Classical Fourier Analysis, Third Edition [3rd Ed] (Instructor Solution Manual, Solutions)

349 Pages·2014·1.562 MB·English

by  Loukas Grafakos

Loukas Grafakos
Classical Fourier Analysis, Third Edition
Solutions of all the exercises
– Monograph –
January22,2014
Springer

v
Iwouldliketoexpressmydeepgratitudetothefollowingpeoplewhohelpedinthepreparationofthesolutionsofthebooks
ClassicalFourierAnalysis,3rdedition,GTM249andModernFourierAnalysis,3rdedition,GTM250.
Mukta Bhandari, Jameson Cahill, Santosh Ghimire, Zheng Hao, Danqing He, Nguyen Hoang, Sapto Indratno, Richard
Lynch,DiegoMaldonado,HanhVanNguyen,PeterNguyen,JessePeterson,SharadSilwal,BrianTuomanen,XiaojingZhang,
Iremainsolelyresponsibleforanyerrorcontainedintheenclosedsolutions.

Contents
vii

1. Lp Spaces and Interpolation
Section1.1.Lp andWeakLp
Exercise1.1.1
Suppose f and f aremeasurablefunctionson(X,µ).Provethat
n
(a)d isrightcontinuouson[0,∞).
f
(b)If|f|≤liminf |f |µ-a.e.,thend ≤liminf d .
n→∞ n f n→∞ fn
(c)If|f |↑|f|,thend ↑d .
(cid:2) n fn f
Hint:Part(a):Lett beadecreasingsequenceofpositivenumbersthattendstozero.Showthatd (α +t )↑d (α )using
n f 0 n f 0
a convergence theorem. Part (b): Let E ={x∈X : |f(x)|>α} and E ={x∈X : |f (x)|>α}. Use that µ(cid:0)(cid:84)∞ E (cid:1)≤
n n n=m n
liminfµ(E )andE(cid:106)(cid:83)∞ (cid:84)∞ E µ-a.e.(cid:3)
n m=1 n=m n
n→∞
Solution.
(a)Forallα >0letE ={x∈X : |f(x)|>α}.Fixα ≥0andlett beadecreasingsequenceofpositivenumbersthattends
α 0 n
to0.ObservethatE (cid:106)E foreveryn.Thereforeχ ↑χ asn→∞,sobytheLebesguemonotoneconvergence
α0+tn α0+tn+1 Eα0+tn Eα0
theoremitfollowsthat
µ(E )↑µ(E )
α0+tn α0
whichestablishesthedesiredconclusion.
(b)FirstnotethatE⊆∪∞ ∩∞ E µ−a.e.,sowehave
m=1 n=m n
d (α)=µ(E)≤µ(∪∞ ∩∞ E )≤liminfµ(E )=liminfd (α).
f m=1 n=m n n→∞ n n→∞ fn
(c)LetE andE beasabove.TheclearlywehaveE ⊆E andlim E =E,sotheconclusionfollowsfromtheLebesgue
n n n+1 n→∞ n
monotoneconvergencetheoremasinpart(a). (cid:4)
Exercise1.1.2
(Ho¨lder’sinequality)Let0<p,p1,...,pk≤∞,wherek≥2,andlet fj beinLpj =Lpj(X,µ).Assumethat
1 1 1
= +···+ .
p p p
1 k
(a)Showthattheproduct f ···f isinLpandthat
1 k
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
(cid:13)f1···fk(cid:13)Lp ≤(cid:13)f1(cid:13)Lp1···(cid:13)fk(cid:13)Lpk.
1
2 Contents
(b)Whenno pj isinfinite,showthatifequalityholdsinpart(a),thenitmustbethecasethatc1|f1|p1 =···=ck|fk|pk µ-a.e.
forsomec ≥0.
j
(c) Let 0<q<1 and q(cid:48) = q−q1. For r<0 and g>0 almost everywhere, define (cid:107)g(cid:107)Lr =(cid:13)(cid:13)g−1(cid:13)(cid:13)−L|1r|. Show that if g is strictly
positiveµ-a.e.andliesinLq(cid:48) and f ismeasurablesuchthat fgbelongstoL1,wehave
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
(cid:13)fg(cid:13)L1 ≥(cid:13)f(cid:13)Lq(cid:13)g(cid:13)Lq(cid:48).
Solution.
(a)Wemayassumethatall pj<∞,sinceotherwise,onecanfactorfromthe(cid:107)f1···fk(cid:107)Lp theL∞ normofthe fj’swith pj=∞
andtheinequalityreducestotheremaining p ’swhicharenotinfinite.
j
Ifone(cid:107)fj(cid:107)Lp1 is0or∞,thentheconclusionisobvious.Let
f
j
F =
j
(cid:107)fj(cid:107)Lp1
Theconvexityofet impliestheinequality
1 1 a a
ap1 ···apk ≤ 1 +···+ k
1 k p p
1 k
when1<p <∞and 1 +···+ 1 =1.Usingthisinequalityweobtain
j p1 pk2
(|F|p1)p11 ···(|F|pk)p1k ≤ |f1|p1 +···+ |fk|pk .
p1(cid:107)f1(cid:107)Lp1 pk(cid:107)fk(cid:107)Lpk
Integratingandusing p11+···+p1k =1,weobtain(cid:107)f1···fk(cid:107)L1≤(cid:107)f1(cid:107)Lp1···(cid:107)fk(cid:107)Lpk.Forp(cid:54)=1reducetheinequalitytothecase
p=1asfollows:
1 (cid:16) (cid:17)1
(cid:107)f1···fk(cid:107)Lp =(cid:107)|f1···fk|p(cid:107)Lp1 ≤ (cid:107)|f1|p(cid:107)Lpp1 ···(cid:107)|fk|p(cid:107)Lppk p =(cid:107)f1(cid:107)Lp1···(cid:107)fk(cid:107)Lpk
usingthefactthat1/p +···+1/p =1/pisequivalentto(p /p)−1+···+(p /p)−1=1.
1 k 1 k
(b)Wenotethatequalityintheinequality
1 1 a a
ap1 ···apk ≤ 1 +···+ k
1 k p p
1 k
holdsifandonlyifa =···=a .Indeed,thisinequalityisarestatementofthefactthatt(cid:55)→et isconvex.Andsinceet isstrictly
1 k
convex,itfollowsthatequalityholdsintheprecedinginequalityifandonlyifallthea ’sareequaltoeachother.Usingthis
j
aj=|fj|pj/(cid:107)fj(cid:107)Lpj,weobtaintheclaimwithcj=(cid:107)fj(cid:107)−Lp1j.
(c)Wehave
(cid:107)f(cid:107)Lq =(cid:107)fgg−1(cid:107)Lq ≤(cid:107)fg(cid:107)L1(cid:107)g−1(cid:107)L|q(cid:48)| =(cid:107)fg(cid:107)L1(cid:107)g(cid:107)−Lq1(cid:48),
since1= 1+ 1 implies 1 = 1+ 1 .Theconclusionfollows. (cid:4)
q q(cid:48) q 1 |q(cid:48)|
Exercise1.1.3
Let(X,µ)beameasurespace.
(a)If f isinLp0(X,µ)forsome p0<∞,provethat
(cid:13) (cid:13) (cid:13) (cid:13)
lim(cid:13)f(cid:13) =(cid:13)f(cid:13) .
p→∞ Lp L∞
(b)(Jensen’sinequality)Supposethatµ(X)=1.Showthat
Contents 3
(cid:18)(cid:90) (cid:19)
(cid:13) (cid:13)
(cid:13)f(cid:13) ≥exp log|f(x)|dµ(x)
Lp
X
forall0<p<∞.
(c)Ifµ(X)=1and f isinsomeLp0(X,µ)forsome p0>0,then
(cid:18)(cid:90) (cid:19)
(cid:13) (cid:13)
lim(cid:13)f(cid:13) =exp log|f(x)|dµ(x)
p→0 Lp X
withtheinterpretatione−∞=0.
(cid:2)Hint:Part(a):If0<(cid:107)f(cid:107)L∞ <∞,usethat(cid:107)f(cid:107)Lp ≤(cid:107)f(cid:107)(Lp∞−p0)/p(cid:107)f(cid:107)Lp0p0/p toobtainlimsupp→∞(cid:107)f(cid:107)Lp ≤(cid:107)f(cid:107)L∞.Conversely,let
Eγ ={x∈X : |f(x)|>γ(cid:107)f(cid:107)L∞} for γ in (0,1). Then µ(Eγ)>0, (cid:107)f(cid:107)Lp0(Eγ) >0, and (cid:107)f(cid:107)Lp ≥(cid:0)γ(cid:107)f(cid:107)L∞(cid:1)(p−p0)/p(cid:107)f(cid:107)Lp0p0/(pEγ),
1
hence liminfp→∞(cid:107)f(cid:107)Lp ≥γ(cid:107)f(cid:107)L∞. If (cid:107)f(cid:107)L∞ =∞, set Gn ={|f|>n} and use(cid:82)that (cid:107)f(cid:107)Lp ≥(cid:107)f(cid:107)L(cid:0)p((cid:82)Gn) ≥nµ(cid:1)(Gn)p to obtain
liminfp→∞(cid:107)f(cid:107)Lp ≥n. Part (b) is a direct consequence of Jensen’s inequality Xlog|h|dµ ≤log X|h|dµ . Part (c): Fix a
sequence0<p <p suchthat p ↓0anddefine
n 0 n
1 1
hn(x)= (|f(x)|p0−1)− (|f(x)|pn−1).
p p
0 n
Usethat 1(tp−1)↓logt as p↓0forallt>0.TheLebesguemonotoneconvergencetheoremyields(cid:82) h dµ ↑(cid:82) hdµ,hence
p X n X
(cid:82) 1 (|f|pn−1)dµ ↓(cid:82) log|f|dµ,wherethelattercouldbe−∞.Use
X pn X
exp(cid:18)(cid:90) log|f|dµ(cid:19)≤(cid:18)(cid:90) |f|pndµ(cid:19)p1n ≤exp(cid:18)(cid:90) 1 (|f|pn−1)dµ(cid:19)
p
X X X n
(cid:3)
tocompletetheproof.
Solution.
(a)If(cid:107)f(cid:107)L∞ <∞usethat
(cid:18)(cid:90) (cid:19)1p (cid:18)(cid:90) (cid:19)1p p−p0 p0
|f|pdµ = |f|p0|f|p−p0dµ ≤(cid:107)f(cid:107) p (cid:107)f(cid:107) p
L∞ Lp0
X X
from which it follows that limsupp→∞(cid:107)f(cid:107)Lp ≤(cid:107)f(cid:107)L∞. Conversely, let Eγ ={x∈X : |f(x)|>γ(cid:107)f(cid:107)L∞} for γ ∈(0,1). Then
µ(Eγ)>0andtheintegralof|f|p0 overEγ ispositive.For p>p0wehave
(cid:107)f(cid:107)Lp ≥(cid:18)(cid:90) |f|pdµ(cid:19)1p ≥γp−pp0(cid:107)f(cid:107)Lp∞−pp0(cid:18)(cid:90) |f|p0dµ(cid:19)pp0
Eγ Eγ
andfromthisitfollowsthatliminfp→∞(cid:107)f(cid:107)Lp ≥γ(cid:107)f(cid:107)L∞.Sinceγ <1wasarbitrary,liminfp→∞(cid:107)f(cid:107)Lp ≥(cid:107)f(cid:107)L∞.
Nowsupposethat(cid:107)f(cid:107)L∞ =∞.ThenifGn={|f|>n},wemusthaveµ(Gn)>0foralln=1,2,....Then
1
(cid:107)f(cid:107)Lp ≥(cid:107)f(cid:107)Lp(Gn)≥nµ(Gn)p
andsoliminfp→∞(cid:107)f(cid:107)Lp ≥n.Thusliminfp→∞(cid:107)f(cid:107)Lp =∞=(cid:107)f(cid:107)L∞.
(b)ByJensen’sinequalitywehavethatifφ usconcave,then
(cid:18)(cid:90) (cid:19) (cid:90)
φ gdµ ≥ φ◦gdµ, g≥0,a.e.
X X
Hence
(cid:18)(cid:90) (cid:19) (cid:90)
log |h|dµ ≥ log(|h|)dµ.
X X
Let|h|=|f|p,thenwehave
4 Contents
(cid:18)(cid:90) (cid:19)1p (cid:90) (cid:90)
log |f|pdµ ≥ log(|f|)dµ ⇒(cid:107)f(cid:107)Lp ≥exp( log(|f|))dµ.
X X X
(c)Bypart(b)wehave
(cid:18)(cid:90) (cid:19)
limsup(cid:107)f(cid:107)Lp ≥exp log|f|dµ .
p→0 X
Fixasequence0<p <p suchthat p ↓0.BytheLebesguemonotoneconvergencetheoremweknowthat
n 0 n
(cid:90) |f|p0−1 |f|pn−1 (cid:90) |f|p0−1
− dµ ↑ −log(|f|)dµ,
p p p
X 0 n X 0
since f ∈Lp0.Therefore
(cid:90) |f|pn−1 (cid:90)
dµ ↓ log|f|dµ.
p
X n X
ae≤ea,∀a>0,so(cid:82) |f|pn ≤exp((cid:82) |f|pn−1).Therefore
X pn X pn
(cid:90) |f|p (cid:18)(cid:90) (cid:19)
liminf dµ ≤exp log|f|dµ ≤limsup(cid:107)f(cid:107)Lp,
p→0 X p X p→0
andtheconclusionfollows. (cid:4)
Exercise1.1.4
Leta beasequenceofpositivereals.Showthat
j
(a)(cid:0)∑∞j=1aj(cid:1)θ ≤∑∞j=1aθj,forany0≤θ ≤1.
(b)∑∞j=1aθj ≤(cid:0)∑∞j=1aj(cid:1)θ,forany1≤θ <∞.
(c)(cid:0)∑Nj=1aj(cid:1)θ ≤Nθ−1∑Nj=1aθj,when1≤θ <∞.
(d)∑Nj=1aθj ≤N1−θ(cid:0)∑Nj=1aj(cid:1)θ,when0≤θ ≤1.
Solution.
(a)Considerthenumbers
a
j
∑∞k=1ak
whicharelessthan1.Thenwehave
(cid:18) a (cid:19)θ a
j j
≥
∑∞k=1ak ∑∞k=1ak
since0≤θ ≤1.Summingover jweobtain
∞ (cid:18) a (cid:19)θ
∑ j ≥1.
j=1 ∑∞k=1ak
Thisisjusttheclaimedinequality
(b)Thisisprovedinawaysimilartothatin(a),withtheonlydifferencebeingthat≥isreplacedwith≤.
(c)Wehave
N N N N
∑aj≤(cid:0)∑1(cid:1)θ−θ1(cid:0)∑aθj(cid:1)θ1 =Nθ−θ1(cid:0)∑aθj(cid:1)θ1
j=1 j=1 j=1 j=1
therefore
N N
(cid:0)∑a (cid:1)θ ≤Nθ−1∑aθ.
j j
j=1 j=1