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whichwhich gives gives \noindentt to the powerwhich of 1 minus i alpha u open parenthesis closing parenthesis bar t = 0 = .. period t1−α u ( t ) | t = 0 = b . Now operating1− byα I to the power of 1 minus alpha on both sides of open parenthesis 2 closing parenthesis comma then Now\ centerline , operating{ $ by tI ˆ{ 1on− both \ sidesalpha of} ($ 2 ) u , then( t $ ) \mid $ t $ = 0 = $ \quad b . } I to the power of 1 minus alpha1 u−α open parenthesis t closing parenthesis = b sub 1 plus .. I open parenthesis t comma u open parenthesis phi open parenthesis I u ( t ) = b1 + I f ( t , u (φ( t ) ) ) . Differentiatingt closing parenthesis both closing sides parenthesis we get closing parenthesis period \noindentDifferentiatingNow both , operating sides we get by $ I ˆ{ 1 − \alpha }$ on both sides of ( 2 ) , then Dα u ( t ) = f ( t , u (φ( t ) ) ) . ConverselyD to the power, let u of ( alpha t ) b u e open a solution parenthesis of ( t 1 closing ) , integrate parenthesis both = .. sides f open , then parenthesis t comma u open parenthesis phi open parenthesis t closing parenthesis closing\ centerline parenthesis{ $ closing I ˆ{ parenthesis1 − \ periodalpha }$ u ( t $ ) = b { 1 } + $ \quad I f ( t , u $ ( \phi ( $ t ) ) ) . } I1−α u ( t ) − I1−α u ( t ) | t = 0 = I f ( t , u (φ( t ) ) ) , Conversely comma let u open parenthesis t closing parenthesis b e a solution of open parenthesis 1 closing parenthesis comma integrate both sides comma then op erating by Iα on both sides of the last equation , then \noindentI to the powerDifferentiating of 1 minus alpha uboth open sides parenthesis we gett closing parenthesis minus I to the power of 1 minus alpha u open parenthesis t closing parenthesis bar t = Iu ( t ) − Iα = I1+α f ( t , u (φ( t ) ) ) , differentiate0 = .. I f open both parenthesis sides , t then comma u open parenthesis phi open parenthesis t closing parenthesis closing parenthesis closing parenthesis comma \ centerlineop erating by{ I$ to the ˆ{\ poweralpha of alpha}$ on u both ( t sides $ ) of the = last $ equation\quad commaf ( t then , u $ ( \phi ( $ t ) ) ) . } u ( t ) − C tα−1 = Iα f ( t , u (φ( t ) ) ) , Iu open parenthesis t closing parenthesis minus1 I to the power of alpha C .. = I to the power of 1 plus alpha f open parenthesis t comma u open parenthesis phi f\−noindentNro−om− tConversely− dh − ee−fin ,− i letn−e i ut (− tti− )ha b−e el ac − solutionop − on−ed of− a (it− 1to− )ir− , integratea − ws both− e find sides that ,C then1 = b , open parenthesis t closing parenthesis closing parenthesis closing parenthesis commanT −comma thendifferentiate we obtain both ( 2 sides) , i .comma e . then Problem ( 1 ) and equation\ centerlineu open ( parenthesis 2 ){ are$ equivalentI t ˆ closing{ 1 parenthesis− t o each \alpha minusother}$ C . sub u (1 t t to the $ ) power− of alphaI ˆ{ minus1 1− = I to \alpha the power}$ of u alpha ( t f open $ ) parenthesis\mid $ t comma t $ u = open 0 parenthesis = $ phi\quad openI parenthesis f ( t , t u closing $ ( parenthesis\phi closing( $ parenthesisTu t ) t )) ) = closing , } b parenthesisα1 comma 0 Γ(f-Nα) subf ( o-o s , sub u ( -wφ( s t-d) ) -e ) sub , e-fi t -i∈ sub(0, 1) n-e. To i t-t solve sub i-hequation a-e c-o ( -o 2 ) sub it is n-e necessary d-a sub i t-t to o-i find sub a r-o fixed sub point n sub of T-comma a-w s-e find that C sub 1 = b thecomma\noindent op erator then weop T obtain erating. open byparenthesis $ I ˆ 2{\ closingalpha parenthesis}$ on comma both isides periodof e period the .... last Problem equation open parenthesis , then 1 closing parenthesis .... and Nowequation , we present open parenthesis our main 2 result closing by parenthesis proving are some equivalent local and t o each global other existence period theorems for \ centerline { Iu ( t $ ) − I ˆ{\alpha }$ C \quad $ = I ˆ{ 1 + \alpha }$ f ( t , u $ ( \phi ( $ t ) ) ) , } theTu integral t closing equation parenthesis ( 2 = ) ..in bL alpha1. 1 To facilit ate our discussion , let us first state the following 0 Capital Gamma open parenthesis alpha closing parenthesis .. f open parenthesis s comma u open parenthesis phi open parenthesis s closing parenthesis closing parenthesis\noindent closingdifferentiate parenthesis ds both comma sides .. t in , open then parenthesis 0 comma 1 closing parenthesis period To solve equation open parenthesis 2 closing parenthesisassumptions it is necessary: to find a fixed point of the op erator T period \ centerlineNow comma{ weu ( present t $ our ) main− resultC { by1 proving} t some ˆ{\ localalpha and global− existence1 } = theorems I ˆ{\ foralpha }$ f ( t , u $ ( \phi ( $ t ) ) ) , } : (0, 1)× → the integral equation( i ) open f parenthesisR 2 closingR b e parenthesis a function in withL sub the 1 period following .... To properties facilit ate our : discussion comma let us first state the following ∈ (0, 1), \noindentassumptions$ : f−N { r( o a− )o { form− eachw }} t t−d hf− (e t ,{ . )e− isf i continuous} n−i { , n−e }$ i $ t−t { i−h a−e }$ l $ c−o p−o { n−e } ∈ d−aopen{ parenthesisi t−t i closing o−i parenthesis{(r b− )o { for ..n f each :{ openT− ucomma parenthesisR ,}}}} f ( .0 comma, ua− )w i s1 measurableclosing s−e $ parenthesis find , that times R$C right{ arrow1 } R=$ b e a function b , thenweobtain with the following ( properties 2 ) , i: . e . \ h f i l l Problem ( 1 ) \ h f i l l and → → open parenthesis( c a ) closing there parenthesis exist two .. real for each functions t in open t parenthesisa ( t ) , t 0 commab ( t 1 ) closing such that parenthesis comma f open parenthesis t comma period closing | ) | ≤ ) + ) | |, ∈ (0, 1), ∈ parenthesis\noindent isequation continuousf ( t , u comma ( 2 ) area (equivalent t b (t t o eachu otherfor . each t u R, (.) ∈ L (0, 1) ) φ : (0, 1) → (0, 1) whereopen a parenthesis1 b closingand b parenthesis ( . ) is measurable .. for each u and in R b comma ounded f open . ( parenthesisii period commai s u nondecreasing closing parenthesis and i s measurable comma > 0 φ0 ≥ there\ centerlineopen i s parenthesis a constant{Tu t c M closing $ )such parenthesis = $ that\quad .. therebM $exist\alpha two real functions1 $ } t right arrow a open parenthesis t closing parenthesis comma t right arrow b open parenthesis t closing parenthesis such that a . e . on ( 0 , 1 ) . \noindentbar f open parenthesis$ 0 \Gamma t comma u( closing\alpha parenthesis) $ bar\ lessquad or equalf ( .. s a , open uEJQTDEparenthesis $ ( ,\phi 2007 t closing No(. $parenthesis s 30 ) , ) p plus ). ds .. 3 b , open\quad parenthesist $ \ tin closing( parenthesis 0 , bar1 u ) bar comma . $ .. for each .. t in open parenthesis 0 comma 1 closing parenthesis comma .. u in R comma Towhere solve a open equation parenthesis ( 2period ) it closing is necessary parenthesis in to L sub find 1 open a fixed parenthesis point 0 comma of the 1 closing op erator parenthesis T . and b open parenthesis period closing parenthesis is measurable and b ounded period \noindentopen parenthesisNow , ii we closing present parenthesis our main phi : open result parenthesis by proving 0 comma some 1 closing local parenthesis and global right arrowexistence open parenthesis theorems 0 forcomma 1 closing parenthesis i s nondecreasing and there i s a constant M greater 0 such that phi to the power of greater equal M \noindenta period ethe period integral .. on open equation parenthesis (0 comma 2 ) in 1 closing $ L parenthesis{ 1 } . period $ \ h f i l l To facilit ate our discussion , let us first state the following EJQTDE comma 2007 No period .. 30 comma p period .. 3 \ begin { a l i g n ∗} assumptions : \end{ a l i g n ∗}

\ centerline {( i ) \quad f $ : ( 0 , 1 ) \times $ R $ \rightarrow $ R b e a function with the following properties : }

\ centerline {( a ) \quad f o r each t $ \ in ( 0 , 1 ) ,$ f(t,.)iscontinuous, }

\ centerline {( b ) \quad f o r each u $ \ in $ R, f ( . ,u) i s measurable , }

\ centerline {( c ) \quad there exist two real functions t $ \rightarrow $ a ( t ) , t $ \rightarrow $ b ( t ) such that }

\ centerline { $ \mid $ f ( t , u $ ) \mid \ leq $ \quad a ( t $ ) + $ \quad b ( t $ ) \mid $ u $ \mid , $ \quad f o r each \quad t $ \ in ( 0 , 1 ) , $ \quad u $ \ in $ R , }

\noindent where a $ ( . ) \ in L { 1 } ( 0 , 1 )$ andb( . ) is measurableandbounded . ( i i $ ) \phi : ( 0 , 1 ) \rightarrow ( 0 , 1 ) $ i s nondecreasing and there i s a constantM $ > 0 $ such that $ \phi ˆ{\prime }\geq $ M

\ centerline {a . e . \quad on ( 0 , 1 ) . }

\ hspace ∗{\ f i l l }EJQTDE , 2007 No . \quad 30 , p . \quad 3 L-N sub o-e sub t-w sub comma-t f-e sub o a-r s-t h-s sub u-e l-m sub o-p t-c a-i l-o n-e sub -s i-parenleft i-s t-parenright sub e n-a n-c d-e o-parenleft f-i L\−noindentNo−et−w $ Lf−−N eoa{−ros−−eth −{ sut−ewl − m{ o−commapt−ca−−ilt−}}}on − exf−−sie−parenleft{ oi − a−str−}parenrights−ten h−ans −cd{−euo−−eparenleftf} l−m −{ o−p t−c a−i } l−o n−e { x−s parenright-tcomma a-h− subt e-r s-e o-s a-l sub u-t i-t sub i-s fi-o sub n-e d-s we have the following theorem : iparenrighti−p a r e n l e− fta t −} he−i−rss− eo t− psa a− r el nu− r iti g− ht t { e nwe−a have n− thec following d−e } o theorem−p a r e n l : e f t f−i parenright −t a−h { e−r } s−e o−s Theorem 3 period 1 i−sfi−on−ed−s Theorema−lhe ..{ nthu 3−t .fraciona 1 i−t .. lo{ rdi− rnt e f gr i − ao leq{ uation−e ..} n opend−s parenthesis}}$ we 2 have the following theorem : heb nth fraciona lo rd e rnt e gr a leq uatio n ( 2 \noindentLine 1 r lessTheorem or equal 3 Line . 1 2 1 alpha minus Capital Gammab open parenthesis 1 plus alpha closing parenthesis from supremum to plus Capital Gamma open parenthesis 1 to the power of 1 plus alpha closing parenthesis 1 bar sub bar bar a b open parenthesis bar t closing parenthesis to the power of bar bar 1 M period \noindentProof periodhe ....\quad Let unth b e an fraciona arbitrary element\quad inlo B sub rd r e period rnt e.... gr Thenr afrom leq≤ the uatio assumptions\quad n open ( 2 parenthesis i closing parenthesis open parenthesis ii closing parenthesis comma we have 1 α Γ(1 + α)sup | | ab(|| | 1M. − + 1 1 | t) \ centerlinebar bar Tu bar{b bar} = integral sub 0 to the power of 1 barΓ(1 Tu+α) open parenthesis t closing parenthesis bar dt Proofless . or equal Let integral u b sube an 0 arbitraryto the power element of 1 bar in b t-zeroBr. to theThen power from of alpha the minus assumptions 1 bar zero-d ( i sub) - ( t iiplus ) , integral we have sub 0 to the power of 1 bar integral sub 0 \ [ \ begin { a l i g n e d } r \ leq \\ 1 to the power of t open parenthesis t .. minus|| Tu Capital|| = GammaR sub| Tu open ( t parenthesis) | dt alpha sub closing parenthesis to the power of s closing parenthesis to the power 1 \alpha { − } \Gamma ( 1 + \0alpha ) ˆ{\sup } { + {\Gamma ( 1 ˆ{ 1 } + \alpha ) } 1 }\mid {\mid } of alpha minus1 1 .. f open parenthesis s comma u open1 parenthesist phi open parenthesiss)α−1 s closing parenthesis closing parenthesis closing parenthesis ds bar dt ≤ R | b t − zeroα−1 | zero − d + R | R ( t −Γ f ( s , u (φ( s ) ) ) ds | dt \midless ora{ equal0b } parenleftbigg( \mid to{ thet power ) of}ˆt b{\ submid alpha0 }\ to themid0 power1{ ofM t} to(α the. )\ powerend{ a of l i alpha g n e d parenrightbigg}\ ] 1 Capital Gamma open parenthesis alpha closing b tα parenthesis .. dt bar f open parenthesis≤ ( sα comma)1 Γ( uα open) dt parenthesis| f ( s , u phi(φ open( s ))) parenthesis| ds s closing parenthesis closing parenthesis closing parenthesis bar ds less or equal b alpha plus integral sub 0 to the power of1 1 open parenthesis Capital Gamma sub open parenthesis 1 to the power of t minus s plus sub alpha α closing\noindent parenthesisProof to the. \ powerh f i l l ofLet closing u b parenthesis e an arbitrary to the powert element of alpha) barin1 to $ the B power{ r of} 1 . $ \ h f i l l Then from the assumptions ( i ) − ( ii ) , we have ≤ bα + (Γ(1 − s+α) | alpha plus integral sub 0 to the power of 1 open parenthesis0 Capital Gamma sub open parenthesis 1 to the power of 1 minus plus s sub alpha sub closing \ centerline { $ \midα \mid $ Tu $ \mid \mid = \ int ˆ{ 1 } { 0 }\mid $ Tu ( t $ ) \mid $ dt } parenthesisR 1 to1 the power) of closing parenthesis to the power of alpha open parenthesis bar a open parenthesis s closingR 1 parenthesis bar plus bar b open parenthesis α + ( (1 − +sα ) (| a ( s ) | + | b ( s ) | | u (φ( s )) |) ds ≤ bα + Γ(11+α) (| a ( s closing0 parenthesisΓ bar bar u open parenthesis phi open parenthesis s closing parenthesis closing parenthesis bar closing0 parenthesis .. ds s ) | + | b ( s ) | | u (φ( s )) |) ds \ centerlineless or equal{ b$ alpha\ leq plus Capital\ int ˆ Gamma{ 1 } open{ 0 parenthesis}\mid $ 1 1 b sub $ plus t− alphazero closing ˆ{\alpha parenthesis− integral1 }\ submid 0 to thezero power−d of 1{ opent } parenthesis+ \ int barˆ a{ open1 } { 0 } parenthesis\mid \ int s closingˆ{ t parenthesis} { 0 } bar( plus $ bar t \ bquad open parenthesis$−{ \Gamma s closing} { parenthesis( \alpha bar bar}ˆ{ us open ) parenthesis ˆ{\alpha phi open− parenthesis1 }} { s) closing}$ \ parenthesisquad f ( closing s , u ≤ b + Γ(11 α) || a || + 1 parenthesis$ ( \phi bar closing( $ parenthesis s ) ) ) .. ds ds $ \midα $ dt } + less or equal b alpha plus Capital Gamma open parenthesis 1 1 sub plus alpha closing parenthesis bar bar a bar bar plus 1 \ centerline { $ \ leq ( ˆ{ b } {\alpha }ˆ{ t ˆ{\alpha }} ) 1 \Gamma ( \alpha ) $ \quad dt $ \mid $ f ( s , u $ ( \phi ( $ s $ ) ) ) \mid $ ds }

\ [ \ leq b {\alpha } + \ int ˆ{ 1 } { 0 } ( {\Gamma }ˆ{ t } { ( 1 } − s { + }ˆ{ ) ˆ{\alpha }} {\alpha ) }\mid ˆ{ 1 }\ ]

\noindent $ \alpha + \ int ˆ{ 1 } { 0 } ( {\Gamma }ˆ{ 1 } { ( 1 } −{ + } s {\alpha }ˆ{ ) ˆ{\alpha }} { ) } ( \mid $ a ( s $ ) \mid + \mid $ b ( s $ ) \mid \mid $ u $ ( \phi ( $ s $ ) ) \mid ) $ \quad ds $ \ leq b {\alpha } + \Gamma ( 1 1 { + }\alpha ) \ int ˆ{ 1 } { 0 } ( \mid $ a ( s $ ) \mid + \mid $ b ( s $ ) \mid \mid $ u $ ( \phi ( $ s $ ) ) \mid ) $ \quad ds

\ [ \ leq b {\alpha } + \Gamma ( 1 1 { + }\alpha ) \mid \mid a \mid \mid + 1 \ ] whichwhich implies implies that that \noindentb alpha pluswhich Capital implies Gamma openthat parenthesis 1 1 sub plus alpha closing parenthesis bar bar a bar bar plus Capital Gamma open parenthesis 1 1 sub plus alpha b + Γ(11 α) || a || + Γ(11 α) sup | b ( t ) | .1M r ≤ r . Therefore closingα parenthesis+ .. sup bar b open parenthesis+ t closing parenthesis bar period 1 M r .. less or equal .. r period \noindent $ b {\alpha } + \Gamma ( 1 1 { + }\alpha ) \mid \mid $ a $ \mid \mid + \Gamma Therefore − sup | ( 1 1 { + }\alpha )r $ ≤\quad1 αsupb Γ(1 + $α)\+mid $1 || b| ab ((| t| 1M. $ ) \mid . 1{ M }$ r \quad $ \ leq $ \quad r . r less or equal 1 alpha from minus to b Capital Gamma open parenthesisΓ(11+α) 1 plust) alpha closing parenthesis from supremum to plus Capital Gamma open parenthesis Therefore Using1 to the inequality power of 1 ( plus 3 ) alpha we deduce closing that parenthesis r > 0. 1 barMoreover sub bar bar , we a b have open parenthesis bar t closing parenthesis to the power of bar bar 1 M period Using inequality open parenthesis 3 closing parenthesis1 we deduce that r greater 0 period .. Moreover comma we have || f || = R | f ( s , u (φ( s ))) | ds \ [bar r bar\ leq f bar bar1 =\ integralalpha subˆ{ 0 − to } the{ b power}\0 ofGamma 1 bar f open( parenthesis 1 + \ salpha comma u open) ˆ{\ parenthesissup } { phi+ open{\Gamma parenthesis( s closing 1 ˆ{ parenthesis1 } + closing\alpha ) } 1 }\mid {\mid R}\1 mid a{ b } ( \mid { t ) }ˆ{\mid }\mid 1{ M } . \ ] parenthesis closing parenthesis≤ bar0 ds(| a ( s ) | + | b ( s ) | | u (φ( s )) |) ds less or equal integral sub 0 to the≤ power || a of|| 1 open+ sup parenthesis| b ( t bar) a| open.1M parenthesis|| u || . s closing parenthesis bar plus bar b open parenthesis s closing parenthesis Thisbar bar estimation u open parenthesis shows that phi open f in parenthesisL1(0, 1). s closing parenthesis closing parenthesis bar closing parenthesis .. ds \noindent Using inequality ( 3 ) we deduce that r $ > 0α . $ \quad Moreover , we have Furtherless or , equal f is bar continuous bar a bar bar in plus u .. ( sup assumption bar b open parenthesis ( a ) ) t and closingI parenthesismaps barL period1(0, 1) 1continuously M bar bar into bar period α itselfThis,I estimationf ( t , u shows(φ( t that ) ) ) f isin continuousL sub 1 open in parenthesis u . 0 comma Since 1 u closing i s an parenthesis arbitrary period element in Br, T maps Br \ centerline { $ \mid \mid $ f $ \mid \mid = \ int ˆ{ 1 } { 0 }\mid $ f ( s , u $ ( \phi ( $ s $ ) ) continuouslyFurther comma into ....L1(0 f, is1) ..... Now continuous , we will in showu .... open that parenthesis T is compact assumption , t o achieve .... open this parenthesis goal we will a closing apply parenthesis Theorem closing parenthesis .... and .... I to the 2power) . 3 .\ ofmid alpha So$ , .... ds maps} .... L sub 1 open parenthesis 0 comma 1 closing parenthesis .... continuously into let itselfΩ b comma e a bounded I to the subset power of alpha of Br f. openThen parenthesis T (Ω) ti comma s bounded u open in parenthesisL1(0, 1), phii . open e . parenthesis condition t closing ( i ) parenthesis closing parenthesis closing \ centerline { $ \ leq \ int ˆ{ 1 } { 0 } ( \mid $ a ( s $ ) \mid + \mid $ b ( s $ ) \mid \mid $ u $ ( ofparenthesis Theorem is 2 continuous . 3 i s satisfied in u period . .... It Since remains u i s an t arbitrary o show that element in ( B Tu sub ) rh comma→ TTu maps in BL1 sub(0, 1) r as h → 0, uniformly\phicontinuously( with $ sinto resp $ L ect sub) to 1 ) open Tu \ parenthesismid∈ T Ω. ) $We 0 comma\ havequad 1 the closingds following} parenthesis estimation period : R 1 Now comma we|| will( Tu show ) h that− T isTu compact|| = comma t| o achieve( Tu ) thish ( goal t ) we− will( applyTu ) Theorem ( t ) | 2dt period 3 period .. So comma \ centerline { $ \ leq \mid \mid $ a0 $ \mid \mid + $ \quad sup $ \mid $ b ( t $ ) \mid . 1{ M }\mid let .... Capital Omega= b e aR bounded1 | 1h subsetR t+h ....( ofTu B sub) (s r period) ds ....− Then( Tu T ) open ( t ) parenthesis| dt Capital Omega closing parenthesis .... i s bounded in L sub 1 \mid $ u $ \mid \mid0 . $ } t open parenthesis 0 comma≤ 1R closing1 ( 1 parenthesish R t+h comma| ( Tu .... ) i ( period s ) e− period( Tu ) .... ( t condition) | ds ....) opendt parenthesis i closing parenthesis of Theorem 2 period 3 i s satisfied0 periodt .... It remains t o show that .... open parenthesis Tu closing parenthesis h .... right arrow Tu in L sub 1 open parenthesis \noindent This estimation shows that f in $ L { 1 } ( 0 , 1 ) . $ 0 comma 1 closing parenthesis .... as h right arrow 0 commaZ 1 uniformly with resp ect to Tu in T Capital Omega period≤ .. We1 have the following estimation : \noindentbar bar openFurther parenthesis , \ Tuh f i closing l l fparenthesis i s \ h f i l l hcontinuous .. minus0 .. Tu barin ubar\ =h f integral i l l ( sub assumption 0 to the power\ h f of i l 1l bar( a open ) ) parenthesis\ h f i l l and Tu closing\ h f i parenthesisl l $ I ˆ h{\ openalpha }$ parenthesis\ h f i l l maps t closing\ h f parenthesis i l l $ L minus{ 1 open} ( parenthesis 0 , Tu 1closing ) parenthesis $ \ h f i l lopencontinuously parenthesis t closing into parenthesis bar dt = integral sub 0 to the power of 1 bar 1 h integral sub t to the power of t plus h open parenthesis Tu closing parenthesis open parenthesis s closing parenthesis ds\noindent .. minus openi t s parenthesis e l f $ , Tu closing I ˆ{\ parenthesisalpha }$ open f parenthesis( t , u t $ closing ( \ parenthesisphi ( $bar dtt ) ) ) is continuous in u . \ h f i l l Since u i s an arbitrary element in $ Bless{ orr equal} , integral $ T sub maps 0 to $ the B power{ r of}$ 1 open parenthesis 1 h integral sub t to the power of t plus h bar open parenthesis Tu closing parenthesis open parenthesis s closing parenthesis minus open parenthesis Tu closing parenthesis open parenthesis t closing parenthesis bar ds closing parenthesis .. dt \noindentless or equalcontinuously integral sub 0 to into the power $ L of{ 1 11 } ( 0 , 1 ) . $ Now , we will show that T is compact , t o achieve this goal we will apply Theorem 2 . 3 . \quad So ,

\noindent l e t \ h f i l l $ \Omega $ b e a bounded subset \ h f i l l o f $ B { r } . $ \ h f i l l Then T $ ( \Omega ) $ \ h f i l l i s bounded in $ L { 1 } ( 0 , 1 ) , $ \ h f i l l i . e . \ h f i l l c o n d i t i o n \ h f i l l ( i )

\noindent of Theorem 2 . 3 i s satisfied . \ h f i l l It remains t o show that \ h f i l l ( Tu ) h \ h f i l l $ \rightarrow $ Tu in $ L { 1 } ( 0 , 1 ) $ \ h f i l l as h $ \rightarrow 0 , $

\noindent uniformly with resp ect to Tu $ \ in $ T $ \Omega . $ \quad We have the following estimation :

\ centerline { $ \mid \mid ( $ Tu ) h \quad $ − $ \quad Tu $ \mid \mid = \ int ˆ{ 1 } { 0 }\mid ( $ Tu ) h ( t $ ) − ( $ Tu ) ( t $ ) \mid $ dt }

\ centerline { $ = \ int ˆ{ 1 } { 0 }\mid 1{ h }\ int ˆ{ t + h } { t } ( $ Tu ) ( s ) ds \quad $ − ( $ Tu ) ( t $ ) \mid $ dt }

\ centerline { $ \ leq \ int ˆ{ 1 } { 0 } ( 1{ h }\ int ˆ{ t + h } { t }\mid ( $ Tu ) ( s $ ) − ( $ Tu ) ( t $ ) \mid $ ds $ ) $ \quad dt }

\ [ \ leq \ int ˆ{ 1 } { 0 } 1 \ ]