Functional Analysis I

Transcription

1 Fuctioal Aalysis I Term 1, Vassili Gelfreich

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3 Cotets 1 Vector spaces Defiitio Examples of vector spaces Hamel bases Normed spaces Norms Four famous iequalities Examples of orms o a space of fuctios Equivalece of orms Liear Isometries Covergece i a ormed space Defiitio ad examples Topology o a ormed space Closed sets Compactess Baach spaces Completeess: Defiitio ad examples The completio of a ormed space Examples Lebesgue spaces Itegrable fuctios Properties of Lebesgue itegrals Lebesgue space L 1 (R) L p spaces Hilbert spaces Ier product spaces Natural orms Parallelogram law ad polarisatio idetity Hilbert spaces: Defiitio ad examples Orthoormal bases i Hilbert spaces Orthoormal sets Gram-Schmidt orthoormalisatio Bessel s iequality Covergece Orthoormal basis i a Hilbert space iii

4 8 Closest poits ad approximatios Closest poits i covex subsets Orthogoal complemets Best approximatios Weierstrass Approximatio Theorem Separable Hilbert spaces Defiitio ad examples Isometry to l Liear maps betwee Baach spaces Cotiuous liear maps Examples Kerel ad rage Liear fuctioals Defiitio ad examples Riesz represetatio theorem Liear operators o Hilbert spaces Complexificatio Adjoit operators Self-adjoit operators Itroductio to Spectral Theory Poit spectrum Ivertible operators Resolvet ad spectrum Compact operators Spectral theory for compact self-adjoit operators Sturm-Liouville problems 82 iv

5 Preface These otes follow the lectures o Fuctioal Aalysis give i the Autum Term of If you fid a mistake or misprit please iform the author by sedig a to v.gelfreich@warwick.ac.uk. The author thaks James Robiso for his set of otes ad selectio of exercises which sigificatly facilitated the preparatio of the lectures. The author also thaks all studets who helped with proofreadig the otes. 1 Vector spaces 1.1 Defiitio. A vector space V over a field K is a set equipped with two biary operatios called vector additio ad multiplicatio by scalars. Elemets of V are called vectors ad elemets of K are called scalars. The sum of two vectors x,y V is deoted x + y, the product of a scalar α K ad vector x V is deoted αx. It is possible to cosider vector spaces over a arbitrary field K, but we will cosider the fields R ad C oly. So we will always assume that K deotes either R or C ad refer to V as a real or complex vector space respectively. I a vector space, additio ad multiplicatio have to satisfy the followig set of axioms: Let x,y,z be arbitrary vectors i V, ad α,β be arbitrary scalars i K, the Associativity of additio: x + (y + z) = (x + y) + z. Commutativity of additio: x + y = y + z. There exists a elemet 0 V, called the zero vector, such that x + 0 = x for all x V. For all x V, there exists a elemet y V, called the additive iverse of x, such that x + y = 0. The additive iverse is deoted x. Associativity of multiplicatio: 1 α(βx) = (αβ)x. Distributivity: α(x + y) = αx + αy ad (α + β)x = αx + βx. There is a elemet 1 K such that 1x = x for all x V. This elemet is called the multiplicative idetity i K. 1 The purist would ot use the word associativity for this property as it icludes two differet operatios: αβ is a product of two scalars ad βx ivolves a vector ad scalar. 1

6 It is coveiet to defie two additioal operatios: subtractio of two vectors ad divisio by a (o-zero) scalar are defied by 1.2 Examples of vector spaces 1. R is a real vector space. 2. C is a complex vector space. 3. C is a real vector space. x y = x + ( y), x/α = (1/α)x. 4. The set of all polyomials P is a vector space: P = { α k x k : α k K, N k=0 }. 5. The set of all bouded sequeces l (K) is a vector space: { } l (K) = (x 1,x 2,...) : x k K for all k N, sup x k < k N. For two sequeces x,y l (K), we defie x + y by For α K, we set x + y = (x 1 + y 1,x 2 + y 2,...). αx = (αx 1,αx 2,...). We will always use these defiitios of additio ad multiplicatio by scalars for sequeces. 6. Let 1 p <. The set l p (K) of all p th power summable sequeces is a vector space: { } l p (K) = (x 1,x 2,...) : x k K, x k p <. The defiitio of the multiplicatio by scalars ad vector additio is the same as i the previous example. Let us check that the sum x + y l p (K) for ay x,y l p (K). Ideed, x k + y k p ( x k + y k ) p 2 p ( x k p + y k p ) 2 p x k p + 2 p y k p <. 2

7 7. The space C[0, 1] of all real-valued cotiuous fuctios o the closed iterval [0, 1] is a vector space. The additio ad multiplicatio by scalars are defied aturally: for f,g C[0,1] ad α R we defied by f + g the fuctio whose values are give by ( f + g)(t) = f (t) + g(t), t [0,1], ad α f is the fuctio whose values are (α f )(t) = α f (t), t [0,1], We will always use these defiitios o all spaces of fuctios to be cosidered later. 8. The set L 1 (0,1) of all real-valued cotiuous fuctios f o the ope iterval (0,1) for which 1 f (t) dt < is a vector space. 0 If f C[0,1] the f L 1 (0,1). Ideed, sice [0,1] is compact f is bouded (ad attais its lower ad upper bouds). The i.e. f L 1 (0,1). 1 0 f (t) dt max f (t) <, t [0,1] We ote that L 1 (0,1) cotais some fuctios which do ot belog to C[0,1]. For example, f (t) = t 1/2 is ot cotiuous o [0,1] but it is cotiuous o (0,1) ad 1 1 f (t) dt = t 1/2 dx = 2t 1/2 1 = 2 <, 0 so f L 1 (0,1). 0 We coclude that C[0,1] is a strict subset of f L 1 (0,1). 1.3 Hamel bases 0 Defiitio 1.1 The liear spa of a subset E of a vector space V is the collectio of all fiite liear combiatios of elemets of E: { } Spa(E) = x V : x = α j e j, N, α j K, e j E. We say that E spas V if V = Spa(E), i.e. every elemet of V ca be writte as a fiite liear combiatio of elemets of E. 3

8 Defiitio 1.2 A set E is liearly idepedet if ay fiite collectio of elemets of E is liearly idepedet: α j e j = 0 = α 1 = α 2 = = α = 0 for ay choice of N, e j E ad α j K. Defiitio 1.3 A Hamel basis E for V is a liearly idepedet subset of V which spas V. Examples: 1. Ay basis i R is a Hamel basis. 2. The set E = { 1,x,x 2,... } is a Hamel basis i the space of all polyomials. Lemma 1.4 If E is a Hamel basis for a vector space V the ay elemet x V ca be uiquely writte i the form x = α j e j where N, α j K ad e j E. Exercise: Prove the lemma. Defiitio 1.5 We say that a set is fiite if it cosists of a fiite umber of elemets. Theorem 1.6 If V has a fiite Hamel basis the every Hamel basis for V has the same umber of elemets. Defiitio 1.7 If V has a fiite basis E the the dimesio of V (deoted dimv ) is the umber of elemets i E. If V has o fiite basis the we say that V is ifiitedimesioal. Example: I R ay basis cosists of vectors. Therefore dimr =. Let V ad W be two vector spaces over K. Defiitio 1.8 A map L : V W is called liear if for ay x,y V ad ay α K L(x + αy) = L(x) + αl(y). Defiitio 1.9 If a liear map L : V W is a bijectio, the L is called a liear isomorphism. We say that V ad W are liearly isomorphic if there is a bijective liear map L : V W. 4

9 Propositio 1.10 Ay -dimesioal vector space over K is liearly isomorphic to K. Proof: If E = {e j : 1 j } is a basis i V, the every elemet x V is represeted uiquely i the form x = α j e j. The map L : x (α 1,...,α ) is a liear bijectio V K. Therefore V is liearly isomorphic to K. I order to show that a vector space is ifiite-dimesioal it is sufficiet to fid a ifiite liearly idepedet subset. Let s cosider the followig examples: 1. l p (K) is ifiite-dimesioal (1 p ). Proof. The set E = {(1,0,0,0,...), (0,1,0,0,...), (0,0,1,0,...),...} is ot fiite ad liearly idepedet. Therefore diml p (K) =. Remark: This liearly idepedet subset is ot a Hamel basis. Ideed, the sequece x = (x 1,x 2,x 3,...) with x k = e k belogs to l p (K) for ay p 1 but caot be represeted as a sum of fiitely may elemets of the set E. 2. C[0,1] is ifiite-dimesioal. Proof: The set E = {x k : k N} is liearly idepedet subset of C 0 [0,1]: Ideed, suppose p(x) = α k x k = 0 for all x [0,1]. Differetiatig the equality times we get p () (x) =!α = 0. Which implies α = 0. Therefore p(x) 0 implies α k = 0 for all k. The liearly idepedet sets provided i the last two examples are ot Hamel bases. This is ot a coicidece: we will see later that l p (K) ad C[0,1] (as well as may other fuctioal spaces) do ot have a coutable Hamel basis. Theorem 1.11 Every vector space has a Hamel basis. The proof of this theorem is based o Zor s Lemma. We ote that i may iterestig vector spaces (called ormed spaces), a very large umber of elemets should be icluded ito a Hamel basis i order to eable represetatio of every elemet i the form of a fiite sum. The the basis is too large to be useful for the study of the origial vector space. A atural idea would be to allow ifiite sums i the defiitio of a basis. I order to use ifiite sums we eed to defie covergece which caot be doe usig the axioms of vector spaces oly. A additioal structure o the vector space should be defied. 5

10 2 Normed spaces 2.1 Norms Defiitio 2.1 A orm o a vector space V is a map : V R such that for ay x,y V ad ay α K: 1. x 0, ad x = 0 x = 0 (positive defiiteess); 2. αx = α x (positive homogeeity); 3. x + y x + y (triagle iequality). The pair (V, ) is called a ormed space. I other words, a ormed space is a vector space equipped with a orm. Examples: 1. R with each of the followig orms is a ormed space: (a) x = x k 2 (b) x p = (c) x = max 1 k x k. ( ) 1/p x k p, 1 p < 2. l p (K) is a vector space with the followig orm (1 p < ) x l p = ( ) 1/p x k p. 3. l (K) is a vector space with the followig orm x l = sup x k. k N I order to prove the triagle iequality for the l p orm, we will state ad prove several iequalities. 6

11 2.2 Four famous iequalities Lemma 2.2 (Youg s iequality) If a,b > 0, 1 < p,q <, 1 p + 1 q = 1, the ab ap p + bq q. Proof: Cosider the fuctio f (t) = t p p t + 1 q defied for t 0. Sice f (t) = t p 1 1 vaishes at t = 1 oly, ad f (t) = (p 1)t p 2 0, the poit t = 1 is a global miimum for f. Cosequetly, f (t) f (1) = 0 for all t 0. Now substitute t = ab q/p : f (ab q/p ) = ap b q ab q/p + 1 p q 0. Multiplyig the iequality by b q yields Youg s iequality. Lemma 2.3 (Hölder s iequality) If 1 p,q, 1 p + 1 q = 1, x lp (K), y l q (K), the x j y j x l p y l q. Proof. If 1 < p,q <, we use Youg s iequality to get that for ay N ( x j y j 1 x j p x l p y l q p x p + 1 y j q ) l p q y q 1 l q p + 1 q = 1 Therefore for ay N x j y j x l p y l q. Sice the partial sums are mootoically icreasig ad bouded above, the series coverge ad Hölder s iequality follows by takig the limit as. If p = 1 ad q = : x j y j max y j x j x l 1 y l. 1 j Therefore the series coverges ad Hölder s iequality follows by takig the limit as. Lemma 2.4 (Cauchy-Schwartz iequality) If x,y l 2 (K) the ( ) 1/2 ( ) 1/2 x j y j x j 2 j y 2. 7

12 Proof: This iequality coicides with Hölder s iequality with p = q = 2. Now we state ad prove the triagle iequality for the l p orm. Lemma 2.5 (Mikowski s iequality) If x,y l p (K) for 1 p the x + y l p (K) ad x + y l p x l p + y l p. Proof: If 1 < p <, defie q from 1 p + 1 q = 1. The usig Hölder s iequality (fiite sequeces belog to l p with ay p) we get 2 x j + y j p = x j + y j p 1 x j + y j x j + y j p 1 x j + x j + y j p 1 y j ( ) 1/q ( ) 1/p x j + y j (p 1)q j x p (Hölder s iequality) + Dividig the iequality by 1 q = 1 p, we get for all ( ) 1/q ( ) 1/p x j + y j (p 1)q j y p. ( x j + y j p) 1/q ad usig that (p 1)q = p ad 1 ( ) 1/p x j + y j p ( ) 1/p x j p + ( ) 1/p y j p. The series o the right had side coverge to x l p + y l p. Cosequetly the series o the left had side also coverge, x+y l p (K), ad Mikowski s iequality follows by takig the limit as. Exercise: Prove Mikowski s iequality for p = 1 ad p =. 2.3 Examples of orms o a space of fuctios Each of the followig formulae defies a orm o C[0,1], the space of all cotiuous fuctios o [0,1]: 2 We do ot start directly with = because a priori we do ot kow covergece for some of the series ivolved i the proof. 8

13 1. the sup(remum) orm 2. the L 1 orm 3. the L 2 orm f = sup f (t) ; t [0,1] f L 1 = 1 0 f (t) dt ; ( 1 1/2 f L 2 = f (t) dt) 2. 0 Exercise: Check that each of these formulae defies a orm. For the case of the L 2 orm, you will eed a Cauchy-Schwartz iequality for itegrals. Example: Let k N. The space C k [0,1] cosists of all cotiuous real-valued fuctios which have cotiuous derivatives up to order k. The orm o C k [0,1] is defied by f C k = k j=0 where f ( j) deotes the derivative of order j. 2.4 Equivalece of orms sup t [0,1] f ( j) (t), We have see that various differet orms ca be itroduced o a vector space. I order to compare differet orms it is coveiet to itroduce the followig equivalece relatio. Defiitio 2.6 Two orms 1 ad 2 o a vector space V are equivalet if there are costats c 1,c 2 > 0 such that I this case we write 1 2. c 1 x 1 x 2 c 2 x 1 for all x V. Theorem 2.7 Ay two orms o R are equivalet. 3 Example: The orms L 1 ad o C[0,1] are ot equivalet. 3 You already saw this statemet i Aalysis III ad Differetiatio i Year 2. The proof is based o the observatio that the uit sphere S R is sequetially compact. The we checked that f (x) = x 2 / x 1 is cotiuous o S ad cosequetly it is bouded ad attais its lower ad upper bouds o S. We set c 1 = mi S f ad c 2 = max S f. 9

14 Proof: Cosider the sequece of fuctios f (t) = t with N. Obviously f C[0,1]. We see that f = max t [0,1] t = 1, f L 1 = 1 0 t dt = Suppose the orms are equivalet. The there is a costat c 2 > 0 such that for all f : f f L 1 = + 1 c 2, which is ot possible for all. This cotradictio implies the orms are ot equivalet. 2.5 Liear Isometries Suppose V ad W are ormed spaces. Defiitio 2.8 If a liear map L : V W preserves orms, i.e. L(x) = x for all x V, it is called a liear isometry. This defiitio implies L is ijective, i.e., L : V L(V ) is bijective, but it does ot imply L(V ) = W, i.e., L is ot ecessarily ivertible. Note that sometimes the ivertibility property is icluded ito the defiitio of the isometry. Fially, i Metric Spaces the word isometry is used to deote distace-preservig trasformatios. Defiitio 2.9 We say that two ormed spaces are isometrically isomorphic (or simply isometric), if there is a ivertible liear isometry betwee them. A liear ivertible map ca be used to pull back a orm as follows. Propositio 2.10 Let (V, V ) be a ormed space, W a vector space, ad L : W V a liear isomorphism. The x W := L(x) V defies a orm o W. Proof: For ay x,y V ad ay α K we have: x W = L(x) V 0, αx W = L(αx) V = α L(x) V = α x W. If x W = L(x) V = 0, the L(x) = 0 due to o-degeeracy of the orm V. Sice L is ivertible, we get x = 0. Therefore W is o-degeerate. 10

15 Fially, the triagle iequality follows from the triagle iequality for V : x + y W = L(x) + L(y) V L(x) V + L(y) V = x W + y W. Therefore, W is a orm. Note that i the propositio the ew orm is itroduced i such a way that L : (W, W ) (V, V ) is a liear isometry. Let V be a fiite dimesioal vector space ad = dimv. We have see that V is liearly isomorphic to K. The the propositio implies the followig statemets. Corollary 2.11 Ay fiite dimesioal vector space V ca be equipped with a orm. Corollary 2.12 Ay -dimesioal ormed space V is isometrically isomorphic to K equipped with a suitable orm. Sice ay two orms o R (ad therefore o C ) are equivalet we also get the followig statemet. Theorem 2.13 Let V be a fiite-dimesioal vector space. The all orms o V are equivalet. 11

16 3 Covergece i a ormed space 3.1 Defiitio ad examples The orm o a vector space V ca be used to measure distaces betwee poits x,y V. So we ca defie the limit of a sequece. Defiitio 3.1 A sequece (x ) =1, x V, N, coverges to a limit x V if for ay ε > 0 there is N N such that The we write x x. x x < ε for all > N. We ote that the sequece of vectors x x if ad oly if the sequece of oegative real umbers x x 0. Propositio 3.2 The limit of a coverget sequece is uique. Exercise: Prove it. Propositio 3.3 Ay coverget sequece is bouded. Exercise: Prove it. Propositio 3.4 If x coverges to x, the x x. Exercise: Prove it. It is possible to check covergece of a sequece of real umbers without actually fidig its limit: it is sufficiet to check that it satisfies the followig defiitio: Defiitio 3.5 (Cauchy sequece) A sequece (x ) =1 i a ormed space V is Cauchy if for ay ε > 0 there is a N such that x x m < ε for all m, > N. Theorem 3.6 A sequece of real umbers coverges iff it is Cauchy. Propositio 3.7 Ay coverget sequece is Cauchy. Exercise: Prove it. Propositio 3.8 Ay Cauchy sequece is bouded. Exercise: Prove it. 12

17 Example: Cosider the sequece f C[0,1] defied by f (t) = t. 1. f 0 i the L 1 orm. Proof: We have already computed the orms: Cosequetly, f 0. f L 1 = f does ot coverge i the sup orm. Proof: If m > 2 1 the f (2 1/ ) f m (2 1/ ) = m/ 1 4. Cosequetly ( f ) is ot Cauchy i the sup orm ad hece ot coverget. This example shows that the covergece i the L 1 orm does ot imply the poitwise covergece ad, as a results, does ot imply the covergece i the sup orm (ofte called the uiform covergece). Note that i cotrast to the uiform ad L 1 covergeces the otio of poitwise covergece is ot based o a orm o the space of cotiuous fuctio. Exercise: The poitwise covergece does ot imply the L 1 covergece. Hit: Costruct f with a very small support but make the maximum of f very large to esure that f L 1 >. Therefore f is ot bouded i the L 1 orm, hece ot coverget. We ca also make supp f supp f m = /0 for all m, such that m. The for ay t there is at most oe such that f (t) 0. The last property guaratees poitwise covergece: f (t) 0 for ay t. Propositio 3.9 If f C[0,1] for all N ad f f i the sup orm, the f f i the L 1 orm, i.e., Proof: 0 f f L 1 = f f 0 = f f L Therefore f f L f (t) f (t) dt sup f (t) f (t) = f f 0. 0 t 1 We have see that differet orms may lead to differet coclusios about covergece of a give sequece but sometime covergece i oe orm implies covergece i aother oe. The followig lemma shows that equivalet orms give rise to the same otio of covergece. 13

18 Lemma 3.10 Suppose 1 ad 2 are equivalet orms o a vector space V. The for ay sequece (x ): x x 1 0 x x 2 0. Proof: Sice the orms are equivalet, there are costat c 1,c 2 > 0 such that 0 c 1 x x 1 x x 2 c 2 x x 1 for all. The x x 2 0 implies x x 1 0, ad vice versa. 3.2 Topology o a ormed space We say that a collectio T of subsets of V is a topology o V if it satisfies the followig properties: 1. /0,V T ; 2. ay fiite itersectio of elemets of T belogs to T ; 3. ay uio of elemets of T belogs to T. A set equipped with a topology is called a topological space. The elemets of T are called ope sets. The topology ca be used to defie a coverget sequece ad cotiuous fuctio. We ote that each orm o V ca be used to defie a topology o V, i.e., to defie the otio of a ope set. Defiitio 3.11 A subset X V is ope, if for ay x X there is ε > 0 such that the ball of radius ε cetred aroud x belogs to X: Example: I ay ormed space V : B(x,ε) = {y V : y x < ε} X. 1. The uit ball cetred aroud the zero, B 0 = {x : x < 1}, is ope. 2. Ay ope ball B(x,ε) is ope. 3. V is ope. 4. The empty set is ope. It is ot too difficult to check that the collectio of ope sets defies a topology o V. You ca easily check from the defiitio that equivalet orms geerate the same topology, i.e., ope sets are exactly the same. The otio of covergece ca be defied i terms of the topology. 14

19 Defiitio 3.12 A ope eighbourhood of x is a ope set which cotais x. Lemma 3.13 A sequece x x if ad oly if for ay ope eighbourhood X of x there is N N such that x X for all > N. Proof: ( = ). Let x x. Take ay ope X such that x X. The there is ε > 0 such that B(x,ε) X. Sice the sequece coverges there is N such that x x < ε for all > N. The x B(x,ε) X for the same values of. ( =). Take ay ε > 0. The ball B(x,ε) is ope, therefore there is N such that x B(x,ε) for all > N. Hece x x < ε ad x x. 3.3 Closed sets Defiitio 3.14 A set X V is closed if its complemet V \ X is ope. Example: I ay ormed space V : 1. The uit sphere S = {x : x = 1} is closed 2. Ay closed ball is closed. B(x,ε) = {y V : y x ε } 3. V is closed. 4. The empty set is closed. Lemma 3.15 A subset X V is closed if ad oly if ay coverget sequece i X has its limit i X. Proof: The proof literally repeats the proof give i Differetiatio. Defiitio 3.16 We say that a subset L V is a liear subspace, if it is a vector space itself, i.e., if x 1,x 2 L ad λ K imply x 1 + λx 2 L. Propositio 3.17 Ay fiite dimesioal liear subspace L of a ormed space V is closed. Proof: Sice L is fiite-dimesioal, it has a fiite Hamel basis E = {e 1,e 2,...,e } L such that L = Spa(E). Suppose L is ot closed, the by Lemma 3.15 there is a coverget sequece x k x, x k L but x V \ L. The x is liearly idepedet from E (otherwise it would belog to L). Cosequetly Ẽ = {e 1,e 2,...,e,x } 15

20 is a Hamel basis i L = Spa(Ẽ). I this basis, the compoets of x k are give by (α k 1,...,αk,0) ad x correspods to the vector (0,...,0,1). We get i the limit as k (α k 1,...,αk,0) (0,...,0,1), which is obviously impossible. Therefore L is closed. 4 Example: The subspace of polyomial fuctios is liear but ot closed i C[0, 1] equipped with the sup orm. 3.4 Compactess Defiitio 3.18 (sequetial compactess) A subset K of a ormed space (V, V ) is (sequetially) compact if ay sequece (x ) =1 with x K has a coverget subsequece x j x with x K. Propositio 3.19 A compact set is closed ad bouded. Theorem 3.20 A subset of R is compact iff it is closed ad bouded. Corollary 3.21 A subset of a fiite-dimesioal vector space is compact iff it is closed ad bouded. Example: The uit sphere i l p (K) is closed, bouded but ot compact. Proof: Take the sequece e j such that e j = (0,...,0, }{{} 1,0,...). j th place We ote that e j e k l p = 2 1/p for all j k. Cosequetly, (e j ) does ot have ay coverget subsequece, hece S is ot compact. Lemma 3.22 (Riesz Lemma) Let X be a ormed vector space ad Y be a closed liear subspace of X such that Y X ad α R, 0 < α < 1. The there is x α X such that x α = 1 ad x α y > α for all y Y. Proof: Sice Y X ad Y X there is x X \Y. Sice Y is closed, X \Y is ope ad therefore d := if{ x y : y Y } > 0. 4 This proof implicitly uses equivalece of orms o R +1 to establish that covergece i the orm obtaied by restrictig the origial orm V oto L implies covergece of the compoets of the vectors. 16

21 Sice α 1 > 1 there is a poit z Y such that x z < dα 1. Let x α = x z x z. The x α = 1 ad for ay y Y, x α y = x z x z y x (z + x z y) = x z > d dα 1 = α, as z + x z y Y because Y is a liear subspace. Theorem 3.23 A ormed space is fiite dimesioal iff the uit sphere is compact. Proof: Bolzao-Weierstrass Theorem ad Lemma 3.15 imply that i a fiite dimesioal ormed space the uit sphere is compact (the uit sphere is bouded ad closed). So we oly eed to show that if the uit sphere S B is sequetially compact, the the ormed space V is fiite dimesioal. Ideed, if V is ifiite dimesioal, the Riesz Lemma ca be used to costruct a ifiite sequece of x S such that x x m > α > 0 for all m. This sequece does ot have a coverget subsequece (oe of the subsequeces is Cauchy) ad therefore S is ot compact. We costruct x iductively. Fix α (0,1) ad take ay x 1 S. Suppose that for some 1 we have foud E = {x 1,...,x } such that x k S ad x l x k > α for all 1 k,l, k l (ote that the secod property is automatically fulfilled for = 1). The liear subspace Y = Spa(E ) is -dimesioal ad hece closed (see Propositio 3.17). Sice X is ifiite dimesioal Y X. The Riesz Lemma implies that there is x +1 S such that x +1 x k > α for all 1 k. Repeatig this argumet we geerate x for all N. 17

22 4 Baach spaces 4.1 Completeess: Defiitio ad examples Defiitio 4.1 (Baach space) A ormed space V is called complete if ay Cauchy sequece i V coverges to a limit i V. A complete ormed space is called a Baach space. Theorem 4.2 Every fiite-dimesioal ormed space is complete. Proof: Theorem 3.6 implies that R is complete, i.e., every Cauchy sequece of umbers has a limit. Now let V be a real vector space, dimv = <. Take ay basis i V. The a sequece of vectors i V coverges iff each compoet of the vectors coverges, ad a sequece of vectors is Cauchy iff each compoet is Cauchy. Therefore each compoet has a limit, ad those limits costitute the limit vector for the origial sequece. Hece V is complete. Cosiderig C as a real vector space we coclude that it is also complete. Therefore, ay fiite-dimesioal complex vector space V is also complete. I particular, R ad C are complete. Theorem 4.3 (l p is a Baach space) The space l p (K) equipped with the stadard l p orm is complete. Proof: Suppose that x k = (x1 k,xk 2,...) lp (K) is Cauchy. The for every ε > 0 there is N such that x m x l p = x m j x j p < ε for all m, > N. Cosequetly, for each j N the sequece x k j is Cauchy, ad the completeess of K implies that there is a j K such that x k j a j as k. Let a = (a 1,a 2,...). First we ote that for ay M 1 ad m, > N: Takig the limit as we get M x m j x j p x m j x j p < ε. M x m j a j p ε. 18

23 This holds for ay M, so we ca take the limit as M : x m j a j p ε. We coclude that x m a l p (K). Sice l p (K) is a vector space ad x m l p (K), the a l p (K). Moreover, x m a l p < ε for all m > N. Cosequetly x m a i l p (K) with the stadard orm, ad so l p (K) is complete. Theorem 4.4 (C is a Baach space) The space C[0,1] equipped with the sup orm is complete. Proof: Let f k be a Cauchy sequece. The for ay ε > 0 there is N such that sup f (t) f m (t) < ε t [0,1] for all m, > N. I particular, f (t) is Cauchy for ay fixed t ad cosequetly has a limit. Set f (t) = lim f (t). Let s prove that f (t) f (t) uiformly i t. Ideed, we already kow that f (t) f m (t) < ε for all,m > N ad all t [0,1]. Takig the limit as m we get f (t) f (t) ε for all > N ad all t [0,1]. Therefore f coverges uiformly: f f = sup f (t) f (t) < ε. t [0,1] for all > N. The uiform limit of a sequece of cotiuous fuctios is cotiuous. Cosequetly, f C[0,1] which completes the proof of completeess. Example: The space C[0,2] equipped with the L 1 orm is ot complete. Proof: Cosider the followig sequece of fuctios: { t for 0 t 1, f (t) = 1 for 1 t 2. This is a Cauchy sequece i the L 1 orm. Ideed for ay < m: f f m L 1 = 1 0 (t t m ) dt = m + 1 < 1 + 1, 19

24 ad cosequetly for ay m, > N f f m L 1 < 1 N. Now let us show that f do ot coverge to a cotiuous fuctio i the L 1 orm. Ideed, suppose such a limit exists ad call it f. The Sice implies that f f L 1 = t f (t) dt + 1 f (t) dt 0. 1 f (t) t t f (t) f (t) + t f (t) dt t dt t f (t) dt f (t) dt + t dt, we have 1 0 t f (t) dt 1 0 f (t) dt as ad cosequetly f (t) dt + 1 f (t) dt = 0. 1 As f is assumed to be cotiuous, it follows { 0, 0 < t < 1, f (t) = 1, 1 < t < 2. We see that the limit fuctio f caot be cotiuous. This cotradictio implies that C[0,2] is ot complete with respect to the L 1 orm. 4.2 The completio of a ormed space A ormed space may be icomplete. However, every ormed space X ca be cosidered as a subset of a larger Baach space ˆX. The miimal amog these spaces 5 is called the completio of X. Iformally we ca say that ˆX cosists of limit poits of all Cauchy sequeces i X. Of course, every poit x X is a limit poit of the costat sequece (x = x for all N) ad therefore X ˆX. If X is ot complete, some of the limit poits are ot i X, so ˆX is larger the the origial set X. Defiitio 4.5 (dese set) We say that a subset X V is dese i V if for ay v V ad ay ε > 0 there is x X such that x v < ε. 5 I this cotext miimal meas that if ay other space X has the same property, the the miimal ˆX is isometric to a subspace of X. It turs out that this property ca be achieved by requirig X to be dese i ˆX. 20

25 Note that X is dese i V iff for every poit v V there is a sequece x X such that x v. Theorem 4.6 Let (X, X ) be a ormed space. The there is a complete ormed space (X, X ) ad a liear map i : X X such that i is a isometrical isomorphism betwee (X, X ) ad (i(x), X ), ad i(x) is dese i X. Moreover, X is uique up to isometry, i.e., if there is aother complete ormed space ( X, ) with these properties, the X adx are isometrically isomorphic. X Proof: The proof is relatively log so we break it ito a sequece of steps. Costructio of X. Let Y be the set of all Cauchy sequeces i X. We say that two Cauchy sequeces x = (x ) =1, x X, ad y = (y ) =1, y X, are equivalet, ad write x y, if lim x y X = 0. Let X be the space of all equivalece classes i Y, i.e., it is the factor space: X = Y /. The elemets of X are collectios of equivalet Cauchy sequeces from X. We will use [x] to deote the equivalece class of x. Exercises: Show that X is a vector space. Norm o X. For a η X take ay represetative x = (x ) =1, x X, of the equivalece class η. The the equatio defies a orm o X. Ideed: η X = lim x X. (4.1) 1. Equatio (4.1) defies a fuctio X R, i.e., for ay η X ad ay represetative x η the limit exists ad is idepedet from the choice of the represetative. (Exercise) 2. The fuctio defied by (4.1) satisfies the axioms of orm. (Exercise) Defiitio of i : X X. For ay x X let i(x) = [(x,x,x,x,...)] (the equivalece class of the costat sequece). Obviously, i is a liear isometry, ad it is a bijectio X i(x). Therefore the spaces X ad i(x) are isometrically isomorphic. (η (k)) Completeess of X. Let be a Cauchy sequece i (X, X ). For every k N take a represetative x (k) η (k). Note that x (k) Y is a Cauchy sequece i the space (X, X ). The there is a strictly mootoe sequece of itegers k such that x (k) j x (k) l 1 for all j,l k. (4.2) X k 21

26 Now cosider the sequece x defied by ( ) x = x (k). k Next we will check that x is Cauchy, ad cosider its equivalece class η = [x ] X. The we will prove that η (k) η i (X, X ). The sequece x is Cauchy. Sice the sequece of η (k) is Cauchy, for ay ε > 0 there is M ε such that lim x(k) x (l) X = η (k) η (l) X < ε for all k,l > M ε. Cosequetly, for every k,l > M ε there is N k,l ε x (k) x (l) X < ε such that for all > N k,l ε. (4.3) The fix ay ε > 0. If j,l > ε 3 ad m > max{ j, l,n k,l ε/3 } we have x j xl X = x ( j) j x (l) l X x ( j) j x ( j) m X + x m ( j) < 1 j + ε l < ε x (l) m X + x (l) m x (l) l X where we used (4.3) ad (4.2). Therefore x is Cauchy ad η = [x ] X. The sequece η (k) [x ]. Ideed, take ay ε > 0 ad k > 3ε 1, the η (k) η X = lim x (k) j j ( lim x (k) j j x j X = lim x (k) j j x ( j) j X x (k) k X + x (k) k x ( j) j X ) 1 k + ε < 2ε Therefore η (k) η. We have proved that ay Cauchy sequece i X has a limit i X, so X is complete. Desity of i(x) i X. Take a η X ad let x η. Take ay ε > 0. Sice x is Cauchy, there is N ε such that x m x k X < ε for all k,m > N ε. The η i(x k ) X = lim m x m x k X ε. Therefore i(x) is dese i X. Uiqueess of X up to isometry. I will ot show you the details of the proof. The proof uses the fact that i(x) is isometrically isomorphic to ĩ(x) (sice each oe is isometrically isomorphic to X). Moreover, i(x) is dese i X ad ĩ(x) is dese i X. I order to complete the proof oe has to show that a isometry betwee two dese subsets ca be exteded to a isometry betwee the sets. 22

27 4.3 Examples The theorem provides a explicit costructio for the completio of a ormed space. Ofte this descriptio is ot sufficietly coveiet ad a more direct descriptio is desirable. 1. Example: Cosider the space P[0, 1] of all polyomial fuctios restricted to the iterval [0,1] ad equip this space with the sup orm. This space is ot complete. O the other had ay polyomial is cotiuous, ad therefore P[0, 1] ca be cosidered as a subspace of C[0, 1] which is complete. The Weierstrass approximatio theorem states that ay cotiuous fuctio o [0, 1] ca be uiformly approximated by polyomials. I other words, the polyomials are dese i C[0,1]. The Theorem 4.6 implies that the completio of P[0,1] is isometrically isomorphic to C[0,1] equipped with the sup orm. 2. Example: Let l f (K) be the space of all sequeces which have oly a fiite umber of o-zero elemets. This space is ot complete i the l p orm. The completio of l f (K) i the l p orm is isometric to l p (K). Ideed, we have already see that l p (K) is complete. So i order to prove the claim you oly eed to check that l f (K) is dese i l p (K). We see that the completio of a space depeds both o the space ad o the orm. 3. Example: L 1 (0,1) is the completio of C[0,1] i the L 1 orm. Accordig to this defiitio ay f L 1 (0,1) is a equivalece class of a Cauchy sequece f C[0,1] equipped with the L 1 orm. The orm of f is defied by 1 f L 1 = lim f L 1 = lim f (t) dt. 0 I spite of the fact that f ca be cosidered as a limit of the sequece of cotiuous fuctios f i the L 1 orm, we caot defie the value of f (t) for a give t as the limit of f (t), because the limit may ot exist or may deped o the choice of the represetative i the equivalece class. I spite of that we ca defie the itegral of f by settig f (t)dt := lim 0 f (t)dt. (4.4) Ideed, the limit exists: the sequece of the itegrals is Cauchy as ( f (t)dt f m (t)dt = f (t) f m (t) ) dt f (t) f m (t) dt = f f m L 1,

28 hece coverget. It is ot difficult to check that the limit does ot deped from the choice of a represetative i the equivalece class f. Obviously, if f is a costat sequece, i.e., f (t) = f 0 (t), t [0,1], for all N, the 1 0 f (t) = 1 0 f 0 (t)dt. Therefore this defiitio ca be cosidered as a extesio of the classical Riema s itegral used for cotiuous fuctios. Takig ito accout the ew defiitio of the itegral we ca write f L 1 = 1 0 f (t) dt. The we recall that the orm is o-degeerate, therefore f = g for f,g L 1 (0,1) if ad oly if f g L 1 = 0, i.e., 1 0 f (t) g(t) dt = 0, where the itegral should be iterpreted i the sese of the ew defiitio. Note that f ad g are ot cotiuous ad cosequetly the equality above caot be used to deduce that f (t) = g(t) for all t. Nevertheless a more direct descriptio of L 1 (0,1) is possible: The space L 1 (0,1) is isometrically isomorphic to the space of equivalece classes of Lebesgue itegrable fuctios o (0,1): the fuctios f ad g are equivalet if 1 0 f (t) g(t) dt = 0. Here the itegral should be cosidered as the Lebesgue itegral. We ote that it coicides with the defiitio provided above but its costructio is more direct. Sice the otio of the Lebesgue itegratio is very importat for the fuctioal aalysis ad its applicatios we will discuss it i more details i the ext few lectures. A more detailed study of this topics is a part of MA359 Measure Theory module. 24

29 5 Lebesgue spaces The expositio of the Lebesgue itegral is based o the book H.A.Priesly, Itroductio to itegratio, Oxford Sc.Publ., 1997, 306 p. 5.1 Itegrable fuctios Itegrals of step fuctios We say that ϕ : R R is a step fuctio if it is piecewise costat o a fiite umber of itervals, i.e., it ca be represeted as a fiite sum ϕ(x) = c j χ Ij (x), where c j R, I j R is a iterval, I j I k = /0 if k j, ad χ I is the characteristic fuctio of I: { 1, x I, χ I (x) = 0, x I. We ote that the itervals are allowed to be of ay of the four possible types (e.g. (a,b), [a,b), (a,b] or [a,b]). We defie the itegral of a step fuctio ϕ by ϕ := c j I j, where I j is the legth of I j. We ote that this sum equals to the Riema itegral which you studied i Year 1, i.e., ϕ is the algebraic area uder the graph of the step fuctio ϕ (the area is couted egative o those itervals where ϕ(x) < 0). Sets of measure zero Defiitio 5.1 We say that a set A R has measure zero if for ay ε > 0 there is a (at most coutable) collectio of itervals that cover A ad whose total legth is less tha ε: A [a j,b j ] ad (b j a j ) < ε. Exercise: Show that a coutable uio of measure zero sets has measure zero. Hit: for A choose a cover with ε = ε/2. Examples. The set Q of all ratioal umbers has measure zero. The Cator set has measure zero. Defiitio 5.2 A property is said to hold almost everywhere or for almost every x (ad abbreviated to a.e. ) if the set of poits at which the property does ot hold has measure zero. 25

30 Almost everywhere covergece Theorem 5.3 Let (ϕ (x)) =1 be a icreasig sequece of step fuctios (ϕ +1(x) ϕ (x) for all x) such that ϕ < K. The ϕ (x) coverges for a.e. x. Proof: First ote that a icreasig sequece of umbers has a limit if ad oly if it is bouded from above. So i order to prove the theorem it is sufficiet to show that the set E = {x R : ϕ (x) + } has measure zero. Without loosig i geerality, we ca assume that ϕ (x) 0. 6 Let us defie the set E,m = {x : ϕ (x) > m}. This set is a fiite uio of itervals. Ideed, ϕ (x) = j c () j Let I,m = { j : c () j > m}. The E,m = I () j. j I,m The total legth of those itervals is less tha K/m. Ideed, sice c () j c () j > for j I, K > ϕ = j c () j I () j > m I () j. j I,m χ () I (x) is a step fuctio. j 0 for all j ad Fially, E E m = l=1 E l,m for every m. Sice the sequece ϕ is icreasig, E,m E +1,m. Moreover E +1,m \ E,m cosists of a fiite umber of itervals. The E m = E l,m \ E l 1,m l=1 is at most coutable uio of itervals (we deote E 0,m = /0). Sice E,m = l=1 E l,m \ E l 1,m ad the total legth of the itervals i E,m is less tha K/m for all, the total legth of the itervals i E m is ot larger tha K/m. Sice E m is at most coutable uio of itervals whose total legth is less tha K/m ad E E m for all m, the set E has measure zero. 6 Otherwise replace ϕ by ϕ ϕ 1 which is o-egative. The ew sequece satisfies the assumptio of the theorem (possibly with a differet costat K). Moreover, the covergece of ϕ (x) ϕ 1 (x) is equivalet to the covergece of ϕ (x). 26

31 Lemma 5.4 If ϕ ad ψ are two icreasig sequeces of step fuctios which respectively ted to f ad g a.e. ad f (x) g(x) a.e., the lim ϕ lim ψ. Proof: The followig sets E 1 = E 2 = E 3 = { } x : limϕ (x) f (x) { x : limψ (x) g(x) { } x : f (x) < g(x), }, have measure zero. Let E = E 1 E 2 E 3. Assume x E. Fix arbitrary k N. The sequece ψ k (x) ϕ (x) is decreasig i. The as Cosequetly ψ k (x) ϕ (x) ψ k (x) f (x) g(x) f (x) 0. (ψ k ϕ ) + (x) = max{ψ k (x) ϕ (x), 0} 0 Sice it is a decreasig sequece of o-egative step fuctios which coverges to 0 a.e., 7 (ψ k ϕ ) + 0. Sice ψ k ad ϕ are step fuctios ψ k ϕ = (ψ k ϕ ) (ψ k ϕ ) +. The takig the limit as we get ψ k lim ϕ Fially, take the limit as k to obtai the desired iequality. Corollary 5.5 If ϕ ad ψ are two icreasig sequeces of step fuctios which ted to a fuctio f a.e., the lim ϕ = lim ψ. Proof: Use the previous lemma with g = f. 7 This statemet is provided without a proof. 27

32 Lebesgue itegrable fuctios Defiitio 5.6 If a fuctio f : R R ca be represeted as a a.e. limit of a icreasig sequece of step fuctios ϕ, the the itegral of f is give by f = lim ϕ. If the limit is fiite we write f L ic (R). Note that the limit is idepedet of the choice of the icreasig sequece of step fuctios. Ufortuately L ic (R) is ot a vector space as f L ic (R) does ot imply f L ic (R). Ideed, f is bouded from below by ϕ 1 but ot ecessarily bouded from above. The f is ot bouded from below ad therefore f L ic (R). For example, f (x) = { 1 x, x 0, 0, x = 0, or f = k 1/2 χ [(k+1) 1,k 1 ]. Defiitio 5.7 (Lebesgue itegral) A fuctio f : R R is (Lebesgue) itegrable if f = g h where g,h L ic (R) ad f := This itegral is called the Lebesgue itegral. 8 g Of course i the defiitio the choice of g ad h is ot uique. So we will have to check that the value of the itegral does ot deped from this freedom. 5.2 Properties of Lebesgue itegrals The properties of the Lebesgue itegral are based o the properties of the itegrals for the fuctios from L ic (R). Propositio 5.8 If f,g L ic (R) ad λ R, λ 0, the 1. f + g, λ f, max{ f,g}, mi{ f,g} L ic (R). 2. ( f + g) = f + g. 3. If additioally f (x) g(x) a.e., the f g. 8 This costructio is differet (but equivalet) to the origial defiitio of the Lebesgue itegral. For a alterative approach see Measure Theory module. h. 28

33 Exercise: Prove the propositio. First we state the mai elemetary properties of the Lebesgue itegratio. Theorem 5.9 If f, f 1, f 2 are itegrable ad λ R, the 1. f 1 + λ f 2 is also itegrable ad ( f 1 + λ f 2 ) = f 1 + λ f f (x) is also itegrable ad f f. 3. If additioally f (x) 0 a.e., the f 0. Proof: 1. Sice f 1, f 2 are itegrable, there are fuctios g 1,g 2,h 1,h 2 L ic (R) such that f k = g k h k, k = 1,2. If λ 0 the g 1 + λg 2,h 1 + λh 2 L ic (R) ad ( f 1 + f 2 ) = (g 1 + λg 2 ) (h 1 + λh 2 ) = g 1 + λ g 2 h 1 λ h 2 = f 1 + λ f 2. The case of λ < 0 ca be reduced to the previous oe. Ideed, i this case we ca write f 1 + λ f 2 = f 1 + ( λ)( f 2 ) ad observe that f 2 = h 2 g 2 ad cosequetly is also itegrable. Liearity is proved. 2. Sice f is itegrable, f = g h with g,h L ic (R). Obviously for every x f (x) = max{g(x),h(x)} mi{g(x),h(x)}, the Propositio 5.8 implies that the maximum ad the miimum belog to L ic (R) ad hece f is itegrable. The iequality g(x) + mi{g(x),h(x)} h(x) + max{g(x),h(x)} is valid for every x ad i combiatio with Propositio 5.8 implies g + mi{g,h} h + max{g,h}. Cosequetly f = g h max{g,h} mi{g,h} = f. Applyig this result for f replaced by f we coclude f = ( f ) f = f. Cosequetly, f f. 29

34 3. Sice f (x) = g(x) h(x) with g,h L ic (R) ad f (x) 0 a.e., we coclude g(x) h(x) a.e.. The Propositio 5.8 implies that g h. Cosequetly f = g h 0. We ote that f is itegrable does ot imply that f is itegrable. 9 Exercise: If f is itegrable tha f + = max{ f,0} ad f = mi{ f,0} are itegrable. (Hit: f + = ( f + f )/2 ad f = ( f f )/2.) Itegrals ad limits You should be careful whe swappig lim ad : Examples: lim lim χ ( 0, 1 ) = 1 1 χ (0,) = 1 lim χ( 0, 1 ) = 0. 1 lim χ (0,) = 0. The followig two theorems establish coditios which allow swappig the limit ad itegratio. They play the fudametal role i the theory of Lebesgue itegrals. Theorem 5.10 (Mootoe Covergece Theorem) Suppose that f are itegrable, f (x) f +1 (x) a.e., ad f < K for some costat idepedet of. The there is a itegrable fuctio g such that f (x) g(x) a.e. ad g = lim Corollary 5.11 If f is itegrable ad f = 0, the f (x) = 0 a.e. Proof: Let f (x) = f (x). This sequece satisfies MCT (itegrable, icreasig ad f = 0 < 1), cosequetly there is a itegrable g(x) such that f (x) g(x) for a.e. x. Sice the sequece is icreasig, f (x) g(x) a.e. which implies f (x) g(x)/ for all ad a.e. x. Cosequetly f (x) = 0 a.e. Theorem 5.12 (Domiated Covergece Theorem) Suppose that f are itegrable fuctios ad f (x) f (x) for a.e. x.. If there is a itegrable fuctio g such that f (x) g(x) for every ad a.e. x, the f is itegrable ad f = lim 9 Ideed, we ca sketch a example. It is based o partitioig the iterval [0,1] ito two very asty subsets. So let f (x) = 0 outside [0,1], for x [0,1] let f (x) = 1 if x belogs to the Vitali set ad f (x) = 1 otherwise. The f = χ [0,1] but f is ot itegrable. 30 f. f.

35 It is also possible to itegrate complex valued fuctios: f : R C is itegrable if its real ad imagiary parts are both itegrable, ad f := Re f + i Im f. The MCT has o meaig for complex valued fuctios. The DCT is valid without modificatios (ad ideed follows easily from the real versio). 5.3 Lebesgue space L 1 (R) Defiitio 5.13 The Lebesgue space L 1 (R) is the space of itegrable fuctios modulo the followig equivalece relatio: f g iff f (x) = g(x) a.e. The Lebesgue space is equipped with the L 1 orm: f L 1 = f. It is coveiet to thik about elemets of L 1 (R) as fuctios R R iterpretig the equality f = g as f (x) = g(x) a.e. From the viewpoit of Fuctioal Aalysis, the equivalece relatio is itroduced to esure o-degeeracy of the L 1 orm. Ideed, i the space of itegrable fuctios f = 0 is equivalet to f (x) = 0 a.e. ad therefore does ot imply the latter equality for all x. Theorem 5.14 L 1 (R) is a Baach space. Revise the properties of the Lebesgue itegral which imply that L 1 (R) is a ormed space. The completeess of L 1 (R) follows from the combiatio of the followig two statemets: The first lemma gives a criterio for completeess of a ormed space, ad the secod oe implies that the assumptios of the first lemma are satisfied for X = L 1 (R). Lemma 5.15 If (X, X ) is a ormed space i which y j X < implies the series y j coverges, the X is complete. Proof: Let x j X be a Cauchy sequece. The there is a mootoe icreasig sequece k N such that for every k N x j x l X < 2 k for all k,l k. 31

36 Let y 1 = x 1 ad y k = x k x k 1 for k 2. Sice y k X 2 1 k for k 2, y k X y 1 X + Cosequetly there is x X such that O the other had x = k y j = x 1 + k j=2 2 1 k = y 1 X + 2 <. y j. (x j x j 1 ) = x k ad therefore x k x. Cosequetly x k x ad the space X is complete. Lemma 5.16 If ( f k ) is a sequece of itegrable fuctios such that f k L 1 <, the 1. f k(x) coverges a.e. to a itegrable fuctio, 2. f k(x) coverges a.e. to a itegrable fuctio. Proof: The first statemet follows from MCT applied to the sequece g = f k ad K = f k L 1. So there is a itegrable fuctio g(x) such that g(x) = f k (x) for almost all x. For these values of x the partial sums h (x) = f k(x) obviously coverge, so let h(x) = f k (x). Moreover h (x) = f k (x) f k (x) f k (x) = g(x). Therefore the partial sums h satisfy DCT ad the secod statemet follows. I additio to L 1 (R) we will sometimes cosider the Lebesgue spaces L 1 (I) where I is a iterval. We say that f L 1 (I) if χ I f L 1 (R), i.e., we exted the fuctio by zero outside its origial domai. Propositio 5.17 The space C[0,1] is dese i L 1 (0,1). Proof: First show that step fuctios are dese i L 1 (0,1). The check that every step fuctio ca be approximated by a piecewise liear cotiuous fuctio. Cosequetly the space we costructed i this sectio is isometrically isomorphic to the completio of C[0,1] i the L 1 orm. 32

37 5.4 L p spaces Aother importat class of Lebesgue spaces cosists of L p spaces for 1 p <, amog those the L 2 space is the most remarkable (it is also a Hilbert space, see the ext chapter for details). I this sectio we will sketch the mai defiitios of those spaces otig that the full discussio requires more kowledge of Measure Theory tha we ca fit ito this module). If I = (a,b) is a iterval, the L p (I) ca be defied i terms of the itegratio procedure developed earlier i this chapter. This defiitio is equivalet to the stadard oe which will be give a bit later. The Lebesgue space L p (I) is the space of all itegrable fuctios such that f p L p = f p < modulo the equivalece relatio: f = g iff f (x) = g(x) a.e. We ote that i this case L p (I) L 1 (I). The defiitio of L p (R) is slightly differet. We say that f L p (R) if f is locally itegrable (i.e., f L 1 (I) for ay iterval I) 10 ad its p th power is itegrable. The orm is defied by the same formula: f p L p = f p <. I R We ote that although L 1 (R) L 2 (R) /0 (e.g. both spaces cotai all step fuctios) oe of those spaces is a subset of the other oe. For example, f (x) = 1/(1 + x ) belogs to L 2 (R) but ot to L 1 (R). Ideed, f 2 < but f = so it is ot itegrable o R. O the other had g(x) = χ (0,1)(x) x 1/2 belogs to L 1 (R) but ot to L 2 (R). Theorem 5.18 L p (R) ad L p (I) are Baach spaces for p 1 ad ay iterval I. We will ot give a complete proof but sketch the mai ideas istead. Let 1 p + 1 q = 1 ad f Lp (R), g L q (R). The the Hölder iequality states 11 that f g f p g q. 10 This is a importat requiremet. It is ot sufficiet to defie L p as a set of all fuctios such that f p is itegrable: this space would ot be a vector space. Ideed, let p = 2, g = χ [0,1] ad f be the o-itegrable fuctio from the footote 9. The f 2 = g 2 = χ [0,1] is itegrable. But ( f + g) 2 = f f g + g 2 = f is ot itegrable. Therefore if we followed this defiitio f,g L 2 would ot imply f + g L We will ot discuss the proof of this iequality i these lectures. 33

38 Note that the characteristic fuctio χ I L q (R) for ay iterval I ad ay q 1, moreover χ I L q = I 1 q where I = b a is the legth of I. The Hölder iequality with g = χ I implies that χ I f = f I 1/q f p. The left had side of this iequality is the orm of f i L 1 (I): I f L 1 (I) I 1/q f L p (I). Cosequetly ay Cauchy sequece i L p (I) is automatically a Cauchy sequece i L 1 (I). Sice L 1 is complete the Cauchy sequece coverges to a limit i L 1 (I) (ad cosequetly coverges a.e.). I order to proof completeess of L p it is sufficiet to show that the p th power of this limit is itegrable. This ca be doe o the basis of the Domiated Covergece Theorem. Exercise: The ext two exercises show that L 2 (R) is complete (compare with the proof of completeess for L 1 (R)). 1. Let ( f k ) be a sequece i L2 (R) such that Applyig the MCT to the sequece f k L 2 <. g = ( f k ) 2 show that k f k coverges to a fuctio f with itegrable f Now use the DCT applied to h = f f k 2 to deduce that k f k coverges i the L 2 orm to a fuctio i L 2. If you look ito a textbook, you will probably see a differetly lookig defiitio of the Lebesgue spaces. Traditioally a fuctio f is asked to be measurable istead of locally itegrable. Local itegrability is a stroger property: every locally itegrable fuctio is measurable but there are measurable fuctios which are ot locally itegrable, e.g. x 2 is measurable but ot locally itegrable sice 1 1 x 2 = +. Nevertheless the two alterative defiitios of the Lebesgue space are equivalet. Let us discuss the otio of a measurable fuctio from the perspective of our defiitios. 34