Chater 2 Lebesgue Seuece Saces Abstract I this chater, we will itroduce the so-called Lebesgue seuece saces, i the fiite ad also i the ifiite dimesioal case We study some roerties of the saces, eg, comleteess, searability, duality, ad embeddig We also examie the validity of Hölder, Mikowski, Hardy, ad Hilbert ieuality which are related to the aforemetioed saces Although Lebesgue seuece saces ca be obtaied from Lebesgue saces usig a discrete measure, we will ot follow that aroach ad will rove the results i a direct maer This will highlight some techiues that will be used i the subseuet chaters 2 Hölder ad Mikowski Ieualities I this sectio we study the Hölder ad Mikowski ieuality for sums Due to their imortace i all its forms, they are sometimes called the workhorses of aalysis Defiitio 2 The sace l, with <, deotes the -dimesioal vector sace R for which the fuctioal x l i x i ) 2) is fiite, where x x,,x ) I the case of, we defie l as x l su x i i {,,} From Lemma 24 we obtai i fact that l defies a orm i R Sriger Iteratioal Publishig Switzerlad 206 RE Castillo, H Rafeiro, A Itroductory Course i Lebesgue Saces, CMS Books i Mathematics, DOI 0007/978-3-39-30034-4 2 23
24 2 Lebesgue Seuece Saces Examle 22 Let us draw the uit ball for articular values of for 2, as i Figs 2, 22, ad 23 y e 2 e x Fig 2 Uit ball for l 2 y e 2 e x Fig 22 Uit ball for l 2 Lemma 23 Hölder s ieuality) Let ad be real umbers with < < such that + The x k y k ) / ) / x k y k 22)
2 Hölder ad Mikowski Ieualities 25 y e 2 e x Fig 23 Uit ball for l 2 2 for x k,y k R Proof Let us take α x k x k ), β y k / y k ) / By Youg s ieuality 5) we get x k y k x k ) / y k ) x k / x k + y k y k Termwise summatio gives x k y k x k ) / y k ) / + ad from this we get x k y k ) / ) / x k y k We ca iterret the ieuality 22) i the followig way: If x l ad y l the x y l where stads for comoet-wise multilicatio ad moreover x y l x l y l
26 2 Lebesgue Seuece Saces Lemma 24 Mikowski s ieuality) Let, the ) / ) / ) / k + y k x k x k y + 23) for x k, y k R Proof We have By Lemma 23 we get x k + y k x k + y k x k ) / x k + y k x k + y k x k x k + y k + + y k x k + y k ) / ) / k y k + y k x ) Sice +, the ), from which x k + y k x k ) / + ) / ) / k y k + y k x, the ) ) / ) / k + y k x k x + k y, which etails 23) 22 Lebesgue Seuece Saces We ow wat to exted the -dimesioal l sace ito a ifiite dimesioal seuece sace i a atural way Defiitio 25 The Lebesgue seuece sace also kow as discrete Lebesgue sace) with <, deoted by l or sometimes also by l N), stads for the set of all seueces of real umbers x {x } N such that x k < We edow the Lebesgue seuece sace with the orm, x l {x } N l x k ) /, 24) where x l
22 Lebesgue Seuece Saces 27 We leave as Problem 224 to show that this is ideed a orm i l, therefore l, l ) is a ormed sace We will deote by R the set of all seueces of real umbers x {x } N Examle 26 The Hilbert cube H is defied as the set of all real seueces {x } N such that 0 x /, ie H : {x R : 0 x /} By the hyer-harmoic series we have that the Hilbert cube is ot cotaied i l but is cotaied i all l with > Let us show that l is a subsace of the sace R Letx ad y be elemets of l ad α, β be real umbers By Lemma 24 we have that αx k + βy k ) / α x k ) / + β y k ) / 25) Takig limits i 25), first to the right-had side ad after to the left-had side, we arrive at αx k + βy k ) / α x k ) / + β y k ) /, 26) ad this shows that αx+βy is a elemet of l ad therefore l is a subsace of R The Lebesgue seuece sace l is a comlete ormed sace for all We first rove for the case of fiite exoet ad for the case of it will be show i Theorem 2 Theorem 27 The sace l N) is a Baach sace whe < Proof Let {x } N be a Cauchy seuece i l N), where we take the seuece x as x x ),x) 2,) The for ay ε > 0 there exists a 0 N such that if, m 0, the x x m l < ε, ie / x ) j x m) j j < ε, 27) wheever, m 0 From 27) it is immediate that for all j,2,3, x ) j x m) j < ε, 28) wheever, m 0 Takig a fixed j from 28) we see that x ) j,x 2) j,) is a Cauchy seuece i R, therefore there exists x j R such that lim m x m) j x j
28 2 Lebesgue Seuece Saces Let us defie x x,x 2,) ad show that x is i l ad lim x x From 27) we have that for all, m 0 from which k j k j x m) j x ) j < ε, k,2,3, x j x ) j k j lim x m) m j x ) j ε, wheever 0, This shows that x x l ad we also deduce that lim x x Fially i virtue of the Mikowski ieuality we have x j j / j j x ) j x ) j + x j x ) j / + / x j x ) j j which shows that x is i l N) ad this comletes the roof The ext result shows that the Lebesgue seuece saces are searable whe the exoet is fiite, ie, the sace l admits a eumerable dese subset Theorem 28 The sace l N) is searable wheever < Proof Let M be the set of all seueces of the form, 2,,,0,0,) where N ad k Q We will show that M is dese i l Letx {x k } be a arbitrary elemet of l, the for ε > 0 there exists which deeds o ε such that the x k < ε /2 k+ Now, sice Q R, we have that for each x k there exists a ratioal k such that x k k < ε, 2 which etails x l x k k < ε /2, x k k + k+ x k < ε, ad we arrive at x l < ε This shows that M is dese i l, imlyig that l is searable sice M is eumerable /,
22 Lebesgue Seuece Saces 29 With the otio of Schauder basis recall the defiitio of Schauder basis i Defiitio B3), we ow study the roblem of duality for the Lebesgue seuece sace Theorem 29 Let < < The dual sace of l N) is l N) where + Proof A Schauder basis of l is e k {δ kj } j N where k N ad δ kj stads for the Kroecker delta, ie, δ kj ifk j ad 0 otherwise If f l ), the f x) α k f e k ), x {α k } We defie T f ){ f e k )} We wat to show that the image of T is i l, for that we defie for each, the seuece x ξ ) k ) with The Moreover ξ ) k { f ek ) f e k ) if k ad f e k ) 0, 0 if k > or f e k )0 f x ) ξ ) k f e k ) f x ) f x f f f f e k ) ) ξ ) k ) f e k ) f e k ) ), from which Takig, we obtai ) ) f e k ) f e k ) f f e k ) ) f where { f e k )} l
30 2 Lebesgue Seuece Saces Now, we affirm that: i) T is oto I effect give b β k ) l, we ca associate a bouded liear fuctioal g l ), give by gx) α k β k with x α k ) l the boudedess is deduced by Hölder s ieuality) The g l ) ii) T is - This is almost straightforward to check iii) T is a isometry We see that the orm of f is the l orm of Tf f x) α k f e k ) x ) α k ) f e k ) f e k ) ) Takig the suremum over all x of orm, we have that ) f f e k ) Sice the other ieuality is also true, we ca deduce the euality ) f f e k ), with which we establish the desired isomorhism f {f e k )} The l saces satisfy a embeddig roerty, formig a ested seuece of Lebesgue seueces saces Theorem 20 If 0 < < <, the l N) l N) Proof Let x l, the x < Therefore there exists 0 N such that if 0, the x < Now, sice 0 < <, the 0 < ad x < if > 0, by which x < x if > 0 Let M max{ x, x 2,, x 0,}, the imlyig that x l x x x < M x < +,
23 Sace of Bouded Seueces 3 To show that l N) l N), we take the followig seuece x / for all N with <, ad sice <, the > Now we have x < / The last series is coverget sice it is a hyer-harmoic series with exoet bigger tha, therefore x l N) O the other had x ad we get the harmoic series, which etails that x / l N) 23 Sace of Bouded Seueces The sace of bouded seueces, deoted by l or sometimes l N), isthesetof all real bouded seueces {x } N it is clear that l is a vector sace) We will take the orm i this sace as x x l su x, 29) N where x x,x 2,,x,) The verificatio that 29) is ideed a orm is left to the reader A almost immediate roerty of the l -sace is its comleteess, iheritig this roerty from the comleteess of the real lie Theorem 2 The sace l is a Baach sace Proof Let {x } N be a Cauchy seuece i l, where x x ) for ay ε > 0 there exists 0 > 0 such that if m, 0 the x m x < ε Therefore for fixed j we have that if m, 0, the,x) 2,) The x m) j x ) j < ε 20) resultig that for all fixed j the seuece x ) j,x 2) j,) is a Cauchy seuece i R, ad this imlies that there exists x j R such that lim m x m) j x j Let us defie x x,x 2,) Now we wat to show that x l ad lim x x
32 2 Lebesgue Seuece Saces From 20) we have that for 0, the x j x ) j lim j xm) x ) j ε, 2) sice x {x ) j } j N l, there exists a real umber M such that all j By the triagle ieuality, we have x j x j x ) + j x ) j ε + M x ) j M for wheever 0, this ieuality beig true for ay j Moreover, sice the right-had side does ot deed o j, therefore {x j } j N is a seuece of bouded real umbers, this imlies that x {x j } j N l From 2) we also obtai x x l su x ) j x j < ε j N wheever 0 From this we coclude that lim x x ad therefore l is comlete The followig result shows a atural way to itroduce the orm i the l sace via a limitig rocess Theorem 22 Takig the orm of Lebesgue seuece sace as i 24) we have that lim x l x l Proof Observe that x k x k ), therefore x k x l for k,2,3,,, from which su x k x l, k whece x l limif x l 22) O the other had, ote that x k ) the for all ε > 0, there exists N such that therefore x l N x k + ε ) su x k k ) x l, ) x l N + ε x l N + ε x l ),
23 Sace of Bouded Seueces 33 limsu x l x l 23) Combiig 22) ad 23) results x l limif x l limsu x l x l, ad from this we coclude that lim x l x l Now we study the dual sace of l which is l Theorem 23 The dual sace of l is l Proof For all x l, we ca write x α k e k, where e k δ kj ) j forms a Schauder basis i l, sice ad x α k e k 0,,0,α }{{} +,) x α k e k l k+ α k e k l 0 sice the series α k e k is coverget Let us defie T f ){ f e k )}, for all f l ) Sice f x) α k f e k ), the f e k ) f, sice e k l I coseuece, su f e k ) f, therefore { f e k )} l We affirm that: i) T is oto I fact, for all b {β k } l, let us defie : l R as gx) α k β k if x {α k } l The fuctioal g is bouded ad liear sice gx) α k β k su β k the g l ) Moreover, sice ge k ) j N δ kj β j, α k x l su β k, T g){ge k )} {β k } b ii) T is - If Tf Tf 2, the f e k ) f 2 e k ), for all k Sice we have f x) α k f e k ) ad f 2 x) α k f 2 e k ), the f f 2 iii) T is a isometry I fact, T f su f e k ) f 24)
34 2 Lebesgue Seuece Saces ad f x) α k f e k ) su j N f e k ) α k x l su f e k ) The f su f e k ) T f 25) Combiig 24) ad 25) we get that T f f We thus showed that the saces l ) ad l are isometric Oe of the mai differece betwee l ad l saces is the searability issue The sace of bouded seuece l is ot searable, cotrastig with the searability of the l saces wheever <, see Theorem 28 Theorem 24 The sace l is ot searable Proof Let us take ay eumerable seuece of elemets of l, amely {x } N, where we take the seueces i the form ) x x ),x) 2,x) 3,,x) k, ) x 2 x 3 x 2),x2) 2,x2) 3,,x2) k, ) x 3),x3) 2,x3) 3,,x3) k, ) x k x k),xk) 2,xk) 3,,xk) k, We ow show that there exists a elemet i l which is at a distace bigger tha for all elemets of {x } N, showig the o-searability ature of the l sace Let us take x {x } N as { 0, if x ) ; x x x ) +, if x ) < It is clear that x l ad x x l > for all N, which etails that l is ot searable We ow defie some subsaces of l, which are widely used i fuctioal aalysis, for examle, to costruct couter-examles Defiitio 25 Let x x,x,) By c we deote the subsace of l such that lim x exists ad is fiite By c 0 we deote the subsace of l such that lim x 0 By c 00 we deote the subsace of l such that sux) is fiite
24 Hardy ad Hilbert Ieualities 35 These ewly itroduced saces ejoy some iterestig roerties, eg, c 0 is the closure of c 00 i l For more roerties, see Problem 220 24 Hardy ad Hilbert Ieualities We ow deal with the discrete versio of the well-kow Hardy ieuality Theorem 26 Hardy s ieuality) Let {a } N be a seuece of real ositive umbers such that a < The Proof Let α A ) a k ) a where A a + a 2 + + a, ie, A α, the a + a 2 + + a α, 26) from which we get that a α )α Let us cosider ow α α a α [ ] α )α α I virtue of Corollary 0 we have therefore α ) α α α α α + ) α ) α α ) + α ) + α + )α, α ) α a α α + α + )α ) α α α α + )α + ) )α α α α + )α + + )α [ )α α ],
36 2 Lebesgue Seuece Saces from which N α N α a N Nα N 0 [ )α α ] [ α + α 2α 2 + Nα N ] The N α By Hölder s ieuality we have that N α a α a a ) ) α ) ) ) α, the ad this imlies α ) a ) ) a k ) a We ow wat to study the so-called Hilbert ieuality We eed to remember some basic facts about comlex aalysis, amely siz) z + ) z + + ) 27) z Let us cosider the fuctio f z) zz + ) > ) defied i the regio D {z C : 0 < z < } We wat to obtai the Lauret exasio I fact, if z <, the
24 Hardy ad Hilbert Ieualities 37 therefore + z z) z) 0 f z) By the same reasoig, let us cosider gz) 0 z + 0 ) z, ) z 28) ) + z defied i the regio D 2 {z C : z > } Sice <, the Therefore z + z z ) ) 0 z gz) 0 0 ) z ) z 29) We ow obtai some auxiliary ieuality before showig the validity of the Hilbert ieuality 220) Theorem 27 For each ositive umber m ad for all real > we have Proof Note that m m + ) m ) m + ) si ˆ 0 ˆ 0 ˆ By 28) ad 29) we deduce that m m + ) ˆ 0 0 m x m + x) dx dz z + z) ) z 0 ˆ dz + z + z) ) dz + ˆ z + dz ) + z ) z 0 ) dz
38 2 Lebesgue Seuece Saces 0 0 ˆ ) 0 ) + + 0 ) + + z ˆ dz + ) z dz 0 ) + ) + ) + ) + + ) si This last oe is obtaied by 27) with z Remark 28 I fact the roof of Theorem 27 is a two lie roof if we remember that ˆ x α dx Bα,β) + x) α+β ad the fact that B α,α) 0 siα,0< α <, see Aedix C Before statig ad rovig the Hilbert ieuality we eed to digress ito the cocet of double series Let { } x k, j be a double seuece, viz a real-valued j, fuctio x : N N R We say that a umber L is the limit of the double seuece, deoted by lim x k, j L, k, j if, for all ε > 0 there exists ε) such that x k, j L < ε wheever k > ad j > We ca ow itroduce the otio of double series usig the kow costructio for the series, amely if there exists the double limit x k, j Σ k, j lim k, j Σ k, j Σ
24 Hardy ad Hilbert Ieualities 39 where Σ k, j is the rectagular artial sum give by Σ k, j k j m x m, A otio related to the double series is the otio of iterated series, give by x k, j ad j x k, j ) j We ca visualize the iterated series i the followig way We first rereset the double seuece as umbers i a ifiite rectagular array ad the sum by lies ad by colums i the followig way: x, x,2 x,3 j x, j : L x 2, x 2,2 x 2,3 j x 2, j : L 2 x 3, x 3,2 x 3,3 j x 2, j : L 3 C : x k, C 2 : x k,2 C 3 : x k,3 ad ow the iterated series are give by jc j ad L k It is ecessary some cautio whe dealig with iterated series sice the euality jc j L k is i geeral ot true eve if the series coverges, as the followig examle shows 2 2 0 0 0 0 0 3 4 3 4 0 0 0 7 0 0 8 7 8 0 0 5 0 0 0 6 5 6 0 2 4 8 6 ad clearly the obtaied series are differet Fortuately we have a Fubii tye theorem for series which states that whe a double series is absolutely coverget the the double series ad the iterated series are the same, ie k, j x k, j j x k, j 32 x k, j ) j Not oly that, it is also ossible to show a stroger result, that if the terms of a absolutely coverget double series are ermuted i ay order as a simle series, their sum teds to the same limit
40 2 Lebesgue Seuece Saces Theorem 29 Hilbert s ieuality) Let, > be such that + ad {a } N, {b } N be seueces of oegative umbers such that m am ad b are coverget The a m b m, m + ) ) ) am b 220) si m Proof Usig Hölder s ieuality ad Proositio 27 we get m, m, a m b m + m a m b m + ) m m + ) ) m ) a m, m b m + ) m, m m + ) ) m m ) a m m + ) b m m m + ) ) ) m si am si b ) ) m si am si b ) ) ) ) si si am b m ) ) si am b, m which shows the result 25 Problems 220 Prove the followig roerties of the subsaces of l itroduced i Defiitio 25 a) The sace c 0 is the closure of c 00 i l b) The sace c ad c 0 are Baach saces c) The sace c 00 is ot comlete
25 Problems 4 22 Show that s,ρ) is a comlete metric sace, where s is the set of all seueces x x,x 2,) ad ρ is give by ρx,y) x k y k 2 k + x k yk 222 Let l w), be the set of all real seueces x x,x 2,) such that x k w k < where w w,w 2,) ad w k > 0 Does N : l w) R give by N x) : x k w k ) defies a orm i l w)? 223 As i the case of Examle 22, draw the uit ball for l 3, l3, ad l 3 2 224 Prove that 24) defies a orm i the sace l N) 225 Prove the Cauchy-Buyakovsky-Schwarz ieuality ) 2 x i y i i xi 2 i ) y 2 i i ) without usig Jese s ieuality This ieuality is sometimes called Cauchy, Cauchy-Schwarz or Cauchy-Buyakovsky Hit: Aalyze the uadratic form i x i u + y i v) 2 u 2 i x 2 i + 2uv i x i y i + v 2 i y 2 i 226 Let {a } Z ad {b } Z be seueces of real umbers such that k a < ad b m < m where > Let C m a m b m Prove that a) C k / m a m b m ) / where + b) C ) / k b ) / 227 If a > 0for,2,3, show that a a 2 a e a
42 2 Lebesgue Seuece Saces If a a 2 a k a 0 ad α β > 0 Demostrate that ) /α ak α ) /β a β k 228 Use Theorem 05 to show the Theorem 26 Hit: Choose a seuece {a } N of ositive umbers such that a + a N Cosider A N N a ad defie f a χ,) 229 Demostrate that l is ot the dual sace of l 230 Show that x l x l 22) wheever < < Hit: First, show the ieuality 22) whe x l Use that result ad the homogeeity of the orm to get the geeral case 26 Notes ad Bibliograhic Refereces The history of Hölder s ieuality ca be traced back to Hölder [32] but the aer of Rogers [6] receded the oe from Hölder just by oe year, for the comlete history see Maligrada [48] The Mikowski ieuality is due to Mikowski [5] but it seems that the classical aroach to the Mikowski ieuality via Hölder s ieuality is due to Riesz [58] The Hardy ieuality Theorem 26) aeared i Hardy [26] as a geeralizatio of a tool to rove a certai theorem of Hilbert Accordig to Hardy, Littlewood, ad Pólya [30], the Hilbert ieuality Theorem 29) was icluded by Hilbert for 2 i his lectures, ad it was ublished by Weyl [82], the geeral case > aeared i Hardy [27] The Cauchy-Buyakovsky-Schwarz ieuality, which aears i Problem 225, was first roved by Cauchy [6]
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