exam2011 - solutions.pdf

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MATH31011/41011/61011: FOURIER ANALYSIS &LEBESGUE INTEGRATION - JANUARY 2011 SOLUTIONS

A1. The Fourier series associated to f is the series

a02

+n=1

(an cos(nx) + bn sin(nx)) ,

where

a0 =1

f(x)dx, an =1

f(x)cos(nx)dx, bn =1

f(x)sin(nx)dx, n 1.

Alternative solution:

n=

cneinx,

where

cn =1

2

f(x)einxdx.

[3 marks]

FEEDBACK: Generally answered correctly. A small number of students made slips likereplacing the sum by an integral or forgetting to include the function f(x) in the formulafor the coefficients.

A2. A set E is countably infinite if there exists a bijection : N E.[2 marks]

FEEDBACK: Lots of students lost marks here by writing that a set E was countably

infinite if there was a surjection from N to E. This is the definition of a set being countable,not countably infinite. For example, the set {1} is countable but not infinite and there is asurjection : N {1} given by (n) = 1 for all n N. An acceptable alternative solutionwas to say that there was a surjection from N to E and that E was infinite.

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A3. B is a -algebras if (i) , X B; (ii) E B = X\E B; (iii) if{Ej}j=1 B thenj=1 Ej B.A -algebra containing Q [0, 1] but not [0, 1/2] is the set of E [0, 1] such that either

E or [0, 1]\E is countable (or, alternatively, is a null set).

Two other examples from {, [0, 1]}; Borel sets in [0, 1]; measurable sets in [0, 1];P([0, 1]).[6 marks]

FEEDBACK: Lots of students had difficulties in identifying a -algebra containing Q [0, 1] but not containing [0, 1/2]. Some of these even listed one or both of the solutionsgiven above among the other two examples. An alternative and perfectly correct solutionwas the -algebra {,Q [0, 1], [0, 1]\Q, [0, 1]}.A4. m : A R+ is countably additive if, whenever {Ej}j=1 A are disjoint sets, wehave

m

j=1

Ej =

j=1

m(Ej).

[3 marks]

FEEDBACK: Many students lost marks by forgetting to say the sets had to be disjoint.

A5. We have f d =

ki=1

i(Ei).

[3 marks]

FEEDBACK: Nearly everyone answered this correctly.

A6. A function f : [, ] R is square integrable if |f|2 is integrable, i.e.

|f|2d 0, there exists a countable collection of open intervals{Ij} such that

E

Ij andj

l(Ij) < .

[2 marks]

(ii) IfE is countable then we can write E = {xj}j=1. [I dont expect the simpler finite caseto be written down explicitly.] Given > 0, define open intervals Ij = (xj /2j+2, xj +/2j+2). Clearly E Ij and

j=1

l(Ij) =j=1

2j+1=

2< .

Thus E is a null set.

[6 marks]

(iii) The Middle Third Cantor set is constructed as follows. Start with C0 = [0, 1]. Removethe open middle third (1/3, 2/3) to obtain

C1 = [0, 1/3] [2/3, 1].

Continue in this way by removing the open middle third of each closed interval to obtaina nested sequence of sets

C1 C2 Cn .

Cn will consist on 2n disjoint closed intervals each of length 3n. The Middle Third Cantorset C is defined by

C =n=0

Cn.

[8 marks]

(iv) Call the 2n closed intervals making up Cn C1n, . . . , C

2n

n . Enlarge these to make openintervals by setting

Uin = a 1

2

7n, b +

1

2

7n ,where Cin = [a, b]. Note that

l(Uin) = 3n + 7n c}

is a measurable set (or any similar equivalent criterion).

[3 marks]

(ii) Suppose that

f(x) + g(x) > c.

Then

f(x) > c g(x).Since Q is dense in R, we can find r Q such that

f(x) > r > c

g(x),

which may be rewitten

f(x) > r and g(x) > c r.Now suppose that there exists r Q such that

f(x) > r and g(x) > c r.

This may be rewritten

f(x) > r > c g(x),so

f(x) > c g(x).Rearranging,

f(x) + g(x) > c.

We have shown that f(x) + g(x) > c iff there exists r Q such that f(x) > r andg(x) > c r. Hence

{x [0, 1] : f(x) + g(x) > c} = rQ({x [0, 1] : f(x) > r} {x [0, 1] : g(x) > c r}).

[11 marks]

(iii) Since f is measurable, for each r Q,

Ar = {x R : f(x) > r}

is measurable. Since g is measurable, for each r Q,

Br = {x R : g(x) > c r}5


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is measurable. Since the measurable sets form a -algebra, Ar Br is measurable and sois

rQ

(Ar Br).

By part (ii), the latter shows that

{x R : f(x) + g(x) > c}

is measurable, giving f + g measurable.

[11 marks]

FEEDBACK: Rather few students could give a clear account of the required argumentin part (ii) but part (iii) was answered better. There were a lot of vague and inaccuratesolutions to this question. A lot of students just regurgitated the solution to B9(iii) on themock exam, for which they received no marks.

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B10.(i) f : [0, 1] R is integrable iff+ d,

f d < +

(f+, f are the postive and negative parts of f) or, alternatively, if|f| d < +.

[3 marks]

(ii) Monotone Convergence Theorem: Suppose that fn is an increasing sequence of non-negative integrable functions and that limn+ fn = f (pointwise). Then

limn+

fn

d = f d.[3 marks]

Proof: For each n, choose an increasing sequence of non-negative simple functions fn,kconverging to fn, as k +. Set

gk = maxnk

fn,k.

The gk are simple and form an increasing sequence. Thus we can define

g = limk+

gk.

For 1 n k, we havefn,k gk fk f. ()

Letting k + givesfn g f.

Letting n + givesf g f,

i.e. f = g.Integrating (*) gives, for 1 n k,

fn,k d

gk d

fk d.

Letting k + gives fn d

g d lim

k+

fk d.

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Letting n + gives

limn+

fn d

g d lim

k+

fk d.

Since f = g this completes the proof.(Bookwork)

[12 marks]

(iii) Write

gn =ni=1

fi.

Then gn is an increasing sequence of non-negative measurable functions whose limit is

n=1 fn. So, by the Monotone Convergence Theorem,

gn d

n=1

fn

d,

as n +. But by additivity of the integral,

gn d =ni=1

fi d.

Therefore,n=1

fn d

=

n=1

fn

d.

[7 marks]

FEEDBACK: Part (i) of the question was answered correctly by nearly everyone whoattempted the question. A lot of students could give a complete or nearly complete proofof the MCT. A common slip was to forget to mention that the functions fn,k, and hencethe functions gk, were simple. A number of students answered part (ii) by giving the proofof the DCT (from the mock exam), for which they received no marks. Very few studentsanswered part (iii) correctly. As you can see, it is a straightforward application of the

MCT.

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B11.(i) Let c1, . . . , cn be arbitrary real numbers and set ak = w, vk. Write u =nk=1 akvk and v =

nk=1 ckvk. By orthonormality,

u2 =n

k=1

a2k and v2 =n

k=1

c2k.

Also

w, v =

w,n

k=1

ckvk

=

nk=1

ckw, vk =n

k=1

ckak.

Thusw v2 = w v, w v

= w2 2w, v + v2

=

w

2

2

n

k=1

ckak +n

k=1

c2k

= w2 n

k=1

a2k +n

k=1

(ak ck)2

= w2 u2 +n

k=1

(ak ck)2.

It follows thatw v2 w2 u2

with equality if and only ifnk=1(ak ck)

2 = 0, i.e. if and only if ck = ak =

w, vk

for

all k = 1, . . . , n .[9 marks]

(ii) Riesz-Fischer Theorem: Let f L2([, ], ,R). Then Sn(f, ) converges to f inL2([, ], ,R), i.e,

Sn(f, ) f2 =

1

|Sn(f, ) f|2 d1/2

0, as n +.

[4 marks]

Proof: Suppose f L2([, ], ,R) and let > 0 be given. We can find a continuousfunction g : [, ] R such that f g2 < /2.

By Fejers Theorem, we can choose N sufficiently large that

n N = n(g, ) g < 2

2.

Thus, if n N then

n(g, ) g2

2n(g, ) g < 2

.

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Combing the two estimates, if n N then

f n(g, )2 f g2 + g n(g, )2 < 2

+

2= .

We may writen(g, x) =

nk=n

ckk(x),

for some ck R. By part (i),fn

k=n

f, kk(x)2

f

nk=n

ckk(x)

2

.

Thus, if n

N then

f

Sn(f,

)

2 < , as required.

[12 marks]

FEEDBACK: This question was answered by relatively few students and spome of theseonly made a serious attempt at part of it. Most students who did part (i) give a correctsolution. Also, most people who attempted part (ii) gave a correct statement of the Riesz-Fischer Theorem. There were few serious attempts to give a proof and even fewer thatwere completely correct.

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C12.(i) If (En) = + for some n then the result is immediate.Suppose (En) < + for all n. Choose > 0 and, for each n, a cover {I(n)j } of En by

intervals such that

jl(I

(n)j ) <

(En) +

2n.

Thenn=1{I(n)j } is a cover of

n=1 En by intervals, so

n=1

En

n=1

j

l(I(n)j )

n=1

(En) +

2n

=

n=1

(En) + .

Since > 0 is arbitrary,

n=1

En

n=1

(En).

[8 marks]

(ii) First we show that (a, ) is measurable. Take A R and set

A1 = (a, ) A, A2 = (, a] A.If(A) = + the criterion for measurability is automatically satisfied. Suppose (A) 0. We can find a cover {Ij} of A by intervals such that

j

l(Ij) (A) + .

WriteIj = Ij (a, ), Ij = Ij (, a]

(intervals or empty). Then {Ij} (resp.{Ij }) is a cover of A1 (resp. A2) by intervals. Thus

(A1) + (A2)

j

l(Ij) + l(I

j )

=j

l(Ij) (A) + .

Since > 0 is arbitrary,(A1) +

(A2) (A),11


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which gives (a, ) measurable.Since measurable sets form a -algebra,

(, a] = R\(a, )

is measurable and so is

(, a) =n=1

(, a 1n

].

Hence(a, b) = (a, ) (, b)

is measurable.

[13 marks]

(iii) The Borel sets

Bare the smallest -algebra containing all open intervals in R. By

part (ii), M (measurable sets) is a -algebra containing all open intervals. Thus B M,i.e. all Borel sets are measurable.

[4 marks]

FEEDBACK: This question was generally quite well answered by everyone who made aserious attempt at it.

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C13.(i) Since E1, E2 measurable, we may use the criterion with E = E1 and A = E1 E2.We have E A = E1 and Ec A = E2 and so

(E1) + (E2) = (E1) +

(E2) = (E1 E2) = (E1 E2).

[5 marks]

(ii) LetE1 = E [0, 1 x), E2 = E [1 x, 1).

These are measurable and disjoint. By part (i),

(E) = (E1) + (E2).

We havex

E1 = x + E1

so x E1 is measurable and

(x E1) = (x + E1) = (E1).

Alsox E2 = (x 1) + E2

so x E2 is measurable and

(x

E2) = ((x

1) + E2) = (E2).

Thusx E = (x E1) (x E2)

(disjoint union) is measurable and

(x E) = (x E1) + ((x E2) = (E1) + (E2) = (E).

[7 marks]

(iii) Define an equivalence relation

on [0, 1) by

x y x y Q.

By the Axiom of Choice, we can choose a set E consisting of exactly one element fromeach equivalence class.

Claim: E is not measurable.

Justification: Since Q is countable, we can write

Q [0, 1) = {rn}n=1.13


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Consider x rn E, for some n. Then either

x = rn + en or x = rn + en 1 ()

for some en

E.

Supposex (rn E) (rm E).

By (*),en em Q.

Since E has one element from each equivalence class, en = em. By (*) again, rn = rm(since both are in [0, 1)). Hence

(rn E) (rm E) = = rn E = rm E

(disjointness).Since E contains an element from every equivalence class

[0, 1) =n=1

(rn E)

(disjoint union). If E is measurable then so is every rn E, so

1 = ([0, 1)) =

n=1

(rn E) =

n=1

(rn E).

By part (ii), each summand is equal to (E). If(E) = 0, the sum is zero; if (E) > 0the sum is infinite. Either case gives a contradiction. Therefore E is not measurable.

[13 marks]

FEEDBACK: Some students had problems with giving accurate solutions to parts (i)and (ii) but I was pleased to see that part (iii) was generally well answered.

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