Bernoulli’s inequality: (1 + x)n > 1 + nx for x > −1, we have |fn | = fn =

1 1 1 χ ≤ χ ≤ χ[0,∞) ≡ g(x). [0,n] [0,n] (1 + x2 )n 1 + nx2 1 + nx2

(Or we can consider g = χ[0,n] since fn ≤ χ[0,n] for all n and χ[0,n] √ g ∈ L1 (0, ∞), we let y = nx. So Z Z ∞ ∞ 1 1 1 dy = √ arctan(y) = g(x) dx = √ 2 n 0 1+y n 0

is integrable) Now to show

π √ < ∞. 2 n

By DCT, Z lim

n→∞

0

n

Z ∞ 1 1 dx = lim χ[0,n] dx 2 n n→∞ 0 (1 + x ) (1 + x2 )n Z ∞ 1 = lim χ[0,n] dx n→∞ (1 + x2 )n 0 Z ∞ = 0 dx = 0. 0



MATH 545 and MATH 546 Prelim Problems

21

15. Prove for each y ∈ (0, 1) that 1

Z

x−y (1 − e−nx ) dx

f (y) := lim

n→∞

0

exists and is finite. Find a formula for f (y). SOLUTION: Since −1 < −y < 0 and 0 ≤ x ≤ 1, then x−y ≤ x−1 . Also, we know 1 − e−nx ≤ nx for all n ∈ N. Let fn (x) = x−y (1 − e−nx ), then limn→∞ fn = x−y and |fn | = fn = x−y (1 − e−nx ) ≤ x−1 (nx) = n ≡ g. But Z Z 1

1

n dx = n < ∞.

g dx = 0

0

So by DCT, Z

1 −y

x (1 − e

f (y) := lim

n→∞

−nx

Z ) dx =

0

=

1

lim x−y (1 − e−nx ) dx

n→∞ Z0 1 −y

x

0

dx =

1 . 1−y 

16. Find with proof: Z

1/n

lim

n→∞

0

√ n x enx dx. sin nx

SOLUTION: Let y = nx, then Z lim

n→∞

0

1/n

√ Z 1 √ y ye n x enx √ dx = lim dy. n→∞ 0 sin nx n sin y

√ √ Let fn (y) = y ey /( n sin y)χ[0,1] . Then, limn→∞ fn = 0 and |fn | = fn ≤ 2 e1 y −1/2 / nχ[0,1] ≡ g, since sin y ≥ (y/2) and ex ≤ e1 on [0, 1]. But Z Z 2e1 1 −1/2 4e1 g dy = √ y dy = √ < ∞. n 0 n √

Therefore by DCT, Z lim

n→∞

0

1/n

√ Z 1 n x enx dx = lim n→∞ 0 sin nx Z 1 = lim 0

√ y ye √ dy n sin y Z 1 √ y ye √ dy = 0 dy = 0. n→∞ n sin y 0 

MATH 545 and MATH 546 Prelim Problems

22

17. Let f, g : Rk → Rk be Borel measurable. Show that f ◦ g is Borel measurable. SOLUTION: We need to prove {x ∈ Rk : (f ◦ g)(x) > α} = (f ◦ g)−1 ((α, ∞)) = g −1 (f −1 ((α, ∞))) ∈ B. If we prove that g −1 (f −1 (B)) ∈ B, where B ∈ B, we will be done. To do that, we need to prove A = {A ⊂ Rk : f −1 (A) ∈ B} is a σ-algebra containing the open sets, and repeat the same procedure but with g −1 instead of f −1 . A is not empty since f −1 (∅) = ∅ ∈ B. Let A ∈ A, then B = f −1 (A) ∈ B. Then f −1 (Ac ) = [f −1 (A)]c = B c ∈ B because B is a σ-algebra. Now, let {Ai }i ⊂ A, then Bi = f −1 (Ai ) ∈ B for all i ∈ N. So f −1 (∪i Ai ) = ∪i f −1 (Ai ) = ∪i Bi ∈ B again because B is a σ-algebra. Hence A is a σ-algebra. Now we show A contains the open sets. Let O ⊂ Rk be open, then f −1 (O) ∈ B since f is Borel by definition. Hence A contains the open sets and therefore the Borel sets B. Let f −1 (B1 ) = B2 , where B1 ∈ B, then B2 ∈ B. Repeat the same procedure but with g −1 instead of f −1 , then we will have g −1 (B2 ) ∈ B.  R1 18. Let M > 0, {fn } ⊂ L2 ([0, 1]) such that 0 |fn |2 dm ≤ M and fn (x) → 0 as n → ∞ a.e., m is Lebesgue measure. Show that for all 0 < p < 2, Z 1 |fn |p dm = 0. lim n→∞

0

SOLUTION: Let  > 0. Since m([0, 1]) < ∞, fn are measurable (since fn in L2 ) and fn → 0 (point-wisely) a.e., we can apply Egorov’s Theorem: For  > 0, there exists a measurable set A with m(A) <
2/(2−p) /[2M p/2 ] and fn → 0 uniformly on [0, 1] \ A. This uniform convergence implies that for all x ∈ [0, 1] \ A, there exists an N ∈ N such that for all n ≥ N we have |fn − 0| < (/2)1/p . Then, for n ≥ N: Z 1 Z Z p p |fn | dm = |fn | dm + |fn |p dm 0 [0,1]\A A Z  < m([0, 1] \ A) + |fn |p dm 2 A Z  ≤ m([0, 1]) + |fn |p dm 2 A Z  |fn |p dm = + 2 A R Now we apply Holders inequality on A |fn |p dm with 1/q 0 + 1/q = 1, where q, q 0 to be chosen later. 1/q0 Z 1/q Z Z 0 p pq q |fn | dm ≤ |fn | dm 1 dm A

A

A 1/q

Z

= m(A)

A

1/q0 |fn | dm pq 0

MATH 545 and MATH 546 Prelim Problems

23

Choose q, q 0 such that pq 0 = 2, then q 0 = 2/p and 1/q = 1 − 1/q 0 = 1 − p/2, thus q = 2/(2 − p). So we have Z p/2 Z  p (1−p/2) 2 |fn | dm ≤ m(A) |fn | dm ≤ m(A)(1−p/2) M p/2 < 2 A A Thus, for n ≥ N : 1

Z

|fn |p dm < 

0

Which proves that the limit goes to 0 as n goes to infinity. 19. Find the following limit: Z

∞

lim

n→∞

−1



√

nf (x) dx, 1 + nx2

if f is continuous and integrable on R. SOLUTION: √ Let y = nx, then we have √ Z ∞√ Z ∞ f (y/ n) nf (x) lim dx = lim dy n→∞ −1 1 + nx2 n→∞ −√n 1 + y 2 √ √ Z ∞ Z √n f (y/ n) f (y/ n) dy + lim √ dy. = lim n→∞ n→∞ −√n 1 + y 2 1 + y2 n For the second integral, we have √ √ Z ∞ Z ∞ |f (y/ n)| |f (y/ n)| C lim √ dy ≤ lim √ dy ≤ lim 2 n→∞ n→∞ n→∞ 1 + n 1+y 1+n n n

since f ∈ L1 (R)

= 0. So we have

√ f (y/ n) lim dy. 2 √ n→∞ −1 − n 1+y √ √ Now notice that f is continuous, hence it is bounded on [− n, n]. So if we let √ fn = f (y/ n)χ[−√n,√n] /(1 + y 2 ), then fn → f (0)/(1 + y 2 )χ(−∞,∞) and √ |fn | = |f (y/ n)|χ[−√n,√n] /(1 + y 2 ) ≤ supt∈[−1,1] |f (t)|χ[−√n,√n] ≡ g, where g is clearly integrable. Hence by DCT, √ Z ∞√ Z √n Z ∞ nf (x) f (y/ n) dy lim dx = lim dy = f (0) = π f (0). 2 2 2 √ n→∞ −1 1 + nx n→∞ − n 1 + y −∞ 1 + y Z

∞

√

√

nf (x) dx = lim n→∞ 1 + nx2

Z

n

 20. Let (X, A, µ) be a measure space. For f ∈ L1 (X, µ) define Z P (f ) = |f |(1 + log+ |f |) dµ X

MATH 545 and MATH 546 Prelim Problems

24

where log+ x = 0 if 0 ≤ x < 1 and log+ x = log x if x ≥ 1. Suppose {fn } ⊂ L1 (X, µ) is a sequence of functions such that P (fn ) < ∞ for each n and for all  > 0 there is an N ∈ N such that P (fn − fm ) <  for all n, m > N . Show there is an f ∈ L1 (X, µ) with P (f ) < ∞ and P (fn − f ) → 0 as n → ∞. SOLUTION: Let  > 0. So for fn ∈ L1 (X, µ) let Z

|fn |(1 + log+ |fn |) dµ

P (fn ) = X

with P (fn ) < ∞. Let N be chosen such that for all n, m > N we have Z P (fn − fm ) = |fn − fm |(1 + log+ |fn − fm |) dµ < . X

Then we have

Z

Z |fn − fm | dµ ≤

|fn − fm |(1 + log+ |fn − fm |) dµ < .

X

X 1

Hence {fn } is a Cauchy sequence in L and since L1 is complete, there exists a function f ∈ L1 (X, µ) such that fn → f in L1 . Now we need to prove that P (f ) < ∞. Let {fnk }k be a subsequence of {fn } such that limk→∞ fnk (x) = f (x) a.e.. By Fatou’s Lemma Z Z Z |f | dµ = lim |fnk | dµ ≤ lim inf |fnk | dµ < ∞. X k→∞

X

k→∞

X

Now for for all n, nk > N Z

|fn − f |(1 + log+ |fn − f |) dµ

P (fn − f ) = ZX =

lim |fn − fnk |(1 + log+ |fn − fnk |) dµ

X k→∞Z

|fn − fnk |(1 + log+ |fn − fnk |) dµ

≤ lim inf k→∞

X

= lim inf P (fn − fnk ) k→∞

< .  21. Evaluate the following limit: Zπ/2  √ n lim n 1 − sin x dx.

n→∞

0

SOLUTION: Note that

  √ (sin x)1/n − 10 n n 1 − sin x = − . 1/n

MATH 545 and MATH 546 Prelim Problems

25

Then if we let f (y) = ((sin x)y − 10 )/(y − 0) where y = 1/n, then (sin x)y − 10 (sin x)1/n − 10 = lim = f 0 (y) = (sin x)y ln(sin x) = (sin x)1/n ln(sin x). y→0 n→∞ 1/n y−0     √ √ So let fn = n 1 − n sin x , then fn → f = − ln(sin x), and |fn | = n 1 − n sin x ≤ n ≡ g, which is clearly integrable on [0, π/2]. By DCT lim

Zπ/2 Zπ/2  √ n lim n 1 − sin x dx = − ln(sin x) = π ln(2)/2.

n→∞

0

0

 22. Find

Z lim

n→∞

0

∞

sin xn dx. xn

SOLUTION: We split the integral into two parts: Z 1 Z ∞ Z ∞ sin xn sin xn sin xn dx = lim dx + lim dx. lim n→∞ 0 n→∞ 1 n→∞ 0 xn xn xn The second integral on the right is zero because if we let fn = sin xn /xn , then fn → 0 and we can use DCT to prove its limit is zero. Let u = xn , then dx = du/(nxn−1 ) = du/(nu1−1/n ) and Z ∞ Z ∞ sin u sin xn dx = lim du. lim n n→∞ n→∞ 1 x nu2−1/n 1 Let fn = sin u/(nu2−1/n ), then fn → 0 and |fn | ≤ 1/(nu2−1/n ) ≡ g(u). But for all n ≥ 2 Z ∞ 1 u n −1 ∞ 1 g(u) du = . = 1−n 1 n−1 1 Thus by DCT Z lim

n→∞ n

1

∞

sin xn dx = 0. xn

n

For the first integral, let fn = sin x /x , then fn → 1 and |fn | ≤ 1 on [0, 1], since | sin x| ≤ |x| on [0, 1]. Hence again by DCT Z 1 Z 1 sin xn lim dx = 1 dx = 1. n→∞ 0 xn 0  23. Suppose f is Lebesgue integrable on (0, a), a > 0. Show that Z a f (t) dt g(x) = t x is integrable on (0, a).

MATH 545 and MATH 546 Prelim Problems

26

SOLUTION: Ra Ra First notice that |g(x)| ≤ x |f (t)|/t dt ≤ (1/x) x |f (t)| dt < ∞ for all x ∈ (0, a), because f is integrable on (0, a), and hence on any subset of (0, a). That means the function f (t)/t is integrable on (x, a) for x ∈ (0, a). So we can use Fubini’s Theorem as follows, to prove that g(x) is integrable on (0, a): Z aZ a Z aZ a |f (t)| |f (t)| dt dx = χ{0≤x≤t≤a} (t) dt dx t t 0 x 0 0 Z a Z |f (t)| a = χ{0≤x≤t≤a} (x) dx dt t 0 0 Z a Z |f (t)| t = 1 dx dt t 0 0 Z a Z a |f (t)| |f (t)| dt < ∞. t dt = = t 0 0  24. Suppose f and g are nonnegative measurable functions on the interval [0, 1], with the properties Z 1 Z 1 Z 1 f (x) dx = 2, g(x) dx = 1, and [f (x)]2 dx ≤ C 0

0

0

for some constant C > 4. Let E = {x ∈ [0, 1] : f (x) > g(x)}. Show that E has measure m(E) ≥ 1/C. SOLUTION: Suppose not, i.e. m(E) < 1/C. Then Z

1

Z

Z f (x) dx +

f (x) dx =

2=

Z ≤

f (x) dx [0,1]/E 1/2 Z

E

0

2

1/2 1 dx

[f (x)] dx E

Z [0,1]/E

Z 1/2 Z p 2 ≤ m(E) [f (x)] dx + E 1 p m(E) [f (x)]2 dx p √ 0 ≤ m(E) C + 1 1 √ <√ C + 1 = 2. C

Z

≤

f (x) dx, by Holder (note that f ≥ 0)

+

E

g(x) dx

since f (x) ≤ g(x) on E c

[0,1]/E

1/2

Z +

1

g(x) dx 0

So we have 2 < 2 which is a contradiction. So m(E) ≥ C. 

MATH 545 and MATH 546 Prelim Problems

27

25. Suppose g is increasing and differentiable on [0, 1]. For every f ∈ L2 (0, 1) define f ∗ (x), for x ∈ [0, 1], by: Z x ∗ 0 f (t) dt. f (x) = g (x) 0

fn∗

2

If fn → f in L (0, 1), then prove that

∗

→ f in L1 (0, 1).

SOLUTION: First notice that since g increasing, g 0 (x) ≥ 0. Now let  > 0. Let Z x 0 ∗ fn (t) dt. fn (x) = g (x) 0

We need to prove that there exists N ∈ N such that for all n ≥ N we have Z 1 |fn∗ − f ∗ | dx < . 0 2

Since fn → f in L (0, 1), there exists N ∈ N such that for all n ≥ N we have 1/2 Z 1  2 < |fn − f | dx . (Notice that g(1) − g(0) > 0 since g increasing.) g(1) − g(0) 0 Then for all n ≥ N , we have Z x Z x Z 1 Z 1 0 0 ∗ ∗ g (x) f (t) dt dx fn (t) dt − g (x) |fn − f | dx = 0 0 0 0 Z 1 Z x ≤ g 0 (x) |fn − f | dt dx notice that g 0 (x) ≥ 0 Z0 1 Z0 1 = g 0 (x) |fn − f | χ{x≥t≥0} (t) dt dx 0

Z ≤

0 1 0

Z

g (x) 0

1

1/2 Z |fn − f | dt 2

0

Z 1 √ 0  = x g (x) dx g(1) − g(0) 0 Z 1  ≤ g 0 (x) dx g(1) − g(0) 0  ≤ (g(1) − g(0)) = . g(1) − g(0)

x

1/2 1 dt dx

Holder’s

0

(By Theorem 1.1.11) 

26. Let {fn } be a sequence of measurable functions on a measure space (X, A, µ) . Suppose ∞ X that the infinite series µ{x ∈ X : |fn (x)| ≥ } converges for each  > 0. Prove that n=1

fn (x) → 0 a.e.

MATH 545 and MATH 546 Prelim Problems

SOLUTION: Let  > 0 be given. Suppose

∞ X

28

µ{x ∈ X : |fn (x)| ≥ } = C < ∞. Notice that

n=1

C=

∞ X

m Z X

µ{x ∈ X : |fn (x)| ≥ } = lim

m→∞

n=1

= lim

Z X m

m→∞

n=1

χ{|fn (x)|≥} (x) dµ

X

χ{|fn (x)|≥} (x) dµ.

X n=1

P Since χ{|fn (x)|≥} (x) is nonnegative, the summation m n=1 χ{|fn (x)|≥} (x) is increasing in m. By MCT Z X Z X m ∞ χ{|fn (x)|≥} (x) dµ = χ{|fn (x)|≥} (x) dµ. lim m→∞

X n=1

Hence,

X n=1

Z X ∞

χ{|fn (x)|≥} (x) dµ = C < ∞.

X n=1

P That implies ∞ n=1 χ{|fn (x)|≥} (x) is finite [µ]-a.e. x ∈ X. That means we have finitely many n such that |fn (x)| ≥  and infinitely many n such that |fn (x)| < , i.e. fn → 0.  27. Let (X, A, µ) be a σ-finite measure space. Let f : X → [0, ∞) be µ-measurable. Show Z Z ∞ f dµ = µ({x ∈ X : f (x) > t}) dt. X

0

SOLUTION: Note that Z∞

∞

Z

Z

µ({x ∈ X : f (x) > t}) dt =

χ{f (x)>t} dµ(x) dt. 0

0

X

Since {x ∈ X : f (x) > t} is measurable, since f is measurable, χ{f (x)>t} is a nonnegative measurable function. By Tonelli’s Z∞

Z

∞

Z

µ({x ∈ X : f (x) > t}) dt =

χ{f (x)>t} dµ(x) dt 0

0

X

Z Z =

∞

χ{f (x)>t} dt dµ(x) X

0

Z Z =

f (x)

dt dµ(x) ZX

=

0

f (x) dµ(x). X



MATH 545 and MATH 546 Prelim Problems

29

28. Let (X, A, µ) be a finite Z measure space and let f : X → [0, ∞] be a µ-measurable. For A ∈ A define σ(A) = f dµ. Show that if {An }n is a sequence of measurable sets such A

that µ(An ) → 0 and σ(An ) ≥ 1 for all n, then σ(∪∞ n=1 An ) = ∞. SOLUTION:

∞=

∞ X

1≤

n=1

∞ X

σ(An ) =

n=1

∞ Z X

f dµ =

An

n=1

∞ Z X n=1

Z

∞ X

=

f X

Z f

= X

where Z f X

∞ Y

f χAn dµ

X

χAn dµ by MCT, since f χAn ≥ 0

n=1 ∞ X

χ An −

n=1

χAn

dµ

n=1

f dµ ≤ ∩∞ n=1 An

n=1

!

Z

Z χAn dµ =

∞ Y

f dµ = 0 Ak

for some k > N , where N as follows: Since µ(An ) → 0, there is an N ∈ N such that for all n > N , µ(An ) <  (Which implies µ(An ) = 0 for n > N ). Therefore, ! Z ∞ ∞ Y X χAn dµ ∞≤ f χ An − X

n=1

n=1

Z = ZX

f χ∪∞ dµ n=1 An

= ∪∞ n=1 An

f dµ = σ(∪∞ n=1 An ).

(Note: We can solve this problem by using disjointness lemma) 29. Guess the value of Z lim

n→∞

0

1

ln(nx + n) dx. π + ln(nx + n)

Prove your guess is correct. SOLUTION: To guess the value of the limit, notice that ln(nx + n) = lim n→∞ π + ln(nx + n) n→∞ lim

1 = 1. π +1 ln(nx + n)

To proof this, notice that fn =

ln(n) + ln(x + 1) π ln(nx + n) = =1− . π + ln(nx + n) π + ln(n) + ln(x + 1) π + ln(n) + ln(x + 1)



MATH 545 and MATH 546 Prelim Problems

30

The point-wise limit of fn is 1 since ln(∞) = ∞. Also, notice that π ≤ π + ln(n) + ln(x + 1) for all n ∈ N and x ∈ [0, 1], hence |fn | ≤ 1 ≡ g for all n. g is integrable on [0, 1], so by DCT we have Z 1 Z 1 ln(nx + n) dx = 1 dx = 1. lim n→∞ 0 π + ln(nx + n) 0  30. Suppose f ∈ Lp [0, ∞) for some 1 ≤ p ≤ ∞. Guess the value of Z ∞ 2 e−nx f (x) dx. lim n→∞

0

Prove your guess is correct. SOLUTION: 2 We expect the limit to be 0 since limn→∞ e−nx = 0 for all x > 0. Let q be conjugate to p (i.e.  1/q π √ 1/p + 1/q = 1). Let  > 0. Choose N ∈ N such that kf kpp < . Then for all n ≥ N we 2 Nq have Z ∞ Z ∞ 2 −nx2 −nx e f (x) dx ≤ e f (x) dx 0 Z 0∞ 2 e−nx |f (x)| dx ≤ 0

≤

kf kpp

∞

Z

e

−nqx2

1/q dx

(by Holder’s)

0

1/q



π √ 2 nq



π √ 2 Nq

= ≤

kf kpp

1/q

kf kpp

< . Therefore

Z lim

n→∞

∞

2

e−nx f (x) dx = 0.

0

 31. Suppose µ is a finite Borel measure on R which satisfies µ(A ∩ B) = µ(A)µ(B) for all Borel sets A and B in R. Prove that Z Z Z f g dµ = f dµ g dµ, R

for all f, g ∈ L1 (R).

R

R

MATH 545 and MATH 546 Prelim Problems

31

SOLUTION: We will solve this problem in three steps. P P Step 1. Let f = nk=1 ck χAk and g = m j=1 dj χBj where {Ak }k and {Bj }k are µ-measurable sets, and ck , dj ≥ 0. Then Z X Z Z Z n X m n X m n X m X X f g dµ = ck dj χAk χBj dµ = ck dj χAk χBj dµ = ck dj χ(Ak ∩Bj ) dµ R

=

R k=1 j=1 n X

m X

k=1

j=1

R

k=1 j=1

ck µ(Ak )

Z dj µ(Bj ) =

R

Z f dµ

R

k=1 j=1

g dµ. R

Step 2. Let f ≥ 0 and g ≥ 0. Then by Proposition 1.4, there exist sequences of nonnegative nondecreasing measurable simple functions sfn and sgn such that limn→∞ sfn (x) = f (x) and limn→∞ sgn (x) = g(x) for all x. Hence sfn sgn is a nondecreasing sequence and limn→∞ (sfn sgn )(x) = (f g)(x) for all x. So Z Z f g dµ = lim sfn sgn dµ (by MCT) n→∞ R R   Z Z (by step 1) sgn dµ = lim sfn dµ lim n→∞ n→∞ R R Z Z = f dµ g dµ (by MCT). R

R

Step 3. For general f, g let f = f + − f − and g = g + − g − as in Definition 1.3. Then we can apply step 2 on f + , f − , g + and g − because they all are nonnegative to get Z Z f g dµ = (f + − f − )(g + − g − ) dµ R Z Z Z ZR − + + − + + f g dµ − f g dµ − f g dµ + f − g − dµ = R R ZR Z Z Z Z R Z Z Z + + + − − + − = f dµ g dµ − f dµ g dµ − f dµ g dµ + f dµ g − dµ R R R R R R RZ  Z  ZR Z Z Z = f + dµ g + dµ − g − dµ − f − dµ g + dµ − g − dµ R R R RZ RZ R Z Z = f + dµ − f − dµ g + dµ − g − dµ R R R Z R Z = f dµ g dµ. R

R

 32. Let (X, M, m) be Lebesgue measure space. Suppose f ∈ L1 (X, m). Prove that lim n m({x ∈ X : |f (x)| > n}) = 0.

n→∞

SOLUTION:

MATH 545 and MATH 546 Prelim Problems

32

Let An = {x ∈ X : |f (x)| > n}, then An+1 ⊂ An . Let fn = f χAn , then since f ∈ L1 (X, m), |f | is finite a.e. and hence fn → 0 a.e.. Also we have |fn | ≤ |f |, so by DCT Z Z f χAn dm ≥ lim nχAn dm = lim n m(An ). 0 = lim n→∞

n→∞

X

X

n→∞

 33. Let p ≥ 2 be an extended real number and  > 0. Suppose (X, A, µ) is a finite measure space and fk is a sequence of measurable functions on X such that kfk kp → 0 as k → ∞. If φ is measurable on X and Z φfk dµ ≥ , X

2

prove that φ ∈ / L (X, µ). SOLUTION: Suppose φ ∈ L2 (X, µ). Choose p = 2, then q = 2 is a conjugate to p. Then Z Z 0 <  ≤ φfk dµ ≤ |φ||fk | dµ ≤ kφk2 kfk k2 . X

X

Take the limit as k → ∞ on both sides,we get  ≤ 0 which is a contradiction.  34. If f, g ∈ L1 (R) and g is continuous at 0, prove that Z h 1 lim f (y + t)g(t) dt = g(0)f (y) h→0 2h −h for almost every y ∈ R. SOLUTION: x

Z

f (w) dw. Then by FTC(I), F 0 (x) = f (x) a.e. x.

Let F (x) = 0

Claim. 1 lim h→0 2h

Z

h

f (y + t) dt = f (y) a.e. y. −h

Proof: 1 lim h→0 2h

Z

1 h→0 2h lim

0

h

1 f (y + t) dt = lim h→0 2h

0

Z

y+h

f (w) dw = y y

1 F (y + h) − F (y) 1 1 lim = F 0 (y) = f (y) a.e.. 2 h→0 h 2 2

1 F (y) − F (y − h) 1 1 lim = F 0 (y) = f (y) a.e.. 2 h→0 h 2 2 −h y−h Z h Z 0 Z h 1 1 1 =⇒ lim f (y + t) dt = lim f (y + t) dt + lim f (y + t) dt = f (y) a.e.. h→0 2h −h h→0 2h −h h→0 2h 0

Z

1 h→0 2h

f (y + t) dt = lim

Z

f (w) dw =

MATH 545 and MATH 546 Prelim Problems

33

 Notice that Z h Z h Z h 1 1 1 f (y + t)g(t) dt = lim f (y + t)(g(t) − g(0)) dt + g(0) lim f (y + t) dt lim h→0 2h −h h→0 2h −h h→0 2h −h Since g(x) continuous at x = 0, as h → 0, |h − −h| = 2|h| becomes very small and by continuity there exists δ > 0 such that if 2|h| < δ, then |g(x) − g(0)| <  for all  > 0. Hence sending  to 0 implies g(x) − g(0) = 0. So we have Z h Z h 1 1 lim f (y + t)g(t) dt = g(0) lim f (y + t) dt = g(0)f (y) a.e. y ∈ R. h→0 2h −h h→0 2h −h  35. Suppose that f is integrable and absolutely continuous on [0, ∞). If f 0 ∈ L1 [0, ∞), prove that f (x) → 0 as x → ∞ SOLUTION: Suppose f (x) does not tend to 0 as x tends to ∞. Then there exists an N ∈ N and a sequence of points {xn } such that for all n ≥ N we have xn+1 > xn (i.e. xn → ∞) and |f (xn )| >  for some  > 0. Rx Notice that since f ∈ AC([0, ∞)), by FTC(II) f (x) = f (0) + 0 f 0 (t) dt. And since f 0 ∈ L1 [0, ∞), R∞ we must have limx→∞ x f 0 (t) dt = 0, thus there exists a δ > 0 such that if y > x with |y − x| < δ, R y then we must have x f 0 (x) dx < /2. Therefore, Z x Z y 0 0 f (t) dt f (t) dt − f (0) − |f (y) − f (x)| = f (0) + 0 0 Z y  f 0 (x) dx < . = 2 x Hence if we have |y − xn | < δ for n ≥ N , then |f (xn )| >  and hence |f (y)| > /2. Let xn − δ < y < xn + δ (i.e. |y − xn | < δ), then Z xn +δ |f (y)| dy > δ, xn −δ

and therefore Z

∞

Z

∞

|f (y)| dy ≥ 0

|f (y)| dy = xN −δ

∞ Z X n=N

xn +δ

|f (y)| dy >

xn −δ

∞ X

δ = +∞.

n=N

Which contradicts the fact that f is integrable.  36. Suppose X is a normed linear space and denote its unit sphere by S := {x ∈ X : kxk = 1}. Prove that X is a complete normed linear space (Banach space) ⇐⇒ S is complete.

MATH 545 and MATH 546 Prelim Problems

34

SOLUTION: Suppose X is a complete normed linear space. Let {xn } be a Cauchy sequence in S, i.e. kxn k = 1 for all n, then there is x ∈ X such that xn → x, since S ⊂ X and X is complete. We need to prove that kxk = 1. Notice that since the sequence converges, for all  > 0 there exists an N ∈ N such that for all n ≥ N we have kxn k − kxk ≤ |kxn k − kxk| ≤ kxn − xk < , hence kxk = kxn k +  = 1 + . Since that true for all  > 0, kxk = 1, so S is complete. Now suppose S is complete. Let {xn } be a Cauchy sequence in X. Since kxn k ∈ R for all n and every Cauchy sequence in R is convergent, if we consider the sequence {kxn k} and we prove it is Cauchy in R, then there is α ∈ R such that kxn k → α. Notice that {xn } is Cauchy in X implies for all  > 0 there is N ∈ N such that for all n, m ≥ N we have |kxn k − kxm k| ≤ kxn − xm k <  which implies that {kxn k} is Cauchy in R and hence there is α ∈ R such that kxn k → α. Notice that if the limit of {xn } we are looking for equals 0, then we are done, because 0 ∈ X (since X is vector space). Suppose the limit is not 0, and hence α 6= 0, then there is an N ∈ N such that for all n ≥ N , xn 6= 0. Let n ≥ N and consider the sequence yn = xn /kxn k. Then {yn } ⊂ S since kyn k = 1. Since S is complete, if we prove that {yn } is Cauchy, then there is y ∈ S, yn → y. Let  > 0, and choose M ∈ N such that kkxn k − αk < α/3 and kxn − xm k < α/3 for all n, m ≥ M , then for all n, m ≥ M we have





xn

xn

xm

x x x m n m

≤

+

+ 1 kxn − xm k kyn − ym k = − − −

kxn k kxm k kxn k α kxm k α |α| 1 1 1 = kkxn k − αk + kkxm k − αk + kxn − xm k α α α    < + + = . 3 3 3 So {yn } is Cauchy in S and hence there is y ∈ S, yn → y. Since xn = kxn kyn , xn → αy.  37. Prove that Z 0

∞

e−x − e−xt dx = ln t, x

for all t > 0. SOLUTION: Notice that Z 0

∞

e−x − e−xt dx = x

Z 0

∞

Z 1

t

e−xy dy dx.

MATH 545 and MATH 546 Prelim Problems

35

Let (R+ , M, m) be Lebesgue measure space, where R+ = (0, ∞). Then this space is σ-finite. Let f (x, y) = e−xy , then f > 0, and it is m × m-measurable, since it is continuous. By Tonelli’s Theorem Z tZ ∞ Z t −xy x=∞ Z ∞Z t Z t e 1 −xy −xy e dy dx = e dx dy = dy = ln t. dy = 0 1 1 0 1 −y x=0 1 y  38. Guess the value of Z

 tan

lim

n→∞

π/3

0

x2 + nx − 1 n

 dx.

Prove your guess is correct. SOLUTION:  x2 + nx − 1 , then fn → tan x and |fn | ≤ 2x ≡ g on the interval [0, π/3]. g clearly Let fn = tan n is integrable, hence by DCT we have  2  Z π/3 Z π/3 π/3 x + nx − 1 2 tan tan x dx = ln(sec x) = ln √ . lim dx = n→∞ 0 n 0 3 0 

 39. Guess the value of 1 lim √ n→∞ n

Z

1

1/n

ex 1 sin dx. x x

Prove your guess is correct. SOLUTION: √ 1 ex 1 1 ex Let fn = √ sin χ[1/n,1] , then fn → 0 and |fn | ≤ √ χ[1/n,1] = n ex χ[1/n,1] ≡ g. So x nx n (1/n) Z Z 1 √ √ g dx = n ex dx = n(e1 − e1/n ) < ∞. 1/n

Thus g is integrable and hence by DCT Z 1 x Z e 1 1 lim √ sin dx = lim fn dx = 0. n→∞ n→∞ x n 1/n x  40. Recall that the translate of a set E by y is defined to be E + y = {x : x = e + y for some e ∈ E}. (a) Show that the collection of Borel sets in R is closed under translation.

MATH 545 and MATH 546 Prelim Problems

36

(b) Suppose µ is a Borel measure on R which satisfies µ(E + y) = µ(E) for all Borel sets E and all y ∈ R. Prove that Z Z f (x + y) dµ(x) = f dµ R

R

1

for all f ∈ L (R, µ) and all y ∈ R. SOLUTION: (a) For a fixed y, let C = {B − y : B is Borel}. We need to prove that C is a σ-algebra that contains all open sets and hence all Borel sets. If we do that, then if D ∈ C, then D is Borel and D = B−y where B is also Borel, and hence B = D + y is Borel. Notice that C is not empty, since for example [−y, 0] ∈ C because [−y, 0] = [0, y] − y and [0, y] is Borel. Let D ∈ C, then there is B Borel such that D = B − y and hence Dc = B c − y (Proof: Let a ∈ Dc ⇐⇒ a ∈ / B − y ⇐⇒ a + y ∈ / B ⇐⇒ a + y ∈ B c ⇐⇒ a ∈ B c − y) and B c is Borel, so Dc ∈ C. Let Di ∈ C for all i ∈ N. Then there exists {Bi }i , Bi is Borel for each i, such that Di = Bi − y. ∪i Di = ∪i Bi − y (Proof: Let a ∈ ∪i Di ⇐⇒ ∃i0 3 a ∈ Bi0 − y ⇐⇒ a + y ∈ Bi0 ⇐⇒ a + y ∈ ∪i Bi ⇐⇒ a ∈ ∪i Bi + y.) and ∪i Bi is Borel so ∪i Di ∈ C. Now Let O ∈ R be an open set. Then O+y is open (Proof: Let x ∈ O+y then x−y ∈ O and since O is open there exists an  > 0 such that (x−y −, x−y +) ⊂ O and thus (x−, x+) ⊂ O +y.) Then O + y being open implies it is Borel which implies that O = (O + y − y) ∈ C. C contains all open sets and hence all Borel sets and thus if D ∈ C, then D is Borel and D = B − y where B is also Borel, and hence B = D + y is Borel.  (b) We will solve this part in three steps. P Step 1. Let y ∈ R be fixed and let f (x) = nk=1 ck χEk (x) be an integrable nonnegative simple function, Ek Borel sets and ck ≥ 0. Then Z Z X Z X n n f (x + y) dµ(x) = ck χEk (x) dµ(x + y) = ck χEk −y (x) dµ(x) R

R k=1

= =

R k=1 n X Z

ck

k=1 n X

χEk −y (x) dµ(x)

R

ck µ(Ek − y)

k=1

=

n X

ck µ(Ek )

k=1

=

Z X n R k=1

Z ck χEk dµ =

f dµ. R

MATH 545 and MATH 546 Prelim Problems

37

Step 2. Let f ≥ 0 be an integrable function, hence is measurable, and therefore there exists a sequence of nonnegative measurable simple functions {sn } increasing to f . Then by MCT Z Z Z Z f (x + y) dµ(x) = lim sn (x + y) dµ(x) = lim sn (x) dµ = f dµ. n→∞

R

n→∞

R

R

R

Step 3. For general f let f = f + − f − as in Definition 1.3. Then we can apply step 2 on f + and f − because they all are nonnegative to get Z Z Z Z Z + − + f (x + y) dµ(x) = f (x + y) dµ(x) − f (x + y) dµ(x) = f (x) dµ − f − (x) dµ R R R R ZR = f dµ. R

 41. Suppose (X, A, µ) is a finite measure space and fk a sequence of integrable function on X. Suppose further that to each  > 0 there corresponds an integer N > 0 such that Z |fk | dµ <  E

for all measurable sets E which satisfy N µ(E) < 1 and all integers k which satisfy k > N . If fk → f a.e. µ-on X, prove that f ∈ L1 (X, µ) and fk → f in L1 norm. SOLUTION: First notice since fk → f a.e. µ on X (i.e. point-wise limit a.e.), and hence |fk | → |f | a.e., by Fatou’s Lemma we have Z Z Z |f | dµ = lim |fk | dµ ≤ lim inf |fk | dµ < ∞ (Since fk is integrable for all k). X k→∞

X

k→∞

X

Thus f ∈ L1 (X, µ). Now let  > 0. Since (X, A, µ) is a finite measure space and fk → f a.e. µ-on X, by Egoroff’s Theorem for any δ > 0 there exists a set A with µ(X \ A) < δ such that fk → f uniformly on A. This uniform convergence implies that, there exists an integer M ∈ N such that  |fk − f | < for all k > M and for all x ∈ A. Let N be chosen such that N < 1/δ, then 3µ(A) Z N µ(X \ A) < 1 and |fk | dµ < /3 for all k > N . Let L = max{N, M }, then for all k > L we X\A

have Z

Z

Z

|fk − f | dµ = X

|fk − f | dµ +

|fk − f | dµ Z Z Z ≤ |fk − f | dµ + |f | dµ + |fk | dµ A X\A X\A Z Z Z ≤ |fk − f | dµ + lim inf |fn | dµ + A

A

<

X\A

n→∞ n>L

X\A

|fk | dµ

X\A

   µ(A) + lim inf + = . n→∞ n>L 3 3µ(A) 3 

MATH 545 and MATH 546 Prelim Problems

38

42. (a) Suppose a, b ∈ R, a < b, and f ∈ L1 ([a, b], m), where m is Lebesgue measure. Prove that m-a.e. x ∈ [a, b] has the following property: For any sequence {(an , bn )}∞ n=1 of intervals such that x ∈ (an , bn ) for every n ∈ N and limn→∞ (bn − an ) = 0, we have Z bn 1 lim f dm = f (x). n→∞ bn − an a n (That is, a locally integrable function is a.e. equal to the limit of its averages.) (b) Let E ⊂ R be Lebesgue measurable. A point x ∈ R is called a point of density of E if m(E ∩ (x − h, x + h)) = 1. lim+ h→0 2h Prove that a.e. x ∈ E is a point of density of E. And prove that if x ∈ / E, then lim+

h→0

m(E ∩ (x − h, x + h)) = 0. 2h

Note 1.2.1. If we have E ⊂ Rn instead of R, then a point x ∈ R is called a point of density of E if m(E ∩ B(x, r)) lim+ = 1. r→0 m(B(x, r)) Where B(x, r) ⊂ Rn is a ball of radius r centered at x. (c) Prove that if E ⊂ R is Lebesgue measurable and m(E ∩ I) < 0.99m(I) for every open interval I, then m(E) = 0. SOLUTION: Rx (a) Define F (x) = a f (t) dt, then since f is integrable, F 0 (x) exists and F 0 (x) = f (x) a.e. x ∈ [a, b] by FTC(I). For any such x we have  Z x+h Z Z x 1 x+h 1 lim f (t) dt = lim+ f (t) dt − f (t) dt h→0+ h x h→0 h a a F (x + h) − F (x) = lim+ = F 0 (x) = f (x). h→0 h and 1 lim+ h→0 h

Z x  Z x−h 1 f (t) dt = lim+ f (t) dt − f (t) dt h→0 h x−h a a F (x) − F (x − h) = lim+ = F 0 (x) = f (x). h→0 h

Z

x

Let  > 0. Then since an < x < bn for all n and limn→∞ (bn − an ) = 0, limn→∞ (bn − x) = 0 and limn→∞ (x − an ) = 0. So there exists N ∈ N such that for all n > N we have Z x Z bn 1 1 <  and < . (f (t) − f (x)) dt (f (t) − f (x)) dt x − an bn − x an x

MATH 545 and MATH 546 Prelim Problems

Now for n > N we have Z bn 1 f (t) dt − f (x) b n − an an Z bn 1 (f (t) − f (x)) dt = bn − an an Z x x − an 1 bn − x ≤ (f (t) − f (x)) dt + bn − an x − an an b n − an x − an bn − x < + = . b n − an b n − an

39

Z 1 bn − x

x

bn

(f (t) − f (x)) dt

 1 (b) For n ∈ N let En = E ∩ [−n, n], then E = ∪∞ n=1 En . Let f (x) = χEn (x), then f ∈ L (m) and hence we can apply part (a) on f Z x+h Z 1 1 1 = χEn (x) = lim+ χEn (t) dt = lim+ χEn (t)χ(x−h,x+h) (t) dt h→0 2h x−h h→0 2h Z 1 χ(En ∩(x−h,x+h)) (t) dt = lim+ h→0 2h m(En ∩ (x − h, x + h)) = lim+ . h→0 2h

So a.e. x ∈ En is a point of density for En . But since En ⊂ E, we have 1 = lim+ h→0

m(En ∩ (x − h, x + h)) m(E ∩ (x − h, x + h)) m((x − h, x + h)) ≤ lim+ ≤ lim+ h→0 h→0 2h 2h 2h = lim+ 1 = 1. h→0

Therefore we have a.e. x ∈ En is a point of density of E. Claim. For a.e. x ∈ E is a point of density of E. Proof: Let Bn = {x ∈ En : x is not a point of desnist for E}. Then m(Bn ) = 0 and m(∪∞ n=1 Bn ) = 0. If x ∈ E is not a point of density of E, then there exists n ∈ N such that ∞ x ∈ En is not a point of density of E, since E = ∪∞ n=1 En . So x ∈ Bn , and hence x ∈ ∪n=1 Bn . So B = {x ∈ E : x is not a point of density of E} ⊂ ∪∞ n=1 Bn . Hence m(B) = 0.#  For the c second part, suppose x ∈ / E, i.e. x ∈ E , then a.e. x ∈ E c is a point of density of E c and hence we have m(E c ∩ (x − h, x + h)) lim = 1. h→0+ 2h But since   m((x − h, x + h)) m(E ∩ (x − h, x + h)) m(E c ∩ (x − h, x + h)) 1 = lim+ = lim+ + , h→0 h→0 2h 2h 2h we have lim+

h→0

m(E ∩ (x − h, x + h)) m(E c ∩ (x − h, x + h)) = 1 − lim+ = 1 − 1 = 0. h→0 2h 2h 

MATH 545 and MATH 546 Prelim Problems

40

(c) Suppose for contradiction that m(E) > 0, then there must exist a point of density x of E. By assumption we have m(E ∩ (x − h, x + h)) ≤ 0.99 m((x − h, x + h)) = 0.99(2h). By part b 1 = lim+ h→0

0.99(2h) m(E ∩ (x − h, x + h)) ≤ lim+ = 0.99. h→0 2h 2h

Which is a contradiction.  43. Let L1Loc (R, m) be the set of Lebesgue measurable functions f : R → [−∞, ∞] such that Rn |f | dm < ∞ for each n ∈ N. −n (a) Suppose f ∈ L1Loc (R, m). Prove that for m-a.e. x0 ∈ R, Z x0 +h 1 lim f (x) dm(x) = f (x0 ). h→0+ 2h x0 −h (b) A point x0 is called a Lebesgue point of f if Z x0 +h 1 |f (x) − f (x0 )| dm(x) = 0. lim h→0+ 2h x0 −h Suppose f ∈ L1Loc (R, m). Prove that m-a.e. x0 ∈ R is a Lebesgue point of f . SOLUTION: (a) Since f ∈ L1Loc (R, m), f ∈ L1 (R, m). So by Problem 42, we have for m-a.e. x0 ∈ R, Z x0 +h Z x0 Z x0 +h 1 1 1 lim f (x) dm(x) = lim+ f (x) dm(x) + lim+ f (x) dm(x) = f (x0 ). h→0+ 2h x0 −h h→0 2h x0 −h h→0 2h x0 Another solution. For each n ∈ N let 1 An = {x0 ∈ (−n, n) : lim+ h→0 2h

Z

x0 +h

f (x) dm(x) 6= f (x0 ).} x0 −h

Let fn = f χ(−n,n) . Then 1 An = {x0 ∈ (−n, n) : lim+ h→0 2h

Z

x0 +h

fn (x) dm(x) 6= fn (x0 ), } x0 −h

since f = fn on (x0 − h, x0 + h) ⊂ (−n, n) for h sufficiently small. Since fn ∈ L1 ((−n, n), m), Z x0 +h 1 lim fn (x) dm(x) = fn (x0 ) h→0+ 2h x0 −h for m-a.e. x0 ∈ (−n, n) and hence m(An ) = 0 for all n. Therefore m(∪n An ) = 0 and hence for all x0 ∈ R \ ∪n An ) (i.e., a.e. x0 ∈ R) we have Z x0 +h 1 lim f (x) dm(x) = f (x0 ). h→0+ 2h x0 −h 

MATH 545 and MATH 546 Prelim Problems

41

(b) Let Q = {ri }∞ i=1 be the set of rationals. For each i ∈ N let Z x0 +h 1 Ei = {x0 ∈ R : lim+ |f (x) − ri | dm(x) 6= |f (x0 ) − ri |}. h→0 2h x0 −h Then since |f (x) − ri | ∈ L1 (R, m) for each i, m(Ei ) = 0, by part a, and hence m(∪i Ei ) = 0. Claim. For all x0 ∈ / ∪∞ i=1 Ei , x0 is a Lebesgue point of f . Proof: Let  > 0. Since rationals are dense in R, there exists i ∈ N such that |f (x0 ) − ri | < /2. Then Z x0 +h 1 lim |f (x) − f (x0 )| dm(x) h→0+ 2h x0 −h Z x0 +h Z x0 +h 1 1 ≤ lim+ |f (x) − ri )| dm(x) + lim+ |f (x0 ) − ri | dm(x) h→0 2h x0 −h h→0 2h x0 −h = |f (x0 ) − ri )| + |f (x0 ) − ri )| < . Since this is true for all  > 0, 1 lim+ h→0 2h

Z

x0 +h

|f (x) − f (x0 )| dm(x) = 0.# x0 −h

 Hence m-a.e. x0 ∈ R is a Lebesgue point of f .  44. Let f ∈ L1 (Rk , m) where m is Lebesgue measure. Define g(x) = f (x) if |f (x)| ≤ 1 and g(x) = 0 otherwise. Suppose x0 is a Lebesgue point for f such that f (x0 ) < 1. Show that x0 is a Lebesgue point for g. SOLUTION: We need to prove that 1 lim+ r→0 |B(x0 , r)|

Z |g(x) − g(x0 )| dm(x) = 0 B(x0 ,r)

Where |B(x0 , r)| is Lebesgue measure of a ball of radius r and centered at x0 . Since x0 is a Lebesgue point for f , we have Z 1 lim |f (x) − f (x0 )| dm(x) = 0. r→0+ |B(x0 , r)| B(x0 ,r) Let A = {x ∈ Rk : |f (x)| ≤ 1}. Then g(x) = f (x)χA (x). Since f ∈ L1 (Rk , m), f is Lebesgue measurable and hence A is Lebesgue measurable. By Problem 42 (b), a.e. x ∈ A is a point of density of A, and since |f (x0 )| < 1, x0 ∈ A and therefore it may or may not be a density point of A, but for sure m(Ac ∩ B(x0 , r)) lim+ = 0, r→0 m(B(x0 , r))

MATH 545 and MATH 546 Prelim Problems

42

since x0 ∈ / Ac , by Problem 42 (b). Back to what we need to show Z 1 lim |g(x) − g(x0 )| dm(x) r→0+ |B(x0 , r)| B(x0 ,r)  Z Z 1 |g(x) − g(x0 )| dm(x) + |g(x) − g(x0 )| dm(x) = lim+ r→0 |B(x0 , r)| B(x0 ,r)∩A B(x0 ,r)∩Ac Z  Z 1 = lim+ |f (x) − f (x0 )| dm(x) + |0 − f (x0 )| dm(x) r→0 |B(x0 , r)| B(x0 ,r)∩A B(x0 ,r)∩Ac Z  Z 1 |f (x) − f (x0 )| dm(x) + 1 dm(x) ≤ lim+ r→0 |B(x0 , r)| B(x0 ,r) B(x0 ,r)∩Ac m(B(x0 , r) ∩ Ac ) = lim+ r→0 |B(x0 , r)| = 0.  45. Let 1 ≤ p < ∞ and let f ∈ Lp (R). Show that Z |f (x) − f (x + h)|p dx → 0

as h → 0.

R

SOLUTION: Notice that this is the same as proving Z |f (x) − f (x + 1/n)|p dx → 0

as n → ∞.

R

Let fn (x) = |f (x) − f (x + 1/n)|p . Then limn→∞ fn = 0 and |fn (x)| = |f (x) − f (x + 1/n)|p ≤ (|f (x)| + |f (x + 1/n)|)p ≤ (2 max{|f (x)|, |f (x + 1/n)|})p ≤ 2p max{|f (x)|p , |f (x + 1/n)|p } ≤ 2p (|f (x)|p + |f (x + 1/n)|p ) ≡ g. But Z

p

Z

|f (x)| +

g(x) dx = 2 R

Z

p

R

p

|f (x + 1/n)|

 < ∞.

(Since f ∈ Lp )

R

Then by DCT, Z

|f (x) − f (x + 1/n)| dx =

lim

n→∞

p

R

Z

lim |f (x) − f (x + 1/n)|p dx = 0.

R n→∞

 Another Solution First suppose f is continuous and there exists n ∈ N such that f (x) = 0 for x ∈ / [−n, n], i.e. f is continuous with compact support. Then f is uniformly continuous on [−n − 1, n + 1] (A continuous

MATH 545 and MATH 546 Prelim Problems

43

function on a compact set is uniformly continuous). Note that if |h| < 1, then for x ∈ / [−n − 1, n + 1] we have f (x + h) = 0. Let  > 0. By uniform continuity of f , there exists δ ∈ (0, 1) such that |f (x) − f (x + h)| < /(2n + 2)p for all h such that |h| < δ and for all x ∈ [−n − 1, n + 1]. Hence for all h such that |h| < δ, 1/p Z 1/p Z n+1 p p |f (x) − f (x + h)| dx |f (x) − f (x + h)| dx = −n−1

R

 <

p 2n + 2

Z

n+1

1/p 1 dx = .

−n−1

This establishes the result in the case that f is continuous with compact support. Now let f ∈ Lp . Let  > 0. Then by Theorem 1.1.26 there exists a continuous function g vanishing outside [−n, n] for some n ∈ N such that kf − gkp < /3. By the case above there exists δ > 0 such that for all h with |h| < δ, kg(x) − g(x + h)| < /3. Note that by the translation-invariance of Lebesgue integral, kg(x + h) − f (x + h)kp = kg(x) − f (x)kp < /3. Hence by Minkowski’s inequality kf (x) − f (x + h)kp ≤ kf (x) − g(x)kp + kg(x) − g(x + h)kp + kg(x + h) − f (x + h)kp    < + + = . 3 3 3  46. If h, g ∈ L2 (R, m), prove that Z

1

g(y − x)h(y) dy

F (x) := 0

is a continuous function on R. SOLUTION: Let c > 0, we need to prove that lim |F (x + c) − F (x)| = 0.

c→0

To do that we are going to use Holder’s inequality. For each x ∈ R and y ∈ [0, 1], let G(x) := g(y−x), then G ∈ L2 (R, m) since g is. So Z 1 Z 1 lim |F (x + c) − F (x)| = lim g(y − (x + c))h(y) dy − g(y − x)h(y) dy c→0 c→0 0 0 Z 1 ≤ lim |h(y)||g(y − (x + c)) − g(y − x)| dy c→0

0

Z

1

≤ lim

c→0

2

1/2 Z

1

|h(y)| dy 0

2

1/2

|g(y − (x + c)) − g(y − x)| dy 0

≤ khk2 lim kG(x + c) − G(x)k2 c→0

= 0. Where last limit is zero by the previous problem.



MATH 545 and MATH 546 Prelim Problems

44

47. The Laplace transform of real measurable functions on [0, ∞) is defined by L[f ] = R∞ f (t) e−st dt. If f ∈ L1 [0, ∞), prove that L[f ] exists and is bounded for all s ≥ 0. If 0 fn , f ∈ L1 [0, ∞) for all n, and fn → f in L1 [0, ∞), show that L[fn ] → L[f ] uniformly on [0, ∞). SOLUTION: For all s, t ≥ 0, Z

∞

|L[f ]| ≤

|f (t)| e−st dt ≤

0

Z

∞

|f (t)| dt = kf k1 < ∞. 0

Hence if f ∈ L1 [0, ∞), L[f ] exists and is bounded for all s ≥ 0. Now let  > 0. Then since fn → f R∞ in L1 , there exists an N ∈ N such that for all n > N , 0 |fn (t) − f (t)| dt < . For all s ≥ 0 and n > N , we have Z ∞ −st |L[fn ] − L[f ]| = e (fn (t) − f (t)) dt Z 0∞ ≤ |fn (t) − f (t)| dt < . 0

 48. Let f be a real everywhere differentiable function on (−∞, ∞) show that f 0 is Borel measurable. Further if |f 0 | ≤ M , show that f is absolutely continuous on (−∞, ∞). SOLUTION: Since f is differentiable everywhere, f is continuous everywhere and hence is Borel. Now notice that f (x + 1/n) − f (x) = lim (nf (x + 1/n) − nf (x)). n→∞ n→∞ 1/n

f 0 (x) = lim

But nf (x) and nf (x + 1/n) are continuous everywhere, since f is, and hence Borel. A point-wise limit of Borel functions is also Borel as well. Therefore, f 0 is Borel measurable. Now let  > 0 and P let {(xi−1 , xi )}ni=1 be a collection of disjoint open intervals with ni=1 (xi − xi−1 ) < /M . Then X n n Z xi n Z xi n X X X 0 0 |f (xi ) − f (xi−1 )| = f (x) dx ≤ |f | dx ≤ M (xi − xi−1 ) < . i=1

i=1

xi−1

i=1

xi−1

i=1

Thus f is absolutely continuous.  49. Let {fk } be a non-decreasing sequence of real valued measurable functions on (X, F) R and let µ be a non-negative measure on (X, F). Assume X f1− dµ < ∞, show that Z Z lim fk dµ = lim fk dµ. k→∞

X

X k→∞

MATH 545 and MATH 546 Prelim Problems

45

SOLUTION: This is a generalization of Monotone Convergence Theorem. Notice that if f is non-decreasing, then its positive and negative parts are still non-decreasing. Now consider the sequence gk = fk +f1− , then {gk } is non-decreasing and gk ≥ g1 = f1+ −f1− +f1− = f1+ ≥ 0 for all k. Then we can apply MCT on gk to have Z Z − lim (fk + f1 ) dµ = lim (fk + f1− ) d k→∞ X k→∞ X Z Z Z Z − lim fk dµ + f1 dµ = lim fk d + f1− dµ k→∞ X k→∞ X Z ZX Z X =⇒ lim fk dµ = lim fk dµ (Since f1− dµ < ∞). k→∞

X

X k→∞

X

 50. Let f, g be non-negative real measurable functions on [0, ∞). Let 1 < p ≤ 2 and q = R∞ 1 p/(p − 1). If g ∈ Lp ([0, ∞), m) and 0 f (t) t q −1 dt < ∞, show that Z ∞Z x Z ∞ 1 g(x) −1/q f (t) dt dx ≤ kgkp (q − 1) f (t) t q −1 dt < ∞. x 0 0 0 SOLUTION: Since f, g are measurable and 1/x is continuous on [0, ∞), and hence measurable,

g(x) f (t) is m×mx

g(x) f (t) ≥ 0 for all t, x in [0, ∞). With these information and the x fact ([0, ∞), M, m) is a σ-finite measure space, Tonelli’s Theorem can be applied. Also note that q > 1. So we have Z ∞Z ∞ Z ∞Z x g(x) g(x) f (t) dt dx = f (t)χ{x≥t≥0} dt dx x x 0 0 0 0 Z ∞Z ∞ g(x) = f (t)χ{x≥t≥0} dx dt (Tonelli’s) x 0 0 Z ∞ Z ∞ g(x) = f (t) dx dt x 0 t Z ∞ 1/p Z ∞ 1/q Z ∞ 1 p ≤ f (t) g (x) dx dx dt (Holder’s) xq 0 t t Z ∞ 1/q Z ∞ 1 ≤ kgkp f (t) dx dt xq 0 t ∞ !1/q Z ∞ x1−q = kgkp f (t) dt 1 − q 0 Z ∞ t 1 = kgkp (q − 1)−1/q f (t)t q −1 dt < ∞. measurable on [0, ∞). Moreover,

0

MATH 545 and MATH 546 Prelim Problems

46

 51. Let (X, A, µ) be a measure and let {Ek }k be a sequence of measurable sets in X, such that ∞ X µ(Ek ) < ∞. k=1

Show that µ-a.e. x ∈ X lie in at most finitely many of the sets Ek . SOLUTION: This result called Borel-Cantelli Lemma. Let A = {x ∈ X : x ∈ Ek for infinitely many k}. Then we need to prove that µ(A) = 0. Claim. A=

∞ [ ∞ \

Ek .

n=1 k=n

Proof: Let x ∈ A, then x ∈ Ek for infinitely many k, so for any n ∈ N, there is k ∈ N, k ≥ n such ∞ ∞ that x ∈ Ek . That is x ∈ ∪∞ k=n for every n ∈ N and so x ∈ ∩n=1 ∪k=n Ek . For the other side, let ∞ ∞ x ∈ ∩∞ n=1 ∪k=n Ek , then x ∈ ∪k=n Ek for some n ∈ N and hence x ∈ Ek for all k ≥ n, i.e. infinitely many k, and thus x ∈ A. #  By the claim A ⊂ ∪∞ k=n Ek , and therefor µ(A) ≤

∞ X

µ(Ek )

for each n ∈ N.

(∗)

k=n

P∞

Since k=1 µ(Ek ) < ∞, limn→∞ have µ(A) = 0.

P∞

k=n

µ(Ek ) = 0. Take the limit as n → ∞ on both sides of (∗) we 

52. Let µ, ν, and λ be σ-finite, nonnegative and nontrivial measures on the measure space (X, M) such that for every measurable set A ∈ M, µ(A) = ν(A) + λ(A). Show: (a) If ν ⊥ λ, then ∃ B ∈ M such that ν(A) = µ(A ∩ B). dν (b) Note that v  µ. If ν  λ, then 0 ≤ < 1 a.e. [µ] on X. dµ SOLUTION:

MATH 545 and MATH 546 Prelim Problems

47

(a) Let B such that λ(B) = 0 and ν(B c ) = 0. Then X = B ∪ B c and ν(A) = ν(A ∩ B) + ν(A ∩ B c ) = ν(A ∩ B) since A ∩ B c ⊂ B c and ν(B c ) = 0. Also, λ(A ∩ B) = 0, since A ∩ B ⊂ B and λ(B) = 0. Hence µ(A ∩ B) = ν(A ∩ B) + λ(A ∩ B) = ν(A).  (b) Let A = {x :

dν (x) dµ

≥ 1}, then we need to prove that µ(A) = 0. Since ν  µ, Z

ν(A) = A

dν dµ ≥ µ(A) = ν(A) + λ(A) =⇒ λ(A) = 0 if ν(A) < ∞. dµ

Suppose ν(A) < ∞, then since ν  λ, Z ν(A) = A

dν dλ = 0 (since λ(A) = 0). dλ

Hence µ(A) = 0.  53. Show that the integral Z

∞

Z

0

0

∞

sin πx dx dy (y + ex | sin πx|)2

exists and is finite and evaluate it.

SOLUTION:

Notice that Z [0,∞)×[0,∞)

| sin πx| dm(x × y) = (y + ex | sin πx|)2

Z

1

y ex ( 2 + )2 sin πx | sin πx| Z 1 ≤ dm(x × y) x 2 [0,∞)×[0,∞) (y + e )

dm(x × y)

[0,∞)×[0,∞)

Notice that f (x, y) = 1/(y+ex )2 is continuous for all x, y ∈ [0, ∞), hence m×m-Lebesgue measurable, f (x, y) ≥ 0 and ([0, ∞), M, m) is σ-finite measure space. Thus by Tonelli’s Theorem Z [0,∞)×[0,∞)

1 dm(x × y) = (y + ex )2

Z

∞

0

Z = 0

∞

Z

∞

1 dy dx (y + ex )2 0 Z ∞ −1 ∞ e−x dx = 1 < ∞. dx = y + ex 0 0

MATH 545 and MATH 546 Prelim Problems

48

Hence sin πx/(y + ex | sin πx|)2 is integrable and hence by Fubini theorem Z ∞Z ∞ Z ∞Z ∞ sin πx sin πx dx dy = dy dx x 2 (y + e | sin πx|) (y + ex | sin πx|)2 0 0 0 0 ∞ Z ∞ sin(πx) dx = − y + ex | sin(πx)| y=0 0 Z ∞ sin(πx) −x e dx = | sin(πx)| 0 ∞ Z n+1 X = (−1)n e−x dx =

n=0 ∞ X

n

(e−n − e−(n+1) )(−1)n

n=0

= 1 − e−1 − e−1 + e−2 + e−2 − e−3 − e−3 + e−4 + ... = 2(1 − e−1 + e−2 − e−3 + e−4 + ....) − 1 2e e−1 1 =2 −1= −1= . −1 1 − (−e ) e+1 e+1 Note 1.2.2. We can do the following to check that f (x, y) = sin πx/(y + ex | sin πx|)2 is integrable and hence Fubini can be applied: Notice that f (x, y) is measurable since it is continuous, and |f (x, y)| ≥ 0, then by Tonelli’s Z Z ∞Z ∞ | sin πx| | sin πx| dm(x × y) = dy dx x 2 (y + ex | sin πx|)2 [0,∞)×[0,∞) (y + e | sin πx|) 0 0 Z ∞ Z ∞ 1 dy dx = | sin πx| x (y + e | sin πx|)2 0 0 Z ∞ ∞ −1 = | sin πx| dx x | sin πx| 0 y + e 0 Z ∞ 1 = | sin πx| x dx e | sin πx| 0 Z ∞ = e−x dx = 1 < ∞. 0

 54. Let f ∈ L1 (R). For s > 0 define g(s) :=

1 s

Z (s − r) f (r) dr. (0,s)

(a) Find with proof lims→0+ g(s). (b) For a.e. s > 0 compute g 0 (s). (c) Show, with appropriate definition for g(0), that g ∈ AC([0, 1]). SOLUTION:

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(a) First Solution: Notice that Z lim+ g(s) = lim g(1/n) = lim n→∞

s→0

Let fn =

n→∞

1/n

0

1 (1 − nr) f (r) dr = lim n→∞ n

Z

1

(1 − x) f 0

x n

dx.

x 1 1 (1 − x) f χ[0,1] , then fn → 0 as n → ∞, and |fn | ≤ |f (x/n)|χ[0,1] ≡ g(x). But n n n Z Z 1 1  x  g(x) dx = dx < ∞ (Since f ∈ L1 (R)). f n 0 n R

Hence by DCT 1 lim+ g(s) = lim n→∞ n s→0

Z

1

x

(1 − x) f

n

0

Z

1

x 1 (1 − x) f dx = 0. n→∞ n n lim

dx = 0

Second Solution: Claim. : lims→0+ g(s) = 0. Proof: Since f ∈ L1 (R), by Problem 6, for all  > 0 there is δ > 0 such that if m((0, s)) = s < δ, then Z  |f (r)| dr < . 2 (0,s) Therefore, Z Z Z 1 r |g(s) − 0| = (s − r) f (r) dr ≤ |f (r)| dr + |f (r)| dr s (0,s) (0,s) (0,s) s Z Z ≤ |f (r)| dr + |f (r)| dr (0,s)

(0,s)

  < + = . 2 2 



(b)   Z s+h Z 1 s g(s + h) − g(s) 1 1 g (s) = lim+ = lim+ (s + h − r) f (r) dr − (s − r) f (r) dr h→0 h→0 h s + h 0 h s 0   Z s Z s+h Z 1 1 1 1 s = lim+ (s + h − r) f (r) dr + (s + h − r) f (r) dr − (s − r) f (r) dr h→0 h s + h 0 s+h s s 0 Z s    Z s Z s+h 1 1 1 h 1 = lim+ (s − r) f (r) dr − + f (r) dr + (s + h − r) f (r) dr h→0 h s+h s s+h 0 s+h s 0   Z s+h Z s Z s 1 1 1 = lim+ 2 (r − s) f (r) dr + f (r) dr + (s + h − r) f (r) dr h→0 s + sh 0 s+h 0 sh + h2 s Z s Z s+h r 1 = f (r) dr + lim+ (s + h − r) f (r) dr 2 h→0 sh + h2 s 0 s 0

Claim. 1 lim+ h→0 sh + h2

Z

s+h

(s + h − r) f (r) dr = 0. s

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Proof: Again since f ∈ L1 (R), for all  > 0, there is δ > 0 with m((s, s + h)) = h < δ and Z |f (r)| dr < (s + h). (s,s+h)

Therefore, Z 1 sh + h2

s

 Hence g 0 (s) =

s+h

Z s+h 1 (s + h − s) |f (r)| dr sh + h2 s Z s+h 1 = |f (r)| dr < .# s+h s

(s + h − r) f (r) dr − 0 ≤

1 R r f (r) dr. s2 (0,s)



(c) Clearly we can define g(0) = 0 because lims→0+ g(s) = 0 as shown in part a. Recall from Problem 7, the first step we need to do is proving that g 0 is integrable on an arbitrary interval [0, x] ⊂ [0, 1] and Z x g 0 (s) ds = g(x) − g(0) = g(x). (Note that 0 ≤ s ≤ x) 0 0

So, to prove g is integrable we have to show Z xZ x Z x Z xZ s r r 0 |f (r)| dr ds = |f (r)|χ{x≥s≥r≥0} dr ds < ∞. |g (s)| ds = 2 2 0 0 s 0 0 0 s Notice that (r/s2 )|f (r)| is measurable since f is measurable and r/s2 is continuous and hence measurable. Also (r/s2 )|f (r)| ≥ 0, hence by Tonelli’s Z x Z xZ x r 0 |f (r)|χ{x≥s≥r≥0} ds dr |g (s)| ds = 2 0 0 0 s Z x Z x 1 r |f (r)| = ds dr 2 r s Z0 x  Z x r 1− |f (r)| dr < ∞. (Since f ∈ L1 ) = |f (r)| dr ≤ x 0 0 R x So g 0 is integrable. Now we need to show 0 g 0 (s) ds = g(x). Since g 0 is integrable, by Fubini’s Theorem Z x Z xZ x r 0 g (s) ds = f (r)χ{x≥s≥r≥0} ds dr s2 0 Z0 x 0 Z x 1 = r f (r) ds 2 r s Z0 x  Z r 1 x = 1− f (r) dr = (x − r) f (r) dr = g(x). x x 0 0 Since g 0 is integrable, for all  > 0, there is a δ > 0 such that if E is a measurable set with m(E) < δ, then Z |g 0 (s)| ds < . E

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P Now let E = ∪ni=1 (ai , bi ) with m(E) = ni=1 (bi − ai ) < δ, where {(ai , bi )}ni=1 is a collection of finite disjoint intervals in [0, 1] ordered in such a way bi < ai+1 for i = 1, 2, ..., n − 1. Then Z ai n n Z b i n Z bi X X X 0 0 |g(bi ) − g(ai )| = g (s) ds − g (s) ds ≤ |g 0 (s)| ds i=1

i=1

0

0

ai

Zi=1

|g 0 (s)| ds

= ∪n i=1 (ai ,bi )

Z =

|g 0 (s)| ds < .

E

Thus g ∈ AC([0, 1]).  55. Prove that Z lim

n→∞

1


n 1 − e−x/n dx

0

exists and find its value. SOLUTION:
Let fn = n 1 − e−x/n . Let y = 1/n, then −x y −x 0
(e ) − (e ) d −x/n −xy lim n 1 − e = − lim+ =− e n→∞ y→0 y−0 dy

= x. y=0

(We can find this limit by L’Hopital’s rule, also notice that if n(1 − e−x/n ) = y, then (1 − ny )n = e−x , thus as we know the limit on the left will give us e−y . Compare it to e−x , we have y = x.) Also, |fn | ≤ n ≡ g, which is clearly integrable on [0, 1]. Hence by DCT Z 1 Z 1 Z 1

1 −x/n −x/n n 1−e dx = lim n 1 − e dx = x dx = . lim n→∞ 0 2 0 n→∞ 0  56. Let µn be a sequence of nonnegative measures on a measurable space (X, M) with µn (X) = 1, for n ∈ N ∪ {0}. Construct a measure ν such that ν(X) = 1 and µn  ν, for all n ∈ N ∪ {0}. SOLUTION: µn . We need to prove that ν is a measure (not a signed measure), and it satisfies 2n+1 ν(X) = 1 and µn  ν, for all n ∈ N ∪ {0}.

Let ν =

P∞

n=0

For the first part:

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52

P∞ µn (∅) = 0 since µn n=0 2n+1 is measure for all n = 0, 1, .... Finally, let {Ei } be a countable collection of disjoint sets in M, then

ν is nonnegative since µn is nonnegative for all n = 0, 1, .... Also, ν(∅) =

ν(∪∞ i=1 Ei )

∞ ∞ X ∞ X X µn (∪∞ µn (Ei ) i=1 Ei ) = = . n+1 2 2n+1 n=0 n=0 i=0

But since µn is snonnegative for all n, we can switch the order of summations (We can do that also if we consider the counting measure and then apply Tonelli’s). So we get ν(∪∞ i=1 Ei )

=

∞ ∞ X ∞ X µn (Ei ) X i=0 n=0

2n+1

ν(Ei ).

i=0

For the second part: Note that ν(X) =

∞ X µn (X) n=0

2n+1

∞ X 1 = = 1. n+1 2 n=0

(Geometric series)

Also, if ν(A) = 0 for some measurable set A, then 0=

∞ X µn (A) n=0

2n+1

,

but this is possible if and only if µn (A) = 0 for all n = 0, 1, .... Hence µn  ν for all n. 

Chapter 2

Complex Analysis 2.1

Review

Definition 2.1.1. Let f (z) = u + iv. Define   1 ∂ 1 ∂ ∂ = + ∂z 2 ∂x i ∂y

and

1 ∂ = ∂z 2



1 ∂ ∂ − ∂x i ∂y

 .

Also, we can define the Laplacian ∆ as ∆=

∂2 ∂2 ∂ ∂ ∂ ∂ + =4 =4 . 2 2 ∂x ∂y ∂z ∂z ∂z ∂z

Proposition 2.1.1. If f is holomorphic at z0 , then ∂f ∂f ∂u (z0 ) = 0 and f 0 (z0 ) = (z0 ) = 2 (z0 ) = ux + ivx = ux − ivy . ∂z ∂z ∂z In other words, if f is holomorphic, then it satisfies Cauchy-Riemann equations. Theorem 2.1.2. Suppose f = u + iv is a complex-valued function defined on an open set Ω. If u and v are continuously differentiable and satisfy the Cauchy-Riemann equations on Ω, then f is holomorphic on Ω and f 0 (z) = ∂f /∂z. P n Theorem 2.1.3. Given a power series ∞ n=0 an (z − z0 ) , there exists 0 ≤ R ≤ ∞ such that : (i) If |z| < R, the series converges absolutely. (ii) If |z| > R, the series diverges. R, the radius of convergence, is given by Hadamards formula 1 = lim sup |an |1/n . R n→∞ P n Theorem 2.1.4. f Analytic =⇒ f Holomorphic: The power series f (z) = ∞ n=0 an (z − z0 ) defines a holomorphic function in its disc of convergence R. The derivative of f is also a power series obtained by differentiating term by term the series for f , that is, 0

f (z) =

∞ X

nan (z − z0 )n−1 .

n=0

53

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54

Moreover, f 0 has the same radius of convergence as f , namely R. This theorem says: If f is analytic, then it is holomorphic. Corollary 2.1.5. A power series is infinitely complex differentiable in its disc of convergence, and the higher derivatives are also power series obtained by term-wise differentiation. Definition 2.1.2. Integral of f Along Curve γ: Given a smooth curve γ parametrized by z(t) : [a, b] → C (e.g z(t) = z0 + reit , t ∈ [0, 2π]) and f a continuous function on γ, we define the integral of f along γ by Z Z b

f (z(t))z 0 (t) dt.

f (z) dz = a

γ

By definition, the length of the smooth curve γ is Z length(γ) =

b

|z 0 (t)| dt.

a

Definition 2.1.3. Primitive: Suppose f is a function on the open set Ω. A primitive for f on Ω is a function F that is holomorphic on Ω and such that F 0 (z) = f (z) for all z ∈ Ω. Theorem 2.1.6. If a continuous function f has a primitive F in Ω, and γ is a curve in Ω that begins at w1 and ends at w2 , then Z f (z) dz = F (w2 ) − F (w1 ). γ

Hence, if γ is a closed curve in Ω, then Z f (z) dz = 0. γ

Also, if given a closed curve Ω and the integral Z f (z) dz 6= 0, γ

then f does not have a primitive in Ω. Theorem 2.1.7. Goursat’s Theorem: If f is holomorphic in open set Ω and T ⊂ Ω a triangle whose interior is also contained in Ω. Then Z f (z) dz = 0. T 0

Proof: Let T denote the original T oriented counterclockwise. Let d0 and p0 denote the diameter and perimeter of T 0 , respectively. Bisect each side of the triangle and connecting the midpoints. This creates four new smaller triangles, denoted T11 , T21 , T31 and T41 with the same orientation as T 0 . Hence we have Z Z Z Z Z f (z) dz = f (z) dz + f (z) dz + f (z) dz + f (z) dz. T0

And therefore

T11

T21

Z

T

T31

Z f (z) dz ≤ 4 0

T

f (z) dz , 1

T41

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where T 1 is such that (Z

Z f (z) dz =

max

f (z) dz

j∈{1,2,3,4}

T1

) .

Tj1

Then d1 = d0 /2 and p1 = p0 /2. We now repeat this process for the triangle T 1 , bisecting it into four smaller triangles. Continuing this process, we obtain a sequence of triangles T 0 , T 1 , ..., T n with the properties that Z Z n 0 f (z) dz ≤ 4 n f (z) dz , T

n

0

n

n

0

T

n

n

and d = d /2 and p = p /2 . We also denote by T the solid closed triangle with boundary T n , and observe that our construction yields a sequence of nested compact sets T n ⊂ T n−1 ⊂ ... ⊂ T 1 whose diameter goes to 0. Hence there exists a unique point z0 that belongs to all the solid triangles T n . Since f is holomorphic at z0 we can write f (z) = f (z0 ) + (z − z0 )f 0 (z0 ) + ψ(z)(z − z0 ), where ψ(z) → 0 as z → z0 . Now both the constant and linear functions f (z0 ) and (z − z0 )f 0 (z0 ) have primitive in T 1 and hence in T n . So we can integrate to get Z Z Z n dz = dn pn sup |ψ(z)| = 4−n d0 p0 sup |ψ(z)|. ψ(z)(z − z0 ) dz ≤ d sup |ψ(z)| f (z) dz = Tn

z∈T n

Tn

Tn

z∈T n

z∈T n

Therefore, Z

T

f (z) dz ≤ 4n 4−n d0 p0 sup |ψ(z)| = d0 p0 sup |ψ(z)| → 0 as z → 0. 0 z∈T n z∈T n

 Theorem 2.1.8. A holomorphic function in an open disc has a primitive in that disc. Theorem 2.1.9. Cauchy’s Theorem for a Disc: If f is holomorphic in a disc D, then Z f (z) dz = 0, γ

for any closed curve γ in D. Theorem 2.1.10. Cauchy’s Integral Formulas: Suppose f is holomorphic in an open set Ω that contains the closure of a disc D. If C denotes the boundary circle of this disc with the positive orientation, then Z f (ζ) 1 f (z) = dζ for any point z ∈ D. 2πi C ζ − z Moreover, f has infinitely many complex derivatives in Ω (Regularity Theorem), and Z n! f (ζ) (n) f (z) = dζ for any point z ∈ D. 2πi C (ζ − z)n+1 The above integrals vanish when z is outside D.

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Corollary 2.1.11. Cauchy Inequalities: If f is holomorphic in an open set that contains the closure of a disc D centered at z0 and of radius R, then (n) f (z0 ) ≤ n!kf kC , Rn where kf kC = supz∈C |f (z)| denotes the supremum of |f | on the boundary circle C. Theorem 2.1.12. f Holomorphic =⇒ f Analytic: Suppose f is holomorphic in an open set Ω. If D is a disc centered at z0 and whose closure is contained in Ω, then f has a power series expansion at z0 f (z) =

∞ X f (n) (z0 ) n=0

n!

(z − z0 )n .

Proof: Fix z ∈ D. By Cauchy’s integral formula: Z 1 f (ζ) f (z) = dζ. 2πi C ζ − z

(∗)

Write 1/(ζ − z) as 1 1 1 1 = = . z ζ −z ζ − z0 − (z − z0 ) ζ − z0 1 − − z0 ζ − z0 Since z ∈ D and ζ ∈ C, the boundary of D, we have z − z0 ζ − z0 < 1. Thus we can write

n ∞  X 1 z − z0 z − z0 = ζ − z0 1− n=0 ζ − z0 and this series converges uniformly to its limit, which enables us to interchange the infinite sum with the integral when we substitute it back in (∗). Therefore, after substituting we get Z ∞ ∞ ∞ (n) f (ζ) 1 X 1 X (z0 ) X f (n) (z0 ) n n 2πif (z − z0 ) dζ = (z − z ) = (z − z0 )n . f (z) = 0 n+1 2πi n=0 (ζ − z ) 2πi n! n! 0 C n=0 n=0

 Note 2.1.1. If f is entire, then the theorem implies that f has a power series expansion around 0, say P n (n) f (z) = ∞ (0)/n!, that converges in all of C. n=0 an z with an = f Theorem 2.1.13 (Liouvile’s Theorem). If f is entire and bounded, then f is constant. Proof: We need to prove that f 0 = 0 for all z. Let R > 0 be the radius of a disk D centered at z. Since f is bounded, there is M > 0 such that |f (z)| ≤ M for all z. Since f is entire, f is holomorphic on D, and hence by Cauchy’s inequality Corollary 2.1.11 we have kf kC M ≤ . R R Letting R → ∞, because f is entire), implies |f 0 (z)| = 0 for all z and hence f 0 (z) = 0, so f is constant.  |f 0 (z)| ≤

MATH 545 and MATH 546 Prelim Problems

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Theorem 2.1.14. Suppose f is a holomorphic function in a region Ω, and {wk }∞ k=1 a sequence of distinct points such that wk → z0 ∈ Ω. If f (wk ) = 0 for all k, then f ≡ 0. In other words: If the zeros of a holomorphic function f in the connected open set Ω accumulate in Ω, then f is identically 0. Therefore, if we suppose f and g are holomorphic in a region Ω and f (z) = g(z) for all z in some non-empty open subset of Ω (or more generally for z in some sequence of distinct points with limit point in Ω). Then f (z) = g(z) throughout Ω. Note 2.1.2. By this theorem, if f is analytic in Ω, then we can possibly extend f to outside Ω to get a new holomorphic function F on a new domain Ω0 with F = f on Ω. Such F is called an analytic continuation of f into the region Ω0 . Theorem 2.1.15 (Morera’s Theorem). Suppose f is a continuous function in the open disc D such that for any triangle T ⊂ D, we have Z f (z) dz = 0. T

Then f is holomorphic. Proof: If we prove that f has a primitive F , then since F is holomorphic by Definition 2.1.3, it has infinitely many holomorphic derivatives by regularity theorem Theorem 2.1.10 and hence F 00 = f 0 exists, so f is holomorphic. After a translation, we may assume W.L.O.G that the disc D is centered at the origin. Given a point z ∈ D, consider the piecewise-smooth curve that joins 0 to z first by moving in the horizontal direction from 0 to z˜ where z˜ =Re(z), and then in the vertical direction from z˜ to z. We choose the orientation from 0 to z, and denote this polygonal line, which consists of at most two segments, by γz . Define Z f (w) dw.

F (z) = γz

Then 1 F (z + h) − F (z) = h h

"Z

Z

#

1 f (w) dw − f (w) dw = h γz γz+h 1 = h

"Z

#

Z f (w) dw +

γz+h

f (w) dw −γz

Z f (w) dw, η

where η is the straight line segment between z and z + h. Since f is continuous at z we can write f (w) = f (z) + ψ(w), where ψ(w) → 0 as w → z. (Since f is continuous.) Hence Z Z F (z + h) − F (z) f (z) 1 = dw + ψ(w) dw h h η h η supw∈η |ψ(w)| f (z) ≤ [(z + h) − (z)] + [(z + h) − (z)] h h → f (z) as h → 0.

MATH 545 and MATH 546 Prelim Problems

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 Theorem 2.1.16 (Sequences of holomorphic functions). If {fn } is a sequence of holomorphic functions that converges uniformly to a function f in every compact subset of Ω, then f is holomorphic in Ω. Proof: First recall that a uniform limit of a sequence of continuous functions is continuous. Moreover, if these functions {fn } are integrable on a compact set D ⊂ Ω, then f is integrable and Z Z lim fn (z) dz = f (z) dz. n→∞

D

D

Let D be any disk whose closure is contained in Ω. Let T be a triangle contained in Ω. Then since {fn } are holomorphic on Ω and hence on T , by Goursat’s Theorem Theorem 2.1.7, Z fn (z) dz = 0 T

for all n. Since {fn } converges uniformly to f , f is continuous and because D is a closed set contained in Ω, then it is compact, and thus by the above note f is integrable on D and hence any subset of it and Z Z 0 = lim fn (z) dz = f (z) dz. n→∞

T

T

Thus, by Morera’s Theorem Theorem 2.1.15, f is holomorphic on D. Since that true for all D whose closure contained in Ω, f is holomorphic on Ω.  Theorem 2.1.17 (fn0 converges uniformly to f 0 ). If {fn } is a sequence of holomorphic functions that converges uniformly to a function f in every compact subset of Ω, then the sequence of derivatives {fn0 } converges uniformly to f 0 on every compact subset of Ω. Proof: W.L.O.G. suppose {fn } converges uniformly on all of Ω. For each δ > 0 define Ωδ = {z ∈ Ω : D(z, δ) ⊂ Ω}. Then Ωδ ⊂ Ω and hence by Cauchy’s integral formula Theorem 2.1.10 for any z ∈ Dδ (z) ⊂ Ω Z 1 fn (ζ) − f (ζ) 0 0 0 dζ |(fn − f ) (z)| = |fn (z) − f (z)| = 2π Cδ (ζ − z)2 Z supζ∈Ω |fn (ζ) − f (ζ)| 2π δieiθ ≤ dθ 2π δ 2 e2iθ 0

1 = sup |fn (ζ) − f (ζ)| → 0 (since fn converges uniformly to f ). δ ζ∈Ω  Theorem 2.1.18. Let F (z, s) be defined for (z, s) ∈ Ω × [0, 1] where Ω open set in C. Suppose F satisfies the following: (i) F is holomorphic in z for each s.

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(ii) F is continuous on Ω × [0, 1]. (F is jointly continuous in both arguments.) Then the function f defined on Ω by Z f (z) =

1

F (z, s) ds 0

is holomorphic. Proof: It is enough to prove the result on a disc D whose closure contained in Ω. We can write the previous integral as a Riemann sum. Consider the sequence of functions n 1X fn (z) = F (z, k/n). n k=1 Then fn (z) is holomorphic in Ω by property (i) for each n. We claim that fn converges to f uniformly. To prove the claim, notice first by property (ii) that F is continuous on a compact set D implies it is a uniform continuous function. Hence for all  > 0, there is δ > 0 such that |F (z, s1 ) − F (z, s2 )| < 

whenever |s1 − s2 | < δ.

To prove that fn converges uniformly to f , choose N ∈ N such that N > 1/δ. Then for all n ≥ N , we have Z 1 n 1 X F (z, k/n) − F (z, s) ds |fn − f | = n 0 k=1 n Z k/n n 1 X X F (z, s) ds = F (z, k/n) − n k=1 (k−1)/n n k=1 X Z k/n = [F (z, k/n) − F (z, s)] ds k=1 (k−1)/n n Z k/n X |F (z, k/n) − F (z, s)| ds ≤ k=1 (k−1)/n n Z k/n X

< =

k=1 n X k=1

(k−1)/n

k (k − 1) k−1 k = 1/n < δ) ds (since s ∈ [ , ], so − n n n n

1 = . n

Hence by Theorem 2.1.16, f is holomorphic on D. Since that true for all D in Ω, then f is holomorphic on Ω.  Definition 2.1.4. Let Ω+ denote the part of Ω that lies in the upper half-plane and Ω− that part that lies in the lower half-plane. Let I = Ω ∩ R. Theorem 2.1.19 (Symmetry Principle). If f + and f − are holomorphic functions in Ω+ and Ω− respectively, that extend continuously to I and f + (x) = f − (x) for all x ∈ I, then the function f defined on Ω by  +  if z ∈ Ω+ ,  f (z) f (z) = f + (z) = f − (z) if z ∈ I,   f − (z) if z ∈ Ω−

MATH 545 and MATH 546 Prelim Problems

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is holomorphic on all of Ω. Theorem 2.1.20 (Schwarz Reflection Principle). Suppose that f is a holomorphic function in Ω+ that extends continuously to I and such that f is real-valued on I. Then there exists a function F holomorphic in all of Ω such that F = f on Ω+ . Note 2.1.3. F is defined as follows: For z ∈ Ω− , F (z) = f (z). Theorem 2.1.21 (Runge Theorem). Any function that is holomorphic in a neighborhood of compact set K can be approximated uniformly on K by rational functions whose singularities are in K c . Moreover, if K c is connected, then f can be approximated by polynomials on K. Definition 2.1.5 (Isolated Singularities). A point singularity or isolated singularity of a function f is a complex number z0 such that f is defined in a neighborhood of z0 but not at the point z0 itself (e.g. f (z) = z on C \ {0}, then z = 0 is an isolated singularity.) Isolated singularities are three types: i.) Removable singularity Let f be a function holomorphic in an open set Ω except possibly at one point z0 in Ω. If we can define f at z0 in such a way that f becomes holomorphic in all of Ω, we say that z0 is a removable singularity for f . E.g.: f (z) = z on C \ {0}, then z = 0 is a removable singularity, because we can define f (z) to be 0 at z = 0 and we notice that f (z) = z is holomorphic on all C. ii.) Pole singularity: We say that a function f defined in a deleted neighborhood of z0 has a pole at z0 , if the function 1/f , defined to be zero at z0 , is holomorphic in a full neighborhood of z0 . E.g. : f (z) = 1/z defined on C \ {0} has a pole at z = 0, because g(z) = 1/f (z) = z has a zero at z = 0 and is defined on all C. iii.) Essential singularity: An essential singularity is not a removable singularity or a pole. E.g.: f (z) = e1/z defined on C \ {0} behaves crazy near z = 0. If we reach z = 0 on the real axis from right, f goes to ∞, while if we reach z = 0 on the real axis from left f goes to 0. Also, it oscillates rapidly, yet remains bounded, as z approaches the origin on the imaginary axis. Theorem 2.1.22. If f has a pole of order n at z0 , then f (z) =

a−n a−n+1 a−1 + + ... + + G(z), n n−1 (z − z0 ) (z − z0 ) z − z0

where G is a holomorphic function in a neighborhood of z0 . The sum P (z) =

a−n+1 a−1 a−n + + ... + (z − z0 )n (z − z0 )n−1 z − z0

is called the principal part of f at the pole z0 , and the coefficient a−1 the residue of f at the pole z0 . We write resz0 f = a−1 . And if C is a circle centered at z0 , then we have Z P (z) dz = 2πi a−1 = 2πi resz0 f. C

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Theorem 2.1.23. If f has a pole of order n at z0 , then resz0 f = lim

z→z0

1 dn−1 (z − z0 )n f (z). n−1 (n − 1)! dz

So if f has a simple pole at z0 , then resz0 f = lim (z − z0 )f (z). z→z0

Theorem 2.1.24. Suppose that f is holomorphic in an open set containing a circle C and its interior, except for a pole at z0 inside C. Then Z f (z) dz = 2πi resz0 f. C

Corollary 2.1.25 (Residue Formula). Suppose that f is holomorphic in an open set containing a circle C or a toy contour γ and its interior, except for poles at the points z1 , ..., zN inside C or γ . Then for the circle C Z N X reszk f. f (z) dz = 2πi C

k=1

And for the toy contour Z f (z) dz = 2πi γ

N X

reszk f.

k=1

Theorem 2.1.26 (Riemann’s Theorem on Removable Singularities). Suppose that f is holomorphic in an open set Ω \ {z0 }. If f is bounded on Ω \ {z0 }, then z0 is a removable singularity. Proof: Let D be a disc centered at z0 whose closure is contained in Ω. Let C denote its boundary circle. We need to prove that for z 6= z0 1 f (z) = 2πi

Z C

f (ζ) dζ. ζ −z

(∗)

Why? Because we can consider the parametrization of C defined by z(t) : [0, 1] → C and then rewrite the above integral as Z Z 1 f (ζ) f (ζ(t))ζ 0 (t) dζ = dt. ζ(t) − z C ζ −z 0 And this integral is holomorphic in z for each t and continuous in both z and t, then by Theorem 2.1.18, this integral is holomorphic for all D. Thus for z 6= z0 this integral agrees with f (z), and it is holomorphic on all of D. Then by the definition of removable singularity, this is the continuation we looking for and hence z0 is a removable singularity. To prove (∗), fix z ∈ D, z 6= z0 and consider the toy contour Γ shown in the figure

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Figure 2.1: The multiple keyhole contour Γ in the proof of Riemann’s theorem

Let γ and γ0 be small circles of radius  with negative orientation and centered at z and z0 respectively. Then since f (ζ)/(ζ − z) is holomorphic on Γ and in its interior, then Z Z Z Z f (ζ) f (ζ) f (ζ) f (ζ) dζ + dζ + dζ = dζ = 0. C ζ −z γ ζ − z γ0 ζ − z Γ ζ −z But f (z) is holomorphic on γ and its interior. Hence by Cauchy’s integral formula, Z f (ζ) dζ = −2πif (z), γ ζ − z Z

f (ζ) dζ, as γ0 ζ − z  → 0, there exists δ > 0 such that |ζ − z| > δ, where ζ ∈ γ0 (because z 6= z0 and as  → 0, ζ → z0 ). And from assumption, |f (z)| ≤ M for all z ∈ D \ {z0 }. Hence Z Z f (ζ) |f (ζ)| 2πM dζ 0 ζ − z ≤ 0 |ζ − z| dζ ≤ δ → 0 as  → 0. γ γ and the negative sign comes from negative orientation of γ . For the second integral

This proves that 1 f (z) = 2πi

Z C

f (ζ) dζ. ζ −z

 Corollary 2.1.27. Suppose that f has an isolated singularity at the point z0 . Then z0 is a pole of f if and only if |f (z)| → ∞ as z → z0 . Theorem 2.1.28 (Casorati-Weierstrass). Suppose f is holomorphic in the punctured disc Dr (z0 )\{z0 } and z0 is an essential singularity of f . Then the image of Dr (z0 ) \ {z0 } under f is dense in C. Proof: Suppose not for the purpose of contradiction. Then there exists w ∈ C and δ > 0 such that |f (z) − w| > δ for all z ∈ Dr (z0 ) \ {z0 }. Consider the function g(z) =

1 . f (z) − w

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Then g(z) is holomorphic in Dr (z0 ) \ {z0 } since f (z) − w is, and we have |g(z)| < 1/δ, i.e. bounded. Then by Riemann theorem of removable singularity, g(z) has a removable singularity at z0 , i.e. g can be redefined on all Dr (z0 ) in such a way that makes it holomorphic on all Dr (z0 ). So we let g(z0 ) = c. If c 6= 0, then f (z0 ) − w is finite and f (z) − w is holomorphic at z0 , because g, and hence 1/g, is holomorphic at z0 . But this contradicts the fact that z0 is an essential singularity of f (which means f cannot be defined at z = z0 . If c = 0, then by the definition of the pole, z = z0 is a pole for f (z) − w, i.e. a pole for f (z). But this is also contradicts the fact that z0 is an essential singularity.  Definition 2.1.6. Suppose that f is holomorphic for all large values of z. Let F (z) = f (1/z), then F is holomorphic in a deleted neighborhood of the origin. We say that f has a pole at infinity if F has a pole at z = 0, i.e. if 1/f (1/z) has a zero at z = 0. And we say f has a removable singularity at infinity if F has a removable singularity at z = 0, i.e. f (1/z) is holomorphic at z = 0. And we say that f has an essential singularity at infinity if F has an essential singularity at z = 0. Definition 2.1.7 (Meromorphic). A function f on an open set Ω is meromorphic if there exists a sequence of points {z0 , z1 , z2 , ...} that has no limit points in Ω such that: i) f is holomorphic in Ω \ {z0 , z1 , z2 , ...}, and ii) f has poles at points {z0 , z1 , z2 , ...}. A meromorphic function in the complex plane that is either holomorphic at infinity or has a pole at infinity is said to be meromorphic in the extended complex plane. Note 2.1.4 (Some Topology). . 1. A topological space X is called compact if each of its open covers has a finite subcover. 2. The complex plane C is not compact. For instance, here is an open cover that does not admit any finite subcover: U = {Un }

with

Un = {z ∈ C : |z| < n} .

The reason why U does not admit any finite subcover is easy: if you pick just a finite number of Un ’s, then their union would be equal to the biggest one of them, which is not the whole C. To understand it more deeply: A sequence like {1, 2, 3, ...} which tends to infinity does not have a convergent subsequence in C, whereas any sequence in C has a convergent subsequence (because convergence here accepts infinity as a limit). The addition of infinity of called the one- point compactification (Riemann sphere) 3. The extended complex plane C ∪ {∞} = C is compact. 4. Every closed discrete subset of a compact space is finite. Hence every closed discrete subset of C is finite.

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Theorem 2.1.29. If f is meromorphic in the extended complex plane, then f has finitely many poles. Proof: By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete. Also, The set of poles is closed, since its complement, the set of points at which f is holomorphic, is open, since any point where f is holomorphic has a neighborhood where f is holomorphic on it. Thus by the previous note, the set of poles of f is finite. Also notice from the definition of meromorphic functions, if we suppose we have a sequence of poles {z0 , z1 , z2 , ...} (sequence means: infinite number of points), then by definition of meromorphic functions, this sequence has no limit points in the domain of f . But since f is defined on C, then the sequence will accumulate in C (no other place to accumulate in), hence f is not meromorphic (against the definition), which is a contradiction.  Theorem 2.1.30. The meromorphic functions in the extended complex plane are the rational functions. Proof: From the previous theorem, f has finite number of poles: {z0 , z1 , ..., zn } and by Definition 2.1.7 f could P be holomorphic at infinity or has a pole there. The idea is to write g(z) = f (z) − P∞ (z) − nk=0 Pk (z), where P∞ = f (∞) if f is holomorphic at ∞ or it is the principle part of f if f has a pole at z = ∞, and Pk are the principle parts of f at the pole zk for each k = 0, 1, .., n. Then showing that g is constant and P hence f = P∞ (z) + nk=0 Pk (z)+constant, which is rational. If f is holomorphic at ∞ or it has a pole at ∞, then in both cases f (z) → P∞ (z) as z → ∞. Also, notice that Pk (z) → 0 as z → ∞ for all such k. By the definition of g, g is entire. And since limz→∞ g(z) = P∞ (z) − P∞ (z) − 0 = 0. Hence, there exists an M > 0 such that |g(z)| ≤ M for all z. Hence by Liouville’s theorem Theorem 2.1.13 g is constant. Another proof: As in the previous proof there are a finite number of poles for f , namely {z0 , z1 , ..., zN }. Let n0 , n1 , ..., nN be the orders of these poles, respectively. Then (1)

(1)

(2)

(2)

(N )

(N )

a−nN a−1 a−1 a−1 a−n0 a−n1 + ... + + ... + + ... + f (z) = + + ... + n n n 0 1 N (z − z0 ) (z − z0 ) (z − z0 ) (z − z0 ) (z − z0 ) (z − z0 ) + G1 (z) + G2 (z) + ... + GN (z), G1 , G2 , ..., GN are holomorphic functions. Define g(z) = (z − z0 )n0 (z − z1 )n1 ....(z − zN )nN f (z). Then g(z) is entire and hence analytic throughout C: g(z) =

∞ X n=0

an z n , z ∈ C.

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And ψ(z) = (z − z0 )n0 (z − z1 )n1 ....(z − zN )nN is a polynomial. But since n0  n1  nN      1 1 1 1 1 g = − z0 − z1 ... − zN f z z z z z     1 1 = , (By binomial) + ... f z n0 +n1 +...+nN z g(z) has a pole at ∞ of order : ( n0 + n1 + ... + nN M= n0 + n1 + ... + nN + K Since g has a pole of order M at ∞,

if f is holomorphic at ∞ if f has a pole of order K at ∞

  X ∞ 1 = an z −n g z n=0

has a pole of order M at z = 0. But this implies that an = 0 for n > M (otherwise g(1/z) will have a pole of order > M at 0). Hence g(z) is a polynomial. Since g and ψ are polynomials and g = ψf , then f = g/ψ is rational.  Theorem 2.1.31 (Argument principle). Suppose f is meromorphic in an open set containing a circle C and its interior. If f has no poles and never vanishes on C, then Z 0 1 f (z) dz = (number of zeros of f inside C) − (number of poles of f inside C) 2πi C f (z) where the zeros and poles are counted with their multiplicities. Proof: Suppose f has N zeros {z0 , z1 , ..., zN } inside C with order {n0 , n1 , ..., nN }, respectively. And suppose 0 } inside C with orders {n00 , n01 , ..., n0M }, respectively. Then f can be written it has M poles {z00 , z10 , ..., zM as 0

0

0

0 −nM f (z) = (z − z0 )n0 (z − z1 )n1 ...(z − zN )nN (z − z00 )−n0 (z − z10 )−n1 ...(z − zM ) h(z),

where h(z) is non-vanishing holomorphic function. Since (f1 f2 )0 f10 f20 = + , f1 f2 f1 f2 we have f 0 (z) n0 n1 nN = + + ... + − f (z) z − z0 z − z1 z − zN



n00 n01 n0M + + ... + 0 z − z00 z − z10 z − zM

 +

h0 (z) . h(z)

0 Thus f 0 /f has n0 + n1 + ... + nN + n00 + n01 + ... + n0M simple poles at {z0 , z1 , ..., zN , z00 , z10 , ..., zM }. Therefore by residue formula, Z 0 N M X X 1 f (z) dz = reszk f + reszi0 f = (n0 + n1 + ... + nN ) − (n00 + n01 + ... + n0M ). 2πi C f (z) i=0 k=0



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Theorem 2.1.32 (Rouch´ e’s Theorem). Suppose f and g are holomorphic functions in an open set containing a circle C and its interior. If |f (z)| > |g(z)|

for all z ∈ C,

then f and f + g have the same number of zeros inside C. Definition 2.1.8 (Open Map). A mapping is said to be open if it maps open sets to open sets. Theorem 2.1.33 (Open mapping theorem). If f is holomorphic and non-constant in a region Ω, then f is open. Proof: Let w0 belong to the image of f , say w0 = f (z0 ), where z0 ∈ Ω. We need to prove that if there exists  > 0 such that for all w with |w − w0 | < , then we must have w belongs to the image of f , i.e. there exists z1 ∈ Ω such that f (z1 ) = w. The idea is to show that the function g(z) = f (z) − w has a zero at z1 . Rewrite g as g(z) = (f (z) − w0 ) + (w0 − w) = F (z) + G(z). Choose δ > 0 such that the disc |z − z0 | ≤ δ is contained in Ω and f (z) 6= w0 on |z − z0 | = δ. Select  such that |f (z) − w0 | ≥  on the circle |z − z0 | = δ. So if |w − w0 | < , then on the circle |z − z0 | = δ, F (z) = |f (z) − w0 | ≥  > |w − w0 | = G(z). Thus by Rouch´e’s theorem F and F + G = g have the same number of zeros inside the circle |z − z0 | = δ. Since F (z) has one zero at z0 , then there exists z1 such that |z1 − z0 | < δ and g(z1 ) = f (z1 ) − w = 0, thus w in the image of f . So f is an open mapping.  Theorem 2.1.34 (Maximum Modulus Principle). If f is a non-constant holomorphic function in a region Ω, then f cannot attain a maximum in Ω. Also, if Ω is the compact closure of Ω and f is continuous on Ω, then sup |f (z)| ≤ sup |f (z)|. z∈Ω

z∈∂Ω

and the equality holds only when f is constant, otherwise sup |f (z)| < sup |f (z)|. z∈Ω

z∈∂Ω

(Its converse): Suppose f is holomorphic in an open set Ω. If f attains a maximum in Ω, then f is constant. Definition 2.1.9 (homotopy). Let γ0 and γ1 be two curves in an open set Ω with common end-points. So if γ0 (t) and γ1 (t) are two parametrizations defined on [a, b], we have γ0 (a) = γ1 (a) = α

and

γ0 (b) = γ1 (b) = β.

These two curves are said to be homotopic in Ω if for each 0 ≤ s ≤ 1 there exists a curve γs ⊂ Ω, parametrized by γs (t) defined on [a, b], such that for every s γs (a) = α

and

γs (b) = β,

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and for all t ∈ [a, b] γs (t) s=0 = γ0 (t)

γs (t) s=1 = γ1 (t),

and

and γs (t) should be jointly continuous in s ∈ [0, 1] and t ∈ [a, b]. Definition 2.1.10 (Simply Connected Domain). A region Ω in the complex plane is simply connected if any two pair of curves in Ω with the same end-points are homotopic. Theorem 2.1.35. If f is holomorphic in Ω, then Z Z f (z) dz = f (z) dz γ0

γ1

for any two curves γ0 and γ1 homotopic in Ω. Theorem 2.1.36. Any holomorphic function in a simply connected domain has a primitive. Corollary 2.1.37 (General Cauchy’s Theorem). If f is holomorphic in the simply connected region Ω, then Z f (z) dz = 0 γ

for any closed curve γ in Ω. Theorem 2.1.38. If f is a nowhere vanishing holomorphic function in a simply connected region Ω, then there exists a holomorphic function g on Ω such that f (z) = eg(z) . In fact g(z) = log f (z) (a branch of the logarithm) Theorem 2.1.39. If u is harmonic function on a simply connected domain Ω, then there exists a holomorphic function f on Ω such that u =Re(f ). Moreover, Im(f ) is uniquely defined up to an additive real constant. ∂ 1 ∂  + u(x, y) = ux − iuy . Proof: Let g be the complex function defined by g(z) = 2∂u/∂z = ∂x i ∂y Let uˆ be the real part of g and vˆ be the imaginary part of g. Then, by harmonicity uˆx = uxx = −uyy = vˆy . Also, uˆy = uxy = uyx = −ˆ vx . Hence, since uˆ and vˆ are continuously differentiable and satisfy the CauchyRiemann equations on Ω, g(z) is holomorphic on Ω. Then by Theorem 2.1.36 g(z) has a primitive F in Ω such that F 0 = g. Claim. Re(F ) − u = constant ∈ R. Proof of claim: Let F = Re(F ) + iIm(F ). Since F is a primitive of g, F 0 = (Re(F ))x + i(Im(F ))x = (Re(F ))x − i(Re(F ))y = g = uˆ + iˆ v = ux − iuy . Hence, (Re(F ))x = ux , so u(x, y) = Re(F )(x, y) + c(y) for some functions c which only depends on y. Likewise, by comparing the imaginary parts, (Re(F ))y = uy , so u(x, y) = Re(F )(x, y) + c(x), where c is a function depends only on x. Hence, c has to be constant, and therefore Re(F ) − u = c(constant). End of the proof of claim.

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Now let f = F − (Re(F ) − u) = u + iIm(F ), then Re(f ) = u and f is holomorphic on D, since F and Re(F ) − u = (constant) are holomorphic. Now suppose h is another holomorphic function on Ω such that Re(f ) =Re(h) = u, then Re(f − h) = 0. Therefore, (f − h)0 = 2∂[Re(f − h)]/∂z = 0 and hence f = h + c where c ∈ C, but since Re(f ) =Re(h)+Re(c) = u =Re(h), Re(c) = 0 and hence c is purely imaginary. Thus Im(f ) is uniquely defined up to an additive real constant.  Definition 2.1.11 (Conformal Mapping). A bijective holomorphic function f : U → V is called a conformal map. Given such a mapping f , we say that U and V are conformally equivalent or simply biholomorphic. An important fact is that the inverse of f is then automatically holomorphic. Proposition 2.1.40. If f : U → V is holomorphic and injective (one-to-one), then f 0 (z) 6= 0 for all z ∈ U . In particular, the inverse of f defined on its range is holomorphic, and thus the inverse of a conformal map is also holomorphic. Lemma 2.1.41. Let V and U be open sets in C and F : V → U a holomorphic function. If u : U → C is a harmonic function, then u ◦ F is harmonic on V . Lemma 2.1.42 (Schwarz Lemma). Let f : D → D be holomorphic with f (0) = 0. Then i) |f (z)| ≤ |z| for all z ∈ D ii) If for some z0 6= 0 we have |f (z0 )| = |z0 |, then f is a rotation, i.e. f (z) = eiθ z, θ ∈ R. iii) |f 0 (0)| ≤ 1, and if equality holds, then f is a rotation. Proof: i) Let f (z) = a0 + a1 z + a2 z 2 + ... be the power series expansion of f in D. Since f (0) = 0, we have a0 = 0. Consider the function f (z)/z. Then f (z)/z is holomorphic on D \ {0}, and since f (0) = 0, there exists a natural integer n and a non-vanishing holomorphic function h in a neighborhood U of z = 0 such that f (z) = z n h(z). Hence f (z)/z = z n−1 h(z) is holomorphic on U , since f and z n−1 holomorphic on D. Therefore it is holomorphic at z = 0 and so it is holomorphic throughout D. (In fact z = 0 is a removable singularity) If |z| = r < 1, then since |f (z)| ≤ 1, we have f (z) 1 1 z ≤ |z| = r , and by the maximum modulus principle, we can conclude that this is true whenever |z| ≤ r. Let r → 1 we conclude (i). ii) f (z)/z attains its maximum at z when |f (z)/z| = 1, since |f (z)| ≤ |z| for all z ∈ D. By assumption |f (z0 )/z0 | = 1, hence f /z attains a maximum at z0 . But since z0 ∈ D, by maximum modulus principle converse ??, f /z is constant, i.e. f (z) = cz. But since |z0 | = |f (z0 )| = |c||z0 |, c = eiθ for some θ ∈ R. Thus f is a rotation.

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iii) Let g(z) = f (z)/z, then f (z) − f (0) f (z) = lim = f 0 (0). z→0 z→0 z z−0

g(0) = lim

But by (i), |g(z)| ≤ 1, hence |g(0)| = |f 0 (0)| ≤ 1. If |f 0 (0)| = 1, then |g(0)| = 1, but since |g(z)| ≤ 1, we notice that g(z) attains a maximum at z = 0, but by maximum modulus principle g is a constant, i.e. f = cz. Argue as in (ii) we have f is a rotation.  Definition 2.1.12 (Automorphisms). A conformal map from an open set Ω to itself is called an automorphism of Ω. The set of all automorphisms of Ω is denoted by Aut(Ω), and carries the structure of a group. Note 2.1.5. An important automorphism on D is ψα (z) =

α−z , 1 − αz

where α, z ∈ D. We use this map mostly as a helping tool to map a point α ∈ D to another point β ∈ D by considering the map f (z) = (ψβ ◦ ψα )(z). (since f (α) = ψβ (ψα (α)) = ψβ (0) = β.) Theorem 2.1.43. Any automorphism f ∈ Aut(D) can be written as f (z) = eiθ

α−z 1 − αz

for some θ ∈ R and α ∈ D. i−z is a conformal map from H into D, and Note 2.1.6. Notice that F : H → D given by F (z) = i+z 1−z . Let ψ = eiθ ψα . Then any automorphism f ∈ Aut(H) is defined hence F −1 : D → H is F −1 (z) = i 1+z as f = F −1 ◦ ψ ◦ F : H → H and given by f (z) =

az + b cz + d

with a, b, c, and d ∈ R and ad − bc = 1.

Theorem 2.1.44 (Riemann Mapping Theorem). Suppose Ω is proper (non-empty and not the whole C) and simply connected. If z0 ∈ Ω, then there exists a unique conformal map F : Ω → D such that F (z0 ) = 0

and

F 0 (z0 ) > 0.

Corollary 2.1.45. Any two proper simply connected open subsets in C are conformally equivalent. Definition 2.1.13 (Normal Family). Let Ω be an open subset of C. A family F of holomorphic functions on Ω is said to be normal if every sequence in F has a subsequence that converges uniformly on every compact subset of Ω (the limit need not be in F). Definition 2.1.14 (Uniform Boundedness on Compact Sets). The family F is said to be uniformly bounded on compact subsets of Ω if for each compact set K ⊂ Ω there exists B > 0, such that |f (z)| ≤ B

for all z ∈ K and f ∈ F.

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Definition 2.1.15 (Equicontinuity). The family F is equicontinuous on a compact set K if for every  > 0 there exists δ > 0 such that whenever z, w ∈ K and |z − w| < δ, then |f (z) − f (w)| < 

for all f ∈ F.

Theorem 2.1.46 (Montel’s Theorem). Suppose F is a family of holomorphic functions on Ω that is uniformly bounded on compact subsets of Ω. Then: i) F is equicontinuous on every compact subset of Ω. ii) F is a normal family.

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Figure 2.2: Some conformal maps

71

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Definition 2.1.16 (M¨ obius Transformation or Fractional Linear Transformation). A M¨obius transformation is a rational function of the form f (z) =

az + b cz + d

of one complex variable z; here the coefficients a, b, c, d are complex numbers satisfying ad − bc 6= 0. The main using of a fractional linear transformation is when we have a region that is bounded with two curves intersected at one or two points z0 , z1 to send one to ∞, and the other to 0, by f (z) =

z − z0 z − z1

f sends z0 to 0 and z1 to infinity.

2.2

Complex Analysis Problems

1. Suppose   that f is holomorphic on the unit disk |z| < 1. If there is an r ∈ (0, 1) so that f 1 ≤ rn for n ∈ N, prove that f is constant on |z| < 1. n SOLUTION: For contradiction suppose f is not constant. Let n → ∞, then |f (0)| ≤ 0 and hence f (0) = 0. Since f (z) is holomorphic and non-constant, f (z) = z m g(z), where m is the order of the zero, z = 0, and g(z) is a non-vanishing   holomorphic function in a neighborhood of 0. Re plug f (z) in the inequality 1 above to get g ≤ nm rn . Take the limit as n → ∞ again to get g(0) = 0, but this is a n contradiction, g(z) can’t be zero at 0. f is constant and in fact f (z) ≡ 0.  2. Let a ∈ C and let f, g be holomorphic in a neighborhood of a. Suppose that g has a simple zero at a and that f (a) 6= 0. Show that   f (a) f resa = 0 . g g (a) SOLUTION:   f f (z) f (z) f (a) resa = lim (z − a) = lim = 0 z→a g g(z) z→a g(z) − g(a) g (a) z−a since both f and g holomorphic at a. 

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3. Suppose that f is entire and if |f (z) − 1| ≥ 0.01 for all complex z . Prove that f must be constant. SOLUTION: Let g(z) =

1 , then g(z) is entire since f (z) − 1 never zero. Also we have f (z) − 1 |g(z)| =

1 ≤ 100. |f (z) − 1|

By Liouville’s Theorem, g(z) is constant c. Since f is holomorphic and hence doesn’t take infinity, c can’t be zero and therefore f (z) = c−1 + 1.  4. Suppose that f : C → C is holomorphic, m and n integers, and (5 + |z|m )−1 f (n) (z) is bounded on C. Prove that f is a polynomial and find it is possible degree. SOLUTION: (5 + |z|m )−1 f (n) is bounded on C means |f (n) (z)| ≤ M (5 + |z|m ) for all z ∈ C and and some positive integer M. We know that f is entire. Let g(z) = f (n) (z) then and |g(z)| ≤ M (5 + |z|m ). Consider the disc D centered at 0 and has radius R, then by Cauchy’s inequalities |g(0)(k) | ≤

k!M (5 + Rm ) k!kgk∂D ≤ . Rk Rk

Let R → ∞, then the last term goes to 0 when k > m, i.e. g(0)(k) = f (0)(k+n) = 0 when k > m. Being entire, f has a power series expansion around 0, i.e. f (z) =

∞ X

al z l

l=0

But since f (0)(k+n) = 0 for k > m, then f is a polynomial of degree at most m + n.  5. Let f be a holomorphic function in the unit disk D and |f (z)| ≤ M for all z ∈ D. Prove that |f 0 (z)| ≤ M (1 − |z|)−1 for all z ∈ D. SOLUTION: Fix z0 ∈ D and consider the disc D centered at z0 and has radius R = δ − |z0 | where |z0 | < δ < 1 (to make sure D inside D.) By Cauchy’s inequalities, |f 0 (z0 )| ≤

kf k∂D M ≤ . R δ − |z0 |

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Since the last inequality true for all δ such that |z0 | < δ < 1, let δ → 1 to get the desired result.  6. Show that if f is a non-constant entire function, then f (C) is dense in C. SOLUTION: For contradiction, suppose that f (C) is not dense in C. Then there exist some  > 0 and a complex 1 , then g is entire and number w ∈ C such that |f (z) − w| > . Let g(z) = f (z) − w |g(z)| =

1 1 < . |f (z) − w| 

By Liouville’s Theorem g is constant. Since f is entire, and hence doesn’t take infinity, and w 6= ∞, g is not zero and therefore f is constant, which is a contradiction.  7. Determine the number of zeros (counted with multiplicity) of f (z) = ez open unit disk D.

100

− 5z 2 in the

SOLUTION: 100

Let g(z) = e z and h(z) = −5z 2 , then on the circle ∂D, we have |h(z)| = | − 5z 2 | = 5 and 100 100 100 |g(z)| = ez = eRe(z ) ≤ e|z| = e. We note that |h(z)| > |g(z)| on the unit circle and hence by Rouche’s Theorem h and h + g have the same number of zeros in D, but h has only z = 0 as a zero of order 2. So f has 2 zeros in D.  8. Let g be a non-constant holomorphic function in the unit disk D and suppose that g(0) = 0 and |g(z)| ≤ 1. Prove that for any integer n ∈ N, g(z) − 2n z n has exactly n zeros (counted with multiplicity) in D0.5 (0). SOLUTION: Since g is a non-constant holomorphic function on D, by Open Mapping Theorem, g(D) ⊂ D, i.e. |g(z)| < 1 and hence we conclude that g is a holomorphic map from D to D with g(0) = 0. By Schwarz Lemma, |g(z)| ≤ |z| for all z ∈ D. Let f (z) = −2n z n and consider the circle |z| = 1/2. Then |f (z)| = | − 2n z n | = 2n 21n = 1 and |g(z)| ≤ |z| = 1/2. By Rouche’s Theorem f and f + g have the same number of zeros in the disc |z| < 1/2, i.e. −2n z n and g(z) − 2n z n have the same number of zeros in |z| < 1/2. But f has exactly n zeros counted with multiplicity in D0.5 (0), and hence g(z) − 2n z n has exactly n zeros counted with multiplicity in D0.5 (0).  9. Let f (z) be a holomorphic function on C which takes values in the upper half plane H. Show that f (z) is constant.

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SOLUTION: i−z . Then h = g ◦ f : C → D is entire, i+z because f is, and bounded (|h(z)| < 1). By Liouville’s Theorem h is constant and thus f = g −1 ◦ h is constant. Consider the conformal map g : H → D defined by g(z) =

 10. Let f, g be holomorphic and 1-1 on D with f (0) = g(0) and f (D) ⊂ g(D). Show that |f 0 (0)| ≤ |g 0 (0)|. SOLUTION: Notice that f 0 , g 0 6= 0 and f −1 , g −1 exist and they are defined on f and g ranges, respectively, by Proposition 2.1.40. Now consider the function F (z) = (g −1 ◦ f )(z). Note that F is from D to D, it is defined because f (D) in range of g. We have F (0) = g −1 (f (0)) = g −1 (g(0)) = 0. Hence, we can apply Schwarz Lemma on F to get −1  −1 0 0 0 0 |F (0)| = (g ◦ f ) (0) = g (f (0)) f (0) ≤ 1. But  −1 0 (f (0)) = g

1 1 1 = = . g 0 [g −1 (f (0))] g 0 [g −1 (0)] g 0 (0)

Therefore,   −1 0 0 (f (0)) f (0) = g

0 f (0) ≤ 1. 0 g (0) 

11. Find an explicit conformal mapping from the unit disc D onto the plane with the real intervals (−∞, −1] and [1, ∞) removed. SOLUTION: 1−z which maps conformally the unit disc D onto the upper half 1+z plane H. Then let f2 (z) = −iz, then f2 takes H onto the right half-plane (the set of points z ∈ C such that Rez > 0). Next consider f3 (z) = log z which maps the right half-plane onto the horizontal strip {z ∈ C : −π/2 < Imz < π/2}. Next consider the map f4 (z) = iz which maps the horizontal strip onto the vertical strip {z ∈ C : −π/2 < Rez < π/2}. Finally, consider the function f5 (z) = sin z which maps that vertical strip onto the plane with the real intervals (−∞, −1] and [1, ∞) removed. So, our desired map is   1−z F (z) = (f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 )(z) = sin i log . 1+z Consider the function f1 (z) = i



MATH 545 and MATH 546 Prelim Problems

12. Find a conformal map of U := {z ∈ C : |z − i| < plane.

76

√

2 and |z + i| <

√

2} onto the right half

SOLUTION:

i

-1

0

1

2

−i

z+1 Intersection points are z = ±1. Let f1 (z) = . Then f1 (−1+ ) = 0− , f1 (0) = −1, f1 (1− ) = −∞, z−1 √ √ 1 1 i i f1 (i( 2 − 1)− ) = − √ − √ and f1 (−i( 2 − 1)+ ) = − √ + √ . So f1 maps U onto the sector 2 2 2 2 3π 5π {z = reiθ : r > 0 and <θ< } 4 4

i

π/2 0

1

Let f2 (z) = −z. This will map the sector to another sector {z = reiθ : r > 0 and −

π π <θ< } 4 4

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i

π/2 0

1

Let f3 (z) = z 2 . This will map the previous sector to the right half plane. So the conformal map is F = f3 ◦ f2 ◦ f1 .  13. Prove that there is no function f that is analytic on the punctured disk D \ {0} so that f 0 has a simple pole at 0. SOLUTION: Suppose, for contradiction, that there is such a function, call it f . Then there exist Dδ (0) \ {0}, a deleted neighborhood of 0 in which f 0 (z) = a−1 /z + a0 + a1 z + a2 z 2 ... where a−1 6= 0 and g(z) = a0 + a1 z + a2 z 2 + ... is a holomorphic function in Dδ (0). Since Dδ (0) is simply connected, g has a primitive G on it. Hence, f (z) = a−1 log z + G(z) + c, where c is a constant. Since f is holomorphic on D \ {0}, and hence it is holomorphic on Dδ (0) \ {0}, log z must be holomorphic on Dδ (0) \ {0}. But this is impossible, there is no branch of log z that is holomorphic in Dδ (0) \ {0}, which is a contradiction.  14. Suppose f is a non-constant holomorphic function on C such that for all z ∈ C it satisfies the equation f (1 − z) + f (z) = 1. Prove that f omits no points in C. SOLUTION: We can write the above equation as f (1 − z) = 1 − f (z). For contradiction, suppose f omits at least one point w ∈ C, i.e. w ∈ / f (C). From the above equation we notice also 1 − w ∈ / f (C). If w = 1 − w, then w = 1/2 which is not in f (C) as was supposed. But for z = 1/2, we have f (1 − 21 ) + f ( 12 ) = 2f ( 12 ) = 1, and thus f (1/2) = 1/2, therefore 1/2 ∈ f (C). So, w 6= 1 − w, but that means f omits two points, namely, w and 1 − w. Then, by Little Picard’s Theorem f is constant which is the contradiction we wanted. Hence, f omits no points in C. 

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15. Write the explicit formula for the conformal map from U = {z : |z| < 1 and |z − 1| < 1} to the unit disc D.

SOLUTION:

i

0

1

2

√ z− The intersection points are z = 1/2 ± i 3/2. Let f1 (z) = z− √ √ 1 i 3 1−i 3 √ =− − f1 (1) = 2 2 1+i 3

√ 1+i 3 2 √ 1−i 3 2

√ 2z − 1 − i 3 √ . Then = 2z − 1 + i 3

√ √ −(1 + i 3) 1 i 3 √ =− + and f1 (0) = 2 2 −1 + i 3

and f1 (1/2) = −1. Hence f1 maps U to the sector {z = reiθ : r > 0 and

2π 4π <θ< } 3 3

i

2π/3 0

1

Let f2 (z) = z 3/4 . This will map the previous region into the second quadrant π {z = reiθ : r > 0 and < θ < π} 2

MATH 545 and MATH 546 Prelim Problems

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i

-1

0

1

Let f3 (z) = −iz. This will map the second quadrant into the first quadrant π {z = reiθ : r > 0 and 0 < θ < }. 2

i

0

1

Let f4 (z) = z 2 . This will map the first quadrant into the upper half plane H. Finally, let f5 (z) = (i − z)/(i + z) which maps H into D. Let the required map be F = f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 .  16. Prove the converse of Montel’s Theorem, i.e. prove that if F is a normal family of holomorphic functions on Ω, then F is uniformly bounded on compact subsets of Ω. SOLUTION: For contrary suppose that F is normal but not uniformly bounded. Then for a compact subset K in Ω and for each n ∈ N there is a holomorphic function fn ∈ F and a point zn ∈ K such that |fn (zn )| > n. Since F is normal, the sequence of holomorphic functions {fn }∞ n=1 has a subsequence ∞ {fnk }k=1 that converges uniformly on K to a holomorphic function f . That is for some N ∈ N, we have |fnk (z) − f (z)| < 1,

∀k ≥ N,

and z ∈ K.

Since f is holomorphic (hence continuous) on K, f has a maximum, i.e. there is a positive constant M such that M = maxz∈K |f (z)|. Therefore, for each z ∈ K and k ≥ N , we have |fnk (z)| < 1 + |f (z)| ≤ 1 + M but this is impossible, because |fnl (z)| > nk and by choosing k very large we get a contradiction. 

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17. Find a conformal map of the unit disk onto itself with f (1/4) = −1/3 and f 0 (1/4) > 0. SOLUTION: z − 1/4 z + 1/3 Let h(z) = and g(z) = . Then both h and g are from D to D and h(1/4) = 1 − z/4 1 + z/3 13z − 7 g(−1/3) = 0. Let f (z) = (g −1 ◦ h)(z) = . Then f is from D to D, f (1/4) = g −1 (h(1/4)) = 13 − 7z g −1 (0) = −1/3, and f 0 (1/4) = 128/135 > 0.  18. Let Ω = {z ∈ D : Im(z) > 0}. Evaluate sup{Ref 0 (i/2) | f : Ω → D}. SOLUTION: The idea is to use Schwarz Lemma. That is, we want g : D → Ω so that we have f ◦ g : D → D and (f ◦ g)(0) = 0. Since f : Ω → D takes its values in D, amongst all these holomorphic functions f , √ 1+z supf :Ω→D Ref 0 (i/2) = supf :Ω→D |f 0 (i/2)|. The map g1 (z) = i takes D into H, g2 (z) = z maps 1−z z−1 maps the first quadrant to Ω. Let g : D → Ω be H into the first quadrant, and g3 (z) = z+1 r 1+z −1 i 1−z g(z) = (g3 ◦ g2 ◦ g1 )(z) = r . 1+z i +1 1−z √ √ √ Note that g(0) = ( i − 1)/( i + 1) = i( 2 − 1). Let f1 be the automorphism of the unit disc taking g(0) to i/2 , and f2 be the automorphism of the unit disc taking f (i/2) to 0. These functions can be written out explicitly as f1 (z) = (T2−1 ◦ T1 )(z) where T1 (z) = f1 (z) =

(2 − ig(0))z + (i − 2g(0))

(2 + ig(0)) − (i + 2g(0))z z − f (i/2) f2 (z) = . 1 − f (i/2)z

z − g(0) 1 − g(0)z

and T2 (z) =

z − 2i . Hence, 1 + 2i z

.

Now let S(z) = (f2 ◦ f ◦ f1 ◦ g)(z) : D → D and S(0) = 0. By Schwarz Lemma |S 0 (0)| ≤ 1. But we note that "  #     
0 0 0 0 S (0) = f2 f f1 g(0) f f1 g(0) f1 g(0) g 0 (0)   = f20 f (i/2) f 0 (i/2) f10 [g(0)] g 0 (0). Where

√ 2 i g (0) = √ , ( i + 1)2 0

f10 [g(0)] =

3 , 4(1 − |g(0)|2 )

  and f20 f (i/2) =

1 . 1 − |f (i/2)|2

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|S 0 (0)| ≤ 1 implies √ √ 2(2 + 2) 1 − |g(0)|2 1 − |f (i/2)|2 2| i + 1|2 1 − |g(0)|2 1 − |f (i/2)|2 √ = |f 0 (i/2)| ≤ 3 3| i| The equality holds when f is a rotation. Moreover, |f 0 (i/2)| is maximized when f such a rotation with f (i/2) = 0. Hence,

0

sup |f (i/2)| = f :Ω→D

2(2 +

√ 2) 1 − |g(0)|2 3

√ 4 2 = . 3 

19. Find a conformal map from the set Ω = H \ {z : |z − 2i | ≤ 12 } onto H. SOLUTION:

i

This region has two boundaries: The real axis and the circle |z − 2i | = 12 . The intersection between them is at z = 0. Let f1 (z) = 1/z. Then f1 (0) = ∞, f1 (±∞) = 0, so it maps the real axis onto the real axis, and f1 (1/2 + i/2) = 1 − i, f1 (−1/2 + i/2) = −1 − i, and f1 (i) = −i. Thus f1 maps the circle onto the line Im(z) = −1. Hence f1 maps Ω onto the horizontal strip {z : −1 < Im(z) < 0}. Let f2 (z) = πz, then f3 maps the previous strip onto {z : −π < Im(z) < 0}.

0

−iπ Let f3 (z) = −z, then f3 maps the previous strip to the strip {z : 0 < Im(z) < π}.

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iπ

0

Now let f4 (z) = ez , then ez = ex+iy = ex ei y, hence 0 < ex = r < ∞ and 0 < y = θ < π. So f4 maps the previous strip onto H. Let F = f4 ◦ f3 ◦ f2 ◦ f1 , then F : Ω → H is a conformal map.  20. Find a conformal map from the slit unit disc D \ (−1, 0] onto the unit disk such that √ z0 = 2 − 1 is mapped to 0. SOLUTION: The slit disc is {z = reiθ : 0 < r < 1 and − π < θ < π}. Let f1 be a branch of the square root √ iθ function, f1 (z) = z. Then f1 takes the p√slit disc onto the right half disc {z = re : 0 < r < 1 and − π/2 < θ < π/2} and z0 to z1 = p 2 − 1. Now let f2 (z) = iz, this takes the right half disc √ 2 − 1. Let f3 (z) = (1 + z)/(1 − z), this takes the upper onto the upper half disc and z1 to z2 = i half disc onto the first quadrant and z2 to p√ q √ 1+i 2−1 √ p√ z3 = = 2 − 1 + i 2( 2 − 1). 1−i 2−1 p √ √ Let f4 (z) = z 2 , this takes the first quadrant onto H and z3 to z4 = 5 − 4 2 + 2i 10 2 − 14. Let f5 (z) = (i − z)/(i + z), then f5 takes H onto D and z4 to z5 = (i − z4 )/(i + z4 ). Let f6 = ψz5 (z) = (z5 − z)/(1 − z 5 z). Then f6 maps D onto D and z5 to z6 = 0. Now let F = f6 ◦ f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 , √ then F : D → D and F ( 2 − 1) = 0.  21. Let G be a simply connected region which is not the whole plane, and suppose that z¯ ∈ G whenever z ∈ G. Let a ∈ G ∩ R and suppose that f : G → D is a conformal map with f (a) = 0, f 0 (a) > 0. Let G+ = {z ∈ G : Im(z) > 0}. Show that f (G+ ) must lie entirely above or below the real axis. SOLUTION: Notice that f (G+ ) does not lie entirely above or bellow the real axis if and only if there exists z ∈ G+ such that f (z) ∈ D ∩ R. Thus if we prove that f (G ∩ R) = D ∩ R, then f being 1-1 implies that f (G+ ) is either above or below the real axis and hence we are done. What we need to prove is: If z ∈ G ∩ R, then f (z) = f (z) = f (z). If we prove that f (z) = f (z) for all z ∈ G, then we are done. Let g(z) = f (z), then:

MATH 545 and MATH 546 Prelim Problems

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Claim. g(z) is a bijective holomorphic function, i.e. a conformal map. Proof: Proving g is holomorphic: Since f is analytic on G and z, z0 ∈ G implies z, and z 0 are in G, f has a power series expansion around z 0 f (z) =

∞ X

ak (z − z 0 )k

k=0

and hence g(z) = f (z) =

∞ X

ak (z − z0 )k

k=0

which defines a holomorphic function on G#. Proving g is bijective: Since f is bijective, f (G) = D, and both G and D are symmetric with respect to the real axis, we have h1 (z) = z and h2 (z) = f are bijective and hence g(z) = (h2 ◦ f ◦ h1 )(z) is bijective because it is a composition of three bijective functions#. Thus g is a conformal map from G → D.  Claim. g(z) = f (z). Proof: Let a ∈ G ∩ R be as in the question with f (a) = 0 and f 0 (a) > 0. If we prove that g(a) = 0 and g 0 (a) > 0, then by Riemann mapping theorem Theorem 2.1.44 f = g, by uniqueness. Proving g(a) = 0: g(a) = f (a) = f (a) = 0, since a ∈ G ∩ R and f (a) = 0. Proving g 0 (a) > 0: g is holomorphic on G and has a power series expansion, hence 0

h

i0

g (z) = f (z) =

∞ X k=1

k ak (z − z0 )

k−1

=

∞ X

k ak (z − z 0 )k−1 = f 0 (z).

k=1

Thus g 0 (a) = f 0 (a) = f 0 (a) = f 0 (a) > 0 since f 0 (a) ∈ R.   22. Find a conformal mapping from D to D \ [−1/2, 1). SOLUTION: Let f1 (z) = i(1 − z)(1 + z), this maps D onto H. Let f2 (z) = −iz, this maps H onto the right halfplane {z = reiθ : r > 0 and − π/2 < θ < π/2}. Let f3 (z) = z 2 , this maps the right half-plane onto √ C \ (−∞, 0]. Let f4 (z) = z + 9, then this maps C \ (−∞, 0] onto C \ (−∞, 9]. Let f5 (z) = z, then this maps C\(−∞, 9] onto the right half-plane with the slit (0, 3] removed. Let f6 (z) = iz, this maps the right half-plane with the slit (0, 3] removed to H \ {0 < Im(z) ≤ 3}. Let f7 (z) = (i − z)/(i + z), this maps H \ {0 < Im(z) ≤ 3} to D \ [−1/2, 1). Let F = f7 ◦ f6 ◦ f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 , then F is the required conformal mapping.  23. Find a conformal mapping from Ω = {z : |z| < 1 and Re(z) > 0} \ [0, 1/2] to H.

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SOLUTION: Let f1 (z) = z 2 , then f1 maps Ω to D\[−1, 1/4]. Let f2 : D → D be defined by f2 = (z−1/4)/(1−z/4). Then f2 (−1) = −1 and f2 (1/4) = 0 maps D \ [−1, 1/4] to D \ [−1, 0]. Now as in problem Problem 20 √ let f3 (z) = z (defined because negative reals omitted), then f3 (z) takes D \ [−1, 0] onto the right half unit disc. Let f4 (z) = iz, this takes the right half unit disc onto the upper half unit disc. Let f5 (z) = i(1 + z)/(1 − z), this maps the upper half unit disc onto H. Let F = f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1 , then F is the required conformal mapping.  24. Let S denote the class of functions f which are 1-1 and analytic on the open disc D satisfying f (0) = 0 and f 0 (0) = 1. Prove that if {fn } ⊂ S and fn converges to f uniformly on compact sets of D, then f ∈ S. SOLUTION: Suppose {fn } ⊂ S and fn converges to f uniformly on compact sets of D, then by Theorem 2.1.16, f is holomorphic on D. Let K ∈ D be a compact subset contains 0, then fn → f uniformly on K implies fn → f point-wisely on K and hence 0 = fn (0) → f (0) so f (0) = 0. Also, fn and f being holomorphic implies fn0 and f 0 are holomorphic, and by Theorem 2.1.17 fn0 → f 0 uniformly on every compact subset of D and hence point-wisely on every compact subset of D. Let K ∈ D such that K contains 0, then 1 = fn0 (0) → f 0 (0) and thus f 0 (0) = 1. It remains to prove that f is 1-1. Suppose f is not 1-1 on D and hence not 1-1 on some compact set K, then there exists z0 , z1 ∈ K such that z0 6= z1 , fn (z0 ) 6= fn (z1 ) for all n and f (z0 ) = f (z1 ). Since fn (z0 ) 6= fn (z1 ) for all n, then there exists  > 0 and N ∈ N such that for all n > N |fn (z0 ) − fn (z1 )| ≥ . But fn → f point-wisely on K implies that there exists N 0 ∈ N such that for all n > N 0 |fn (z0 ) − f (z0 )| <

 2

and

 |fn (z1 ) − f (z1 )| < . 2

Let N 00 = max{N, N 0 }, then for all n > N 00 , |fn (z0 ) − fn (z1 )| ≤ |fn (z0 ) − f (z0 )| + |fn (z1 ) − f (z1 )| <  which is a contradiction. Thus f is 1-1 and hence f ∈ S.  25. Let Ω ⊂ C be open and a, b ∈ R, a < b and F : Ω × [a, b] → C be continuous such that for Z b every t ∈ [a, b], the function ft = F (z, t) is analytic on Ω. Show that g(z) = F (z, t) dt defines an analytic function on Ω.

a

MATH 545 and MATH 546 Prelim Problems

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SOLUTION: It is enough to prove it for [a, b] = [0, 1], I did that in the proof of Theorem 2.1.18. Or if you want to do it for [a, b], then it is enough to do it for a disc D whose closure in Ω. By changing variables t = x(b − a) + a, we get Z 1 Z b F (z, x(b − a) + a) dx, F (z, t) dt = (b − a) g(z) = 0

a

and hence considering the sequence n

b−aX gn (z) = F n k=0



k z, (b − a) + a n

 for n ∈ N.

Then gn is holomorphic on D for all n. Since F is continuous in t on [a, b], it is uniformly continuous on [a, b], i.e. for all  > 0, there is δ > 0 such that for all t1 , t2 ∈ [a, b]  |t1 − t2 | < δ =⇒ |F (z, t1 ) − F (z, t2 )| < . b−a Claim. {gn } converges uniformly to g on D. Proof: Choose N ∈ N such that N > (b − a)/δ, then for all n > N   Z 1 n b − a X k F (z, x(b − a) + a) dx |gn − g| = F z, (b − a) + a − (b − a) n n 0 k=1   n Z k/n n b − a X X k = F (z, x(b − a) + a) dx F z, (b − a) + a − (b − a) n n k=1 (k−1)/n k=1   n Z k/n X k [F z, (b − a) + a − F (z, x(b − a) + a)] dx = (b − a) n k=1 (k−1)/n   n Z k/n X k dx F z, ≤ (b − a) (b − a) + a − F (z, x(b − a) + a) n k=1 (k−1)/n n Z k/n X < dx =

k=1 n X k=1

(k−1)/n

1 = . n

 k−1 k Where in line (5) we used the fact that F is uniformly continuous on [a, b], since x ∈ , , n n   k(b − a) (k − 1)(b − a) k(b − a) (k − 1)(b − a) hence x(b − a) + a ∈ + a, + a and +a− − a = n n n n (b − a)/n < δ.  

Therefore, by Theorem 2.1.16 g is holomorphic on D. Since that true for all D in Ω, then g is holomorphic on Ω. 

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( 26. Let F =

f holomorphic on D :

2 ∞  (n) X |f (0)|

n! a g ∈ F such that g(0) = sup{Re(f (0)) : f ∈ F}.

) ≤ 1 and f (1/2) = 0 . Show that there is

n=0

SOLUTION: We first prove that F is a normal family. To do that we prove that F is uniformly bounded on compact subsets of D, and hence by Montel’s theorem Theorem 2.1.46 F is a normal family. For contrary suppose that F is not uniformly bounded on compact subsets of D. Then for some compact disc K with radius d in D containing the origin and for each k ∈ N there is a holomorphic function fk ∈ F and a point zk ∈ K such that |fk (zk )| > k. But for all k > 1/(1 − d2 ) we have 1 < k < |fk (zk )| ≤ |fk (zk )|2 1 − d2 2 ∞ (n) X f (0) k = zkn (since fk is holomorphic on D) n=0 n! 2 ∞ ∞  (n) X |f (0)| X |zk |2n (By Cauchy-Shwarz inequality) ≤ n! n=0 n=0 ≤

∞ X

|zk |2n

n=0

1 1 − |zk |2 1 ≤ 1 − d2 which is a contradiction. Hence F is a normal. =

Let E = sup{Re(f (0)) : f ∈ F} and let {fn } ⊂ F such that Re(fn ) → E. Then since F is normal, {fn } has a subsequence {fnk } such that {Re(fnk )} converges uniformly on every compact subset of D, say to g (may not be in F) but it is holomorphic by Theorem 2.1.16. Hence Re(fnk ) → g point-wisely, so Re(fnk (0)) → g(0). But since Re(fn ) → E and {fnk } a subsequence of {fn } such that Re(fnk (0)) → g(0), g(0) = E. It remains to prove that g ∈ F. First, notice since {fnk } in F, fn (1/2) = 0 for all n. Then since Re(fnk ) → g point-wisely, we have 0 = limk→∞ Re(fnk (0)) = g(1/2). Thus g(1/2) = 0. Second, notice that since {fnk } a subsequence such that {Re(fnk )} converges (n) uniformly on every compact subset of D to g, by Theorem 2.1.17 {Re(fnk )} converges uniformly on every compact subset of D to g (n) and hence point-wisely, so   (n)  2   (n)  2   2 ∞ ∞ ∞ Re fnk (0) X |g (n) (0)| X X Re fnk (0)  =≤ lim   = lim  k→∞ k→∞ n! n! n! n=0 n=0 n=0 2 ∞  (n) X |f (0)| ≤ lim ≤ lim 1 = 1. k→∞ k→∞ n! n=0

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Thus g ∈ F.



27. Show that F ⊂ Hol(D) is normal if and only if there is a sequence {Mn } of positive √ P n constants such that lim sup n Mn ≤ 1 and if f (z) = ∞ n=0 an z is in F, then |an | ≤ Mn for all n. SOLUTION:  Let Mn = sup

 |f (n) (0)| |f (n) (0)| : f ∈ F , then it is clear that |an | = ≤ Mn for all f ∈ F. n! n!

Suppose F ⊂ Hol(D) is normal, then by Problem 16 F is uniformly bounded on every compact subset of D, i.e. there is B > 0 such that for any compact disc D in D and all f ∈ F, we have |f (z)| ≤ B for all z ∈ D. Let D be an arbitrary disc with boundary circle C and radius R which contained in D, then by Cauchy’s inequalities |f (n) (0)| kf kC B ≤ ≤ n. n n! R R Hence

p √ 1 1 n lim sup n Mn ≤ lim sup B = R n→∞ R n→∞ √ Sending R to 1, implies lim sup n Mn ≤ 1.

√ P n Now suppose lim sup n Mn ≤ 1 and let g(z) = ∞ n=0 Mn |z| . Then the radius of convergence r of √ g(z) is given by 1/r = lim sup n Mn ≤ 1, hence r ≥ 1 and thus g(z) converges on every compact disc P∞ P n n D in D, say to LD . Therefore, fol all f ∈ F, |f (z)| ≤ ∞ n=0 Mn |z| = LD . Thus F n=0 |an ||z| ≤ is uniformly bounded on compact subsets of D, and hence by Montel’s theorem Theorem 2.1.46 F is normal.  28. Suppose f is analytic on D and satisfies |f (z)| < 1 for |z| = 1. Find the number of solutions, counting multiplicities of the equation f (z) = z n where n is an integer at least 1. SOLUTION: Let g(z) = f (z) − z n , then on |z| = 1, | − z n | = |z|n = 1 > |f (z)| thus by Rouche’s theorem Theorem 2.1.32 −z n and f − z n have the same number of zeros. But −z n has n zeros counted with multiplicities.  29. Let u be harmonic in an open connected domain Ω and set V := {a ∈ Ω : u vanishes in some neighborhood of a}. Show that either V = ∅ or V = Ω.

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SOLUTION: V is open, since if a ∈ V , then a ∈ Ω and u vanishes in some neighborhood of a, call it Dδ (a), where δ some positive real number. Let D (a) be another neighborhood of a such that D (a) subsetneqDδ (a), then for any z ∈ D (a), there is a neighborhood Dδ0 (z) ⊂ Dδ (a). Since u vanishes on Dδ (a), u vanishes on Dδ0 (z), thus z ∈ V . V also closed. Let {zn } ⊂ V such that zn → z ∈ Ω, we need to prove that z ∈ V . Choose  > 0 such that D (z) ⊂ Ω. Then there exists a holomorphic function f on D (z) such that u =Re(f ) on D (z). But since zn gets closer to z as n → ∞, then there is an n ∈ N such that zn ∈ D (z). But since zn ∈ V , then u, and hence f , vanish in a neighborhood of zn . But since f is holomorphic on D (z) and it vanishes on a subset of it, then by Theorem 2.1.14 f , and hence u, vanish on D (z), but that means z ∈ V . Let U be the complement of V in Ω, then both V and U are disjoint open sets and Ω = V ∪ U . But since Ω is connected, either U = ∅ and V = Ω or U = Ω and V = ∅.  30. Let f be analytic in a neighborhood of the closed unit disc D with |f (z)| ≤ 1 for all z with |z| = 1. Show that if there are z1 , z2 ∈ D with f (z1 ) = z1 and f (z2 ) = z2 , then either z1 = z2 or f (z) = z for all z ∈ D. SOLUTION: If z1 = z2 , then there is nothing to prove. Let z1 6= z2 , then f (z1 ) = z1 6= z2 = f (z2 ) and hence f is not constant. Since f is analytic and hence continuous on D, f attains a maximum in D, but since f is non constant in D, then by maximum modulus principle Theorem 2.1.34 f does not attain a maximum in D, thus it attains its maximum on ∂D, i.e. when |z| = 1, but since |f (z)| ≤ 1 when |z| = 1, by maximum modulus principle |f (z)| ≤ sup |f (z)| < z∈D

sup |f (z)| ≤ 1. {z:kz|=1}

Hence f is from D to D. Let ψz1 (z) : D → D be an automorphism of D. Then ψz1 ◦ f ◦ ψz1 : D → D and (ψz1 ◦ f ◦ ψz1 )(0) = 0. Thus by Schwarz lemma Theorem 2.1.42 |ψz1 ◦ f ◦ ψz1 | ≤ |z| for all z ∈ D. Now since ψz1 (z) : D → D and z2 ∈ D, there exists z3 ∈ D such that ψz1 (z3 ) = z2 and z3 6= 0 because if it was 0, then z1 = z2 which is not the case. Therefore, we have (ψz1 ◦ f ◦ ψz1 )(z3 ) = (ψz1 ◦ f )(z2 ) = ψz1 (z2 ) = ψz−1 (z2 ) = z3 . 1

(∗)

Hence by Schwarz lemma since z3 6= 0, ψz1 ◦ f ◦ ψz1 = eiθ z for some θ ∈ R. But by (∗) θ = 0 or 2π and hence ψz1 ◦ f ◦ ψz1 = z, which means f (ψz1 (z)) = ψz1 (z) or f (w) = w for all w ∈ D. 

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31. If f is an analytic function on an open neighborhood G of the unit circle ∂D satisfies |f (z)| = 1 for all z ∈ ∂D. Show that f (z) =

1 f (1/z)

for all z in a neighborhood of ∂D. SOLUTION: Let O ⊂ G be a neighborhood of ∂D such that 0 ∈ / O and O is symmetric with respect to |z| = 1 (i.e., if z ∈ O \ D, then 1/z ∈ D \ O). Let g(z) = f (z) − 1/f (1/z), then if we prove that g is holomrphic on O and vanishes on a sequence of complex numbers zn ∈ O that has a limit in O, we will have g identically zero and hence f (z) = 1/f (1/z). Let zn ∈ ∂D ⊂ O for all n and zn → z0 ∈ O. Then g(zn ) = f (zn ) − 1/f (1/zn ) = f (zn ) − 1/f (zn ) = f (zn ) − f (zn ) = 0.

(Since |z| = 1 and hence 1/z = z) (Since when |z| = 1 we have |f (z)| = 1, hence 1/f (z) = f (z))

It remains to prove that g is holomorphic on O. Since f is holomorphic on O, f has a power series expansion around some z0 ∈ ∂D in O f (z) =

∞ X

an (z − z0 )n .

n=0

But since 0 ∈ / O and O is symmetric with respect to |z| = 1, if z ∈ O, then 1/z ∈ O. Hence f (1/z) =

∞ X

an (1/z − z0 )n .

n=0

Hence f (1/z) =

∞ X

an (1/z − z0 )n ,

(∗)

n=0

but since z0 ∈ ∂D, (∗) defines a convergent series in O, thus f (1/z) is holomorphic in O. Now let O0 ⊂ O such that f (1/z) 6= 0 for all z ∈ O0 (a such neighborhood exists by continuity of f around |z| = 1, since |f (z)| = 1 there), then 1/f (1/z) is holomorphic on O0 and hence g(z) is holomorphic on O0 , then by Theorem 2.1.14 g is identically zero in O0 and hence f (z) = 1/f (1/z) in O0 .