# On the Hermite-Hadamard Inequality and Other Integral Inequalities Involving Several Functions

Authors: Banyat Sroysang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We present some new Hermite-Hadamard-type inequalities and other integral inequalities involving several functions.


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## Body

## 1. Introduction and Preliminaries

In 2003, Pachpatte [1] gave some Hermite-Hadamard-type inequalities involving two convex functions, and then Pachpatte [2] also gave, in 2004, some Hermite-Hadamard-type inequalities involving two log-convex functions. In 2007, Kirmaci et al. [3] gave some Hadamard-type inequalities involving s-convex functions. In 2008, Bakula et al. [4] presented some Hadamard-type inequalities involving m-convex functions and (α,m)-convex functions. In 2010, Set et al. [5] gave some new Hermite-Hadamard-type inequalities and other integral inequalities involving two functions. More details about results proved in [5] will be given in Section 2. In this paper, we present more general Hermite-Hadamard-type inequalities and some integral inequalities involving several functions.Letf:[a,b]→ℝ and p≥1. The p-norm of the function f on [a,b] is defined by
(1)∥f∥p={(∫ab‍|f(x)|pdx)1/p;1≤p<∞,supx∈[a,b]|f(x)|;p=+∞.Below we recall few well-known inequalities that will be useful in the proofs of our results.Hermite-Hadamard’s Inequality (see [6–8]). If f is a convex function on [a,b], then
(2)f(a+b2)≤1b-a∫ab‍f(x)dx≤f(a)+f(b)2.
If f is a concave function on [a,b], then
(3)f(a)+f(b)2≤1b-a∫ab‍f(x)dx≤f(a+b2).Barnes-Gugunova-Levin Inequality (see [9–11]). If f and g are nonnegative concave functions on [a,b], and if p,q>1, then
(4)(∫ab‍fp(x)dx)1/p(∫ab‍gq(x)dx)1/q≤B(p,q)∫ab‍f(x)g(x)dx,
where
(5)B(p,q)=6(b-a)1/p+1/q-1(p+1)1/p(q+1)1/q.The Power-Mean Inequality (see [12]). Let x1,x2,…,xn,p1,p2,…,pn>0 and let r∈ℝ∪{-∞,+∞}. Then
(6)Mn[r]={(∑i=1npixir∑i=1npi)1/r;r≠0,±∞,(∏i=1n‍xipi)1/∑i=1npi;r=0,max{x1,x2,…,xn};r=∞,min{x1,x2,…,xn};r=-∞.
Notice that if -∞≤r<s≤+∞ then
(7)Mn[r]≤Mn[s].A Generalization of Hölder Integral Inequality. For any p1,p2,…,pn>1, ∑i=1n‍(1/pi)=1, if f1,f2,…,fn are nonnegative functions on [a,b] and if f1p1,f2p2,…,fnpn are integrable functions on [a,b], then
(8)∫ab‍(∏i=1n‍fi(x))dx≤∏i=1n‍(∫ab‍fipi(x)dx)1/pi.A Generalization of Minkowski Integral Inequality. If p≥1 and if f1,f2,…,fn are non-negative functions on [a,b] such that 0<∫ab‍fip(x)dx<∞ for all i=1,2,…,n, then
(9)(∫ab‍(∑i=1n‍fi(x))pdx)1/p≤∑i=1n‍(∫ab‍fip(x)dx)1/p.A Generalization of Young Inequality. If x1,x2,…,xn≥0 and p1,p2,…,pn>1, ∑i=1n‍(1/pi)=1, then
(10)∏i=1n‍xi≤∑i=1n‍xpipi.To prove results, we refer to the following lemma.Lemma 1 (see [13]).
Ifp≥1 and if f and g are positive functions on [a,b] such that 0<m≤f(x)/g(x)≤M for all x∈[a,b], then
(11)(∫ab‍fp(x)dx)1/p≤MM+1(∫ab‍(f(x)+g(x))pdx)1/p,(∫ab‍gp(x)dx)1/p≤1m+1(∫ab‍(f(x)+g(x))pdx)1/p.Proof.
Letp≥1. Assume that f and g are positive functions on [a,b] such that 0<m≤f(x)/g(x)≤M for all x∈[a,b]. Then
(12)f≤Mg=M(f+g)-Mf,g≤1mf=1m(f+g)-1mg.
Then
(13)(M+1)pfp≤Mp(f+g)p,(1m+1)pgp≤(1m)p(f+g)p.
Thus,
(14)(∫ab‍fp(x)dx)1/p≤MM+1(∫ab‍(f(x)+g(x))pdx)1/p,(∫ab‍gp(x)dx)1/p≤1m+1(∫ab‍(f(x)+g(x))pdx)1/p.


## 2. Main Results

We start this section with the following.Theorem 2.
Letn be a positive even integer and p1,p2,…,pn>1 and let f1,f2,…,fn be non-negative functions on [a,b] such that f1p1,f2p2,…,fnpn are concave on [a,b]. Then
(15)∏i=1n‍(fi(a)+fi(b)2)≤1(b-a)∑i=1n(1/pi)(∏i=1n/2‍B(p2i-1,p2i)∫ab‍f2i-1(x)f2i(x)dx),
where
(16)B(p2i-1,p2i)=6(b-a)1/p2i-1+1/p2i-1(p2i-1+1)1/p2i-1(p2i+1)1/p2i
for all i=1,…,n/2. Moreover, if ∑i=1n‍(1/pi)=1, then
(17)1b-a∫ab‍(∏i=1n‍fi(x))dx≤∏i=1n‍fi(a+b2).Proof.
Applying the inequality (3) with fipi, for any i=1,…,n, we get
(18)fipi(a)+fipi(b)2≤1b-a∫ab‍fipi(x)dx≤fipi(a+b2),
and, consequently,
(19)(fipi(a)+fipi(b)2)1/pi≤1(b-a)1/pi(∫ab‍fipi(x)dx)1/pi≤fi(a+b2).
By the Barnes-Gudunova-Levin inequality (4), it follows that
(20)∏i=1n‍(fipi(a)+fipi(b)2)1/pi≤1(b-a)∑i=1n(1/pi)∏i=1n‍(∫ab‍fipi(x)dx)1/pi=1(b-a)∑i=1n(1/pi)∏i=1n/2‍(∫ab‍f2i-1p2i-1(x)dx)1/p2i-1×(∫ab‍f2ip2i(x)dx)1/p2i≤1(b-a)∑i=1n(1/pi)(∏i=1n/2‍B(p2i-1,p2i)∫ab‍f2i-1(x)f2i(x)dx),
where
(21)B(p2i-1,p2i)=6(b-a)1/p2i-1+1/p2i-1(p2i-1+1)1/p2i-1(p2i+1)1/p2i
for all i=1,…,n/2.
By the power-mean inequality (7), we have
(22)(fipi(a)+fipi(b)2)1/pi≥fi(a)+fi(b)2
for all i=1,2,…,n.
This implies the inequality (15).
Next, we assume that∑i=1n‍(1/pi)=1. By the inequality (19) and the generalized Hölder inequality, we obtain that
(23)∏i=1n‍fi(a+b2)≥1(b-a)∑i=1n(1/pi)∏i=1n‍(∫ab‍fipi(x)dx)1/pi=1b-a∏i=1n‍(∫ab‍fipi(x)dx)1/pi≥1b-a∫ab‍(∏i=1n‍fi(x))dx.
This proof is completed.It is easy to notice that if we putn=2 in Theorem 2 then we get the following.Corollary 3 (see [5]).
Letp,q>1 and let f,g be non-negative functions on [a,b] such that fp,gq are concave on [a,b]. Then
(24)(f(a)+f(b)2)(g(a)+g(b)2)≤1(b-a)1/p+1/q(B(p,q)∫ab‍f(x)g(x)dx),
where
(25)B(p,q)=6(b-a)1/p+1/q-1(p+1)1/p(q+1)1/q.
Moreover, if 1/p+1/q=1, then
(26)1b-a∫ab‍f(x)g(x)dx≤f(a+b2)g(a+b2).Theorem 4.
Letp≥1 and n be a positive integer such that n≥2 and let f1,f2,…,fn be positive functions on [a,b] such that the functions f1p,f2p,…,fnp are integrable functions on [a,b], 0<∫ab‍fip(x)dx<∞ for all i=1,2,…,n, and
(27)0<mi≤fi(x)fi+1(x)≤Mi
for all x∈[a,b] and for all i=1,…,n-1. Then
(28)(∑i=1n∥fi∥p)n∏i=1n∥fi∥p≥1∏i=1nsi,
where
(29)s1=M1M1+1,sn=1mn-1+1,si=min{1mi-1+1,MiMi+1}
for all 1<i<n.Proof.
By Lemma1, we have
(30)(∫ab‍fip(x)dx)1/p≤MiMi+1(∫ab‍(fi(x)+fi+1(x))pdx)1/p,(∫ab‍fi+1p(x)dx)1/p≤1mi+1(∫ab‍(fi(x)+fi+1(x))pdx)1/p
for all i=1,…,n-1.
Then(31)(∫ab‍fip(x)dx)1/p≤MiMi+1(∫ab‍(∑j=1n‍fj(x))pdx)1/p,(∫ab‍fi+1p(x)dx)1/p≤1mi+1(∫ab‍(∑j=1n‍fj(x))pdx)1/p
for all i=1,…,n-1.
Lets1=M1/(M1+1), sn=1/(mn-1+1), and si=min{1/(mi-1+1),Mi/(Mi+1)} for all 1<i<n.
It follows that(32)(∫ab‍f1p(x)dx)1/p≤s1(∫ab‍(∑j=1n‍fj(x))pdx)1/p,(∫ab‍fnp(x)dx)1/p≤sn(∫ab‍(∑j=1n‍fj(x))pdx)1/p,(∫ab‍fip(x)dx)1/p≤si(∫ab‍(∑j=1n‍fj(x))pdx)1/p
for all 1<i<n.
By multiplying the above inequalities and the generalized Minkowski inequality, we obtain that(33)∏i=1n‍(∫ab‍fip(x)dx)1/p≤(∏i=1n‍si)((∫ab‍(∑j=1n‍fj(x))pdx)1/p)n≤(∏i=1n‍si)(∑j=1n‍(∫ab‍fjp(x)dx)1/p)n.
Then
(34)∏i=1n‍∥fi∥p≤(∏i=1n‍si)(∑j=1n‍∥fi∥p)n.
This implies the inequality (28).Notice that from above theorem one can easily get the following.Corollary 5 (see [5]).
Letp≥1 and let f,g be positive functions on [a,b] such that 0<∫ab‍fp(x)dx<∞, 0<∫ab‍gp(x)dx<∞, and
(35)0<m≤f(x)g(x)≤M
for all x∈[a,b]. Then
(36)∥f∥p2+∥g∥p2∥f∥p∥g∥p≥1s-2,
where s=M/(M+1)(m+1).Proof.
By Theorem4 where n=2, we have
(37)(∥f∥p+∥g∥p)2∥f∥p∥g∥p≥1s1s2,
where
(38)s1=MM+1,s2=1m+1.
Lets=s1s2. Then
(39)1s≤(∥f∥p+∥g∥p)2∥f∥p∥g∥p=∥f∥p2+2∥f∥p∥g∥p+∥g∥p2∥f∥p∥g∥p=∥f∥p2+∥g∥p2∥f∥p∥g∥p+2.
This implies the inequality (36).Theorem 6.
Letn be a positive integer such that n≥2 and p1,p2,…,pn>1 and let f1,f2,…,fn be non-negative functions on [a,b] such that f1p1,f2p2,…,fnpn are concave on [a,b]. Then
(40)∏i=1n‍(fi(a)+fi(b))pi≤2∑i=1npi(b-a)n∏i=1n‍∥fi∥pipi.Proof.
Using the inequality (3) with fipi, for any i=1,…,n, we obtain
(41)fipi(a)+fipi(b)2≤1b-a∫ab‍fipi(x)dx.
Then(42)∏i=1n‍(fipi(a)+fipi(b)2)≤1(b-a)n∏i=1n‍(∫ab‍fipi(x)dx).
By the power-mean inequality (7), we have
(43)(fipi(a)+fipi(b)2)1/pi≥fi(a)+fi(b)2,
so
(44)fipi(a)+fipi(b)2≥(fi(a)+fi(b))pi2pi,
for all i=1,2,…,n.
Then(45)1(b-a)n∏i=1n‍(∫ab‍fipi(x)dx)≥∏i=1n‍((fi(a)+fi(b))pi2pi)=∏i=1n(fi(a)+fi(b))pi2∑i=1npi.
This implies the inequality (40).One can easily check that if we putn=2 in Theorem 6 then we get the following.Corollary 7 (see [5]).
Letp,q>1 and let f,g be non-negative functions on [a,b] such that fp,gq are concave on [a,b]. Then
(46)(f(a)+f(b))p(g(a)+g(b))q2p+q≤1(b-a)2∥f∥pp∥g∥qq.Theorem 8.
Letn be a positive integer such that n≥2 and p1,p2,…,pn>1, ∏i=1n‍(1/pi)=1, and let f1,f2,…,fn be positive functions on [a,b] such that the function fipi is integrable on [a,b], 0<∫ab‍fipi(x)dx<∞ for all i=1,2,…,n, and
(47)0<mi≤fi(x)fi+1(x)≤Mi
for all x∈[a,b] and i=1,…,n-1. Then
(48)∫ab‍(∏i=1n‍fi(x))dx≤∑i=1n‍(sipi(∑j=1n‍∥fj∥pipi)),
where
(49)s1=2p1-1(M1M1+1)p1,sn=2pn-1(1mn-1+1)pn,si=min{2pi-1(1mi-1+1)pi,2pi-1(MiMi+1)pi}
for all 1<i<n.Proof.
By Lemma1, we have
(50)(∫ab‍fipi(x)dx)1/pi≤MiMi+1(∫ab‍(fi(x)+fi+1(x))pidx)1/pi,(∫ab‍fi+1pi+1(x)dx)1/pi+1≤1mi+1(∫ab‍(fi(x)+fi+1(x))pi+1dx)1/pi+1
for all i=1,…,n-1.
Using the elementary inequality(α+β)p≤2p-1(αp+βp) where p>1 and α,β>0, we get
(51)∫ab‍fipi(x)dx≤(MiMi+1)pi∫ab‍(fi(x)+fi+1(x))pidx≤(MiMi+1)pi∫ab‍2pi-1(fipi(x)+fi+1pi(x))dx,∫ab‍fi+1pi+1(x)dx≤(1mi+1)pi+1∫ab‍(fi(x)+fi+1(x))pi+1dx≤(1mi+1)pi+1∫ab‍2pi+1-1×(fipi+1(x)+fi+1pi+1(x))dx
for all i=1,…,n-1.
Then(52)∫ab‍fipi(x)dx≤2pi-1(MiMi+1)pi∫ab‍(∑j=1n‍fjpi(x))dx=2pi-1(MiMi+1)pi∑j=1n‍(∫ab‍fjpi(x)dx),∫ab‍fi+1pi+1(x)dx≤2pi+1-1(1mi+1)pi+1∫ab‍(∑j=1n‍fjpi+1(x))dx=2pi+1-1(1mi+1)pi+1∑j=1n‍(∫ab‍fjpi+1(x)dx)
for all i=1,…,n-1.
Lets1=2p1-1(M1/(M1+1))p1, sn=2pn-1(1/(mn-1+1))pn, and
(53)si=min{2pi-1(1mi-1+1)pi,2pi-1(MiMi+1)pi}
for all 1<i<n.
It follows that(54)∫ab‍f1p1(x)dx≤s1∑j=1n‍(∫ab‍fjp1(x)dx),∫ab‍fnpn(x)dx≤sn∑j=1n‍(∫ab‍fjpn(x)dx),∫ab‍fipi(x)dx≤si∑j=1n‍(∫ab‍fjpi(x)dx)
for all 1<i<n.
By the generalized Young inequality, we obtain that(55)∫ab‍(∏i=1n‍fi(x))dx≤∫ab‍(∑i=1n‍1pifipi(x))dx=∑i=1n‍1pi(∫ab‍fipi(x)dx)≤∑i=1n‍1pi(si∑j=1n‍(∫ab‍fjpi(x)dx))≤∑i=1n‍(sipi(∑j=1n‍∥fj∥pipi)).
This proof is completed.Applying Theorem8 with n=2 and putting there p1=p,p2=q, 2s1/p1=c1, and 2s2/p2=c2, we get the following.Corollary 9 (see [5]).
Letp, q>1, 1/p+1/q=1, and let f and g be positive functions on [a,b] such that 0<∫ab‍fp(x)dx<∞, 0<∫ab‍gq(x)dx<∞, and
(56)0<m≤f(x)g(x)≤M
for all x∈[a,b]. Then
(57)∫ab‍f(x)g(x)dx≤c1(∥f∥pp+∥g∥pp2)+c2(∥f∥qq+∥g∥qq2),
where
(58)c1=2pp(MM+1)p,c2=2qq(1m+1)q.



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