Regulated functions and the regulated integral


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1 Regulted functions nd the regulted integrl Jordn Bell Deprtment of Mthemtics University of Toronto April Regulted functions nd step functions Let = [ b] nd let X be normed spce. A function f : X is sid to be regulted if for ll t [ b) the limit lim s t + f(s) exists nd for ll t ( b] the limit lim s t f(s) exists. We denote these limits respectively by f(t + ) nd f(t ). We define R( X) to be the set of regulted functions X. t is pprent tht R( X) is vector spce. One checks tht regulted function is bounded nd tht R( X) is normed spce with the norm f = sup t [b] f(t). Theorem 1. f is compct intervl in R nd X is normed lgebr then R( X) is normed lgebr. Proof. f f g R( X) then fg R( X) becuse the limit of product is equl to product of limits. For t we hve so fg f g. (fg)(t) = f(t)g(t) f(t) g(t) f g A function f : X where = [ b] is sid to be step function if there re = s 0 < s 1 < < s k = b for which f is constnt on ech open intervl (s i 1 s i ). We denote the set of step functions X by S( X). t is pprent tht S( X) is contined in R( X) nd is vector subspce nd the following theorem sttes tht if X is Bnch spce then S( X) is dense in R( X). 1 Theorem 2. Let be compct intervl in R let X be Bnch spce nd let f X. f R( X) if nd only if for ll ɛ > 0 there is some g S( X) such tht f g < ɛ. We prove in the following theorem tht the set of regulted functions from compct intervl to Bnch spce is itself Bnch spce. 1 Jen Dieudonné Foundtions of Modern Anlysis enlrged nd corrected printing p. 145 Theorem 7.6.1; Rodney Colemn Clculus on Normed Vector Spces p. 70 Proposition 3.3; cf. Robert G. Brtle A Modern Theory of ntegrtion p. 49 Theorem
2 Theorem 3. f is compct intervl in R nd X is Bnch spce then R( X) is Bnch spce. Proof. Let f n R( X) be Cuchy sequence. For ech t we hve f n (t) f m (t) f n f m hence f n (t) is Cuchy sequence in X. As X is Bnch spce this Cuchy sequence converges to some limit nd we define f(t) to be this limit. Thus f X nd f f n 0. We hve to prove tht f R( X). Let ɛ > 0. There is some N for which n N implies tht f f n < ɛ; in prticulr f f N < ɛ. By Theorem 2 there is some g N S( X) with f N g N < ɛ. Then f g N f f N + f N g N < 2ɛ nd by Theorem 2 this implies tht f R( X). The following lemm shows tht the set of points of discontinuity of regulted function tking vlues in Bnch spce is countble. Lemm 4. f is compct intervl in R X is Bnch spce nd f R( X) then {t : f is discontinuous t t} is countble. Proof. For ech n let g n S( X) stisfy f g n 1 n nd let D n = {t : g n is discontinuous t t}. g n is step function so D n is finite nd hence D = n=1 D n is countble. t need not be true tht f is discontinuous t ech point in D but we shll prove tht if t \ D then f is continuous t t which will prove the clim. Suppose tht t \ D let ɛ > 0 nd tke N > 1 ɛ. As t D N the step function g N is continuous t t nd hence there is some δ > 0 for which s t < δ implies tht g N (s) g N (t) < ɛ. f s t < δ then f(s) f(t) f(s) g N (s) + g N (s) g N (t) + g N (t) f(t) 2 f g N + g N (s) g N (t) < 2 N + ɛ < 3ɛ showing tht f is continuous t t. 2
3 2 ntegrls of step functions Let = [ b] nd let X be normed spce. f f S( X) then there is subdivision = s 0 < s 1 < < s k = b of [ b] nd there re c i X such tht f tkes the vlue c i on the open intervl (s i 1 s i ). Suppose tht there is subdivision = t 0 < t 1 < < t l = b of [ b] nd d i X such tht f tkes the vlue d i on the open intervl (t i 1 t i ). One checks tht (s i s i 1 )c i = l (t i t i 1 )d i. We define the integrl of f to be the bove element of X nd denote this element of X by f = b f. Lemm 5. f is compct intervl in R nd X is normed spce then : S( X) X is liner. Lemm 6. f = [ b] nd X is normed spce then : S( X) X is bounded liner mp with opertor norm b. Proof. f f S( X) let = s 0 < s 1 < < s k = b be subdivision of [ b] nd let c i X such tht f tkes the vlue c i on the open intervl (s i 1 s i ). Then f (s i s i 1 ) c i (s i s i 1 ) f = (b ) f. This shows tht b nd if f is constnt sy f(t) = c X for ll t then f = (b )c nd f = (b ) c = (b ) f showing tht = b. Lemm 7. f b c if X is normed spce nd if g S([ c] X) then c g = b g + c b g. 3 The regulted integrl Let be compct intervl in R nd let X be Bnch spce. Theorem 2 shows tht S( X) is dense subspce of R( X) nd therefore if T 0 B(S( X) X) then there is one nd only one T B(R( X) X) whose restriction to S( X) is equl to T 0 nd this opertor stisfies T = T 0. Lemm 6 shows tht : S( X) X is bounded liner opertor thus there is one nd only one bounded liner opertor R( X) X whose restriction to S( X) is equl to nd we denote this opertor R( X) X lso by. With = [ b] we hve = b. We cll : R( X) X the regulted integrl. 3
4 Lemm 8. f b c if X is Bnch spce nd if f R([ c] X) then c f = b f + c b f. Proof. Let 1 = [ b] 2 = [b c] = [ c] nd let f 1 nd f 2 be the restriction of f to 1 nd 2 respectively. From the definition of regulted functions f 1 R( 1 X) nd f 2 R( 2 X). By Theorem 2 for ny ɛ > 0 there is some g S( X) stisfying f g < ɛ. Tking g 1 nd g 2 to be the restriction of g to 1 nd 2 we check tht g 1 S( 1 X) nd g 2 S( 2 X). Then by Lemm 7 f f f 2 = f g + g g g 1 + g 2 f 1 f (f g) (g 1 f 1 ) + (g 2 f 2 ) 1 2 (c ) f g + (b ) g 1 f 2 +(c b) g 2 f 2. 2 g 2 But g 1 f 1 g f nd g 2 f 2 g f hence we obtin f f 1 f 2 < (c )ɛ + (b )ɛ + (c b)ɛ = 2(c )ɛ. 1 2 Since ɛ > 0 ws rbitrry we get f f 1 1 so proving the clim. 2 f 2 f = f f 2 2 = 0 We prove tht pplying bounded liner mp nd tking the regulted integrl commute. 2 Lemm 9. Suppose tht is compct intervl in R nd tht X nd Y re Bnch spces. f f R( X) nd T B(X Y ) then T f R( Y ) nd T f = T f. 2 JenPul Penot Clculus Without Derivtives p. 124 Proposition
5 Proof. Becuse T is continuous we hve T f R( Y ). For ɛ > 0 there is some g S( X) stisfying f g < ɛ. Write = [ b]. Becuse g is step function there is subdivision = s 0 < s 1 < < s k = b of nd there re c i X such tht g tkes the vlue c i on the open intervl (s i 1 s i ). Furthermore T g tkes the vlue T c i on the open intervl (s i 1 s i ) so T g S( Y ) nd then becuse T is liner T g = (s i s i 1 )T c i = T (s i s i 1 )c i = T g. Using this T f T f T f T g + T g T g + T g T f = T (f g) + T (f g) (b ) T (f g) + T (f g) (b ) T f g + T (b ) f g < 2(b ) T ɛ. As ɛ > 0 is rbitrry this mens tht T f T f = 0 nd so T f = T f. 4 Left nd right derivtives Suppose tht is n open intervl in R X is normed spce f X nd t. We sy tht f is rightdifferentible t t if f(t+h) f(t) h hs limit s h 0 + nd tht f is leftdifferentible t t if f(t+h) f(t) h hs limit s h 0. We cll these limits respectively the right derivtive of f t t nd the left derivtive of f t t denoted respectively by f +(t) nd f (t). For f to be differentible t t mens tht f +(t) nd f (t) exist nd re equl. The following is the men vlue theorem for functions tking vlues in Bnch spce. 3 3 Henri Crtn Differentil Clculus p. 39 Theorem
6 Theorem 10 (Men vlue theorem). Suppose tht = [ b] tht X is Bnch spce nd tht f : X nd g : R re continuous functions. f there is countble set D such tht t \ D implies tht f +(t) nd g +(t) exist nd stisfy f +(t) g +(t) then f(b) f() g(b) g(). Corollry 11. Suppose tht = [ b] tht X is Bnch spce nd tht f : X is continuous. f there is countble set D such tht t \ D implies tht f +(t) = 0 then f is constnt on. 5 Primitives Let = [ b] let X be normed spce nd let f g X. We sy tht g is primitive of f if g is continuous nd if there is countble set D such tht t \ D implies tht g is differentible t t nd g (t) = f(t). Lemm 12. Suppose tht is compct intervl in R tht X is Bnch spce nd tht f : X is function. f g 1 g 2 : X re primitives of f then g 1 g 2 is constnt on. Proof. For i = 1 2 s g i is primitive of f there is countble set D i such tht t \ D i implies tht g i is differentible t t nd g i (t) = f(t). Let D = D 1 D 2 which is countble set. Both g 1 nd g 2 re continuous so g = g 1 g 2 : X is continuous nd if t \ D then g is differentible t t nd g (t) = g 1(t) g 2(t) = f(t) f(t) = 0. Then Corollry 11 shows tht g is constnt on i.e. tht g 1 g 2 is constnt on. We now give construction of primitives of regulted functions. 4 Theorem 13. f = [ b] X is Bnch spce nd f R( X) then the mp g : X defined by g(t) = t f is primitive of f on. Proof. For t [ b) nd ɛ > 0 becuse f is regulted there is some 0 < δ < b t such tht 0 < r δ implies tht f(t + r) f(t + ) ɛ. For 0 < r δ nd for ny 0 < η < r using Lemm 8 we hve t+r t t+r t+r t+r f f f(t + ) = f f(t + ) t t t t+η t+r = (f f(t + )) + (f f(t + )) t t+η η sup f(s) f(t + ) t s t+η +(r η) sup f(s) f(t + ) t+η s t+r 2 f η + (r η)ɛ. 4 JenPul Penot Clculus Without Derivtives p. 124 Theorem
7 This is true for ll 0 < η < r so t+r f i.e. This shows tht Similrly t f t+r t f(t + ) rɛ g(t + r) g(t) f(t + ) r ɛ. g +(t) = f(t + ). g (t) = f(t ). Becuse f is regulted Lemm 4 shows tht there is countble set D such tht t \ D implies tht f is continuous t t. Therefore if t \ D then f(t + ) = f(t ) = f(t) so g +(t) = g (t) which mens tht if t \ D then g is differentible t t with g (t) = f(t). To prove tht g is primitive of f on it suffices now to show tht g is continuous. For ɛ > 0 nd t let δ = nd then for s t < δ we hve by Lemm 8 tht s t t g(s) g(t) = f f = f t s f < δ f = ɛ showing tht g is continuous t t completing the proof. s ɛ f Suppose tht X is Bnch spce nd tht f : [ b] X is primitive of regulted function h : [ b] X. Becuse h is regulted by Theorem 13 the function g : [ b] X defined by g(t) = t f is primitive of f on [ b]. Then pplying Lemm 12 there is some c X such tht f(t) g(t) = c for ll t [ b]. But f() g() = f() so c = f(). Hence for ll t [ b] But t h = +η1 f(t) = f() + t t η2 t +η1 t η2 t h + h + h = h + f + h +η 1 t η 2 +η 1 t η 2 h. nd +η1 h η 1 h hence s η nd η t t η 2 h η 2 h t η2 +η 1 f t h 7
8 nd so it mkes sense to write t f = t h nd thus for ll t [ b] f(t) = f() + t f. 8
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