Lecture 1Real and Complex Numbers

Exercise1.1. Show that a bounded monotonic increasing sequence of real numbersconverges (to its least upper bound).

Solution. (This was indicated in class) Let(an) be a bounded monotonic sequence —without loss of generality, say it is monotonic increasing and bounded above byM .Let a = supan. I claim an → a. Let ε > 0 be given. Then there isN such thataN > a − ε; otherwise,a − ε would be an upper bound foran smaller thana.Because(an) is increasing we havea− ε < aN 6 an 6 a for all n > N ; soan → aby definition.

Exercise1.2. Show that any sequence of real numbers either has an increasing sub-sequence or else a decreasing subsequence. Deduce the Bolzano-Weierstrass theo-rem?? (using the previous exercise). Deduce Cauchy’s version of completeness fromBolzano-Weierstrass.

Solution. Let (an) be a sequence of real numbers. Suppose that(an) has no maximalelement (meaning an element of the sequence that is greater than or equal to all otherelements of the sequence). We construct an increasing subsequence of(an) by induc-tion; taken0 = 1 and inductively definenk such thatank

is greater thana1, . . . , ank−1 .We assume then that(an) has no increasing subsequence, and we shall show that

instead it has a (perhaps not strictly) decreasing subsequence. Note that(an), and everysubsequence, has a maximal element. Choosen1 such thatan1 > an for all n (that is,an1 is maximal), and then inductively choosenj such thatanj

> an for all n > nj−1.Theanj then form a decreasing subsequence.

The previous exercise now shows that any bounded sequence contains a conver-gent subsequence (Bolzano-Weierstrass). In particular, any Cauchy sequence containsa convergent subsequence. But, if a Cauchy sequence(an) contains a subsequenceconverging toa, the whole Cauchy sequence converges toa. To prove this formally,let ε > 0. There is anN such that|an − am| < ε for n, m > N . In this inequality letn → ∞ through the subsequencenk. We find that|a − am| 6 ε for m > N , whichproves convergence.

Exercise1.3. (Cantor’s nested interval theorem) Let[an, bn], n = 1, 2, . . . be a se-quence of closed intervals inR that arenestedin the sense that[an+1, bn+1] ⊆ [an, bn]and whose lengthsbn−an tend to zero. Show that the intersection

⋂n[an, bn] contains

one and only one real numberc.

Solution. The (an) are increasing and bounded above byb1, so converge (say toa).Similarly the(bn) converge, say tob. Sincean − bn → 0, a = b = c, say. We havean 6 c 6 bn for all n, soc belongs to the intersection of all the intervals. Any realnumber less thanc is also less than somean, so does not belong to the intersection;any real number greater thanc is also greater than somebn, so does not belong to theintersection. Thus

⋂n[an, bn] = c.

1

Exercise1.4. Use the previous exercise to show that the real numbers areuncountable,i.e. cannot be listed asc1, c2, . . .. Hint: Suppose such a listing is possible. Pick aninterval[a1, b1] of length at most 1 that doesn’t containc1, then a subinterval[a2, b2] oflength at most12 that doesn’t containc2, and so on. Apply the nested interval theorem.What can you say about the limit pointc?

Solution. Follow the construction in the hint. Cantor’s nested interval theorem pro-duces a realc ∈

⋂n[an, bn]. Thusc 6= cn for eachn, contradicting the supposed

enumeration of the reals.

Exercise1.5. Let z andw be complex numbers of absolute value< 1. Show that theabsolute value of

z − w

1− wz

is also less than 1. (Hint: Expand the difference|1 − wz|2 − |z − w|2 in terms ofcomplex conjugates, and factorize the resulting mess.)

Solution. We have|1 − wz|2 = (1 − wz)(1 − wz) = 1 − wz − wz + |w|2|z|2 and|z − w|2 = (z − w)(z − w) = |z|2 − wz − wz + |w|2. Taking the difference

|1− wz|2 − |z − w|2 = 1− |z|2 − |w|2 + |z|2|w|2 = (1− |z|2)(1− |w|2) > 0

and so|z − w| < |1− wz| as required.

2

Lecture 2Metric spaces

Exercise2.1. Show that a functionf as above is continuous atx if and only if, when-ever(xn) is a sequence converging tox, the sequence(f(xn)) converges tof(x).

Solution. Suppose thatf is continuous atx and letε > 0 be given. By definition ofcontinuity there isδ > 0 such thatd(x, x′) < δ implies sd(f(x), f(x′)) < ε. Bydefinition of convergence there isN such that for alln > N , d(x, xn) < δ. Thusd(f(x), f(xn)) < ε andf(xn) converges tof(x).

Suppose thatf is not continuous atx. Then there isε > 0 such that for allδ > 0there isx′ with d(x, x′) < δ andd(f(x), f(x′) > ε. Let xn be a value ofx′ corre-sponding toδ = 1/n. Thenxn → x, butd(f(x), f(xn)) > ε for all n, hencef(xn)does not converge tof(x).

Exercise2.2. Show that the sequence of functions on[0, 1] defined by

fn(t) = nte−nt

doesnotconverge in the metric spaceC[0, 1], even though, for allt ∈ [0, 1], fn(t) → 0asn →∞. (One says that(fn) convergespointwise, but notuniformly.)

Solution. By elementary calculus the functionxe−x has a maximum value atx = 1,and this maximum value is1/e. Thusfn(x) has the same maximum value,1/e, atx = 1/n. It follows thatd(fn, 0) = 1/e for all n, sofn does not converge uniformlyto the zero function. However,fn(0) = 0 for all n, and fort > 0, fn(t) = nte−nt <nt/ 1

2n2t2 → 0 asn →∞; sofn converges pointwise to zero. The figure shows graphsof f1, . . . , f6.

0.5 1 1.5 2

0.05

0.1

0.15

0.2

0.25

0.3

0.35

3

4

Lecture 3Closed sets

Exercise 3.1. Show thata ∈ A is isolated if and only if there isε > 0 such thatB(a; ε) ∩A = a.

Solution. If B(a; ε) ∩ A = a then in any sequence of distinct points ofA, all butat most one member must be at distance> ε from a. Thus no such sequence canconverge toA. Conversely, if every ball arounda meetsA in a point other thana wecan construct a sequencean of distinct points ofA tending toa, by the same inductiveargument as in the proof of Proposition 3.2

Exercise3.2. Show thatA is the collection ofall limits of convergent sequences inA(whether or not the members of the sequence are distinct).

Solution. Any sequence of points ofA either has a subsequence consisting of distinctpoints, or else has a constant subsequence. If the original sequence converges anysubsequence converges to the same limit. Thus the limit is either a limit point ofA (inthe first case) or a member ofA (in the second case). In either instance it belongs toA.

Exercise3.3. A closed ballin a metric space is a set

B(x; r) := x′ ∈ X : d(x, x′) 6 r.

Show that a closed ball is closed. Must every closed ball be the closure of the open ballof the same center and radius?

Solution. If an is a sequence inX converging toa, andd(an, x) 6 r for all r, thend(a, x) 6 r also. To see this, letε > 0; there isn such thatd(an, a) < ε and therefored(x, a) < r + ε by the triangle inequality; this is true for allε sod(x, a) 6 r. Thisshows that a closed ball is indeed a closed set. The two point space0, 1 ⊆ R hasB(0; 1) = 0 which is closed, whereasB(0; 1) = 0, 1. Thus the closure of an openball need not be the corresponding closed ball.

Exercise3.4. Show that the intersection of any (finite or infinite) collection of closedsets is closed.

Solution. Let F =⋂

Fα where eachFα is closed. A limit point ofF is a limit pointof eachFα, so belongs to eachFα, so belongs to their intersection which isF . ThusFcontains all its limit points.

Exercise 3.5. Verify that if f : A → B is onto, andB is uncountable, thenA isuncountable.

Solution. Suppose not. LetA be enumerated asa1, a2, . . .. ThenB = f(a1), f(a2), . . ..EnumerateB as follows:b1 = f(a1), and inductivelybn is f(am) wherem is the leastinteger such thatb1, . . . , bn−1 /∈ f(a1), . . . , f(am) (this construction avoids possi-ble repetitions in our enumeration ofB). ThusB is countable, contradiction.

5

Exercise3.6. Show that any open subset ofR is a union of countably many disjointopen intervals (Lindelof’s theorem). (The example of the Cantor set shows that the‘closed analog’ of this statement is completely false.) Hint: First show the result with-out the countability restriction. To get that, note that each interval must contain arational number.

Solution. LetU ⊆ R be open. Define a relation∼ onU bya ∼ b iff (mina, b,maxa, b) ⊆U . It is easy to check that this is an equivalence relation, soU is partitioned into dis-joint equivalence classes. The equivalence class ofa is a union of open intervals all ofwhich containa, so it is itself an open interval containinga. ThusU is a union of dis-joint open intervals. To see that this is a countable union, note that each open intervalmust contain a rational number, and the rationals are countable.

6

Lecture 4Compactness

Exercise4.1. Let f : X → Y be continuous and surjective. Show that ifX is compactthenY is compact also. What has this got to do with the familiar calculus principlethat ‘a continuous function on a closed bounded interval is itself bounded and attainsits bounds’?

Solution. Let yn be a sequence inY . Becausef is surjective, eachyn = f(xn) forsomexn ∈ X. The sequence(xn) has a convergent subsequencexnk

, becauseX iscompact; the sequenceynk

= f(xnk) is then convergent too, becausef is continuous.

We have shown that an arbitrary sequenceyn in Y has a convergent subsequence, soYis compact.

In the case of the calculus example, letf : [a, b] → R be continuous. Since[a, b] iscompact, so is its imagef([a, b]). In particular the image is bounded (which is what wemean by saying thatf is bounded), and closed. Since the image is closed it containssup f([a, b]) and inf f([a, b]), which are accumulation points off([a, b]); and this iswhat we mean by saying “f attains its bounds”.

Exercise4.2. (Connectedness) Show thatR cannot be written as the disjoint union oftwo nonempty open sets. (Let the sets beU andV . ThenU, V forms an open coverof each closed interval[a, b]. By considering a Lebesgue number for this cover showthat[a, b] lies entirely within one of the two sets.)

Solution. Let the sets beU andV . ThenU, V forms an open cover of each compactinterval[a, b]. Let δ > 0 be a Lebesgue number for this cover. Then two points of[a, b]which are separated by less thanδ are either both inU or both inV . Since any twopoints of[a, b] can be joined by a finite chain of points whose separations are less thanδ, the whole of[a, b] is either inU or in V . Sincethis is true for any interval[a, b], oneof the setsU, V is the whole ofR and the other is empty.

Exercise4.3. A mapf : X → Y between metric spaces isuniformly continuousif foreachε > 0 there isδ > 0 such thatd(f(x), f(x′)) < ε wheneverd(x, x′) < δ. (Theextra information beyond ordinary continuity is thatδ does not depend onx.) Show thatif X is compact, every continuousf is uniformly continuous. Hint: use Theorem??.

Solution. Let ε > 0 be given. For eachx ∈ X there isδx such thatd(f(x), f(x′)) <ε/2 wheneverd(x, x′) < δx. The open setsUx = B(x; δx) form a cover forX. Letδ be a Lebesgue number for this cover. Now, ifd(x′, x′′) < δ, then bothx′ andx′′

belong to the sameUx, and therefore

d(f(x′), f(x)) < ε/2, d(f(x′′), f(x)) < ε/2.

By the triangle inequality, then,d(f(x′), f(x′′)) < ε wheneverd(x′, x′′) < δ. This isuniform continuity.

7

Lecture 5Complete Spaces

Exercise5.1. Banach’s fixed point theorem can be extended as follows: iff : X → Xis a map and some powerfN = f · · · f is a strict contraction, thenf has a uniquefixed point. Prove this.

Solution. Let g = fN . Theng has a unique fixed pointx, by Banach’s theorem. Noticethat

g(f(x)) = fN+1(x) = f(g(x)) = f(x).

Thusf(x) is also a fixed point ofg. By the uniqueness part of Banach’s theorem,f(x) = x; that is,x is a fixed point off .

Exercise5.2. Consider the complete metric spaceCR[0, 1] of continuous real-valuedfunctions on[0, 1]. For a, b ∈ [0, 1] show that the set of functionsf which are non-decreasing on the interval[a, b] is nowhere dense inCR[0, 1]. Using Baire’s theorem,deduce that there exist functionsf ∈ CR[0, 1] that arenowhere monotonic, that is,monotonic on no subinterval of[0, 1].

Solution. Let N [a, b] denote the collection of functionsf which are nondecreasing on[a, b], which is to say thatf(x) 6 f(y) whenevera 6 x 6 y 6 b. If fn is a sequencein N [a, b] which converges to a functionf , then for allx, y as above,

f(x) = lim fn(x) 6 lim fn(y) = f(y)

so f ∈ N [a, b] also. ThusN [a, b] is closed. To show thatN [a, b] is nowhere denseit is therefore enough to show that given anyf ∈ N [a, b] and anyε > 0, there isg ∈ C[0, 1] \N [a, b] with d(g, f) < ε.

Let c = (a+b)/2. Sincef is continuous atc there isδ > 0 such that|f(x)−f(c)| <ε/2 whenever|x−c| < δ; we may assume without loss of generality thatδ < (b−a)/4.Let

g(x) = f(x)− εh((x− c)/δ),

whereh(t) = max0, 1 − |t|. Thend(f, g) = ε, and moreoverg(c) = f(c) − ε,g(c− δ) = f(c− δ) > f(c)− ε/2, sog does not belong toN [a, b].

Note that−N [a, b] is the space of functions that are nonincreasing on[a, b]. Thespace of functions that are monotonic on some subinterval of[0, 1] is⋃

a,b

N [a, b] ∪ (−N [a, b])

where the union may be taken over allrational a, b ∈ [0, 1] with a < b. This is acountable union of nowhere dense sets, so it is of first category. By Baire’s theoremapplied to the complete metric spaceC[0, 1], its complement is dense. In particular thecomplement is nonempty; that is, there exist nowhere monotonic functions.

8

Lecture 6Convergent Series

Exercise6.1. Let Ω be an open subset ofC. Show that a sequence of functions con-verges uniformly on compact subsets ofΩ if, for every pointz ∈ Ω, there is a closeddisk D(z; rz) ⊆ Ω on which the sequence converges uniformly.

Solution. Since the closed disksD(z; rz) are all compact (by the Bolzano-Weierstrasstheorem), uniform convergence on compact sets certainly implies uniform convergenceon any such disk. Conversely, suppose thatK ⊆ Ω is compact and that the conditionof the exercise is satisfied. The open setsD(z; rz), for z ∈ K, form a cover ofK. Thiscover has a finite subcover, sayD(z1; rz1 , . . . , D(zk; rzk

). The sequence convergesuniformly on each closed diskD(zj ; rzj

), and hence on their union, which containsK.

Exercise6.2. Cauchy’s formula for the radius of convergence is

1/r = lim supn→∞

|an|1/n.

Prove this. You need to know the definition of the ‘limit superior’ of a sequence of realnumbers: it is

lim supn→∞

bn := limm→∞

supn>m

bn.

Notice that the limit on the right is the limit of a monotonic sequence, so it alwaysexists (possibly infinite).

Solution. Make use of the following property of the limit superior: ifb = lim sup bn,then for anyε > 0, only finitely manybn are greater thanb + ε, and infinitely manybn

are greater thanb− ε.Applied to Cauchy’s formula, this says that for anyR > r there are infinitely many

n for which |an| > 1/Rn; whereas for anyρ < r, the inequality|an| < 1/ρn issatisfied for all but finitely manyn.

Thus if |z| > R the series∑

anzn has infinitely many terms of absolute valuegreater than 1, so cannot converge. It follows that the radius of convergence is less thanor equal toR. If |z| < ρ then all but finitely many terms of

∑anzn are dominated by

the corresponding terms of the convergent geometric series∑|z|n/ρn. It follows that

the radius of convergence is greater than or equal toρ. Sinceρ,R are arbitrary subjectto the conditionρ < r < R, we conclude that the radius of convergence isr.

Exercise6.3. Let a ∈ C be a constant. Thebinomial seriesis the series

fa(z) =∞∑

n=0

a[n]

n!zn

wherea[n] := a(a − 1) · · · (a − n + 1). Show that the radius of convergence of thebinomial series is at least 1.

9

Solution. We use the ratio test. Letbn = a[n]/n! be the coefficients of the series. Thenbn+1/bn = (a − n)/(n + 1) → −1 asn → ∞. It follows that|bn|ρn is bounded foranyρ < 1, which in turn tells us that the radius of convergence is at leastρ. Sinceρwas arbitrary the radius of convergence is at least 1.

Exercise6.4. Let fa(z) denote the binomial series of the previous exercise. Show thatfa+b(z) = fa(z)fb(z) for |z| < 1. Sincef1(z) = 1 + z, this justifies the notation(1 + z)a for fa(z). Hint: Start by proving, by induction, the formula

(a + b)[n] =n∑

k=0

n!(n− k)!k!

a[k]b[n−k].

Solution. Once we have proved the formula given in the hint, the result follows directlyfrom Proposition 6.7 on multiplying series. To prove the formula by induction, assumeit for n. Then

(a + b)[n+1] =∑

k

(n

k

)a[k]b[n−k](a + b− n).

We can split the last term on the right to get∑k

(n

k

)a[k]b[n−k]((b+n−k)+(a−k)) =

∑k

(n

k

)a[k]b[n+1−k]+

∑k

(n

k − 1

)a[k]b[n+1−k]

where we have substitutedk − 1 for k in the second term. The binomial coefficientidentity (

n + 1k

)=

(n

k

)+

(n

k − 1

)now completes the induction.

Exercise6.5. Fill in the details of the above argument. (Show that for sufficiently largen, the validity of the inequality forn implies its validity forn + 1.)

Solution. Begin with the following observation: if1 < β < α = 1 + a then(1− α

m + 1

) (1 +

1m

)β

6 1, for m large.

There are various ways to see this: direct estimation of the binomial series, L’Hopital’srule, use of logarithms. . . . Apply this to the formulabn+1/bn = (a−n)/(n+1) whichis taken from an earlier exercise. Assume thatn > a and that|bn| 6 Kn−β . Then

|bn+1| 6(

1− α

n + 1

)|bn| 6

(n

n + 1

)β

·Kn−β = K(n + 1)−β

for n sufficiently large. This gives the desired estimate by induction.

10

Lecture 7More about power series

11

Lecture 8The Weierstrass Approximation Theorem

Exercise8.1. In the complex case the Stone-Weierstrass theorem takes the followingform: a ∗-subalgebra ofC(X) which contains the constants and separates points isdense, where the∗-condition means that the algebra is closed under (pointwise) com-plex conjugation. Prove this. Try to give an example to show that the complex theoremis not valid without the extra condition (we shall discuss this in detail later).

Solution. Let A be a∗-subalgebra ofC(X). If f = u + iv with u andv real, thenu = 1

2 (f + f) andv = 12i (f − f). Therefore,u andv belong toA. It now follows

that the collectionB of all real and imaginary parts of members ofA is a subset ofA, closed under addition, and multiplication and containing the (real-valued) constantfunctions; in other words it is a subalgebra ofCR(X). Moreover, sinceA separatespoints, so doesB (if f(x) 6= f(y) then eitheru(x) 6= u(y) or v(x) 6= v(y)). ThusB is dense inCR(X) by the real form of the Stone-Weierstrass theorem, and it easilyfollows thatA is dense inC(X).

The desired example may be given by takingX to be the unit circleS1 = z ∈ C :|z| = 1, andA the algebra of (complex) polynomial functions onX. This separatespoints and contains the constants but it is not a∗-algebra; and in fact it can be shownthat the functionz−1 = z is not in the closure ofA.

12

Lecture 9Compact Sets inC(X)

Exercise9.1. Prove the converse of the Arzela-Ascoli theorem: a compact subset ofC(X) is closed, bounded and equicontinuous.

Solution. All that needs to be proved is equicontinuity, since any compact set is closedand bounded. LetF be a compact subset ofC(X) and letε > 0 be given. Then, bycompactness,F has a finite cover byε/3-balls, say centered atf1, . . . , fn. Moreover,for anyx ∈ X there isδ > 0 such that, ifd(x, x′) < δ, thend(fj(x), fj(x′)) < ε/3for each of the (finitely many) functionsfj , j = 1, . . . , n. For anyf ∈ F there is somefj such thatd(f, fj) < ε/3 and it follows that

d(f(x), f(x′)) 6 d(f(x), fj(x)) + d(fj(x), fj(x′)) + d(fj(x′)), f(x′)) < ε

as required.

13

Lecture 10Normed Vector Spaces

Exercise10.1. Let a andb be positive real numbers and letp > 1. Show that

inft1−pap + (1− t)1−pbp : t ∈ (0, 1) = (a + b)p.

Hence (or otherwise) show that the expression

‖x‖p = (|x1|p + · · ·+ |xn|p)1/p

is also a norm onRn or Cn.

Solution. To prove the first statement, one can use standard calculus methods to findthe minimum of the function att = a/(a + b). Alternatively, argue as follows: thefunction x 7→ xp, p > 1, is convex (the graph always lies below its chords) andtherefore

(a + b)p =[ta

t+ (1− t)

b

1− t

]p

6 t(a

t

)p

+ (1− t)(

b

1− t

)p

= t1−pap + (1− t)1−pbp

with equality att = a/(a + b).Now to prove that‖·‖p is a norm, the only nontrivial thing is the triangle inequality.

Let x andy be vectors. We have fort ∈ (0, 1)

‖x + y‖pp 6

∑i

(|xi|+ |yi|)p

6∑

i

(t1−p|xi|p + (1− t)1−p|yi|p

)6 t1−p‖x‖p

p + (1− t)1−p‖y‖pp.

Taking the infimum over allt we find that

‖x + y‖pp 6 (‖x‖p + ‖y‖p)

p,

and takingp’th roots gives the result. (See L. Maligranda,Simple proof of the Holderand Minkowski inequalities, American Mathematical Monthly102(1995), 256–259.)

Exercise10.2. Show that, with the above norm, the collectionB(V,W ) of boundedlinear maps fromV to W is a normed vector space, and that it is a Banach space ifWis a Banach space.

14

Solution. We prove the triangle inequality for the norm onB(V,W ) (the other axiomsare easier). LetS, T ∈ B(V,W ). Then

‖S + T‖ = sup‖Sv + Tv‖ : ‖v‖ 6 16 sup‖Sv‖+ ‖Tv‖ : ‖v‖ 6 16 sup‖Sv‖ : ‖v‖ 6 1+ sup‖Tv‖ : ‖T‖ 6 1= ‖S‖+ ‖T‖.

As for completeness, let us re-examine our proof thatC(X) is complete forX compact.Notice that this proof really shows that the space ofboundedcontinuous functions fromanyX to acompletespaceY is itself complete. But now,B(V,W ) is a closed subspaceof the space of bounded functions from the unit ball ofV to W , so it is complete ifWis complete.

Exercise10.3. Show that the norm of linear maps is submultiplicative under composi-tion, i.e.‖S T‖ 6 ‖S‖‖T‖.

Solution. We have‖STv‖ 6 ‖S‖‖Tv‖ 6 ‖S‖‖T‖‖v‖,

which gives the result.

15

Lecture 11Differentiation

Exercise11.1. Suppose thatT : V → W is a linear map. Show thatT (h) = o(‖h‖) ifand only ifT = 0.

Solution. By definition of the norm, if‖T‖ > 0 then there is a vectorv ∈ V such that‖Tv‖ > 1

2‖T‖‖v‖. By linearity, any nonzero multiple ofv has this same property.Thus it cannot be the case that‖T (h)‖/‖h‖ tends to zero as‖h‖ → 0. We concludeby contraposition that ifT (h) = o(‖h‖) then‖T‖ = 0 and thusT = 0.

Exercise11.2. Show that the derivative of alinear map is always equal to the mapitself.

Solution. Let T : V → W be continuous (thus, bounded) and linear. Then

T (x + h) = T (x) + T · h

which, by definition, shows that the derivative ofT atx is T itself.

Exercise 11.3. (Open-ended) Formulate precisely the advanced calculus result that“the mixed derivatives are symmetric”, that is∂2f/∂x∂y = ∂2f/∂y∂x, and prove it(under suitable hypotheses) for maps between normed vector spaces.

Solution. Let f : V → W be a twice-differentiable map fromV toW , whereV andWare normed vector spaces. The first derivativeDf therefore takes values in the vectorspaceB(V,W ), and the second derivativeD2f takes values inB(V,B(V,W )), whichcan be regarded as the space ofbilinear maps

S : V × V → W

which are bounded in the sense that‖S(v, v′)‖ 6 c‖v‖‖v′‖ for some constantc. Thestatement is that the second derivative, so regarded, issymmetricin the sense thatS(v, v′) = S(v′, v).

Here is a sketch proof. Suppose thatf is twice differentiable at 0, thatf(0) = 0,and thatDf(0) = 0 also. (The general case can easily be reduced to this.) I claim thenthat

f(h + k)− f(h)− f(k) = D2f(0) · h · k + o((‖h‖+ ‖k‖)2).

The result follows from this, since one the one hand a version of exercise 1 shows thatthis determines the bilinear mapD2f(0) uniquely, and on the other hand the left sideis obviously symmetric inh andk.

To prove the claim let us look at the expression

g(k) = f(h + k)− f(h)− f(k)−Df(h) · k

as a function ofk. Clearly it vanishes whenk = 0. Its derivative is

Dg(k) = Df(h + k)−Df(k)−Df(h)

16

differentiating the first, third and fourth terms in the definition ofg (the second term,considered as a function ofk, is constant.) Now we have

Df(p) = D2f(0) · p + o(‖p‖)

by definition of the second derivative and the assumption thatDf(0) = 0. Substitutingthese into the expression forDg above we find thatDg(k) = o((‖h‖ + ‖k‖)). Fromthe mean value theorem, sinceg(0) = 0, we deduce that

‖g(k)‖ 6 ‖k‖o((‖h‖+ ‖k‖)) 6 o((‖h‖+ ‖k‖)2).

On the other hand we have

Df(h) · k −D2f(0) · h · k = o((‖h‖+ ‖k‖)2)

from the definition of the second derivative. Putting that together with the previousinequality and the definition ofg gives

f(h + k)− f(h)− f(k) = D2f(0) · h · k + o((‖h‖+ ‖k‖)2)

as asserted.

17

Lecture 12The inverse function theorem

Exercise12.1. Let V be a normed space and letW be the normed spaceB(V, V ).Show that the mappingi : T 7→ T−1 is differentiable (where defined) onW , and thatits derivative is

Di(T ) ·H = −T−1HT−1.

Solution. This follows from the algebraic identity

S−1 − T−1 = −S−1(S − T )T−1.

That is

i(T + H)− i(T ) = −i(T + H)Hi(T ) = −i(T )Hi(T ) + o(‖H‖).

Theo-estimate follows from the continuity ofi nearT , which we have already proved.

18

Lecture 13Measure Spaces

Exercise 13.1. Show that theσ-algebra of Borel subsets ofR is generated by theintervals(a,∞), with a ∈ Q.

Solution. Let B be theσ-algebra generated by the intervals mentioned in the proposi-tion. ClearlyB contains all the intervals(a, b) with rational endpoints. But any openset inR is the union of all the intervals with rational endpoints contained in it. Sincethis union is necessarily countable, any open subset ofR belongs toB; the result fol-lows.

Exercise13.2. Let An be a monotonedecreasingsequence of measurable sets,A1 ⊇A2 ⊇ · · · , and suppose thatν(A1) is finite. Show that thenν (

⋂An) = lim ν(An).

Give an example to show that the finiteness hypothesis is necessary.

Solution. Supposing thatA1 has finite measure, letBn = A1 \ An. Then theBn area monotone increasing sequence of sets and

⋃Bn = A1 \ A, whereA =

⋂An. By

previous resultsν(A1 \A) = lim ν(Bn), that is

ν(A1)− ν(A) = limn

(ν(A1)− ν(An)

which gives the result.For a counterexample in the infinite measure case takeν to be the counting measure

onN andAn = n, n + 1, . . ..

Exercise13.3. Let An be a sequence of measurable subsets ofX. The limit superiorlim supAn is the set of thosex ∈ X that belong to infinitely many of theAn; the limitinferior lim inf An is the set of thosex ∈ X that belong to all but finitely many of theAn.

(a) Show thatlim inf An andlim sup An are measurable.

(b) Show thatν(lim inf An) 6 lim inf ν(An).

(c) Show thatν(lim sup An) > lim sup ν(An) provided that⋃

An has finite mea-sure.

(See exercise 6.2 for the definitions oflim inf andlim sup for sequences of real num-bers.)

Solution. We can writelim inf An =⋃

k

⋂n>k An; the unions and intersections are

countable, solim inf An is measurable. Similarly forlim sup An =⋂

k

⋃n>k An.

Let Bk =⋂

n>k An. Then theBk increase monotonically tolim inf An, and there-fore ν(lim inf An) = lim ν(Bk). But Bk ⊆ Ak so ν(Bk) 6 ν(Ak) and thereforelim ν(Bk) 6 lim inf ν(Ak). Similarly for the supremum.

19

Lecture 14Integration

Exercise14.1. (Consistency) Verify that a nonnegative simple function is integrable,and that the two senses of “integral” agree for such functions.

Solution. Since the integral is a positive linear functional on the space of simple func-tions, if g 6 f with g andf simple, then

∫gdν 6

∫fdν, both integrals being taken in

the sense of simple functions. On the other hand,f itself is a simple function6 f . Itfollows that

sup∫

gdν : g simple,g 6 f =∫

fdν

as asserted.

Exercise14.2. Verify that if f, g are nonnegative measurable functions withf 6 g,then

∫fdν 6

∫gdν. Verify also that

∫cfdν = c

∫fdν for a positive constantc.

Solution. If f 6 g, then every simple function6 f is also6 g; this gives the firstresult. As for the second,g is a simple function6 f if and only if cg is a simplefunction6 cf .

Exercise14.3. Verify that the integral of the characteristic function of any measurableA ⊆ X is equal to the measure ofA. (With our definitions, you will need to assumethat the measure isσ-finite.)

Solution. If A has finite measure, its characteristic function is a simple function andthe statement is obvious. IfA has infinite measure, letXn be an increasing sequenceof finite measure subsets whose union isX. ThenA ∩ Xn increases toA, soν(A ∩Xn) →∞. The characteristic functions of theA∩Xn are nonnegative simple functionsdominated by the characteristic function ofA, and their integrals areν(A∩Xn) →∞;so the integral ofχA is∞ also.

20

Lecture 15Integrable functions and the convergence theorems

Exercise 15.1. Show that the integrable functions form a vector space and that theintegral is a linear functional on this vector space.

Solution. If f1 = g1 − h1 andf2 = g2 − h2 are integrable, so isf1 + f2 = (g1 +g2)− (h1 + h2). Also, if c > 0 is a constant,cf1 = cg1 − ch1 is integrable; ifc < 0,cf1 = (−c)h1 − (−c)g1 is still integrable. In each case it is easy to check that theintegral behaves linearly.

Exercise15.2. Give a sensible definition of the integral forcomplex-valued functions.

Solution. The only sensible definition is that iff = u + iv, with u, v real-valued, then∫f =

∫u + i

∫v.

Exercise15.3. Show that ∣∣∣∣∫ fdν

∣∣∣∣ 6∫|f |dν

for any integrable functionf .

Solution. Write f = f+ − f−, as in the proof of lemma 15.2, and notice then that theleft side of the inequality is|

∫f+ −

∫f−| and the right side is

∫f+ +

∫f−. The

ordinary triangle inequality now gives the result.It is worth noticing that this inequality is also true forcomplexf . Suppose that

f = u + iv with u, , v real. Neither side of the inequality is affected if we multiplyf by a complex numberw of absolute value 1, and by multiplying by a suitable suchw we may arrange that

∫fdν is real and positive, which is to say that

∫udν > 0 and∫

vdν = 0. Under this hypothesis∣∣∣∣∫ fdν

∣∣∣∣ =∫

udν 6∫|f |dν

sinceu 6 |f | pointwise.

Exercise15.4. Let fn be a sequence of nonnegative integrable functions on a measurespace. Show that if

∞∑n=1

∫fndν < ∞,

thenfn(x) → 0 for almost allx.

Solution. Let gm =∑m

n=1 fn be the partial sums. By the monotone convergencetheoremgm(x) converges a.e. tog(x) (andg is an integrable function, but that doesnot matter here). Thusfn(x) = gn+1(x)− gn(x) converges a.e. to zero.

Exercise15.5. If a nonnegative integrable functionf has∫

fdν = 0, show thatf(x) =0 for almost allx. (Apply the MCT to the sequencenf.)

21

Solution. The hint is the solution, or equivalently apply the preceding exercise to∑∞n=1 f(x) (a sum of infinitely many copies of the same function).

22

Lecture 16Functions and Spaces

Exercise16.1. Give counterexamples to other possible implications among notions ofconvergence. That is, give examples to show that neither convergence in measure, norconvergence a.e., necessarily imply convergence inL1, and that convergence a.e. doesnot imply convergence in measure in caseν(X) = ∞.

Solution. Let X be the set of natural numbers equipped with the measure that assignsmass2−k to the numberk. The functionsfn defined by

fn(k) =

2k (k = n)0 (k 6= n)

then tend to 0 in measure, and pointwise (everywhere), but not inL1.Let Y be the set of natural numbers with the counting measure. Then the functions

gn defined by

gn(k) =

1 (k = n)0 (k 6= n)

tend to zero pointwise, but not in measure.

Exercise16.2. Show that iffn → f in measure and|fn| 6 g for some integrableg,thenfn → f in L1.

Solution. Sincefn → f in measure, any subsequence converges in measure tof ,which means that it has a sub-subsequence converging almost everywhere. By thedominated convergence theorem, this sub-subsequence converges inL1. Thus, everysubsequence of(fn) has a sub-subsequence converging inL1. Now use the fact (truein any metric spaceX) that if xk is a sequence for which any subsequence has a sub-subsequence converging tox, thenxk itself converges tox. (Proof of fact: Supposexk does not converge tox. Then there is someδ > 0 for which there exist arbitrarilylargek with d(xk, x) > δ. The correspondingxk form a subsequence with no sub-subsequence converging tox.)

23

Lecture 17Constructing Measures I

Exercise17.1. Show that any set of outer measure 0 is measurable.

Solution. Let A have outer measure zero. Then, for anyB, B ∩ A ⊆ A has outermeasure zero by monotonicity. Thus

µ∗(B) 6 µ ∗ (B ∩A) + µ∗(B \A) = µ∗(B \A) 6 µ∗(B);

we must have equality all through, soB is measurable.

Exercise17.2. LetX be an uncountable set and forA ⊆ X defineµ∗(A) =

0 if A is countable

1 if A is uncountable.

Show thatµ∗ is an outer measure. Which subsets ofX areµ∗-measurable and what istheir measure?

Solution. It is easy to check thatµ∗ is an outer measure (use the fact that a countableunion of countable sets is countable). By the previous exercise any countable set ismeasurable (with measure 0), and the complement of a countable set is also measur-able (with measure 1). If bothA and its complement are uncountable, thenA is notmeasurable, sinceµ∗(A) + µ∗(X \A) = 2 6= µ∗(X).

Exercise17.3. Let (X,B, ν) be aσ-finite measure space. Show thatµ∗(A) = infν(B) :A ⊆ B,B ∈ B defines an outer measure onX, and that measure produced fromµ byCaratheodory’s construction agrees withν onB. Deduce thatX can be extended to acomplete measure space.

Solution. Clearlyµ∗ is monotonic. It is also countably subadditive because ifAn ⊆Bn, Bn ∈ B, thenA =

⋃An is a subset ofB =

⋃Bn, andν(B) 6

∑ν(Bn). If

B ∈ B thenµ∗(B) = ν(B) from the definition. We show thatB is measurable. LetCbe any set and letε > 0. There existsB′ ∈ B such thatC ⊆ B′ andµ∗(C)+ε > ν(B′).Now

ν(B′) = ν(B′ ∩B) + ν(B′ \B);

the first set on the right containsC ∩B and the second containsC \B, so we deduce

µ∗(C) + ε > ν(B) > µ∗(C ∩B) + µ∗(C \B).

Let ε → 0 to obtainµ∗(C) > µ∗(C ∩B) + µ∗(C \B).

Together with subadditivity, this gives the result.

24

Lecture 18Lebesgue Measure

Exercise18.1. Show that for ameasurablesubsetA of R (having finite measure) andany ε > 0, there is a closed setF ⊆ A with λ(A) < λ(F ) + ε. (The measurabilityrequirement cannot be dropped — this is tricky to see.)

Solution. SinceA is measurable and the setsAn = A ∩ [−n, n] form an increasingsequence whose union isA, there is somen such thatλ(An) > λ(A) − ε/2. LetB = [−(n+1), (n+1)]\An. By the previous proposition there is an open setU ⊇ Bsuch thatλ(U) < λ(B) + ε/2. ThenF = [−(n + 1), (n + 1) \ U is a closed setcontained inAn, and

λ(F ) > (2n + 2)− λ(U) > (2n + 2)− λ(B)− ε/2 > λ(An)− ε/2 > λ(A)− ε

as required.

Exercise18.2. Show that a subsetA ⊆ [a, b] is Lebesgue measurable if and only if itsouter measureλ∗(A) is equal to itsinner measureλ∗(A) := (b− a)− λ∗([a, b] \ A).(This connects the definition of Lebesgue measure to the classical ‘exhaustion method’of Eudoxus and Archimedes.)

Solution. Suppose thatA satisfies the given condition. By the previous proposition,there is a Borel setG with A ⊆ G andλ∗(A) = λ(G). If we can prove thatN = G\Ahas (outer) Lebesgue measure zero, we shall be finished. PutX = [a, b].

Write the definition of measurability forG, usingX \A as a test set. This gives

λ∗(X \A) = λ∗(N) + λ∗(X \G).

But λ∗(X \A) = λ(X)−λ∗(A) because outer and inner measures agree forA; and thecorresponding fact forG is true becauseG is measurable. Soλ∗N) = λ(G)−λ∗(A) =0, as required.

25

Lecture 19Lebesgue Integration

Exercise19.1. Let f(x) = sin xx . Use the alternating series test from calculus to show

that limR→∞∫ R

0f(x)dx exists (later, we shall see that the value of this limit isπ/2).

Nevertheless, show thatf is not Lebesgue integrable.

Solution. Let

an =∫ (n+1)π

nπ

sinx

xdx.

Then the signs of thean are alternately positive and negative, and we also have|an| <frm[o]−−/n. Thus, by the alternating series test, the series

∑an converges. Since

the partial sums of this series are the integrals∫ nπ

0(sinx)/x dx, it follows that these

integrals converge asn → ∞. Now for anyR we can find ann (the greatest integer< R/π) such that ∣∣∣∣∣

∫ R

0

sinx

xdx−

∫ nπ

0

sinx

xdx

∣∣∣∣∣ <1

R− π

and it follows that∫ R

0(sinx)/x dx converges asR →∞.

If the function(sinx)/x were Lebesgue integrable, however, then the series∑

an

would beabsolutelyconvergent. The estimate

|an| >1

(n + 1)π

∫ (n+1)π

nπ

| sinx|dx =2

(n + 1)π

shows that this is not the case.

Exercise19.2. (“Differentiating under the integral sign”) Suppose that the functionF (x, y) is defined onR × R, is integrable (as a function ofx) for each fixedy, andis differentiable (as a function ofy) for each fixedx; suppose also that its (partial)derivativeFy satisfies

|Fy(x, y)| 6 g(x)

for some integrable functiong and ally. Show that then the function

G(y) =∫

F (x, y)dx

is differentiable and thatG′(y) =∫

Fy(x, y)dx. (Same method as Example??, to-gether with the mean value theorem.)Exercise19.3. (Normal numbers) Define functionsfn on [0, 1] as follows:

fn(x) =

−1 if b2nxc is even

+1 if b2nxc is odd

26

Show that ∫ 1

0

fn(x)fm(x)dx =

1 (m = n)0 (m 6= n)

and deduce that ifgn = (f1 + · · · + fn)/n, then∫

g2n(x)dx = 1/n. Conclude using

Exercise 15.4 thatgk2(x) → 0 for almost allx ask → ∞, and hence thatgn(x) → 0for almost allx asn →∞. This conclusion is usually expressed by saying that ‘almostall x ∈ [0, 1] are normal in base 2’, i.e. have on average as many 0’s as 1’s in their binaryexpansion. Can you show that almost allx are normal ineverybase simultaneously?(No explicit example of such anx is known.)

Solution. If n = m thenfnfm = 1 everywhere and so its integral is equal to 1. Ifn > m then the integral offn is zero over every subinterval[k2−m, (k + 1)2−m] onwhichfm is constant; so

∫fnfm = 0. Similarly if m > n.

Now we can write ∫g2

n =1n2

n∑i,j=1j

∫fifj

.

The terms withi 6= j vanish and then terms withi = j yield 1, so∫

g2n = 1/n. Thus∑

k

∫(gk2)2 < ∞, sogk2(x) → 0 for almost allx by the Borel-Cantelli lemma.

To deduce thatgn → 0 almost everywhere letk be the greatest integer such thatk2 6 n. Then

|gn(x)− gk2(x)| 6 (1− k2n) |gk2(x)|+ 2k + 1k2

.

This tends to zero asn → ∞, andg2k(x) tends to zero (a.e.) asn → ∞, so we are

done.A generalization of the same argument shows that almost allx are normal to any

given basem. Since a countable union of null sets is null, it follows that almost allxareperfectly normal, i.e. normal in every base simultaneously.

27

Lecture 20Set-Theoretic Questions

28

Lecture 21The Riesz Representation Theorem

Exercise21.1. Show that Lebesgue outer measure is metric.

Solution. Let A,B ⊆ R be separated and letU, V be disjoint open sets containingU, V respectively. By regularity, givenε > 0 there is an open setW containingA ∪Bwith λ(W ) < λ∗(A∪B)+ε. ThenW ∩U andW ∩V are disjoint open sets containingA, B respectively; so

λ∗(A) + λ∗(B) 6 λ(W ∩ U) + λ(W ∩ V ) 6 λ(W ) 6 λ∗(A ∪B) + ε.

Let ε → 0 to getλ∗(A ∪B) > λ∗(A) + λ∗(B).

The opposite inequality follows from subadditivity, so Lebesgue outer measure is met-ric.

Exercise21.2. The measuresν constructed by the Riesz representation theorem havevarious ‘regularity’ properties. Verify that for everyν∗-measurable setE and everyε > 0 there exist an open setU and a closed setK with K ⊆ E ⊆ U andν(U \K) < ε.Deduce that everyν∗-measurable set differs from a Borel set by a set ofν∗-measurezero. Prove also that the continuous functions form a dense subspace of the normedvector spaceL1(X, ν).

Solution. The first part is a general theorem for finite Borel measures, see ExtendedHW 2.

If now E is measurable there exist open setsUn ⊇ E with ν(Un \ E) 6 1/n. TheintersectionG of all theUn is a Borel set containingE andν(G \ E) = 0.

Let E be a Borel set. Forε > 0 chooseK ⊆ E ⊆ U , K closed,U open withν(U \K) < ε. Choose a continuous,[0, 1]-valued functionφ supported inU and equalto 1 onK. Then∫

X

|φ− χE |dν =∫

U\K|φ− χE |dν 6 ν(U \K) 6 ε.

It follows thatχE is anL1 limit of continuous functions. This extends to measurableEsince any suchE differs from a Borel set by a set of measure zero. But now it followsthat every simple function is anL1 limit of continuous functions; and simple functionsare dense inL1 by construction.

Exercise21.3. Show that we have uniqueness in the Riesz representation theorem: twoBorel measures that determine the same linear functional are the same. (Suggestion:LetA be the class of sets on which the two measures agree. Show thatA is aσ-algebraand contains the open sets.

29

Lecture 22Lebesgue-Stieltjes Measures

Exercise22.1. TakeI = [−1, 1] and letg(x) =

0 (x < 0)1 (x > 0)

. What is the corre-

sponding Stieltjes measure? Does the value ofg at 0 affect your answer?

Solution. The Stieltjes measure associated with the distribution functiong is the unitmass at the origin (that is,

∫fdg = f(0)). To see this, notice that from the definition,

if f1 = f2 in a neighborhood of 0, then∫

f1dg =∫

f2dg. Now given anyε > 0 wecan find a constant functionc = f(0) such that|f − c| < ε in a neighborhood of theorigin. Then ∫

fdg =∫

cdg + O(ε) = c + O(ε)

since integration is a continuous linear functional. Becauseε is arbitrary we get∫

fdg =c = f(0). The choise ofg at 0 does not affect the answer.

Exercise22.2. With dg a Stieltjes measure as above, show that

dg(c, d) = g(d−)− g(c+)

for every open interval(c, d) ⊆ [a, b]. (Hereg(d−) = limxd g(x) andg(c+) =limxc g(x); the limits exist becauseg is monotone.) Find corresponding formulae forclosed and half-open intervals. Hence show that ifg is continuous from the right, orfrom the left, thendg determinesg uniquely up to an additive constant.

Solution. Let fn be the continuous function defined as follows:fn(x) = 0 for x /∈(c + 1/n, d − 1/n), fn(x) = 1 for x ∈ [c + 2/n, d − 2/n], andfn is linear on[c+1/n, c+2/n] and[d−2/n, d−1/n]. Thenfn is a monotone sequence of continuousfunctions increasing to the characteristic function of(c, d), so

∫fndg → dg(c, d).

From the definition of the Stieltjes integral we have

g(d− 2/n)− g(c + 2/n) 6∫

fndg 6 g(d− 1/n)− g(c + 1/n),

and asn → ∞ both sides tend tog(d−) − g(c+). This gives the result. A similarargument for other intervals givesdg[c, d] = g(d+)−g(c−), dg(c, d] = g(d+)−g(c+),and so on.

If g is right continuous theng(x) = g(x+) for all x. Thusdg(c, d] = g(d)− g(c),sodg determinesg up to an additive constant. Similarly for left continuity.

Exercise22.3. Conversely, suppose thatν is a finite Borel measure on[a, b], such thatνa = 0. Defineg(x) = ν(a, x] (with g(a) = 0). Show thatν = dg. Thus everyBorel measure on[a, b] arises from the Stieltjes construction.

30

Solution. The functiong so defined is right continuous (by the monotone convergencetheorem). Thusdg(A) = ν(A) wheneverg is a half-open interval(a, x]. The collectionof setsA on which the two measures agree contains[a, b] and is closed under countableunions and set-theoretic difference, so it is aσ-algebra. Since the half-open intervalsabove generate theσ-algebra of all Borel sets, sodg(A) = ν(A) for all Borel setsA.

Exercise22.4. Show that ifg is continuous, thendg(E) = λ(g(E)) for any BorelsubsetE of [a, b].

Solution. By the last exercise but one,dg(E) = λ(g(E)) wheneverE is an interval.Now both sides are measures and we can argue as in the previous exercise.

Exercise22.5. Let f : [0, 1] → R be a function. One says thatf is absolutely contin-uousif, for any ε > 0, there isδ > 0 with the following property: given any (finite orcountable) collection[ak, bk] of disjoint closed subintervals of[0, 1],

if∑

k

|ak − bk| < δ, then∑

k

|f(ak)− f(bk)| < ε.

Show that every absolutely continuous function is continuous and of bounded varia-tion. Show that the Cantor function?? is continuous and of bounded variation, butnotabsolutely continuous.

31

Lecture 23Signed measures

Exercise23.1. Let ν1, ν2, andµ be finite measures (on the sameσ-algebra). Supposethatν1 andµ are mutually singular, and thatν2 andµ are mutually singular. Prove thatthenν1 + ν2 andµ are mutually singular.

Solution. Suppose that we have disjoint decompositions

X = A1 tB1 = A2 tB2

with µ(A1) = µ(A2) = ν1(B1) = ν2(B2) = 0. PutA = A1 ∪ A2, B = B1 ∩ B2.ThenX = A ∪ B is a disjoint decomposition,µ(A) 6 µ(A1) + µ(A2) = 0, and(ν1 + ν2)(B) = ν1(B) + ν2(B) = 0. Thusµ andν1 + ν2 are mutually singular.

32

Lecture 24Absolute continuity

Exercise24.1. Let A be theσ-algebra of Lebesgue measurable subsets of[0, 1], letµ be the counting measure onA, and letλ be Lebesgue measure. Show thatλ isabsolutely continuous with respect toµ. Does the Radon-Nikodym theorem apply? Ifit does, find the R–N derivativedλ/dµ; if it does not, say what goes wrong.

Solution. The only set ofµ-measure zero is the empty set, which certainly has Lebesguemeasure zero. However, the Radon-Nikodym theorem does not work here. If it did,there would be a (positive) real-valued functionf such that

λ(E) =∑x∈E

f(x).

This is impossible: since every one-point set has zero Lebesgue measure, we wouldhave to havef(x) = 0 for all x. The problem is that the measureµ is not finite or evenσ-finite.

Exercise24.2. Show that ifν is bothabsolutely continuousandmutually singular withrespect toµ, then it is zero. (This gives the uniqueness in (b) above.)

Solution. Takeν to be a positive measure (otherwise, apply the following argument toits absolute value). LetX = A ∪B with ν(A) = µ(B) = 0. If ν µ thenν(B) = 0also, and soν(X) = 0.

Exercise24.3. Show that every signed measure is absolutely continuous with respectto its own absolute value. What can you say about the Radon-Nikodym derivative inthis case? How does it relate to the Hahn decomposition?

Solution. Let ν be a signed measure and letX = P ∪ N be a Hahn decomposition,so thatν+(E) = ν(E ∩ P ) andν−(E) = −ν(E ∩ N) are positive measures and|ν| = ν+ + ν−. Then

ν(E) = |ν|(P ∩ E)− |ν|(N ∩ E) =∫

E

(χP − χN )d|ν|

which shows thatν is absolutely continuous wrt|ν| and the Radon-Nikodym derivativedν/d|ν| equalsχP − χN .

33

Lecture 25Inner Products andL2(X, µ)

Exercise25.1. Show that the simple functions form a dense subspace ofL2(X, µ) forany (σ-finite) measure space(X, µ).

Solution. Let f ∈ L2(X, µ). Definefn(x) to equal±2−nb2n|f(x)|c (choosing thesame sign as that off(x)) if −n 6 f(x) 6 n, and otherwise to equal 0. Thenfn isa simple function,|fn| 6 |f |, andfn → f pointwise. By the dominated convergencetheorem, then,

‖fn − f‖2 =∫|fn − f |2dµ → 0

since the integrand is dominated by the integrable function4|f |2.

Exercise25.2. If µ is a finite Borel measure on the compact spaceX, show that thecontinuous functions are dense inL2(X, µ).

Solution. Let E be a Borel set. By regularity, given anyε > 0 there exist a compactK and an openU with K ⊆ E ⊆ U andµ(U \K) < ε. Pick a[0, 1]-valued functionf equal to 1 onK and supported withinU . Thenf − χE vanishes except on a set ofmeasureε, where it has absolute value6 1, and thus‖f − χE‖ 6 ε1/2. We concludethat every characteristic function of a Borel set, and therefore every simple function, isa limit of continuous functions. Since the simple functions are dense by the precedingexercise, the result follows.

34

Lecture 26The Geometry of Hilbert Space

Exercise26.1. In the Hilbert spaceL2[−1, 1] (with respect to Lebesgue measure) showthat the setV of functionsf that satisfyf(x) = f(−x) a.e. is a closed subspace, andfind its orthogonal complement.

Solution. The setV of functionsf satisfyingf(x) = f(−x) almost everywhere iscertainly a subspace. To see that it is closed, consider the expression

Φ(f) =∫ 1

−1

|f(x)− f(−x)|2dx.

By Cauchy-Schwarz,Φ(f) depends continuously onf , and the subspace we want isexactlyΦ−10, so it is closed.

The orthogonal complement ofV is the setW of functionsg such thatg(x) =−g(−x) almost everywhere. To prove this note thatW is a closed subspace (by thesame reasoning) and thatV andW are orthogonal (iff ∈ V andg ∈ W then

∫ 0

−1fg =

−∫ 1

0fg, so 〈f, g〉 = 0. Finally, every function is the sum of an even and an odd

function, soH = V ⊕W and thusV andW are orthogonal complements.

Exercise26.2. Let S be any subset of a Hilbert space. Show thatS⊥⊥ is theclosedlinear spanof S, that is, the smallest closed subspace that containsS.

Solution. For any subsetS, S⊥ is a subspace, and it is closed because it is the inter-section of the closed setsx : 〈x, s〉 = 0 for all thes ∈ S. MoreoverS⊥⊥ containsS, because anything inS is orthogonal to anything that is orthogonal to everything inS. ThusS⊥⊥ is a closed subspace containingS. To see that it is the smallest suchsubspace, supposeV is another one. ThenV containsS, soV ⊥ is contained inS⊥

and thusV ⊥⊥ containsS⊥⊥. But, becauseV is a closed subspace by assumption, theprojection theorem shows thatV ⊥⊥ = V . The result follows.

Exercise26.3. Show that‖T ∗T‖ = ‖T‖2 for any bounded operatorT on a Hilbertspace.

Solution. Write

‖Tx‖2 = 〈Tx, Tx〉 = 〈T ∗Tx, x〉 6 ‖T ∗T‖‖x‖2

using Cauchy-Schwarz at the last step. Taking the supremum overx in the unit bal;lgives

‖T‖2 6 ‖T ∗T‖.

On the other hand‖T ∗T‖ 6 ‖T ∗‖‖T‖ = ‖T‖2

so we have equality all through, which is what is needed.

35

Exercise26.4. Show that a bounded operatorU on a Hilbert spaceH gives an isometryfrom H onto itself if and only ifU∗U = I = UU∗ (in this case we sayU is unitary).What is the corresponding condition for an operatorV to give an isometry ofH into(but not necessarilyonto) itself?

Solution. Let V be an isometry. Then‖V x‖2 = ‖x‖2 for all x, which is to say that〈(V ∗V − I)x, x〉 = 0 for all x. Let T = V ∗V − I. Thepolarization identity

〈T (x + y), x + y〉 − 〈T (x− y), x− y〉 = 2〈Tx, y〉+ 2〈x, Ty〉 = 4<〈Tx, y〉

(together with the similar identity withy replaced byiy) shows that〈Tx, y〉 = 0 forall x, y and thus thatT = 0. ThusV ∗V = I if V is an isometry. IfV is in additiononto, then it is bijective and its inverse is also an isometry. But the identityV ∗V = Ishows that its inverse (if it exists) must beV ∗; and thenV V ∗ = I sinceV ∗ is anisometry.

Exercise 26.5. Let (X, µ) be a measure space and letf be a bounded measurablefunctionX → C. Show that themultiplication operatorMf defined by

Mf (g) = fg

is bounded onL2(X, µ). What is the adjoint ofMf ?

Solution. Suppose that|f | is bounded by the constantC. Then

‖Mf (g)‖2 =∫|f |2|g|2 6 C2

∫|g|2 = C2‖g‖2

soMf is bounded, with norm at mostC. The adjoint ofMf is Mf , since

〈Mfg, h〉 =∫

f gh =∫

gfh = 〈g,Mfh〉.

36

Lecture 27Proof of the Radon-Nikodym Theorem

Exercise27.1. Let g be a real-valued function inL1(X, ν). If∫

Ag > 0 for every

measurable setA, theng > 0 almost everywhere.

Solution. Let An be the set of points whereg 6 −1/n. Then∫

Angdν 6 1ν(An)/n.

Since by hypothesis this integral is non-negative,ν(An) = 0. But then the union⋃

An

also has measure 0, and this is the set of points whereg < 0.

Exercise27.2. Formulate and prove a version of the Radon-Nikodym theorem that isvalid when bothµ andν areσ-finite.

Solution. Let µ andν beσ-finite (positive) measures and suppose thatnu µ. Thenthere is a locallyµ- integrable functionf such that

ν(E) =∫

E

fdµ

for all measurable setsE. Here ‘locally integrable’ means thatf is µ-integrable overeach set of finiteν-measure.

For the proof, representX as a countable disjoint union of setsXn of finite νmeasure and apply the previous version of the Radon-Nikodym theorem to each one.One obtains functionsfn ∈ L1(Xn, µ) such that

ν(E) =∫

E

fndµ (E ⊆ Xn).

Let f(x) = fn(x) if x ∈ En. Then for every measurableE,

ν(E) =∑

n

ν(E ∩Xn) =∑

n

∫E∩Xn

fdµ =∫

E

fdµ

using the Monotone Convergence Theorem for the last step.

37

Lecture 28Orthonormal Bases and Fourier Series

Exercise28.1. Prove that a Hilbert space is separable if and only if it admits a count-able complete orthonormal set.

Solution. Suppose thatH is separable and letx1, x2, . . . be a countable dense set.We define an inductive process for producing an orthonormal set as follows:S0 = ∅,and, given an orthonormal setSn (having at mostn elements), letm be the least indexsuch thatxm does not belong to the span ofSn (if no suchm exists, stop), letym bethe orthogonal projection ofxm ontoSm, and letsn = (xm − ym)/‖xm − ym‖; thensn is a unit vector orthogonal to the span ofSn, and thusSn+1 = Sn ∪ sn is anorthonormal set. Having completed this induction, letS =

⋃Sn; S is an orthonormal

set whose span includes all thexn, so its closed span is all ofH.Conversely, ifS admits a countable complete orthonormal set it is either finite-

dimensional or isomorphic to2(N), both of which spaces are separable (the sequenceshaving rational values only finitely many of which are nonzero form a countable densesubset of 2(N)).

Exercise28.2. Part of the complication of the above proof arises from the fact that wehave not yet shown that the continuous functions are dense inL2. Prove this (for anyfinite Borel measure on a compact metric space).

Solution. I’ve now put this in as exercises for Lecture 25.

38

Lecture 29Pointwise results about Fourier Series

Exercise29.1. Show that the Fourier series of the functionf(x) = |x|1/3 convergesto zero at the origin (despite the fact that it is not differentiable there).

Solution. The quotient|f(x)/x| = |x|−2/3 is integrable, so the previous propositionapplies.

Exercise29.2. Suppose thatf is aC2 function on some subinterval(a, b) of [−π, π].Show that the Fourier series off converges tof uniformly on compact subsetsof (a, b).

Solution. For any compact subintervalI of (a, b) we can writef = f1 + f2, wheref1 is a globallyC2 function supported in(a, b) andf2 vanishes on some open setUcontainingI. By integrating by parts twice we find that the Fourier coefficients off1

are of orderO(n−2), so the Fourier series off1 converges uniformly by Weierstrass’M -test. What aboutf2? The proof that its Fourier series converges onI depends onthe Riemann-Lebesgue lemma. Now, if one examines the proof of the R–L lemma, onesees that uniformity holds in the following sense: for anycompactsubsetF of L1, theFourier coefficients tend to zerouniformly for f ∈ F . (For, givenε > 0, one can findfinitely manytrigonometric polynomials such that everyf ∈ F lies within ε of one ofthem.) In our situation we want to apply the R–L lemma to functions of the form

f2(x)sin( 1

2 (x− y), y ∈ I.

By the Arzela-Ascoli theorem these functions form a compact set inC[−π, π], andthusa fortiori in L1.

Exercise29.3. The Cesaro meansof the Fourier series forf are the averages of thepartial sums,1n (S1(f) + · · · + Sn(f)). Show that for any continuous functionf theCesaro means converge uniformly tof . (Construct a kernel, called theFejer kernelFn,such that

12π

∫Fn(x− y)f(y)dy

represents then’th Cesaro mean of the Fourier series forf . Show thatFn is apositivefunction.)

Solution. The same kind of reasoning as before shows that the Cesaro means of theFourier series are represented by the expression above, where

FN (t) =N∑

n=−N

(1− |n|

N + 1

)eint.

Recall that

sin2(t/2) =−e−it + 2− eit

4.

39

Using this we compute

sin2(t/2)FN (t) =1

N + 1−e−i(N+1)t + 2− ei(N+1)t

4= sin2((N + 1/2)t).

Thus the Fejer kernels are indeed positive, and the integral of the Fejer kernel is equalto 2π. Moreover, for anyδ > 0, ∫

|t|>δ

FN (t)dt → 0

asN →∞.Sincef is continuous, for anyε > 0 there isδ > 0 such that|x − y| < δ implies

|f(x) − f(y)| < ε. There is alsoN > 0 such that∫|t|>δ

FN (t)dt < ε. Thus, splittingthe range of integration,∣∣∣∣∫ FN (x− y)(f(x)− f(y))dy

∣∣∣∣ 6 2ε sup ‖f‖+ 2πε

where the first term estimates the integral over|x − y| > δ and the second estimatesthe integral over|x− y| < δ. Thus

∫FN (x− y)(f(x)− f(y))dy tends to zero, which

is to say that the Cesaro means of the Fourier series tend tof(x).

40

Lecture 30Coverings and Maximal Inequalities

Exercise30.1. Show that the Hardy-Littlewood maximal function is measurable. (Hint:it suffices to consider intervals with rational endpoints.) Show that it is usually not in-tegrable.

Solution. For each pair of rationalsa, b with a < b, definefa,b(x) to be(b−a)−1∫ b

a|f(y)|dy

if a < x < b, and 0 otherwise. Eachfa,b is measurable (it’s a multiple ofχ(a,b)) andMf is the supremum of thefa,b. ThusMf is measurable.

It is not integrable in general. In fact, iff is nonzero then there is some interval[−r, r] such that

∫ r

−r|f(y)|dy = a > 0. Then forx > r, Mf(x) > a/(r + x + 1) by

considering the interval(a, b) = (r − 12 , x + 1

2 ). But the functionx 7→ 1/(r + x + 1)is not integrable.

Exercise30.2. Show that the functions of weakL1 type form a vector space.

Solution. The only difficulty is to show that the sum of two functions of weakL1 typeis again of weakL1 type. For a functionf let Ar(f) = x : |f(x)| > r. If both |f |and|g| are less than or equal tor/2, then|f + g| is less than or equal tor. This gives

Ar(f + g) ⊆ Ar/2(f) ∪Ar/2(g).

Suppose thatf andg are of weakL1 type, with constantsC andD respectively. Then

λ(Ar(f + g)) 6 λ(Ar/2(f)) + λ(Ar/2(g)) 6 2C/r + 2D/r,

showing thatf + g is also of weakL1 type.

Exercise30.3. Let E be a subset ofR. A collectionI of closed, nondegenerate inter-vals is called aVitali coverof E if for everyx ∈ E and everyε > 0 there is an intervalI ∈ I of length less thanε and containingx. Prove that, from any Vitali cover, one canselect a finite or countable disjoint subfamily that covers all ofE except possibly for aset of Lebesgue measure zero. (Apply the previous result inductively.)

Solution. From the previous result, one can findfinitelymany disjoint intervalsI01 , . . . , I0

n

fromI whose total measure is at leastλ(E)/6. Consider nowE1 = E\(I01∪· · ·∪I0

n0).

Because the union of finitely many closed sets is closed, each point ofE1 has a neigh-borhood that does not meet(I0

1 ∪· · ·∪I0n0

). Thus, the collection of allI ∈ I that do notmeet(I0

1 ∪ · · · ∪ I0n0

) is a Vitali cover ofE1. Now there are finitely many ofthesein-tervals, sayI1

0 , . . . , I1n1

whose total measure is at leastλ(E1)/6. Continue inductivelyin this way producing families on intervalsIn

j . At then’th stage the total measure yetuncovered is at most(5/6)nλ(E); so the whole family ofIn

j cover all ofE except aset of measure zero.

Exercise 30.4. Prove that an arbitrary (even uncountable) union of nondegenerateclosed intervals is Lebesgue measurable. (Use the previous exercise.)

41

Solution. Let E be such a union. By the previous result it is comprised of a countableunion of closed intervals (which is a Borel set) together with a set of Lebesgue measurezero. Therefore, it is Lebesgue measurable.

42

Lecture 31The Lebesgue Differentiation Theorem

Exercise31.1. Let E be a (measurable) subset ofR. A point p is apoint of densityofE if the limit

limI3p

λ(I ∩ E)λ(I)

exists and equals 1. Show that ifE has positive measure, then almost every point ofEis a point of density ofE.

Solution. Apply the preceding proposition to the characteristic functionf = χE . Ifx ∈ E then ∫

I

|f(y)− f(x)|dy = λ(I \ E) = λ(I)− λ(I ∩ E).

Thus1

λ(I)

∫I

|f(y)− f(x)|dy = 1− λ(I ∩ E)λ(I)

and the result follows.

43

Lecture 32Product Measures

Exercise32.1. Prove associativity for the product measure construction: if(Xi,Ai, µi),i = 1, 2, 3, are finite (orσ-finite) measure spaces, then

(µ1 × µ2)× µ3 = µ1 × (µ2 × µ3).

As a result, it makes unambiguous sense to speak of the product of any finite numberof measure spaces.

Solution. LetAi be theσ-algebra ofµi-measurable subsets ofXi, i = 1, 2, 3. A set ofthe formA1×A2×A3, whereAi isµi-measurable, will be called ameasurable cuboid.I claim that(A1 × A2) × A3 is precisely theσ-algebra generated by the measurablecuboids. On the one hand, every measurable cuboid belongs to(A1 × A2) × A3; onthe other hand, ifA3 is fixed then the cuboidsA1 × A2 × A3 generate precisely theσ-algebra of sets of the formB × A3, B ∈ A1 ×A2, and sets of this form (now withvarying A3) generate(A1 × A2) × A3; so anyσ-algebra containing all the cuboidscontains(A1 ×A2)×A3.

It follows that(A1×A2)×A3 = A1×(A2×A3). The measures(µ1×µ2)×µ3 andµ1 × (µ2 × µ3) clearly agree on any ‘measurable cuboid’ of the formA1 × A2 × A3,whereAi is µi-measurable. Therefore they agree on any element of theσ-algebragenerated by the measurable cuboids, and the result is proved.

Exercise32.2. Assume the continuum hypothesis for this exercise. According to thecontinuum hypothesis, one can well-order[0, 1] in such a way that each element hasonly countably many predecessors. Let≺ be such a well-ordering, and define a func-tion f(x, y) on [0, 1]×[0, 1] to be the characteristic function of the set(x, y) : x ≺ y.Show that eachx-section and eachy-section off is bounded and Lebesgue measurable,but that nevertheless∫∫

f(x, y)dλ(x)dλ(y) 6=∫∫

f(x, y)dλ(y)dλ(x).

(The point here is thatf is nonmeasurableas a function of two variables.)

Solution. For anyy there are only countably manyx such thatx ≺ y. Thus, for eachfixedy, we havef(x, y) = 0 for all but countably manyx, and for each fixedx we havef(x, y) = 1 for all but countably manyy. Eachf is therefore Lebesgue measurable,∫

f(x, y)dx = 0 for eachy, and∫

f(x, y)dy = 1 for eachx.

44

Lecture 33Applications of Fubini’s Theorem

Exercise33.1. (Long) Show that the following procedures yield an identical definitionof Lebesgue measure onRn.

• Define a functionalτ on the class of rectangular parallelepipeds in the obviousway, use it as a premeasure and extend by Caratheodory’s procedure.

• The same, but start with the class of measurable rectangles.

• Define a linear functional on the class of (compactly supported) continuous func-tions by iterated integration,

f 7→∫· · ·

∫f(x1, . . . , xn)dx1 . . . dxn,

and apply the Riesz representation theorem.

Exercise33.2. Show that∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dxdy = −π

4,

∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dydx =

π

4.

What went wrong here? Why did Fubini’s theorem not ‘work’?

Solution. Observe that

∂

∂x

x

x2 + y2=

1x2 + y2

− x · 2x

(x2 + y2)2=

y2 − x2

(x2 + y2)2.

Therefore∫ 1

0

∫ 1

0

x2 − y2

(x2 + y2)2dxdy =

∫ 1

0

[−x

x2 + y2

]x=1

x=0

dy

=∫ 1

0

−dy

1 + y2= −π

4.

A corresponding argument withx andy interchanged givesπ/4.The ‘failure’ of Fubini’s theorem must be due to the fact thatf(x, y) = (x2 −

y2)/(x2 + y2)2 is not an integrable function over[0, 1] × [0, 1]. One can see thisdirectly be integrating only the positive part off , that is∫ 1

0

∫ 1

y

x2 − y2

(x2 + y2)2dxdy.

Working out the inner integral as above we get∫ 1

0

(1− y)2

2y(1 + y2)dy,

which diverges logarithmically neary = 0.

45

Exercise33.3. Prove that the operation of convolution is commutative and associative.

Solution. Before beginning the proof notice that for any integrable functionh on R,and any constantc, ∫

h(t)dt =∫

h(u + c)du =∫

h(c− v)dv

(in other words, Lebesgue measure is invariant under translations and reflections). Nowto see that convolution is commutative, write

f ∗ g(x) =∫

f(t)g(x− t)dt =∫

f(x− v)g(v)dv = g ∗ f(x).

To see that it is associative, letf, g, h be three integrable functions. Essentially thesame argument as above shows that

(x, t, u) 7→ f(t)g(u− t)h(x− u)

is an integrable function onR3. We have then for almost allx,

(f ∗ g) ∗ h(x) =∫ (∫

f(t)g(u− t)dt

)h(x− u)du

=∫∫

f(t)g(u− t)h(x− u)dtdu

=∫∫

f(t)g(u− t)h(x− u)dudt

=∫

f(t)(∫

g(u− t)h(x− u)du

)dt

=∫

f(t)(∫

g(v)h(x− t− v)dv

)dt = f ∗ (g ∗ h)(x)

where the interchange of integration (between lines 2 and 3) is justified by Fubini’stheorem.

46

Lecture 34Change of variable in multiple integrals

Exercise34.1. Let I =∫∞−∞ e−x2

dx. By writing

I2 =∫∫

e−x2−y2dxdy

in polar coordinates, show thatI =√

π.

Solution. Sincee−x2< 1/(1+x2), it is integrable, and soe−x2−y2

is integrable onR2

(because it is positive and its iterated integrals are finite). Change to polar coordinatesto get

I2 =∫ 2π

0

∫ ∞

0

e−r2rdrdθ = 2π

[−e−r2

2

]∞0

= π.

47

Lecture 35Fourier Transforms

Exercise35.1. Show that the Fourier transform ‘converts convolution into multiplica-tion’: the Fourier transformf ∗ g is equal to a constant timesfg.Exercise35.2. Let gn be an approximate identity. Show that for everyf ∈ L1, f ∗gn → f in L1 asn → ∞. (Hint: First prove it for continuous functions of compactsupport.)

48