foll and real analysis

8/14/2019 Foll and Real Analysis

1/28

Chapter 4.1, Page 117

Problem 10: A topological space X is called disconnected if there exist

nonempty open sets U, V such that U V = and U V =X; otherwiseX is connected. When we speak of connected or disconnected subsets ofX, we refer to the relative topology on them.(a) X is connected iff and X are the only subsets ofX that are both

open and closed.(b) If{E}Ais a collection of connected subsets ofXsuch that AE=

, thenAE is connected.(c) IfA X is connected, then A is connected.=(d) Every pointx Xis contained in a unique maximal connected subset

ofX, and this subset is closed. (It is called the connected compo-nent ofx.)

Solution.

(a) IfXis disconnected, there exist nonempty disjoint open sets U andV with U V =X. Then U andVare both closed as well since theyare complements of open sets, so there exist clopen sets (both U andV) other than and X. Conversely, supposeF is a clopen set withF = and F =X. Let U=F and V =X\ F. Then V is nonemptysinceF =Xand open because Fis closed, so X is disconnected.

(b) LetX= Eand supposeXis disconnected, soX=U V whereUand V are disjoint, open, and nonempty. For eachE, we must haveeitherE U or E V since otherwise E= (E U) (E V)is a disconnection ofE. BecauseU andVare both nonempty, thereexist E1 U and E2 V . But thenE1 E2 =, contradicting thefact thatE is nonempty. HenceXmust be connected.

(c) Suppose A is disconnected. Then disjoint open U, V X withA UV and A U and AV nonempty. Then A U must be

nonempty since otherwise X\ Uwould be a closed set containing Aso that A U would be empty. Similarly, A V must be nonempty.ThenA = (A U) (A V) is a disconnection ofA.

(d) Givenx X, let

Cx=

xUUconnected

U.

ThenCxis connected by part (b), sincex U. Since every connectedset containingx is a subset ofCx, it is the maximal connected subsetcontaining x. Now supposey Cx. By the maximality ofCx,Cx {y}

is disconnected, so it is contained in disjoint open sets UandV. Thesecannot both intersect Cx since this would give a disconnection ofCx,so there exist U and V open and disjoint with Cx U and y V.But then y V Cx, so C

x is open, so Cx is closed.

Probem 13: IfXis a topological space, U is open in X, and A is dense inX, then U=U A.

1


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Solution. Obviously U A U since U A U. Conversely, since A =(A U) (A U) whereU =X\ U, we have

X= A= A U A U.Now U is a closed set containing A U, so A U U. ThereforeUA U. But this implies UA U.


Problem 26: LetX andY be topological spaces.(a) IfX is connected and fC(X, Y), then f(X) is connected.(b) X is called arcwise connected if for all x0, x1 X there exists f

C([0, 1], X) with f(0) = x0 and f(1) = x1. Every arcwise connectedspace is connected.

(c) LetX={(s, t) R2 :t = sin(s1)} {(0, 0)}, with the relative topol-ogy induced from R2. ThenXis connected but not arcwise connected.

Solution.

(a) Suppose f(X) = U V where U, Vare disjoint, nonempty, and open.Then X = f1(f(X)) = f1(U) f1(V) which are clearly disjointand nonempty, and are open because fis continuous.

(b) Let Xbe arcwise connected, and suppose X is disconnected, so X=U V where U, V are disjoint, nonempty, and open. Letx1 U andx2 V. LetCbe a curve from x1 tox2. ThenC= (C U) (C V)and both are nonempty because x1 C U and x2 C V. Thiscontradicts the fact that C is connected by part (a) because it is theimage of the connected set [0, 1] under a continuous function.

(c) First, we show thatX is connected: SupposeU, Vis a disconnection ofX. Suppose WLOG that (0, 0) U. SinceVis nonempty it intersects

either {(s, sin(s

1

)) : s > 0} or {(s, sin(s

1

)) : s < 0}, WLOG theformer, which we callX. ButU intersectsX as well since it contains aneighborhood of the origin. Then (XU)(XV) is a disconnectionofX. But X is the image of the connected set (0, ) R underthe continuous function x (x, sin(x1)) and hence connected, acontradiction. Therefore Xmust be connected.To show that X is not path connected, suppose : [0, 1] X is apath with (0) = (0, 0) and (1) = ( 12 , 0). Then I = x((t)) isa connected subset of the real line, where x((a, b)) = a, since xand are both continuous. Since I contains 0 and 12 , it containseverything in between. This implies that the image of includes thepointspn= (

12n+/2 , 1) forn = 1, 2, . . . . Butpn (0, 1) which is not

inXand hence not in the image of. Thus, the image ofis not closed

in R2

; but in fact it must be compact and therefore closed because iscontinuous and [0, 1] is compact. This contradiction proves that thereis no path from (0, 0) to ( 12 , 0), soXis not path connected.

Problem 27: If X is connected for each A, then X =

A X isconnected.

Solution. We first solve the problem for finite products:


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Lemma 1. The product of finitely many connected spaces is connected.

Proof. By induction, it is sufficient to prove this for the product of two

spaces, sincenj=1 Xj = (n1j=1 Xj )Xn. SupposeXand Yare connected.IfX Y is disconnected, letU andVbe disjoint open nonempty sets withU V = XY. Let (x0, y0) U and (x1, y1) V. If (x0, y1) U,let Z = {(x, y0) : x X}, which is homeomorphic to Xby Exercise 18.Then ZU and ZV are both nonempty, since (x0, y1) UZ and(x1, y1) V Z. But thenZ = (U Z) (V Z) is a disconnection ofZ, a contradiction because X is connected. Similarly, if (x0, y1) V, letZ= {(x0, y) : y Y} which is homeomorphic to Yby Exercise 18, and weobtain a disconnection ofY by the same argument. HenceX Y must beconnected.

Now letX=

Xbe an arbitrary product of connected spaces. Choose

anyx X. For any finite set{1, . . . , n}of indices, let

U1,...,n ={y X :(y) = (x) for all =1, . . . , n}.

Then U1,...,n is homeomorphic to X1 Xn by Exercise 18, andtherefore connected by the lemma. Since each U1,...,n contains x, theirunion, which we call U, is connected by Exercise 10b. Therefore U is alsoconnected by Exercise 10c. Finally, we will show that U=X. Let V Xbe any open set. Using the basis for the product topology, Vmust containa set of the form

V =U1 Um

=1,...,m

X

for some open sets Ui Xi . Let z U1 Um be any point, anddefine a point z Xby

(z) =i(z) = 1, . . . , m

(x) else.

Then z U1,...,m U since it equals x in all but finitely many coordi-

nates; but z V V as well. Hence every open set V X contains apoint ofU, so U=Xwhich must be connected.


Problem 39: Every sequentially compact space is countably compact.

Solution. LetXbe a sequentially compact space and {U1, U2, . . . }a count-able open cover. Suppose there is no finite subcover. Construct a sequence

xn as follows: Let j1 = 1 and x1 be any point of Uj1 = U1. Next, letj2 be the smallest n 2 such that Uj2 \U1 = , and let x2 Uj2 \U1.Then let x3 Uj3 \ (Uj1 Uj2) where j3 is the smallest possible value forwhich the latter set is nonempty. This process cannot terminate becausethat would imply that Uj1 , . . . , U jm would be a finite subcover. Note thatxn has the property that xn / Ui for any i < jn. Now let x be the limitof some subsequence xnk . Let UM be any open set containing x. ThenUMcontains all but finitely many of the xnk . Butxnk cannot be inUi for


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i < nk, a contradiction. Hence the process must terminate, i.e. there is afinite subcover.

Problem 43: Forx [0, 1), let 1 an(x)2n (an(x) = 0 or 1) be the base-2decimal expansion of x. (If x is a dyadic rational, choose the expansionsuch thatan(x) = 0 for n large.) Then the sequencean in {0, 1}

[0,1) hasno pointwise convergent subsequence. (Hence {0, 1}[0,1) with the producttopology arising from the discrete topology on {0, 1} is not sequentiallycompact. It is, however, compact.)

Solution. Suppose some subsequence anj converges to a function a {0, 1}[0,1).

Letz [0, 1) be a number such that anj (z) is 0 for j even and 1 forj odd.

Then {b {0, 1}[0,1) : b(z) = a(z)} is a neighborhood of a in {0, 1}[0,1).However, it does not contain all but finitely many of the anj ; it only con-tainsanj forj even ifa(z) = 0, and for j odd ifa(z) = 1. Hence anj doesnot converge to a, a contradiction.


Problem 49: LetXbe a compact Hausdorff space and E X.(a) IfEis open, then E is locally compact in the relative topology.(b) IfE is dense in Xand locally compact in the relative topology, then

Eis open. (Use Exercise 13.)(c) Eis locally compact in the relative topology iffEis relatively open in

E.

Solution.

(a) Let x E. By Proposition 4.30, x N E for some compactneighborhood N ofx. ThenNis also a compact neighborhood in the

relative topology, so E is locally compact.(b) For anyx E, we have x U K Ewhere U is open in E and

Kcompact in E in the relative topology. This means that U=U Efor some U open in X. Moreover, Kmust be compact as a subset ofX: If{U}is any open cover, {U E} is an open cover for Kin therelative topology, which has a finite subcover U1 E, . . . , U n E,from which it follows that U1 , . . . , U n is a finite subcover. Then

U U=U E= U K= K E,

where all the closures are taken inX, by Exercise 13. Sox is containedin an open set contained inE. Therefore Eis open.

(c) Suppose E is locally compact in the relative topology. Now E is acompact Hausdorff space in which Eis dense, soEis open in Eby part

(b). Conversely, suppose E is open in E. For any x E, Proposition4.30 applied to the LCH space E implies that there is a compactneighborhood NE E, soE is locally compact.

Problem 50: LetUbe an open subset of a compact Hausdorff spaceXandU its one-point compactification. If : XU is defined by (x) =x ifx U and(x) = ifx Uc, then is continuous.


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Solution. Let V U be any open set. We wish to show that1(V) isopen in X. Suppose first that V is an open subset ofU. (Vis open in therelative topology, but sinceU is open this implies it is also open as a subsetofX.) Then 1(V) = V since is the identity map on Uand everythingelse is sent to /V. Thus 1(V) is open in X. Now consider the othercase, where V ={} (U\ K) whereKU is compact. Now

1(V) = 1() 1(U\ K) = (X\ U) (U\ K) = X\ K

which is open in X. Hence is continuous.

Problem 56: Define : [0, ] [0, 1] by (t) = t/(t + 1) fort [0, ) and() = 1.(a) is strictly increasing and (t + s) (t) + (s).(b) If (Y, ) is a metric space, then is a bounded metric on Y that

defines the same topology as .(c) IfX is a topological space, the function (f, g) = (sup

xX|f(x)

g(x)|) is a metric on CX whose associated topology is the topology ofuniform convergence.

(d) IfX is a -compact LCH space and {Un}1 is as in Proposition 4.39,the function

(f, g) =1

2n

sup

xUn

|f(x) g(x)|

is a metric on CX whose associated topology is the topology of uniformconvergence on compact sets.

Solution.

(a) That is strictly decreasing follows from the fact that

1 (t) = 1

t+1 0 t


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Finally,

((x, y)) ((x, z) + (z, y)) (since is increasing)

((x, z)) + ((z, y)) (since (t + s) (t) + (s))

so satisfies the triangle inequality. Thus, it is a metric, andis obviously bounded since its values are in [0, 1]. The collection ofpositive-diameter balls is the same in both metrics, since (x, x0) < ((x, x0)) 0, or f(0) = 0andy [0, 1] with f(y)= 0, in which case

f|f(y) f(0)|

|y 0| =

|f(y)|

|y| >0.

On the other hand, iff= 0 then clearly f= 0.Also,

cf= |cf(0)| + sup|cf(x) cf(y)|

|x y|

=|c||f(0)| + sup|c||f(x) f(y)|

|x y|

=|c||f(0)| + |c| sup|f(x) f(y)|

|x y|

=|c|f.

Finally,

f+ g= |f(0) + g(0)| + sup|f(x) + g(x) f(y) g(y)|

|x y|

|f(0)| + |g(0)| + sup|f(x) f(y)| + |g(x) g(y)|

|x y|

|f(0)| + |g(0)| + sup|f(x) f(y)|

|x y| + sup

|g(x) g(y)|

|x y|

=f+ g.

Hence, is a norm. To show that this normed linear space is com-plete, supposefi ([0, 1]) with

fi< . Then

fi<

by the triangle inequality, so if we define f=

fi thenf([0, 1]).

Let gN =fN

i=1 fi =

i=N+1 fi. Then gN

i=N+1 fi 0. Since every absolutely convergent sequence converges, ([0, 1]) isBanach.

(b) Since

|af(x) + bg(x) af(y) bg(y)|

|x y| |a|

|f(x) f(y)|

|x y| + |b|

|g(x) g(y)|

|x y| ,

iff, g thenaf+bg . To show this subspace is closed, supposefn f in the norm. Since

fn f= |fn(0) f(0)| + sup|fn(x) f(x) fn(x) + f(y)|

|x y| ,

this implies

sup|fn(x) f(x) fn(x) + f(y)|

|x y| 0.


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Given > 0, choose N such that this supremum is less than 2 forn= N. For fixed y , we can then choose >0 such that

|fN(x) fN(y)||x y|

< 2

for 0< |x y|< . Then for 0< |x y|< ,

|f(x) f(y)|

|x y|

|fN(x) fN(y)|

|x y| +

|f(x) fN(x) f(y) + fN(y)|

|x y| 0;however,p(x) = 0 because 0 Mand 0 (x) P. This contradictsthe construction of F as satisfying F(x) p(x). So F(x) 0 forx P.


Problem 28: The Baire category theorem remains true ifXis assumed to bean LCH space rather than a complete metric space. (The proof is similar;the substitute for completeness is Proposition 4.21.)

Solution. Let Xbe LCH and {Un} open and dense. Let W Xbe open.We wish to show W intersectsUn. SinceU1 is dense,U1 W is nonempty,and of course its open. By Proposition 4.30, every open set in an LCH spacecontains a compact set which contains another open set. LetV1 K1 (U1 W) withV1 open andK1 compact. Now U2 is dense, so it intersectsV1. Let V2 K2 (V1 U2) with V2 open and K2 compact. Continuing,we get a nested sequence Kn of nonempty compact sets. By Proposition4.21,Kn is nonempty; but Kn Un W, soW (Un) is nonempty.

Problem 29: Let Y /L1() where is counting measure on N, and let X=

{fY : 1 n|f(n)|< }, equipped with the L1 norm.(a) Xis a proper dense subspace ofY ; henceX is not complete.(b) Define T : X Y by T f(n) = nf(n). Then T is closed but not

bounded.(c) Let S = T1. ThenS : Y X is bounded and surjective but not

open.

Solution.

(a) Xis a subspace because

n|f(n)|<

n|cf(n)|= |c|

n|f(n)|0, letA ={x:|f(x)|> 1 }. Then 0< (A) (1 )A, so fq ((A))

1/q(1 ).

Becausex1/q

1 asq for every positive number x, this implies thatfq is eventually greater than 1 2. On the other hand, |f|q |f|p a.e.

for all q > p, so fq fp/qp which tends to 1 as q , so fq is

eventually less than 1 + .

Problem 8: Suppose (X) = 1 and f Lp for some p > 0, so that f Lq

for 0< q < p.(a) log fq

log |f|.

(b) (

|f|q 1)/q log fq, and (

|f|q 1)/q

log |f| as q 0.(c) limq0 fq = exp(

log |f|).

Solution.

(a) Becauseex is convex, Jensens inequality implies

e log |f|

q

elog |f|q(e

log |f|)q

|f|q

e log |f| fq

log |f| log fq.

(b) Because y 1 log y for all positive real numbers y (this is easilychecked by noting their equality at y = 1 and comparing derivatives),

|f|q 1 log

|f|q = log fqq = qlog fq.By LHopitals Rule,

limq0 |f|q 1q = limq0 |f|q log |f|= log |f|where the differentiation under the integral is justified by Theorem2.27. The question then becomes how to justify the final equality. If|f|= 0 on a set of positive measure then we dont even need LH opitalsRule because the numerator

|f|q 1 approaches a nonzero negative

number, so |f|q1

q =

log |f|. Hence, we assume |f| = 0

a.e. and writeX =j=0Ej where E0 ={|f| 1} and Ej ={ 1j+1


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|f| < 1j } for j 1. Now on E0, we can choose any p < p; then

|f|p

log |f|= O(|f|p) L1 and |f|q log |f|< |f|p

log |f| for q < p, so

E0 |f|q log |f| E0 log |f| by the Dominated Convergence Theorem.For the rest of the domain, if

X\E0

log |f|=

j=1

Ej

log |f|

diverges, then for any M there exists k such thatk

j=1

Ej

log |f| k , soj=k+1Ej

|f|q log |f|

j=k+1

Ej

log |f|

< .This implies that

X\E0

|f|q log |f| and

X\E0log |f| are eventually

within 2.

(c) By parts (a) and (b) and the Sandwich Theorem, log fq log |f|.Exponentiating both sides gives the result.

Problem 11: Iffis a measurable function on X, define theessential rangeRf offto be the set of all z Csuch that{x: |f(x) z|< }has positivemeasure for all >0.(a) Rf is closed.(b) IffL, then Rfis compact and f = max{|z|: z Rf}.

Solution.

(a) Letzn z wherezn Rf. Given >0, choose zNwith|zN z|< 2 .

Then{x:|f(x)zN|< 2} {x: |f(x)z|< }, and since the former

set has positive measure, the latter does as well. Hence z Rf.

(b) Let M =f. Then Rf BM(0) so its bounded as well as closed,hence compact by the Heine-Borel theorem.Clearly maxzRf|z| M. If it were strictly less than M, say ,choose with < < M. Then every pointz in the closed annulusA= {z : |z| M} has a neighborhood Nz such that {x: f(x)Nz} has measure zero. Since this annulus is compact, there is a finitesubcover, which implies that{x: f(x) A}has measure zero as well.But then({x:|f(x)|> }) = 0, contradicting the fact that f> .


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Problem 13: Lp(Rr, m) is separable for 1 p


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. By Lusins theorem, there exists a subsetB A with(A \ B)< and fn funiformly on B . Then

|fn f|g= A\B

|fn f|g+ B

|fn f|g+ X\A

|fn f|g.

We use the notation hr,E =hEr = (

E|h|r)1/r; note that then

hr,E hr which is the usual Lr norm. Then the first integral isat most fn fp,A\Bgq,A\B by Holders inequality. This in turn

is at most 2M1/q where M = sup fnp < . The second integralis at most fn fp,A\Bgq,A\B, which tends to 0 as n sincefn f uniformly on B. Finally, the third integral is at mostfnfp,X\Agq,X\A 2M

1/q. Putting the pieces together, we see that|fn f|g 0 asn . Hencefn f weakly.

(b) Consider fn = [n,n+1]. Then fn 0 pointwise, and fnp = 1 isbounded. However, fn 0 weakly: Consider the functiong(x) = 1,

which is in L = (L1). Then fng = 1 0 = 0g. Similarly,in 1 = L1(Z) we can let fn = n be the nth unit sequence; thenfn 0 pointwise and fn= 1 is bounded, but the constant functiong= 1 (1) shows that fn0 weakly.Now consider the case p = with -finite. Since L = (L1),weak-* convergence means that

gfn

gffor all g L1. But |fn f|g |fn f||g| 0

by the Dominated Convergence Theorem, since |fn f||g| 2M|g| L1 whereM= sup fn< .

Problem 21: If 1< p < , fn fweakly inp

(A) iff supn fnp< andfn f pointwise.

Solution. If supn fnp< and fn fpointwise, thenfn fweakly inp(A) by Problem 20. For the converse, supposefn fweakly. WLOG wemay assumef= 0. To show fn 0 pointwise, let gy(x) = xy forx, y A.Then gq = 1 so g q(A), and

x gy(x)fn(x) = fn(y); by hypothesis,

this tends to 0 asn withy fixed. To showfnp is bounded, supposeit is not. By taking a subsequence if necessary, we may assumefnp . Let fn1 be such that fn1p > 2, and choose a finite subset A1 Asuch that

xA1

|fn1(x)|p = M1 > 1 and

xA\A1

|fn1(x)|p < . Now

because fn 0 pointwise and A1 is finite, fn 0 uniformly on A1,so

xA1|fn(x)|

p < 2 for all sufficiently large n. Take n2 large enough

that this obtains and that fn2p = M2 > 4. Let A2 (A\ A1) be afinite set withxA2 |fn2(x)|p =M2 >2 andxA\A2 |fn2(x)|p < . Wethen take n3 large enough that

xA1A2

|fn3(x)|p < 2 and fn3p > 8;

let A3 (A\ A1 \A2) be finite with

xA3|fn3(x)|

p = M3 > 4 andxA\A3

|fn3(x)|p < , etc. We now define g : A C by

g(x) =

|fnk (x)|

p2fnk (x)

Mkx Ak

0 x /Ak for any k.


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Then

gqq =

k=1 xAk |g(x)|q=

k=1

xAk

|fnk(x)|q(p1)Mqk

=

k=1

1

Mqk

xAk

|fnk(x)|p

=

k=1

1

MqkMk

=

k=1

1

Mq1k

k=1

12k(q1)

1

2 . This contradicts

the fact thatfn 0 weakly in p

. Hencefnp must be bounded.

Problem 22: LetX= [0, 1] with Lebesgue measure.(a) Letfn(x) = cos2nx. Then fn 0 weakly in L

2, but fn0 a.e. orin measure.

(b) Let fn(x) =n(0,1/n). Then fn 0 a.e. and in measure, but fn0weakly in Lp for anyp.

Solution.


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(a) Since cos 2nx = e2inx+e2inx

2 , it suffices to prove that

f(n) =

1

0 f(x)e2inx

dx 0

as |n| for fL2([0, 1]). But f2 by Plancherels theorem, so

certainly f(n) 0 as |n| . However, since nx is dense in [0, 1]for irrational x, it is in{y : cos 2y > 12} for infinitely many n. Hencecos2nx 0 for irrational x, so it certainly does not converge to 0 a.e.Moreover, by symmetry, {x : cos2nx 12} consists of 4n intervals

of length 12narccos(12) (some of these intervals are adjacent to each

other, so they could actually be viewed as 2n 1 intervals of length1

narccos12 and two intervals of length

12narccos

12 , but this doesnt

matter.) Hence({x : | cos2nx > 12}) is a constant and thereforedoes not tend to 0, so cos 2nx 0 in measure.

(b) Clearly fn(x) 0 for all x = 0, which is certainly a.e., and ({x :

|fn(x) }) = 1n 0 for all > 0, so fn 0 in measure. However,letg(x) = [0,1] which is inL

p for all 1 p . Then

fn(x)g(x) =10, which proves that fn0 weakly in L

p for 1 p


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Problem 29: Suppose that 1 p < , r > 0, and h is a nonnegativemeasurable function on (0, ). Then

0

xr1 x0

h(y)dyp dx prp

0

xpr1h(x)pdx,

0

xr1

x

h(y)dy

pdx

pr

p 0

xp+r1h(x)pdx.

Solution. Let K(x, y) = x1y y>x. Then K(x,y) = 1K(x, y).

Withf(x) = xh(x) andg (y) = yh(y), we have

T f(y) =

0

K(x, y)f(x)dx=

0

x+1y y>xh(x)dx= y

y0

x+1h(x)dx

and

Sg(x) = 0 K(x, y)g(y)dy=

0 x1

y

y>xh(y)dy = x1 x y h(y)dy.

By Theorem 6.20,0

yp y

0

x+1h(x)dx

pdy =

0

|T f|ppdy Cpfpp= C

p

0

xph(x)pdx

and 0

xq(1)

x

y h(y)dy

qdx= Sgqq C

qgqq = Cq

0

yq h(y)qdy

where

C=

0

|K(x, 1)|x1/pdx=

0

x11>xx1/pdx=

1

0

x11/pdx= 1

1/p.

Letting= 1 and= r+1p in the first, and letting= , = 1 + r1

p

with the role of p and q interchanged in the second, yields the desiredequalities.

Problem 31 (A Generalized Holder Inequality): Suppose that 1 pj and

n1p

1j =r

1 1. Iffj Lpj forj = 1, 2, . . . , n, thenn

1fj Lr

andn

1 fj r n

1fj pj .

Solution. By induction on n, it is sufficient to consider the case n = 2.Let g1 = |f1|r and g2 = |f2|r. Then g1 Lp1/r and g2 Lp2/r. Since1

p1/r+ 1p2/r = 1, Holders inequality says that g1g2 L

1 and g1g21

g1p1/rg2p2/r. Translating these back into the corresponding statements

about f1 andf2, we have f1f2 Lr andf1f2r f1p1f2p2 .


Problem 38: fLp iff

2kpf(2k)< .

Solution. Let

En= {x: 2n


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Then the productf g: X[0, 1] is continuous, (f g)1({1}) = K, andsupp(f g) F is compact.


Problem 11: Suppose thatis a Radon measure onXsuch that({x}) = 0for all x X, and A BX satisfies 0 < (A) < . Then for any suchthat 0< < (A) there is a Borel set B A such that(B) = .

Solution. Let > 0. By inner regularity, there exists compact K0 Asuch that < (K) (A). Now for each x K, ({x}) = 0, so byouter regularity on sets of finite measure, A Gx x open such that(Gx) < . (We may take Gx to be open in the relative topology, since

the restriction of to A is still Radon.) By compactness, there is a finitesubcoverGx1 , . . . , Gxn ofK. For eachj = 1, . . . , n, let G(j) = Gx1

Gxj . Then for somej, < (G(j))< +, because(G(j))< (G(j1))+

and(G(n)) (K0)> . LetG0 be equal to the G(j) with this property.Thus G0 A and < (G0) < + . By inner regularity, K1 G0compact with (K1)> . Then we can cover K1 with finitely many opensets of measure at most/2, which we can also require to be subsets ofG0;an appropriate union of these yields G1 G0 with < (G1)< +/2.Continuing, we have a nested sequence G0 G1 . . . with < (Gj )< + 2j . Then E= Gj is a subset ofA with measure .

Problem 12: Let X= RRd, where Rddenotes R with the discrete topology.Iff is a function on X, let fy(x) = f(x, y); and ifE X, let Ey = {x :(x, y) E}.(a) f CC(X) iff fy CC(R) for all y and fy = 0 for all but finitely

many y .(b) Define a positive linear functional on CC(X) by I(f) =

yR

f(x, y)dx,

and let be the associated Radon measure on X. Then(E) = for any Esuch thatEy = for uncountably many y .

(c) Let E = {0} Rd. Then(E) = but (K) = 0 for all compactK E.

Solution.

(a) SupposefCC(X). Then for any fixedy,fy =e fwheree : R X

is the continuous embedding x (x, y). So fy is continuous. More-

over, supp(fy) (supp(f)) where : X Ris defined by (x, y) =x. So supp(fy) is a closed subset of the continuous image of a compactset, hence compact. Thus, fy CC(R) for all y. Suppose there areinfininitely many pairs (x, y) with y distinct and f(x, y) = 0.Consider the open cover {R {y} : y Rd} of supp(f). Any finitesubcover will only include finitely many of the (x, y) and hence can-not cover the whole set, contradicting its compactness. Hence fy = 0for all but finitely many y .


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Conversely, supposefy = 0 for all but finitely many y and fy CC(R)for all y . Then

supp(f) = {y:fy=0}

supp(fy) {y}

is a finite union of compact sets, hence compact. Also, for any opensubsetUof the range,

f1(U) =

yRd

(fy)1(U)

is a union of open sets and therefore open. Thus, fCC(X).(b) Let Ebe a set such that Ey = for uncountably many y, and let G

be any open set with E G. ThenG contains an intervalIy {y}foreachy withEy =, since these intervals are a basis for the topology ofR Rd. Since there are uncountably many intervals, there must exist

N such that uncountably many of the intervals have length greaterthan 1N. For each such interval Iy , there exists a function f Iywith

Iy

f > 12N (e.g. take a subinterval of length at least 12Nand use

Urysohns lemma to construct a function which is 1 on this subintervaland 0 outside Iy), so

(G) =yR

f(x, y)dx

1

2N =.

By regularity, since (G) = for any open set G containing E, then(E) = .

(c) By part (b), (E) = . However, ifK E is compact, then K ={0} y1, . . . , yn for some finite subset {y1, . . . , yn} Rd. Then, given >0, let f(x) : R R be a continuous function with values in [0, 1]which is 1 at the origin and 0 for |x|> . Then

g(x, y) =

f(x) y = y1, . . . , yn

0 else

is inCC(X), g K, and

I(g) =n

i=1

f(x)dx 2n.

Thus,

(K) = inf{I(g) : g CC(X), g K}= 0.


Problem 20: Some examples of nonreflexivity ofC0(X):(a) If M(X), let () =

xX({x}). This sum is well defined,

and M(X). If there exists a nonzero M(X) such that({x}) = 0 for all x X, then is not in the image of C0(x) inM(X) C0(X).


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Problem 22: A sequence {fn} in C0(X) converges weakly to f C0(X) iffsup fnu < and fn f pointwise.

Solution. Suppose sup fn = M 0.By Egorovs theorem, EXwith (E)< 4M and fn funiformly onX\ E. Then for any C0(X) =M(X),

X

(fn f)d=

E

(fn f)d +

X\E

fn f d

where the first integral is at most /2 since fn f 2M and(E)

foll and real analysis

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