C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 67, N umb er 2, Pages 252–263 (2018) D O I: 10.1501/C om mua1_ 0000000879 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
SOME NEW SIMPSON TYPE INEQUALITIES FOR THE p-CONVEX AND p-CONCAVE FUNCTIONS
·
IMDAT ·I¸SCAN, N·IHAN KALYONCU KONUK, AND MAH·IR KADAKAL
Abstract. In this paper, we establish some new Simpson type inequalities for the class of functions whose derivatives in absolute values at certain powers are p-convex and p-concave.
1. INTRODUCTION
A function f : I R ! R is said to be convex if the inequality f (tx + (1 t)y) tf (x) + (1 t)f (y)
is valid for all x; y 2 I and t 2 [0; 1]. If this inequality reverses, then f is said to be concave on interval I 6= ;. This de…nition is well known in the literature.
It is well known that theory of convex sets and convex functions play an impor-tant role in mathematics and the other pure and applied sciences. In recent years, the concept of convexity has been extended and generalized in various directions using novel and innovative techniques. For some inequalities, generalizations and applications concerning convexity see [1, 2, 4, 5, 6, 16, 20].
In [9], the author gave de…nition harmonically convex and concave functions as follow.
De…nition 1. Let I Rn f0g be a real interval. A function f : I! R is said to be harmonically convex, if
f xy
tx + (1 t)y tf (y) + (1 t)f (x)
for all x; y 2 I and t 2 [0; 1]. If this inequality is reversed, then f is said to be harmonically concave.
Received by the editors: June 21, 2017, Accepted: July 21, 2017. 2010 Mathematics Subject Classi…cation. 26A51, 26D10, 26D15.
Key words and phrases. Convex function, p-Convex function, Simpson type inequalities. c 2 0 1 8 A n ka ra U n ive rsity. C o m m u n ic a tio n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a tic s a n d S t a tis tic s . C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra -S é rie s A 1 M a t h e m a tic s a n d S t a tis t ic s .
De…nition 2. Let I (0; 1) be a real interval and p 2 Rn f0g. A function f : I! R is said to be a p-convex function, if
f [txp+ (1 t)yp]1=p tf (x) + (1 t)f (y)
for all x; y 2 I and t 2 [0; 1]. If this inequality is reversed, then f is said to be p-concave.
According to De…nition 2, It can be easily seen that for p = 1 and p = 1, p-convexity reduces to ordinary convexity and harmonically convexity of functions de…ned on I (0; 1), respectively.
Hermite-Hadamard inequality for the p-convex function is following:
Theorem 1. Let f : I (0; 1) ! R be a p-convex function, p 2 Rn f0g, and a; b 2 I with a < b. If f 2 L [a; b] then we have
f a p+ bp 2 1 p! p bp ap b Z a f (x) x1 pdx f (a) + f (b) 2 :
These inequalities are sharp [5, 8]. If these inequalities are reversed, then f is said to be p-concave.
Many papers have been written by a number of mathematicians concerning in-equalities for di¤erent classes of harmonically convex and p-convex functions see for instance the recent papers [3, 7, 8, 9, 10, 11, 12, 17, 18, 19, 21, 22, 24] and the references within these papers.
The following integral inequality, named Simpson’s integral inequality, is one of the best known results in the literature.
Theorem 2. (Simpson’s Integral Inequality). Let f : I = [a; b] R ! R be a four time continuously di¤ erentiable on I , where I is the interior of I and f(4)
1< 1. Then 1 3 f (a) + f (b) 2 + 2f a + b 2 1 b a b Z a f (x)dx 1 2880 f (4) 1(b a) 4 :
There are substantial literature on Simpson type integral inequalities. Here we mention the result of [13, 14, 15] and the corresponding references cited therein.
Throughout this paper we will use the following notations. Let 0 < a < b and p 2 Rn f0g. Ap = Ap(a; b) = ap+ bp 2 ; A1= A = A (a; b) = a + b 2 Mp = Mp(a; b) = A 1 p p = ap+ bp 2 1 p ; H = H (a; b) = 2ab a + b It(x; Ap; u; v) = t 13 u [(1 t) xp+ tA p]v v p; Jt(x; Ap; u; v) = t 13 u(1 t) [(1 t) xp+ tA p]v v p; Kt(x; Ap; u; v) = t 1 3 u t [(1 t) xp+ tA p]v v p: where t 2 [0; 1] and u; v 0. 2. MAIN RESULTS
In this section, we will use the following Lemma for we obtain the main results: Lemma 1. Let f : I (0; 1) ! R be a di¤erentiable mapping on I (interior of I) and a; b 2 I with a < b and p 2 Rn f0g. If f0 2 L[a; b], then the following equality
holds: 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx = b p ap 4p "Z 1 0 t 13 [(1 t) ap+ tA p]1 1 p f0 [(1 t) ap+ tAp] 1 p dt + Z 1 0 t 23 [(1 t) Ap+ tbp]1 1 p f0 [(1 t) Ap+ tbp] 1 p dt # : Proof. Firstly, let’s calculate the following integral:
bp ap 4p "Z 1 0 t 13 [(1 t) ap+ tA p]1 1 p f0 [(1 t) ap+ tAp] 1 p dt + Z 1 0 t 23 [(1 t) Ap+ tbp]1 1 p f0 [(1 t) Ap+ tbp] 1 p dt # :
For shortness, we will use the notations I1 = Z 1 0 t 1 3 df [(1 t) a p+ tA p] 1 p ; I2 = Z 1 0 t 2 3 df [(1 t) Ap+ tb p]p1 :
Using the partial integration method and the method of changing variables re-spectively for the integrals I1 and I2as following, we get
I1 = Z 1 0 t 1 3 df [(1 t) a p+ tA p] 1 p = t 1 3 f [(1 t) a p+ tA p] 1 p 1 0 Z 1 0 f [(1 t) ap+ tAp] 1 p dt = 2 3f (Mp) + 1 3f (a) 2p bp ap Z Ap a f (x) x1 pdx (2.1) I2 = Z 1 0 t 2 3 df [(1 t) Ap+ tb p]1p = t 2 3 f [(1 t) Ap+ tb p]1p 1 0 Z 1 0 f [(1 t) Ap+ tbp] 1 p dt = 1 3f (b) + 2 3f (Mp) 2p bp ap Z b Ap f (x) x1 pdx: (2.2)
Summing up side by side (2.1) and (2.2), we have I1+ I2 = 1 3[f (a) + f (b)] + 4 3f (Mp) 2p bp ap Z b a f (x) x1 pdx; I1+ I2 2 = 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx:
Theorem 3. Let f : I (0; 1) ! R be a di¤erentiable mapping on I (the interior of I) and a; b 2 I with a < b and p 2 Rn f0g. If f0 2 L[a; b] and jf0jq
is p-convex on I for q 1, then the following inequality holds:
1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p [Cp(a; b)] 1 1 q jf0(a)jqD p(a; b) + jf0(Mp)j q Ep(a; b) 1 q +b p ap 4p [Fp(a; b)] 1 1 q jf0(M p)jqGp(a; b) + jf0(b)jqHp(a; b) 1 q
where Cp(a; b) = Z 1 0 It(a; Ap; 1; 1) dt; Dp(a; b) = Z 1 0 Jt(a; Ap; 1; 1) dt; Ep(a; b) = Z 1 0 Kt(a; Ap; 1; 1) dt; Fp(a; b) = Z 1 0 I1 t(b; Ap; 1; 1) dt; Gp(a; b) = Z 1 0 K1 t(b; Ap; 1; 1) dt; Hp(a; b) = Z 1 0 J1 t(b; Ap; 1; 1) dt:
Proof. Using Lemma 1 and the power mean inequality, we have 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p "Z 1 0 t 13 [(1 t) ap+ tA p]1 1 p f0 [(1 t) ap+ tAp] 1 p dt # +b p ap 4p "Z 1 0 t 23 [(1 t) Ap+ tbp]1 1 p f0 [(1 t) Ap+ tbp] 1 p dt # bp ap 4p Z 1 0 t 1 3 [(1 t) ap+ tA p]1 1 p dt !1 1 q Z 1 0 t 1 3 [(1 t) ap+ tA p]1 1 p f0 [(1 t) ap+ tAp] 1 p q! 1 q dt +b p ap 4p Z 1 0 t 23 [(1 t) Ap+ tbp]1 1 p dt !1 1 q Z 1 0 t 23 [(1 t) Ap+ tbp]1 1 p f0 [(1 t) Ap+ tbp] 1 p q! 1 q dt bp ap 4p Z 1 0 It(a; Ap; 1; 1) dt 1 1 q Z 1 0 t 13 (1 t) jf0(a)jq + t jf0(Mp)jq [(1 t) ap+ tA p]1 1 p !1 q dt +b p ap 4p Z 1 0 I1 t(b; Ap; 1; 1) dt 1 1 q Z 1 0 t 23 (1 t) jf0(Mp)jq+ t jf0(b)jq [(1 t) Ap+ tbp]1 1 p !1 q dt
= b p ap 4p Z 1 0 It(a; Ap; 1; 1) dt 1 1 q jf0(a)jq Z 1 0 Jt(a; Ap; 1; 1) dt + jf0(Mp)jq Z 1 0 Kt(a; Ap; 1; 1) dt 1 q +b p ap 4p Z 1 0 I1 t(b; Ap; 1; 1) dt 1 1 q jf0(Mp)j qZ 1 0 K1 t(b; Ap; 1; 1) dt + jf0(b)j qZ 1 0 J1 t(b; Ap; 1; 1) dt 1 q bp ap 4p [Cp(a; b)] 1 1 q jf0(a)jqD p(a; b) + jf0(Mp)j q Ep(a; b) 1 q +b p ap 4p [Fp(a; b)] 1 1 q jf0(M p)jqGp(a; b) + jf0(b)jqHp(a; b) 1 q:
This completes the proof of theorem. Corollary 1. Under conditions of Theorem 3
i. If we take p = 1, then we obtain the following inequality for convex function: 1 6 f (a) + 4f a + b 2 + f (b) 1 b a Z b a f (x) dx b a 4 5 18 1 1 q 8 81jf 0(a)jq + 29 162 f 0 a + b 2 q 1q +b a 4 5 18 1 1 q 29 162 f 0 a + b 2 q + 8 81jf 0(b)jq 1 q
which is the same of the inequality [6, Corollary 10] for s = 1.
ii. If we take p = 1, then we obtain the following inequality for harmonically convex function: 1 6 f (a) + 4f 2ab a + b + f (b) ab b a Z b a f (x) x2 dx b a 4ab [C 1(a; b)] 1 1 q jf0(a)jqD 1(a; b) + f0 2ab a + b q E 1(a; b) 1 q +b a 4ab [F 1(a; b)] 1 1 q f0 2ab a + b q G 1(a; b) + jf0(b)jqH 1(a; b) 1 q : Theorem 4. Let f : I (0; 1) ! R be a di¤erentiable mapping on I (the interior of I) and a; b 2 I with a < b and p 2 Rn f0g. If f0 2 L[a; b] and jf0jq
on I for q > 1, 1 r+ 1 q = 1, then 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p h Nr1 p;r(a; b) A 1 q jf0(a)jq; jf0(M p)jq + O 1 r p;r(a; b) A 1 q jf0(M p)jq; jf0(b)jq i ; where Np;r(a; b) = Z 1 0 It(a; Ap; r; r) dt; Op;r(a; b) = Z 1 0 I1 t(b; Ap; r; r) dt:
Proof. From Lemma 1, Hölder’s integral inequality and the p-convexity of jf0jq
on [a; b], we have, 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p N 1 r p;r(a; b) Z 1 0 f0 ((1 t) ap+ tAp) 1 p q dt 1 q +b p ap 4p O 1 r p;r(a; b) Z 1 0 f0 ((1 t) Ap+ tbp) 1 p q dt 1 q bp ap 4p " N 1 r p;r(a; b) Z 1 0 (1 t) jf0(a)jq+ t jf0(Mp)jq dt 1 q + O1r p;r(a; b) Z 1 0 (1 t) jf0(Mp)jq+ t jf0(b)jq dt 1 q# = b p ap 4p N 1 r p;r(a; b) jf 0(a)jq + jf0(Mp)jq 2 + O 1 r p;r(a; b) jf 0(M p)jq+ jf0(b)jq 2 = b p ap 4p h N 1 r p;r(a; b) Mq(jf0(a)j ; jf0(Mp)j) + O 1 r p;r(a; b) Mq(jf0(Mp)j ; jf0(b)j) i : This completes the proof of theorem.
Corollary 2. Under conditions of Theorem 4,
i. If we take p = 1, then we obtain the following inequality for convex function: 1 6 f (a) + 4f a + b 2 + f (b) 1 b a Z b a f (x) dx b a 12 1 + 2r+1 3 (r + 1) 1 r Mq jf0(a)j ; f0 a + b 2 + Mq f 0 a + b 2 ; jf 0(b)j
which is the same as the inequality in [22, Theorem 8] for s = 1.
ii. If we take p = 1, then we obtain the following inequality for harmonically convex function: 1 6 f (a) + 4f 2ab a + b + f (b) ab b a Z b a f (x) x2 dx b a 4ab h N 1 r 1;r(a; b) Mq(jf0(a)j ; jf0(H)j) + O 1 r 1;r(a; b) Mq(jf0(H)j ; jf0(b)j) i :
Theorem 5. Let f : I (0; 1) ! R be a di¤erentiable mapping on I (the interior of I) and a; b 2 I with a < b and p 2 Rn f0g. If f0 2 L[a; b] and jf0jq
is p-convex on I for q > 1, 1 r+ 1 q = 1, then 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 12p 1 + 2r+1 3 (r + 1) 1 r n Qp;q(a; b) jf0(a)j q + Rp;q(a; b) jf0(Mp)j q 1q + Sp;q(a; b) jf0(b)j q + Tp;q(a; b) jf0(Mp)j q 1qo; where Qp;q(a; b) = Z 1 0 Jt(a; Ap; 0; q) dt; Sp;q(a; b) = Z 1 0 K1 t(b; Ap; 0; q) dt; Rp;q(a; b) = Z 1 0 Kt(a; Ap; 0; q) dt; Tp;q(a; b) = Z 1 0 J1 t(b; Ap; 0; q) dt:
Proof. From Lemma 1, Hölder’s integral inequality and the p-convexity of jf0jq on [a; b], we obtain, 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p Z 1 0 t 1 3 1 [(1 t) ap+ tA p]1 1 p f0 [(1 t) ap+ tAp] 1 p dt +b p ap 4p Z 1 0 t 2 3 1 [(1 t) Ap+ tbp]1 1 p f0 [(1 t) Ap+ tbp] 1 p dt bp ap 12p 1 + 2r+1 3 (r + 1) 1 r Z 1 0 1 [(1 t) ap+ tA p]q q p f0 (1 t) ap+ tMpp 1 p q dt !1 q +b p ap 12p 1 + 2r+1 3 (r + 1) 1 r Z 1 0 1 [(1 t) Ap+ tbp]q q p f0 (1 t) Mpp+ tbp 1 p q dt !1 q bp ap 12p 1 + 2r+1 3 (r + 1) 1 rn
Qp;q(a; b) jf0(a)jq+ Rp;q(a; b) jf0(Mp)jq
1 q + Sp;q(a; b) jf0(b)jq+ Tp;q(a; b) jf0(Mp)jq 1 q o : This completes the proof of theorem.
Corollary 3. Under conditions of Theorem 5,
i. If we take p = 1, then we obtain the following inequality for convex function: 1 6 f (a) + 4f a + b 2 + f (b) 1 b a Z b a f (x) dx b a 12 1 + 2r+1 3 (r + 1) 1 r Mq jf0(a)j ; f0 a + b 2 + Mq f 0 a + b 2 ; jf 0(b)j
which reduce the inequality in Corollary 2 (i).
ii. If we take p = 1, then we obtain the following inequality for harmonically convex function: 1 6 f (a) + 4f 2ab a + b + f (b) ab b a Z b a f (x) x2 dx b a 12ab 1 + 2r+1 3 (r + 1) 1 r n
Q 1;q(a; b) jf0(a)jq+ R 1;q(a; b) jf0(H)jq
1 q + S 1;q(a; b) jf0(b)j q + T 1;q(a; b) jf0(H)j q 1qo:
Theorem 6. Let f : I (0; 1) ! R be a di¤erentiable mapping onI (the interior of I) and a; b 2 I with a < b and p 2 Rn f0g. If f0 2 L[a; b] and jf0jq
is p-concave on I for q > 1, 1 r+ 1 q = 1, then 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p " N 1 r p;r(a; b) f0 3ap+ bp 4 1 p ! + O 1 r p;r(a; b) f0 ap+ 3bp 4 1 p ! # : Proof. From Lemma 1, Hölder’s integral inequality and the p-concavity ofjf0jq on
[a; b], we have, 1 6[f (a) + 4f (Mp) + f (b)] p bp ap Z b a f (x) x1 pdx bp ap 4p N 1 r p;r(a; b) Z 1 0 f0 ((1 t) ap+ tAp) 1 p q dt 1 q +b p ap 4p O 1 r p;r(a; b) Z 1 0 f0 ((1 t) Ap+ tbp) 1 p q dt 1 q bp ap 4p " N1r p;r(a; b) f0 3ap+ bp 4 1 p! + O1r p;r(a; b) f0 ap+ 3bp 4 1 p! # : This completes the proof of theorem.
Corollary 4. Under conditions of Theorem 6,
i. If we take p = 1, then we obtain the following inequality for concave function: 1 6 f (a) + 4f a + b 2 + f (b) 1 b a Z b a f (x) dx b a 12 1 + 2r+1 3 (r + 1) 1 r f0 3a + b 4 + f 0 a + 3b 4 This is the same of the inequality in [6, Corollary 28] for s = 1.
ii. If we take p = 1, then we obtain the following inequality for harmonically concave function: 1 6 f (a) + 4f 2ab a + b + f (b) ab b a Z b a f (x) x2 dx b a 4ab N 1 r 1;r(a; b) f0 4ab a + 3b + O 1 r 1;r(a; b) f0 4ab 3a + b :
3. CONCLUSION
The paper deals with Simpson type inequalities for p-convex and p-concave func-tions. Firstly, we give a new identity for di¤erentiable functions and get some new integral inequalities for the p-convex and p-concave functions by using this identity.
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Current address : ·IMDAT ·I¸SCAN: Department of Mathematics, Faculty of Arts and Sciences, Giresun University, 28100, Giresun, Turkey.
E-mail address : [email protected]
ORCID Address: http://orcid.org/0000-0001-6749-0591
Current address : N·IHAN KALYONCU KONUK: Institute of Sciences and Arts, Giresun University-Giresun-Turkey
E-mail address : [email protected]
ORCID Address: http://orcid.org/0000-0003-1287-3370
Current address : MAH·IR KADAKAL: Department of Mathematics, Faculty of Arts and Sci-ences, Giresun University, 28100, Giresun, Turkey.
E-mail address : [email protected]
