ψ ψ -Caputo Fractional Iyengar's Type Inequalities
aa r X i v : . [ m a t h . G M ] S e p ψψψ -CAPUTO FRACTIONAL IYENGAR’S TYPE INEQUALITIES BHAGWAT R. YEWALE AND DEEPAK B. PACHPATTE
Abstract.
In this paper, we establish Iyengar type inequalities utilizing ψ -Caputo fractional derivatives that is, fractional derivative of a function withrespect to another function, which is generalization of some known fractionalderivatives such as Riemann-Liouville, Hadamard, Erd´elyi-Kober. The inequal-ities in this article are with respect to L p norms, 1 ≤ p ≤ ∞ . The tools used inthe analysis are based on Taylor’s formulae for ψ -Caputo fractional derivatives. Introduction
In the past several years, the development of the theories of fractional calculusis well contributed by many researchers. The field of fractional calculus mainlyfocuses on the study of integrals and derivatives of non integer order. Also, thisfield has been attracting many researchers due to the scope of research as well asproviding significant applications in various fields, for example control theory [29],bioengineering [23], viscoelasticity [20]. Development of the theory of fractionalcalculus leads to the various fractional operators (i.e. fractional integrals andfractional derivative), such as the Riemann-Liouville, Caputo, Hadamard, Erd´elyi-Kober, Riesz (see [19]).In literature, there are numerous approaches acquiring a generalization of frac-tional operators that authorize the exhibition of distinct kernel. For instance,Prabhakar [30], introduced a new fractional integral operator with the help ofthree parameter Mittag-Leffler function. In [37], Srivastava with Tomovski inves-tigated generalized Mittag-Leffler function and defined corresponding fractionalintegral operator known as the generalized Prabhakar fractional integral. Almeida[4] have introduced one more dimension to the study of fractional calculus by pre-senting a fractional derivative of a function with respect to another function inCaputo sense known as ψ -Caputo fractional derivative. In [5, 6, 10] authors pro-vides applications and various results for ψ -Caputo fractional derivative. Recently,motivated by the definition of Hilfer fractional dervative Sousa and Oliveira [38],established new fractional derivative called ψ -Hilfer fractional derivative.A wide variety of such operators led researchers to use them, however someauthors studied fractional differential problems and some authors established frac-tional inequalities for various fractional operators. For instance, Pachpatte [27],studied ˇCebyˇsev like inequalities for functions of two and three variables using ψ -Caputo fractional derivative. In [16], Eshaghi with Ansari established Lyapunov Mathematics Subject Classification.
Key words and phrases.
Iyengar Inequality, ψ -Caputo fractional derivative, Taylor’s formulae. inequality for fractional differential equations for Prabhakar derivative. Rashidet al. [32, 33] investigated new fractional integral operator known Generalizedproportional fractional integral operators with respect to another function andestablished certain Gr¨uss type and the reverse Minkowski type inequalities forthis operators. Aljaaidi with Pachpatte [3], presented Minkowski inequalities bymean of ψ -Riemann Liouville fractional integral operators. A large variety of in-equalities have been proposed and studied for different fractional operators, see[11, 31, 35, 36] and the references therein. In last few decades, a number ofmathematician have devoted their efforts to generalize, extends and give numer-ous variants of the inequalities associated with the names of Minkowski, H¨older’s,Hardy, Trapezoid, Ostrowski, ˇCebyˇsev, for more details, see [25, 28]. In 1938,Iyengar proved the following useful inequality (see [26], p. 471):Let g be a differentiable function on [a, b]. If | g ′ ( z ) | ≤ M , ∀ z ∈ [a, b], thenfollowing inequality holds: (cid:12)(cid:12)(cid:12)(cid:12) Z ba g ( z ) dz −
12 ( b − a )( g ( a ) + g ( b )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ M( b − a ) − ( g ( b ) − g ( a )) . (1.1)Due to its usefulness, over the past four decades, inequality (1.1) has beenextensively studied and generalized in various directions as well as some of itsapplications can be found in [2, 12, 24, 39]. Agarwal with Dragomir [1], presentedan extension of the inequality (1.1) by using Steffensen’s inequality as a maintool. However it has been seen that the Iyengar type inequalities established byElezovi´c with Peˇcari´c [15] and ˇCuljak with Elezovi´c [13] had used the same tooland also provided more improvements in it. Furthermore, in [21], Liu obtained theIyengar type inequalities for certain weaker conditions that is, the function in theIyengar type inequality is not necessarily differentiable. Anastassiou [7, 8], studiedthe above mentioned inequality by employing Caputo fractional derivative andCanavati fractional derivative. In recent study, significant contribution have beenmade to which inequality (1.1) leads, and numerous extensions and improvementscan be found related to it in the literature [9, 14, 17, 18, 22, 34].Motivated by the aforementioned work, in this paper, we derive the Iyengar typeinequalities involving Caputo operator with respect to the new function ψ . Theoverall structure of this paper take the form of three sections with an introduc-tion. The remaining paper is organized as follows: In section 2, we present somepreliminaries, essential definitions and mathematical tools which are used to carryout our work. Main results of the Iyengar type inequalities are established in thesection 3. 2. Preliminaries
In this section we give some basic definitions, preliminaries and weighted spaceswhich are useful for our subsequent discussions. Let [a, b] with ( −∞ < a
Definition 2.1 . [19] Let α > , f be an integrable function on [a, b] and ψ ∈C [ a, b ] be an increasing function such that ψ ′ ( t ) = 0 , ∀ t ∈ [ a, b ] then left sided ψ -Riemann-Liouvlle fractional integral of a function f is given by I α,ψa + f ( t ) = 1Γ( α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − f ( s ) ds. Definition 2.2 . [19] Let α > , f be an integrable function on [a, b] and ψ ∈C [ a, b ] be an increasing function such that ψ ′ ( x ) = 0 , ∀ t ∈ [ a, b ] then right sided ψ -Riemann-Liouville fractional integral of a function f is given by I α,ψb − f ( t ) = 1Γ( α ) Z bt ψ ′ ( s )( ψ ( s ) − ψ ( t )) α − f ( s ) ds. Definition 2.3 . [4] Let α > , n ∈ N . Let f , ψ ∈ C n [ a, b ] with ψ an increasingfunction such that ψ ′ ( t ) = 0 , ∀ t ∈ [ a, b ] then left sided ψ -Caputo fractionalderivative of a function f is defined as D α,ψa + f ( t ) = 1Γ( n − α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) n − α − (cid:18) ψ ′ ( t ) ddt (cid:19) n f ( s ) ds. Where n = ⌈ α ⌉ + 1, ⌈ α ⌉ is the integer part of α . Definition 2.4 . [4] Let α > , n ∈ N . Let f , ψ ∈ C n [ a, b ] with ψ an increasingfunction such that ψ ′ ( t ) = 0 , ∀ t ∈ [ a, b ] then right sided ψ -Caputo fractionalderivative of a function f is defined as D α,ψb − f ( t ) = 1Γ( n − α ) Z bt ψ ′ ( s )( ψ ( s ) − ψ ( t )) n − α − (cid:18) − ψ ′ ( t ) ddt (cid:19) n f ( s ) ds. Where n = ⌈ α ⌉ + 1, ⌈ α ⌉ is the integer part of α . Lemma 2.1 . [4] The left sided and right sided ψ -fractional Taylor’s formulae aregiven as follows: f ( t ) = n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k + I α,ψa + D α,ψa + f ( t ) BHAGWAT YEWALE AND DEEPAK PACHPATTE and f ( t ) = n − X k =0 ( − k f [ k ] ψ ( b ) k ! ( ψ ( b ) − ψ ( t )) k + I α,ψb − D α,ψb − f ( t ) , where f [ k ] ψ ( t ) = (cid:0) ψ ′ ( t ) ddt (cid:1) k f ( t ).3. Iyengar Inequalities using ψ -Caputo Fractional Operators This section deals with the Iyengar type inequalities via ψ -Caputo fractionalderivative operators. Theorem 3.1
Let α > f ∈ A C n ([ a, b ]) and ψ ∈ C n ([ a, b ]) with ψ is an increasingand ψ ′ ( t ) = 0, ∀ t ∈ [ a, b ]. Suppose D α,ψa + f , D α,ψb − f ∈ L ∞ ([ a, b ]). Then for a ≤ s ≤ b following inequalities hold:(i) (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:0) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 + ( − k f [ k ] ψ ( b )( ψ ( b ) − ψ ( s )) k +1 (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) [( ψ ( s ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( s )) α +1 ] . (3.1)(ii) The right hand side of the inequality (3.1) is minimized at ψ ( s ) = ψ ( a )+ ψ ( b )2 ,with the value max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α +2) ( ψ ( b ) − ψ ( a )) α +1 α , that is (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! ( ψ ( b ) − ψ ( a )) k +1 k +1 (cid:0) f [ k ] ψ ( a ) + ( − k f [ k ] ψ ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) ( ψ ( b ) − ψ ( a )) α +1 α . (3.2)(iii) When f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0, for k = 0 , , , ..., n − , from (3.2) we get a sharpinequality as follows: (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) ( ψ ( b ) − ψ ( a )) α +1 α . (3.3)(iv) In general, for i = 0 , , , ...m ∈ N , we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) k +1 (cid:2) i k +1 f [ k ] ψ ( a ) + ( − k ( m − i ) k +1 f [ k ] ψ ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α +1 (cid:2) i α +1 + ( m − i ) α +1 (cid:3) . (3.4) YENGAR’S INEQUALITY 5 (v) If f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0 , k = 1 , , ..., n −
1, then from (3.4) we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a ) m (cid:19)(cid:2) if ( a ) + ( m − i ) f ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α +1 (cid:2) i α +1 + ( m − i ) α +1 (cid:3) . (3.5)(vi) For i=1, m=2 from (3.5) we get (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a )2 (cid:19)(cid:2) f ( a ) + f ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ ([ a,b ]) , kD α,ψb − f k L ∞ ([ a,b ]) (cid:3) Γ( α + 2) (cid:0) ψ ( b ) − ψ ( a ) (cid:1) α +1 α . (3.6) Proof (i)From the left ψ -Caputo fractional Taylor’s formula we have f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k = 1Γ( α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − D α,ψa + f ( s ) ds. (3.7)It follows that (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − (cid:12)(cid:12) D α,ψa + f ( s ) (cid:12)(cid:12) ds ≤ α ) (cid:18) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − ds (cid:19) kD α,ψa + f k L ∞ = kD α,ψa + f k L ∞ Γ( α + 1) ( ψ ( t ) − ψ ( a )) α . (3.8)Similarly, from the right ψ -Caputo fractional Taylor’s formula we have (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) Z bt ψ ′ ( s )( ψ ( s ) − ψ ( t )) α − (cid:12)(cid:12) D α,ψb − f ( s ) (cid:12)(cid:12) ds ≤ kD α,ψb − f k L ∞ Γ( α + 1) ( ψ ( b ) − ψ ( t )) α . (3.9)Set ρ := max (cid:18) kD α,ψa + f k L ∞ Γ( α + 1) , kD α,ψb − f k L ∞ Γ( α + 1) (cid:19) . (3.10)Then from (3.8), (3.9) and (3.10) we get (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( ψ ( t ) − ψ ( a )) α (3.11) BHAGWAT YEWALE AND DEEPAK PACHPATTE and (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( ψ ( b ) − ψ ( t )) α . (3.12)It follows that n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k − ρ ( ψ ( t ) − ψ ( a )) α ≤ f ( t ) ≤ n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k + ρ ( ψ ( t ) − ψ ( a )) α (3.13)and n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k − ρ ( ψ ( b ) − ψ ( t )) α ≤ f ( t ) ≤ n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k + ρ ( ψ ( b ) − ψ ( t )) α . (3.14)Integrating (3.13) with respect to t from a to s , we obtain n − X k =0 f [ k ] ψ ( a )( k + 1)! ( ψ ( s ) − ψ ( a )) k +1 − ρ ( α + 1) ( ψ ( s ) − ψ ( a )) α +1 ≤ Z sa f ( t ) dt ≤ n − X k =0 f [ k ] ψ ( a )( k + 1)! ( ψ ( s ) − ψ ( a )) k +1 + ρ ( α + 1) ( ψ ( s ) − ψ ( a )) α +1 . (3.15)Integrating (3.14) with respect to t from s to b , we obtain − n − X k =0 f [ k ] ψ ( b )( k + 1)! ( ψ ( s ) − ψ ( b )) k +1 − ρ ( α + 1) ( ψ ( b ) − ψ ( s )) α +1 ≤ Z bs f ( t ) dt ≤ − n − X k =0 f [ k ] ψ ( b )( k + 1)! ( ψ ( s ) − ψ ( b )) k +1 + ρ ( α + 1) ( ψ ( b ) − ψ ( s )) α +1 . (3.16)Adding (3.15) and (3.16), we get n − X k =0 k + 1)! (cid:2) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 − f [ k ] ψ ( b )( ψ ( s ) − ψ ( b )) k +1 (cid:3) − ρ ( α + 1) (cid:2) ( ψ ( s ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( s )) α +1 (cid:3) ≤ Z ba f ( t ) dt ≤ n − X k =0 k + 1)! (cid:2) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 − f [ k ] ψ ( b )( ψ ( s ) − ψ ( b )) k +1 (cid:3) + ρ ( α + 1) (cid:2) ( ψ ( s ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( s )) α +1 (cid:3) . (3.17) YENGAR’S INEQUALITY 7
From above it follows that (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:2) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 + ( − k f [ k ] ψ ( b )( ψ ( s ) − ψ ( b )) k +1 (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( α + 1) (cid:2) ( ψ ( s ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( s )) α +1 (cid:3) . (3.18)(ii) Suppose u( ψ ( s )) = ( ψ ( s ) − ψ ( a )) α +1 +( ψ ( b ) − ψ ( s )) α +1 , for all ψ ( s ) ∈ [ ψ ( a ) , ψ ( b )].Then u ′ ( ψ ( s )) =( α + 1) (cid:2) ( ψ ( s ) − ψ ( a )) α − ( ψ ( b ) − ψ ( s )) α (cid:3) = 0 ⇒ ( ψ ( s ) − ψ ( a )) α = ( ψ ( b ) − ψ ( s )) α ⇒ ψ ( s ) − ψ ( a ) = ψ ( b ) − ψ ( s ) ⇒ ψ ( s ) = ψ ( a ) + ψ ( b )2Therefore ψ ( s ) = ψ ( a )+ ψ ( b )2 is the critical point of u. If we put ψ ( s ) = ψ ( a )+ ψ ( b )2 in(3.1) we get the inequality (3.2).(iii) When f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0, for k = 0 , , , ..., n − , from (3.2) we get a sharpinequality as follows: (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ∞ , kD α,ψb − f k L ∞ (cid:3) Γ( α + 2) ( ψ ( b ) − ψ ( a )) α +1 α . (3.19)(iv) Let ψ ( t i ) = ψ ( a ) + i (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) , i = 0 , , , ..., m .That is ψ ( t ) = ψ ( a ) , ψ ( t ) = ψ ( a ) + (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) , ... , ψ ( t m ) = ψ ( b ) . (3.20)Therefore( ψ ( t i ) − ψ ( a )) = i (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) , ( ψ ( b ) − ψ ( t i )) = ( m − i ) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) . (3.21)Therefore by (3.21) we can write( ψ ( t i ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( t i )) α +1 = (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α +1 (cid:0) i α +1 + ( m − i ) α +1 (cid:1) . (3.22)Also by using (3.21) for k=0,1,2,...,n-1 and i=0,1,2,...m we have f [ k ] ( a )( ψ ( t i ) − ψ ( a )) k +1 +( − k f [ k ] ( b )( ψ ( b ) − ψ ( t i )) k +1 = (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) k +1 [ f [ k ] ( a ) i k +1 + ( − k f [ k ] ( b )( m − i ) k +1 ] . (3.23) BHAGWAT YEWALE AND DEEPAK PACHPATTE
Hence from (3.18), (3.22) and (3.23), we get (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:0) ψ ( b ) − ψ ( a ) m (cid:1) k +1 (cid:2) f [ k ] ψ ( a ) i k +1 + ( − k f [ k ] ψ ( b )( m − i ) k +1 (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( α + 1) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α +1 [ i α +1 + ( m − i ) α +1 ] . (3.24)From above we get required inequality.(v) If f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0 , k = 1 , , ..., n −
1, then from (3.24) we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a ) m (cid:19)(cid:2) if ( a ) + ( m − i ) f ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( α + 1) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α +1 (cid:2) i α +1 + ( m − i ) α +1 (cid:3) . (3.25)(vi) For i=1, m=2 from (3.25) we get (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a )2 (cid:19)(cid:0) f ( a ) + f ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ρ ( α + 1) (cid:0) ψ ( b ) − ψ ( a ) (cid:1) α +1 α . (cid:3) Theorem 3.2
Let α ≥ f ∈ A C n ([ a, b ]) and ψ ∈ C n ([ a, b ]) with ψ is an increasingand ψ ′ ( t ) = 0, ∀ t ∈ [ a, b ]. Suppose D α,ψa + f , D α,ψb − f ∈ L ([ a, b ] , ψ ). Then for a ≤ s ≤ b following inequalities hold:(i) (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:0) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 + ( − k f [ k ] ψ ( b )( ψ ( b ) − ψ ( s )) k +1 (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 2) [( ψ ( s ) − ψ ( a )) α +1 + ( ψ ( b ) − ψ ( s )) α +1 ] . (3.26)(ii) The right hand side of the inequality (3.26) is minimized at ψ ( s ) = ψ ( a )+ ψ ( b )2 ,with the value max (cid:2) kD α,ψa + f k L a,b ] ,ψ ) , kD α,ψb − f k L a,b ] ,ψ ) (cid:3) Γ( α +1) ( ψ ( b ) − ψ ( a )) α α − , that is (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! ( ψ ( b ) − ψ ( a )) k +1 k +1 (cid:0) f [ k ] ψ ( a ) + ( − k f [ k ] ψ ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 1) ( ψ ( b ) − ψ ( a )) α α − . (3.27) YENGAR’S INEQUALITY 9 (iii) When f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0, for k = 0 , , , ..., n − , from (3.27) we get a sharpinequality as follows: (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 1) ( ψ ( b ) − ψ ( a )) α α − . (3.28)(iv) In general, for i = 0 , , , ...m ∈ N , we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) k +1 (cid:2) i k +1 f [ k ] ψ ( a ) + ( − k ( m − i ) k +1 f [ k ] ψ ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 1) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α (cid:2) i α + ( m − i ) α (cid:3) . (3.29)(v) If f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0 , k = 1 , , ..., n −
1, then from (3.29) we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a ) m (cid:19)(cid:2) if ( a ) + ( m − i ) f ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 1) (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α (cid:2) i α + ( m − i ) α (cid:3) . (3.30)(vi) For i=1, m=2 from (3.30) we get (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a )2 (cid:19)(cid:0) f ( a ) + f ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L ([ a,b ] ,ψ ) , kD α,ψb − f k L ([ a,b ] ,ψ ) (cid:3) Γ( α + 1) (cid:0) ψ ( b ) − ψ ( a ) (cid:1) α α − . (3.31) Proof
From the left ψ -Caputo fractional Taylor’s formula we have f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k = 1Γ( α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − D α,ψa + f ( s ) ds. (3.32)Then (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − (cid:12)(cid:12) D α,ψa + f ( s ) (cid:12)(cid:12) ds. (3.33)Here α ≥ ψ is increasing for a ≤ s ≤ t , we have (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ ( ψ ( t ) − ψ ( a )) α − Γ( α ) Z ta kD α,ψa + f k L dψ ( s ) = kD α,ψa + f k L ([ a,b ] ,ψ ) Γ( α ) ( ψ ( t ) − ψ ( a )) α − . (3.34)Similarly, from the right ψ -Caputo fractional Taylor’s formula we obtain (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ kD α,ψb − f k L ([ a,b ] ,ψ ) Γ( α ) ( ψ ( b ) − ψ ( t )) α − . (3.35)Set η := max (cid:18) kD α,ψa + f k L ([ a,b ] ,ψ ) Γ( α ) , kD α,ψb − f k L ([ a,b ] ,ψ ) Γ( α ) (cid:19) . (3.36)Hence from (3.34), (3.35) and (3.36) we have (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ η ( ψ ( t ) − ψ ( a )) α − (3.37)and (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ η ( ψ ( b ) − ψ ( t )) α − . (3.38)By using similar arguments as in the proof of theorem (3.1), we can write (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:2) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 + ( − k f [ k ] ψ ( b )( ψ ( s ) − ψ ( b )) k +1 (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ηα (cid:2) ( ψ ( s ) − ψ ( a )) α + ( ψ ( b ) − ψ ( s )) α (cid:3) . The rest of the proof similar to the proof of theorem (3.1). (cid:3)
Theorem 3.3
Let p, q > p + q = 1 and α > q >
0. Let f ∈ A C n ([ a, b ]) and ψ ∈ C n ([ a, b ]) with ψ is an increasing and ψ ′ ( t ) = 0, ∀ t ∈ [ a, b ]. Suppose D α,ψa + f , D α,ψb − f ∈ L q ([ a, b ] , ψ ). Then for a ≤ s ≤ b following inequalities hold:(i) (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:0) f [ k ] ψ ( a )( ψ ( s ) − ψ ( a )) k +1 + ( − k f [ k ] ψ ( b )( ψ ( b ) − ψ ( s )) k +1 (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p [( ψ ( s ) − ψ ( a )) α + p + ( ψ ( b ) − ψ ( s )) α + p ] . (3.39)(ii) The right hand side of the inequality (3.39) is minimized at ψ ( s ) = ψ ( a )+ ψ ( b )2 ,with the value max (cid:2) kD α,ψa + f k Lq ([ a,b ] ,ψ ) , kD α,ψb − f k Lq ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α − p ( ψ ( b ) − ψ ( a )) α + 1 p α − q , that is (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! ( ψ ( b ) − ψ ( a )) k +1 k +1 (cid:0) f [ k ] ψ ( a ) + ( − k f [ k ] ψ ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) YENGAR’S INEQUALITY 11 ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p ( ψ ( b ) − ψ ( a )) α + p α − q . (3.40)(iii) When f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0, for k = 0 , , , ..., n − , from (3.40) we get a sharpinequality as follows: (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p ( ψ ( b ) − ψ ( a )) α + p α − q . (3.41)(iv) In general, for i = 0 , , , ...m ∈ N , we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − n − X k =0 k + 1)! (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) k +1 (cid:2) i k +1 f [ k ] ψ ( a ) + ( − k ( m − i ) k +1 f [ k ] ψ ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α + p [ i α + p + ( m − i ) α + p ] . (3.42)(v) If f [ k ] ψ ( a ) = f [ k ] ψ ( b ) = 0 , k = 1 , , ..., n −
1, then from (3.42) we obtain (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a ) m (cid:19)(cid:2) if ( a ) + ( m − i ) f ( b ) (cid:3)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p (cid:18) ψ ( b ) − ψ ( a ) m (cid:19) α + p (cid:2) i α + p + ( m − i ) α + p (cid:3) . (3.43)(vi) For i=1, m=2 from (3.43) we get (cid:12)(cid:12)(cid:12)(cid:12) Z ba f ( t ) dt − (cid:18) ψ ( b ) − ψ ( a )2 (cid:19)(cid:0) f ( a ) + f ( b ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ max (cid:2) kD α,ψa + f k L q ([ a,b ] ,ψ ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) (cid:3) Γ( α )( α + p )( p ( α −
1) + 1) p (cid:0) ψ ( b ) − ψ ( a ) (cid:1) α + p α − q . (3.44) Proof
From left sided ψ -Caputo fractional Taylor’s formula we have (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) Z ta ψ ′ ( s )( ψ ( t ) − ψ ( s )) α − (cid:12)(cid:12) D α,ψa + f ( s ) (cid:12)(cid:12) ds. By using H¨older’s inequality, we obtain (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) (cid:18) Z ta ( ψ ( t ) − ψ ( s )) p ( α − dψ ( s ) (cid:19) p (cid:18) Z ta (cid:12)(cid:12) D α,ψa + f ( s ) dψ ( s ) (cid:12)(cid:12) q (cid:19) q ≤ α ) ( ψ ( t ) − ψ ( a )) p ( α − p ( p ( α −
1) + 1) p kD α,ψa + f k L q ([ a,b ] ,ψ ) = 1Γ( α ) kD α,ψa + f k L q ([ a,b ] ,ψ ) ( p ( α −
1) + 1) p ( ψ ( t ) − ψ ( a )) α − q . (3.45)Similarly, from right sided ψ -Caputo fractional Taylor’s formula we have (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ α ) kD α,ψb − f k [ L q ,ψ ] ( p ( α −
1) + 1) p ( ψ ( b ) − ψ ( t )) α − q . (3.46)Set δ := max (cid:18) kD α,ψa + f k L q ([ a,b ] ,ψ ) Γ( α ) , kD α,ψb − f k L q ([ a,b ] ,ψ ) Γ( α ) (cid:19) . (3.47)From (3.45), (3.46) and (3.47) it follows that (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( a ) k ! ( ψ ( t ) − ψ ( a )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ δ ( ψ ( t ) − ψ ( a )) α − q and (cid:12)(cid:12)(cid:12)(cid:12) f ( t ) − n − X k =0 f [ k ] ψ ( b ) k ! ( ψ ( t ) − ψ ( b )) k (cid:12)(cid:12)(cid:12)(cid:12) ≤ δ ( ψ ( b ) − ψ ( t )) α − q . The remaining proof follows by using the similar arguments as in the proof oftheorem (3.1). (cid:3)
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Bhagwat R. YewaleDept. of Mathematics, Dr. B. A. M. University, Aurangabad, Maharashtra431004, India
E-mail address : [email protected] Deepak B. PachpatteDept. of Mathematics, Dr. B. A. M. University, Aurangabad, Maharashtra431004, India
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