Product integration. Its history and applications

Transcription

1 Product integrtion. Its history nd pplictions Leesgue product integrtion In: Antonín Slvík uthor: Product integrtion. Its history nd pplictions. English. Prh: Mtfyzpress, pp Persistent URL: Terms of use: Antonín Slvík Institute of Mthemtics of the Czech Acdemy of Sciences provides ccess to digitized documents strictly for personl use. Ech copy of ny prt of this document must contin these Terms of use. This document hs een digitized, optimized for electronic delivery nd stmped with digitl signture within the project DML-CZ: The Czech Digitl Mthemtics Lirry

2 Chpter 3 Leesgue product integrtion While it is sufficient to use the Riemnn integrl in pplictions, it is rther unstisfctory from the viewpoint of theoreticl mthemtics. The generliztion of Riemnn integrl due to Henri Leesgue is sed on the notion of mesure. The prolem of etending Volterr s definition of product integrl in similr wy hs een solved y Ludwig Schlesinger. Volterr s nd Schlesinger s works differ in yet nother wy: Volterr did not worry out using infinitesiml quntities, nd it is not lwys esy to trnslte his ides into the lnguge of modern mthemtics. Schlesinger s proofs re rther precise nd cn e red without greter effort ecept for occsionlly strnge nottion. The foundtions of mthemticl nlysis in 1930 s were firmer thn in 1887; moreover, Schlesinger inclined towrds theoreticl mthemtics, s opposed to Volterr, who lwys kept pplictions in mind. Ludwig Schlesinger 1 Schlesinger s iogrphies cn e found in [Le, McT]: Ludwig Ljos in Hungrin Schlesinger ws orn on the 1st Novemer 1864 in Hungrin town Trnv Ngyszomt, which now elongs to Slovki. He studied mthemtics nd physics t the universities of Heidelerg nd Berlin, where he received doctorte in Photo from [McT] 65

3 The dvisors of his thesis which ws concerned with homogeneous liner differentil equtions of the fourth order were Lzrus Fuchs who lter ecme his fther-in-lw nd Leopold Kronecker. Two yers lter Schlesinger ecme n ssocite professor in Berlin nd in 1897 n invited professor t the University of Bonn. During the yers 1897 to 1911 he served s n ordinry professor nd lso s the hed of the deprtment of higher mthemtics t the University of Kolozsvár now Cluj in Romni. In 1911 he moved to Giessen in Germny where he continued to tech until his retirement in Ludwig Schlesinger died on the 16th Decemer Schlesinger devoted himself especilly to comple function theory nd liner differentil equtions; he lso mde vlule contriutions to the history of mthemtics. He trnslted Descrtes Geometrie into Germn, nd ws one of the orgnizers of the centenry festivities dedicted to the hundredth nniversry of János Bolyi, one of the pioneers of non-eucliden geometry. The most importnt works of Schlesinger include Hnduch der Theorie der lineren Differentilgleichungen , J. Bolyi in Memorim 1902, Vorlesungen üer linere Differentilgleichungen 1908 nd Rum, Zeit und Reltivitätstheorie Schlesinger s pper on product integrtion clled Neue Grundlgen für einen Infinitesimlklkul der Mtrizen [LS1] ws pulished in The uthor links up to Volterr s theory of product integrl. He strts with the Riemnn-type definition nd estlishes the sic properties of the product integrl. His proofs re nevertheless originl while Volterr proved most of his sttements using the Peno series epnsion, Schlesinger prefers the ε δ proofs. He then proceeds to define the Leesgue product integrl s it of product integrls of step functions nd eplores its properties. A continution of this pper ppered in 1932 under the title Weitere Beiträge zum Infinitesimlklkul der Mtrizen [LS2]. Schlesinger gin studies the properties of Leesgue product integrl nd is lso concerned with contour product integrtion in R 2 nd in C. This chpter summrizes the most importnt results from oth Schlesinger s ppers; the finl section then presents generliztion of Schlesinger s definition of the Leesgue product integrl. 3.1 Riemnn integrle mtri functions When deling with product integrl we need to work with sequences of mtrices nd their its. Volterr ws minly working with the individul entries of the mtrices nd convergence of sequence of mtrices ws for him equivlent to convergence of ll entries. Schlesinger chooses different pproch: He defines the norm of mtri A = { ij } n i, y [A] = n m 1 i,j n ij. He lso mentions nother norm Ω A = m{ λ ; λ is n eigenvlue of A} 66

4 nd sttes tht n Ω A ij 2 [A]. i, The second inequlity is ovious, the first is proved in [LS1] 1. Schlesinger s norm [A] hs the nice property tht [A B] [A] [B] for every A, B R n n, ut its disdvntge is tht [I] = n. In the following tet we will use the opertor norm A = sup{ A ; 1}, where A nd denote the Eucliden norms of vectors A, R n. This simplifies Schlesinger s proofs slightly, ecuse I = 1 nd A B A B still holds for every A, B R n n. It should e noted tht the spce R n n is finite-dimensionl, therefore it doesn t mtter which norm we choose since they re ll equivlent. The convergence of sequence of mtrices nd the it of mtri function is now defined in stndrd wy using the norm introduced ove. For n ritrry mtri function A : [, ] R n n nd tgged prtition D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = of intervl [, ] with division points t i nd tgs ξ i we denote where t k = t k t k 1. P A, D = I + Aξ k t k, Schlesinger is now interested in the it vlue of P A, D s the lengths of the intervls [t k 1, t k ] pproch zero if the it eists independently on the choice of ξ k [t k 1, t k ]. Clerly, the it is nothing else thn Volterr s right product integrl. Definition Consider function A : [, ] R n n. In cse the it P A, D νd 0 eists, it is clled the product integrl of function A on intervl [, ] nd denoted y the symol I + At dt. 1 [LS1], p

5 Remrk Schlesinger in fct defines the product integrl s the it of the products P A, D = Y 0 m I + Aξ k t k, where Y 0 is n ritrry regulr mtri which plys the role of n integrtion constnt. In the following tet we ssume for simplicity tht Y 0 = I. Also, insted of Schlesinger s nottion {{ I + A d we use the symol I + A d to denote the product integrl. Lemm Let A 1, A 2,..., A m R n n e ritrry mtrices. Then m I + A 1 I + A 2 I + A m ep A k. Proof. A simple consequence of the inequlities I + A k 1 + A k ep A k. Corollry If A M for every [, ], then P A, D e M for every tgged prtition D of intervl [, ]. Corollry If the function A : [, ] R n n is product integrle nd A M for every [, ], then I + A d em. Schlesinger s first tsk is to prove the eistence of product integrl for Riemnn integrle mtri functions, i.e. functions A : [, ] R n n whose entries ij re Riemnn integrle on [, ]. The proof is sustntilly different from the proof given y Volterr; the technique is similr to Cuchy s proof of the eistence of f for continuous function f see [CE, SŠ]. Definition Consider function A : [, ] R n n nd let [c, d] [, ]. The oscilltion of A on intervl [c, d] is the numer 1 [LS1], p [LS1], p. 38 osca, [c, d] = sup{ Aξ 1 Aξ 2 ; ξ 1, ξ 2 [c, d]}. 68

6 The following chrcteriztion of Riemnn integrle function will e needed in susequent proofs: Lemm If A : [, ] R n n is Riemnn integrle function, then νd 0 m osca, [t k 1, t k ] t k = 0. Proof. The sttement follows esily from Drou s definition of the Riemnn integrl which is sed on upper nd lower sums; it is in fct equivlent to Riemnn integrility of the given function see e.g. [Sch2]. Definition We sy tht tgged prtition D is refinement of tgged prtition D we write D D, if every division point of D is lso division point of D no condition eing imposed on the tgs. Lemm Let the function A : [, ] R n n e such tht A M for every [, ]. Then for every pir of tgged prtitions D, D of intervl [, ] such tht D D we hve P A, D P A, D e M m osca, [t k 1, t k ] t k + M t k 2 e M tk, where t i, i = 0,..., m re division points of the prtition D. Proof. Let the prtition D consist of division points nd tgs D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m =. First, we refine it only on the suintervl [t k 1, t k ], i.e. we consider prtition D which contins division points nd tgs t k 1 = u 0 η 1 u 1 u l 1 η l u l = t k nd coincides with the prtition D on the rest of intervl [, ]. Then k 1 P A, D P A, D I + Aξ i t i We estimte l I + Aη j u j I Aξ k t k 1 [LS1], p k 1 I + Aξ i t i i=k+1 i=k+1 I + Aξ i t i. I + Aξ i t i em 69

7 nd l I + Aη j u j I Aξ k t k l Aη j Aξ k u j + l + p=2 1 r 1< <r p l Aη r1 Aη rp u r1 u rp osca, [t k 1, t k ] t k + l p=2 1 r 1< <r p l M p u r1 u rp = = osca, [t k 1, t k ] t k + l 1 + M u j 1 l M u j osca, [t k 1, t k ] t k +e M tk 1 M t k osca, [t k 1, t k ] t k +M t k 2 e M tk. Therefore we conclude tht P A, D P A, D e M osca, [t k 1, t k ] t k + M t k 2 e M tk. Now, since the given prtition D cn e otined from D y successively refining the suintervls [t 0, t 1 ],..., [t m 1, t m ], we otin P A, D P A, D e M m osca, [t k 1, t k ] t k + M t k 2 e M tk. Corollry Consider Riemnn integrle function A : [, ] R n n. Then for every ε > 0 there eists δ > 0 such tht whenever νd < δ nd D D. P A, D P A, D < ε Proof. The sttement follows from the previous lemm, Lemm nd the estimte m m M t k 2 e M tk νdm 2 e MνD t k = νdm 2 e MνD. 1 [LS1], p

8 Theorem The product integrl I+A d eists for every Riemnn integrle function A : [, ] R n n. Proof. Tke ε > 0. Corollry gurntees the eistence of δ > 0 such tht P A, D P A, D < ε/2 whenever νd < δ nd D D. Consider pir of tgged prtitions D 1, D 2 of intervl [, ] stisfying νd 1 < δ nd νd 2 < δ. These prtitions hve common refinement, i.e. prtition D such tht D D 1, D D 2 the tgs in D cn e chosen ritrrily. Then P A, D 1 P A, D 2 P A, D 1 P A, D + P A, D P A, D 2 < ε. We hve proved tht every Riemnn integrle function A : [, ] R n n stisfies certin Cuchy condition nd this is lso the end of Schlesinger s proof; the eistence of product integrl follows from the Cuchy condition in the sme wy s in the nloguous theorem for the ordinry Riemnn integrl see e.g. [Sch2]. Theorem Consider Riemnn integrle function A : [, ] R n n. If c [, ], then I + A d = I + A d c I + A d. c Proof. As Schlesinger remrks, the proof follows directly from the definition of product integrl see the proof in Chpter Mtri eponentil function Let A : [, ] R n n e constnt function. If D m is prtition of [, ] to m suintervls of length /m, then P A, D m = I + m m A. Since νd m 0 s m, we hve I + A d = I + m m m A = e A. The lst equlity follows from the fct tht e A = m I + A/m m for every A R n n ; recll tht the mtri eponentil ws defined in Chpter 2 using the series e A A m = m! [LS1], p [LS1], p. 41 m=0 71

9 Lemm If A 1,..., A m R n n nd B 1,..., B m R n n, then m i 1 A i B i = B j A i B i. j=i+1 A j Proof. m A i B i = B 1 B i 1 A i A m B 1 B i A i+1 A m = m i 1 = B j A i B i j=i+1 A j. Theorem Consider Riemnn integrle function A : [, ] R n n. Then νd 0 e Aξk tk = νd 0 I + Aξ k t k = I + At dt. Proof. Since every Riemnn integrle function is ounded, we hve A M for some M R nd for every [, ]. The definition of mtri eponentil implies e Aξk tk I + Aξ k t k Aξ k t k 2 e Aξk tk M t k 2 e M tk for k = 1,..., m. According to Lemm 3.2.1, e Aξk tk I + Aξ k t k = m j 1 = I + Aξ k t k e Aξj tj I Aξ j t j e M 1 [LS1], p. 42 m e Aξj tj I Aξ j t j e M M 2 e M M 2 νde MνD m k=j+1 m e Aξk tk t j 2 e M tj t j = e M M 2 νde MνD. 72

10 By choosing sufficiently fine prtition D of [, ], the lst epression cn e mde ritrrily smll. Definition The trce of mtri A = { ij } n i, is the numer Tr A = n ii. Theorem If A : [, ] R n n is Riemnn integrle function, then det I + A d = ep Tr A d. Proof. det = I + A d νd 0 = det νd 0 e Tr Aξk tk = νd 0 ep e Aξk tk = νd 0 det e Aξk tk = m Tr Aξ k t k = ep Tr A d we hve used theorem from liner lger: det ep A = ep Tr A. Remrk This formul sometimes clled the Jcoi formul ppered lredy in Volterr s work. Schlesinger employs different proof nd his sttement is lso more generl it requires only the Riemnn integrility of A, in contrst to Volterr s ssumption tht A is continuous. Corollry If A : [, ] R n n is Riemnn integrle function, then the product integrl I + A d is regulr mtri. Recll tht Volterr hs lso ssigned mening to product integrls whose lower it is greter thn the upper it; his definition for the right integrl ws I + At dt = νd 0 k=m 1 I Aξ k t k. If A is Riemnn integrle, we know tht this is equivlent to I + At dt = νd 0 k=m 1 e Aξk tk. 1 [LS1], p

11 Thus I = = νd 0 νd 0 m e Aξk tk e Aξk tk 1 e Aξk tk = k=m νd 0 k=m 1 e Aξk tk = I + At dt I + At dt, which proves tht I +At dt is the inverse mtri of I +At dt ; compre with Volterr s proof of Theorem The indefinite product integrl Schlesinger now proceeds to study the properties of the indefinite product integrl, i.e. of the function Y = I + At dt Theorem If A : [, ] R n n is Riemnn integrle, then the function Y = I + At dt is continuous on [, ]. Proof. We prove the right-continuity of Y t 0 [, ; continuity from left is proved similrly. Let h. The function A is ounded: A M for some M R. We now employ the inequlity from Lemm Let D e prtition of intervl [ 0, 0 + h]. Then I + A 0 h P A, D e Mh osca, [ 0, 0 + h]h + Mh 2 e Mh. Pssing to the it νd 0 we otin I + A 0+h 0h I + At dt emh osca, [ 0, 0 + h]h + Mh 2 e Mh, which implies Therefore 0. 0+h I + At dt = I. h 0+ 0 Y 0 + h Y 0 = Y 0 h 0+ I + At dt 0+h 0 I = 0. 1 [LS1], p

12 Theorem If A : [, ] R n n is Riemnn integrle, then the function Y = I + At dt, stisfies the integrl eqution Y = I + Y tat dt, [, ]. Proof. It is sufficient to prove the sttement for =. Let D : = t 0 ξ 1 t 1 ξ 2 t m 1 ξ m t m = e tgged prtition of intervl [, ]. We define k Y k = I + Aξ i t i, k = 0,..., m. Then Y k Y k 1 = Y k 1 Aξ k t k, k = 1,..., m Since Y 0 = I nd Y m = P A, D, dding the equlities for k = 1,..., m yields m P A, D I = Y k 1 Aξ k t k. The function A is ounded: A M for some M R. We estimte Y I + P A, D I Y tat dt Y P A, D + Y tat dt Y P A, D + m + Y k 1 m Y t k 1 Aξ k t k + Y t k 1 Y ξ k Aξ k t k + 1 [LS1], p m + Y ξ k Aξ k t k Y tat dt

13 Using the inequlities m m Y k 1 Y t k 1 Aξ k t k M Y k 1 Y t k 1 t k M m m e M t k osca, [t j 1, t j ] t j + M t j 2 e M tj m Me M osca, [t j 1, t j ] t j + M 2 νde MνD we hve used Lemm nd m m Y t k 1 Y ξ k Aξ k t k M oscy, [t k 1, t k ] t k, we see tht ll terms on the right-hnd side of cn e mde ritrrily smll if the prtition D is sufficiently fine. Corollry If A : [, ] R n n is continuous, then the function Y = I + At dt provides solution of the differentil eqution Y = Y A, [, ] nd stisfies the initil condition Y = I. Remrk The function Y is therefore the fundmentl mtri of the system n y i = ji y j, i = 1,..., n. Schlesinger uses the nottion D Y = A, where D Y = Y 1 Y, i.e. D is ectly Volterr s right derivtive of mtri function. 3.4 Product integrl inequlities In this section we summrize vrious inequlities tht will e useful lter. 1 [LS1], p

14 Lemm If A : [, ] R n n is Riemnn integrle function, then I + A d ep A d. Proof. Lemm implies tht m I + Aξ i t i ep Aξ i t i for every tgged prtition D of intervl [, ]; the proof is completed y pssing to the it νd 0. Lemm Let m N, A k, B k R n n for every k = 1,..., m. Then m I + B k I + A k ep m A k ep B k A k 1. Proof. Define Y k = k I + A i, Z k = k I + B i, k = 0,..., m where the empty product for k = 0 equls the identity mtri. Then for k = 1,..., m. This implies where Y k Y k 1 = Y k 1 A k, Z k Z k 1 = Z k 1 B k, Z k Y k = Z k 1 Y k 1 I + B k + E k, E k = Y k 1 B k A k. Applying the equlity m times on the difference Z m Y m we otin 1 [LS1], p [LS1], p Z m Y m = m 1 E k I + B k+1 I + B m + E m. 77

15 We lso estimte E k ep k 1 A i B k A k the empty sum for k = 0 equls zero, m 1 k 1 m Z m Y m ep A i B k A k ep B i A i + A i + Since = m 1 + ep m 1 A i ep A i B k A k ep i k + ep m 1 A i i=k+1 B m A m = m i=k+1 B m A m i=k+1 B i A i m m m ep A i B k A k ep B i A i. B k A k ep B k A k 1, we conclude tht I + B k I + A k = Zm Y m m m m ep A i ep B k A k 1 ep B i A i = i=k i=k+1 m m m m = ep A i ep B i A i ep B i A i = i=k+1 m m = ep A i ep B i A i 1. + Corollry If A, B : [, ] R n n re Riemnn integrle functions, then I + B d I + A d 1 [LS1], p

16 ep A d ep B A d 1. Proof. The previous lemm ensures tht for every tgged prtition D of intervl [, ] we hve P B, D P A, D = I + Bξ k t k I + Aξ k t k m m ep Aξ k t k ep Bξ k Aξ k t k 1. The proof is completed y pssing to the it νd 0. Remrk Lemm is not present in Schlesinger s work, he proves directly the Corollry 3.4.3; our presenttion is perhps more redle. 3.5 Leesgue product integrl The most vlule contriution of Schlesinger s pper is his generlized definition of product integrl which is pplicle to ll mtri functions with ounded nd mesurle i.e. ounded Leesgue integrle entries. From historicl point of view, such generliztion certinly wsn t strightforwrd one. Recll the originl Leesgue s definition: To compute the integrl f of ounded mesurle function f : [, ] [m, M], we choose prtition then form the sets D : m = m 0 < m 1 < < m p = M, E 0 = { [, ]; f = m}, E j = { [, ]; m j 1 < f m j }, j = 1,..., p, nd compute the lower nd upper sums sf, D = m 0 µ 0 + p m j 1 µ j, Sf, D = m 0 µ 0 + p m j µ j, where µ j = µe j is the Leesgue mesure of the set E j. Since Sf, D sf, D = p m j m j 1 µ j νd, 79

17 the sums in pproch common it s νd 0 nd we define f d = sf, D = Sf, D. νd 0 νd 0 Similr procedure cnnot e used to define product integrl of mtri function A : [, ] R n n, ecuse R n n is not n ordered set. Schlesinger ws insted inspired y n equivlent definition of Leesgue integrl which is due to Friedrich Riesz see [FR, KZ]: A ounded function f : [, ] R is integrle, if nd only if there eists uniformly ounded sequence of step i.e. piecewise-constnt functions {f n } n=1 such tht f n f lmost everywhere on [, ]; in this cse, f d = n f n d. To proceed to the definition of product integrl we first recll tht see Theorem I + A d = e Aξk tk νd 0 for every Riemnn integrle function A : [, ] R n n. The product on the right side might e interpreted s e Aξk tk where A D is step function defined y = I + A D t dt, A D t = Aξ k, t t k 1, t k the vlues At k, k = 0,..., m, might e chosen ritrrily. If {D k } is sequence of tgged prtitions of [, ] such tht νd k = 0, it is esily proved tht A D k t = At t every point t [, ] t which A is continuous. Since Riemnn integrle functions re continuous lmost everywhere, the Eqution holds.e. on [, ]. We re therefore led to the following generlized definition of product integrl: I + A d = I + A k d where {A k } is suitly chosen sequence of mtri step functions tht converge to A lmost everywhere. Definition A function A : [, ] R n n is clled step function if there eist numers = t 0 < t 1 < < t m = 80,

18 such tht A is constnt function on every intervl t k 1, t k, k = 1,..., m. Clerly, mtri function A = { ij } n i, is step function if nd only if ll the entries ij re step functions. Definition A sequence of functions A k : [, ] R n n, k N, is clled uniformly ounded if there eists numer M R such tht A k M for every k N nd every [, ]. Definition A function A : [, ] R n n is clled mesurle if ll the entries ij re mesurle functions. Lemm Let A k : [, ] R n n, k N, e uniformly ounded sequence of mesurle functions such tht A k = A.e. on [, ]. Then A k A in the norm of the spce L 1, i.e. A k A d = 0. Proof. Choose ε > 0. As A k M for every k N nd every [, ], we cn estimte A k A d ε + 2Mµ{; A k A ε}. The convergence A k A.e. implies convergence in mesure 1, i.e. for every ε > 0 we hve µ{; A A k ε} = 0. Therefore for every ε > 0. A k A d ε Theorem Let A k : [, ] R n n, k N, e sequence of step functions such tht Then the it 1 [IR], Proposition 8.3.3, p [LS1], p A k A d = 0. I + A k d 81

19 eists nd is independent on the choice of the sequence {A k }. Proof. We verify tht I + A k d is Cuchy sequence. Corollry we hve I + A l d I + A m d ep A m d ep A l A m d 1. According to The ssumption of our theorem implies tht the sequence of numers A m d is ounded nd tht l,m A l A m d = 0, which proves the eistence of the it. To verify the uniqueness consider two sequences of step functions {A k }, {B k } tht stisfy the ssumption of the theorem. We construct sequence {C k }, where C 2k 1 = A k nd C 2k = B k. Then C k A.e. nd C k A d = 0, which mens tht I + C k d eists. Every susequence of {C k} must hve the sme it, therefore I + A k d = I + B k d. Definition Consider function A : [, ] R n n. Assume there eists uniformly ounded sequence of step functions A k : [, ] R n n such tht A k = A.e. on [, ]. Then the function A is clled product integrle nd we define I + A d = I + A k d. We use the symol L [, ], R n n to denote the set of ll product integrle functions. Remrk The correctness of the previous definition is gurnteed y Lemm nd Theorem Every function A L [, ], R n n is clerly ounded 82

20 nd mesurle step functions re mesurle nd the it of mesurle functions is gin mesurle. Assume on the contrry tht A : [, ] R n n is mesurle function on [, ] such tht ij M, [, ], i, j = 1,..., n. There eists 1 sequence of step functions {A k } which converge to A in the L 1 norm. This sequence contins 2 susequence {B k } of mtri functions B k = { k ij }n i, such tht B k A.e. on [, ]. Without loss of generlity we cn ssume tht the sequence {B k } is uniformly ounded otherwise consider the functions minm M, k ij, M. We hve thus found uniformly ounded sequence of step functions which converge to A.e. on [, ]. This mens tht L [, ], R n n = { A : [, ] R n n ; A is mesurle nd ounded }. Schlesinger remrks tht it is possile to further etend the definition of product integrl to encompss ll mtri functions with Leesgue integrle not necessrily ounded entries, ut he doesn t give ny detils. We return to this question t the end of the chpter. 3.6 Properties of Leesgue product integrl After hving defined the Leesgue product integrl in [LS1], Schlesinger crefully studies its properties. Interesting results my e found lso in [LS2]. Lemm Assume tht {A k } is uniformly ounded sequence of functions from L [, ], R n n, nd tht A k A. e. on [, ]. Then A d = A k d. Proof. According to the Leesgue s dominted convergence theorem, A k d = we hve used continuity of the norm. A k d = A d Corollry Inequlities nd re stisfied for ll step functions. As consequence of the previous lemm we see they re vlid even for functions from L [, ], R n n. The net sttement represents dominted convergence theorem for the Leesgue product integrl. 1 [RG], Corollry 3.29, p [IR], Theorem , p. 267, nd Theorem 8.3.6, p

21 Theorem Assume tht {A k } is uniformly ounded sequence of functions from L [, ], R n n such tht A k A. e. on [, ]. Then I + A d = I + A k d. Proof. The function A is mesurle nd ounded, therefore A L [, ], R n n. To complete the proof we use Corollry in the form I + A d I + A k d nd Lemm ep A d ep A k A d 1 Remrk The previous theorem holds lso for Riemnn product integrl in cse we dd n etr ssumption tht the it function A is Riemnn product integrle. Definition If M is mesurle suset of [, ] nd A L [, ], R n n, we define I + A d = I + χ M A d M where χ M is the chrcteristic function of the set M. The previous definition is correct, ecuse the product χ M A is oviously mesurle ounded function. Remrk The following theorem is proved in the theory of Leesgue integrl 2 : For every f L 1 [, ] nd every ε > 0 there eists δ > 0 such tht f d < ε M whenever M is mesurle suset of [, ] nd µm < δ. Schlesinger proceeds to prove n nloguous theorem for the product integrl he speks out totl continuity. Theorem δ > 0 such tht 1 [LS1], p [RG], theorem 3.26, p [LS1], p. 59 For every A L [, ], R n n nd every ε > 0 there eists I + A d I < ε M 84

22 whenever M is mesurle suset of [, ] nd µm < δ. Proof. Sustituting B = 0 to Corollry we otin I + A d I ep A d ep A d 1, M M M which completes the proof see Remrk Schlesinger now turns his ttention to the indefinite product integrl. Recll tht if f L 1 [, ], then the indefinite integrl F = ft dt, [, ] is n solutely continuous function nd F = f. e. on [, ]. Before looking t product nlogy of this theorem we stte the following lemm. Lemm If A L [, ], R n n, then for lmost ll,. 1 h 0 h Proof. If f L 1 [, ], then 1 1 h 0 h +h h 0 At A dt = 0 f + t f dt = 0 for lmost ll, every such is clled the Leesgue point of f. Applying this equlity to the entries of A we otin 1 h 0 h +h for lmost ll,. 1 At A dt = h 0 h h 0 A + t A dt = 0 Theorem If A L [, ], R n n, then the indefinite integrl Y = I + At dt. stisfies Y 1 Y = A for lmost ll [, ]. Proof. According to the definition of derivtive, 1 [IR], Theorem 6.3.2, p [LS1], p Y 1 Y Y 1 Y + h I =. h 0 h 85

23 We now prove tht Y 1 +h Y + h I 1 = I + At dt I = A h 0+ h h 0+ h for lmost ll [, ]; the procedure is similr for the it from left. We estimte +h 1 +h I + At dt I A h 1 I + At dt e Ah + h 1 A k h k + h k! 1 I + At dt h k=2 +h e Ah + A 2 h e A h Since A M for some M R, the Corollry yields +h 1 +h I + At dt e Ah 1 +h = I + At dt I + A dt h h 1 +h +h h ep A dt ep At A dt 1 = ep A h 1 k 1 +h At A dt h k! epmh 1 h +h At A dt + 2M 2 h ep2mh Equtions 3.6.2, 3.6.3, nd Lemm imply Eqution = Remrk In the previous theorem we hve tcitly ssumed tht the mtri Y = I + At dt is regulr for every [, ]. Schlesinger proved it only for A R[, ], R n n see Corollry 3.2.6, ut the proof is esily djusted to A L [, ], R n n : If {A k } is uniformly ounded sequence of step functions such tht A k A. e. on [, ], then using nd Leesgue s dominted convergence theorem deti + At dt = det I + A kt dt = deti + A kt dt = ep Tr A k t dt = ep Tr At dt > =

24 Theorem If A L [, ], R n n, then I + A d = I + k 2 A 1 A k d 1 d k where the integrls on the right side re tken in the sense of Leesgue. Proof. Let {A k } e uniformly ounded sequence of step functions such tht A k A. e. on [, ]. Every function A k is ssocited with prtition D k : = t k 0 < t k 1 < < t k mk = such tht A k = A k j, t k j 1, t k j. According to the definition of Leesgue product integrl, I + A d = I + A k d mk = epa k j t k j. Schlesinger proves 2 first tht the product integrl might e lso clculted s I + A d mk = I + A k i t k i, provided tht νd k = which cn e ssumed without loss of generlity; note tht if is not stisfied, need not hold consider A = A k = I nd the prtitions = t k 0 < t k 1 = for every k N. Schlesinger s proof of seems too complicted nd even fulty; we insted rgue similrly s in the proof of Theorem 3.2.2: Tke positive numer M such A k M for every k N nd [, ]. Then epa k j t k j I A k j t k j M t k j 2 e M tk j for every k N nd j = 1,..., mk. According to Lemm 3.2.1, mk mk epa k j t k j I + A k j t k j = mk j 1 = I + A k l t k l epa k j t k j I A k j t k j l=1 1 [LS2], p [LS2], p mk l=j+1 epa k l t k l 87

25 mk mk e M epa k j t k j I A k j t k j e M M 2 t k j 2 e M tk j mk e M M 2 νd k e MνDk t k j = e M M 2 νd k e MνDk. This completes the proof of Schlesinger now sttes tht mk I + A k i t k i = I + mk nd concludes the proof sying tht 1 i 1< <i s mk s=1 1 i 1< <i s mk A k i 1 A k i s t k i 1 t k i s A k i 1 A k i s t k i 1 t k i s = s 2 = A 1 A s d 1 d s The lst step perhps deserves etter eplntion: Denote nd X s = { 1,..., s R s ; 1 < 2 < < s }, X s k = 1 i 1< <i s mk [t i1 1, t i1 ] [t i2 1, t i2 ] [t is 1, t is ], where s nd k re ritrry positive integers. If χ s nd χ s k denote the chrcteristic functions of X s nd Xk s, then χs k χs for k. Consequently A k i 1 A k i s t k i 1 t k i s = = = 1 i 1< <i s mk = s A k 1 A k s χ s k 1,..., s d 1 d s = A 1 A s χ s 1,..., s d 1 d s = 2 A 1 A s d 1 d s we hve used the dominted convergence theorem. Remrk The deficiency in the previous proof is tht Schlesinger didn t justify the equlity mk I + A k i 1 A k i s t k i 1 t k i s = s=1 1 i 1< <i s mk 88

26 I + s=1 1 i 1< <i s mk A k i 1 A k i s t k i 1 t k i s. We hve lredy encountered similr inccurcy when discussing Volterr s proof of the Peno series epnsion theorem for product integrl; see lso Msni s proof of Theorem Remrk Recll tht, ccording to Theorem 2.3.5, the right derivtive of mtri function stisfies CD 1 d d = D C d d D d D 1. d Consider two continuous mtri functions A, B defined on [, ]. Using the previous formul nd lso the convention tht 1 y I + At dt = I + At dt for y >, we infer the equlity d I + Bt dt I + At dt d = y = I + At dt B AI + At dt for every [, ]. Denoting S = I + At dt we otin d I + Bt dt I + At dt d = SB AS 1, nd consequently since the left hnd side is equl to I for = I + Bt dt I + At dt = I + StBt AtS 1 t dt Sustituting = nd inverting oth sides of the eqution yields. I + At dt I + Bt dt = I + StBt AtS 1 t dt A similr theorem concerning the left product integrl ws lredy present in Volterr s work 1. Schlesinger proves 2 tht the sttement remins true even if A, B L [, ], R n n. The proof is rther technicl nd we don t reproduce it here. 1 [VH], p [LS2], p

27 Theorem Let A : [, ] [c, d] R n n e such tht the integrl P t = I + A, t d eists for every t [c, d] nd tht A, t t M, [, ], t [c, d], for some M R. Then P d dt = P 1 tp t = S, t A t, ts 1, t d, where S, t = I + Au, t du. Proof. The definition of derivtive gives P 1 tp 1 t = I + A, t d I + A, t + h d I. h 0 h Using Eqution we convert the ove it to 1 I + S, ta, t + h A, ts 1, t d h 0 h I. Epnding the product integrl to Peno series see Theorem we otin 1 h 0 h k 2 1, t, h k, t, h d 1 d k, where, t, h = S, ta, t + h A, ts 1, t. As the Peno series converges uniformly the Weierstrss M-test, see Theorem 2.4.5, we cn interchnge the order of it nd summtion. According to the men vlue theorem there is ξh [t, t + h] such tht A, t + h A, t h = A, ξh t M. The dominted convergence theorem therefore implies 1 [LS2], p h 0 h 1, t, h d 1 = h 0 1, t, h h d 1 = 90

28 nd for k 2 = 1 h 0 h k = k 2 S 1, t A t 1, ts 1 1, t d 1, 2 1, t, h k, t, h d 1 d k = h 0 which completes the proof. h k 1 1, t, h k, t, h h h d 1 d k = 0, The following sttement generlizes Theorem ; Schlesinger replces Volterr s ssumption A C[, ], R n n y weker condition A L [, ], R n n. Theorem If A L [, ], R n n nd C R n n is regulr mtri, then I + C 1 AC d = C 1 I + A d C. Proof. Since C 1 AC k = C 1 A k C for every k N, we hve If A is step function, then epc 1 AC = C 1 epac. I + C 1 AC d = e C 1 Aξ ic t i = m = C 1 e Aξi ti C = C 1 I + A d C. In the generl cse when A L [, ], R n n, we rewrite the ove eqution with simple functions A k in plce of A, nd then pss to the it k. 3.7 Doule nd contour product integrls A considerle prt of the pper [LS2] is devoted to doule nd contour product integrls in R 2 s well s in C. Proly the most remrkle chievement is Schlesinger s proof of the Green s theorem for product integrl, which is reproduced in the following tet. Definition Let G e the rectngle [, ] [c, d] in R 2 nd A : G R n n mtri function on G. The doule product integrl of A over G is defined s I + A, y d dy d = I + A, y d dy, G c 1 [LS2], p

29 provided oth integrls on the right hnd side eist in the sense of Leesgue. Definition Let G e the rectngle [, ] [c, d] in R 2 nd P, Q : G R n n continuous functions on G. We denote U, y = I + P t, c dt I + Q, t dt y, c y T, y = I + Q, t dt I + P t, y dt for every [, ], y [c, d]. The contour product integrl over the oundry of rectngle G is defined s the mtri c I + P, y d + Q, y dy G = U, dt, d Remrk Schlesinger refers to the mtrices U, d nd T, d s to the integrl over the lower step nd integrl over the upper step of the rectngle G. They re clerly specil cse of the contour product integrl s defined y Volterr see definition 2.6.8; the mtri corresponds to the vlue of contour product integrl long the nticlockwise oriented oundry of G. Theorem Let G e the rectngle [, ] [c, d] in R 2 nd P, Q : G R n n continuous mtri functions on G. Assume tht the derivtives P y, Q eist nd re continuous on G. Then I + P, y d + Q, y dy =I + T P, Q T 1 d dy, G G where P, Q = Q P y + P Q QP, T, y = I + Q, t dt y I + P t, y dt c. Proof. A simple clcultion revels tht compre to Lemm [LS2], p T P, Q T 1 = T Q T d T 1. dy 92

30 Tking the product integrl over G we otin I + T P, Q T 1 d dy = G [ = I + T, y Q, y T, y d ] d T, y 1 dy dy c According to the rules for differentiting product of functions see Theorem 2.3.2, T d dy = I + P t, y dt d Q, yi + P t, y dt + I + P t, y dt dy. Theorem on differentiting the product integrl with respect to prmeter yields d I + P t, y dt dy = 0, nd consequently The equlities nd imply T d = Q, y dy I + T P, Q T 1 d dy G = = I + T, y Q, y T, y d d T, y 1 dy = dy c = I + T, y Q, y T, y d d T, y 1 dy dy c we hve used Theorem on interchnging the order of it nd integrl. For every [, ] we hve nd lso T, d 1 T, y = T, y d dy = T, d 1 T, y d dy I + T, u d y du du Using Theorem nd Eqution we rrive t d = I + T, d 1 T, u y d du du. I + T, y Q, y T, y d d T, y 1 dy = T, d dy c 93 d

31 I + T, d 1 T, y Q, y T, d 1 T, y d d T, y 1 T, d dy dy c T, d 1 = T, d I + T, d 1 T, y c d dy dy d d I + Q, y dy T, d 1 = T, d T, d 1 T, ci + Q, y dy Finlly, Eqution gives c T, d 1 = T, ci + Q, y dy I + T P, Q T 1 d dy G = d T, d 1. T, ci + Q, y dy c d d T, d 1 = c c = T, ci + Q, y dy d T, d 1 = I + P, y d + Q, y dy. c G Remrk The previous theorem represents n nlogy of Green s theorem for the product integrl; we hve lredy encountered similr sttement when discussing Volterr s work. Volterr s nlogy of the curl opertor ws P, Q = Q P y while Schlesinger s curl hs the form + QP P Q, P, Q = Q P y + P Q QP. The reson is tht Volterr stted his theorem for the left product integrl, while Schlesinger ws concerned with the right product integrl see Theorem nd Remrk Wheres Volterr worked with simply connected domin G see definition , Schlesinger considers only rectngles. Consider functions P, Q tht stisfy ssumptions of Theorem nd such tht everywhere in G. Then P, Q = I + P, y d + Q, y dy G = I, 94

32 which in consequence mens tht the vlues of contour product integrl over the lower step nd over the upper step re the sme. Schlesinger then denotes the common vlue of the mtrices U, y nd T, y see definition y the symol,d I + P, y d + Q, y dy. Clerly,y I + P u, v du + Qu, v dv d d,c,c d = T, y = P, y, d,y I + P u, v du + Qu, v dv d dy = U, y d = Q, y. dy,c Schlesinger now proceeds to define product integrl long contour nd shows tht in simply connected domin the condition implies tht the vlue of product integrl depends only on the endpoints of the contour. His method is lmost the sme s Volterr s nd we don t repet it here. At the end of pper [LS2] Schlesinger trets mtri functions of comple vrile. He defines the contour product integrl in comple domin nd recpitultes the results proved erlier y Volterr theorems 2.7.4, 2.7.7, nd Generliztion of Schlesinger s definition Thnks to the definition proposed y Ludwig Schlesinger it is possile to etend the clss of product integrle functions nd to work with ounded mesurle functions insted of Riemnn integrle functions. At this plce we remind the nottion L [, ], R n n = { A : [, ] R n n ; A is mesurle nd ounded }. Schlesinger ws wre tht his definition might e etended to ll mtri functions with Leesgue integrle not necessrily ounded entries, i. e. to the clss { } L[, ], R n n = A : [, ] R n n ; L At dt < where the symol L emphsizes tht we re deling with the Leesgue integrl. Clerly L L. If {A k } is uniformly ounded sequence of functions which converge to A lmost everywhere, then ccording to lemm A k A 1 = 95 A k A d = 0,,

33 i.e. A k converge to A lso in the norm of spce L[, ], R n n. Tking ccount of Theorem it is nturl to stte the following definition. Definition A function A : [, ] R n n is clled product integrle if there eists sequence of step functions {A k } such tht A k A 1 = 0. We define I + At dt = I + A kt dt. eists iff A L[, ], R n n, i. e. iff the integrl L At dt eists. Interested reders re referred to the ook [DF] for more detils out the theory of product integrl sed on definition As n interesting emple we present the proof of theorem on differentiting the product integrl with respect to the upper ound of integrtion. We strt with preinry lemm which follows lso from Theorem 3.3.2, ut we don t wnt to use it s we re seeking nother wy to prove it. Remrk The correctness of the previous definition is ensured y theorem Since step functions elong to the spce L[, ], R n n, which is complete, every product integrle function lso elongs to this spce. Moreover, step functions form dense suset in this spce 1, nd therefore I + At dt Lemm If A : [, ] R n n is step function, then Y = I + Y tat dt, [, ]. Proof. There eist prtition = t 0 < t 1 < < t m = nd mtrices A 1,..., A m R n n such tht Then A = A k, t k 1, t k. Y = I + At dt = e A1t2 t1 e Ak 1tk 1 tk 2 e Ak tk 1 for every [t k 1, t k ]. The function Y is continuous on [, ] nd differentile ecept finite numer of points; we hve 1 [RG], Corollry 3.29, p. 47 Y = Y A

34 for [, ]\{t 0, t 1,..., t m }. This implies Y = I + Y tat dt, [, ]. Theorem Consider function A L[, ], R n n. For every [, ] the integrl Y = I + At dt eists nd the function Y stisfies the eqution Y = I + Y tat dt, [, ] Proof. Let A k : [, ] R n n, k N e sequence of step functions such tht Then clerly A k A 1 = A k t At dt = 0, i. e. the definition is correct. Denote Y k = I + A k t dt Becuse A k re step functions, Lemm implies Y k = I + A k t At dt = [, ], Y k ta k t dt, [, ] According to Corollry 3.4.3, Y l Y m ep A l t dt ep A l t A m t dt 1 = = ep A l 1 ep A l t A m t 1 1. From Eqution we see tht {A k } is ounded nd Cuchy sequence with respect to the norm 1. The previous inequlity therefore implies tht Y k converge uniformly to Y, i. e. 1 [DF], p Y k Y = sup Y k Y 0 pro k. [,] 97

35 We now estimte Y k ta k t dt Y tat dt Y ka k Y A 1 Y k Y A k 1 + Y A k A 1 A k 1 Y k Y + A k A 1 Y, nd consequently Y k ta k t dt = Y tat dt. The equlity is otined y pssing to the it in eqution Corollry If A L[, ], R n n, then the function Y = I + At dt, [, ], is solutely continuous on [, ] nd lmost everywhere on [, ]. Y 1 Y = A Remrk In our proof of the previous theorem we hve employed Schlesinger s estimte from Corollry 3.4.3, whose proof is somewht lorious. The uthors of [DF] insted mke use of different inequlity, which is esier to demonstrte. Let A, B : [, ] R n n e two step functions. Denoting Y = I + At dt, Z = I + Bt dt, we see tht the function Y Z 1 is continuous on [, ] nd differentile ecept finite numer of points. Using Eqution we clculte Y Z 1 = Y Z 1 + Y Z 1 = Y Y 1 Y Z 1 Y Z 1 Z Z 1 = Y A BZ 1, nd consequently Y Z 1 = I + Y Z 1 t dt = I + Y tat BtZ 1 t dt. Multiplying this eqution y Z from right nd sustituting = we otin I + At dt I + Bt dt = Y Z Y t At Bt Z 1 t dt Z e 2 B 1 e A 1 A B 1 we hve used Lemm to estimte Y t, Z 1 t nd Z. The mening of the lst inequlity is similr to the mening of inequlity from Corollry 3.4.3: If two step functions A, B re close with respect to the norm 1, then the vlues of their product integrls re lso close to ech other. 98