# Trapezoidal Type Fejér Inequalities Related to Harmonically Convex Functions and Application

Authors: Sercan Turhan
Journal: Journal of Function Spaces
 (2019)

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## Abstract

Some authors introduced the concepts of the harmonically arithmetic convex functions and establish some integral inequalities of Hermite Hadamard Fejér type related to the harmonically arithmetic convex functions. In this paper, a mappingM(t) is considered to get some preliminary results and a new trapezoidal form of Fejér inequality related to the harmonically arithmetic convex functions. By using a mapping M(t), the new theorems and corollaries are obtained. Taking advantage of these, applications were given for some real number averages.


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## Body

## 1. Introduction

In 1906, the Hungarian mathematician L. Fejér proved the following integral inequalities known in the literature as Fejér inequality [1, 2]:(1)fa+b2∫abgxdx≤∫abfxgxdx≤fa+fb2∫abgxdxwhere f:[a,b]→R is convex and g:[a,b]→R+ is integrable and symmetric to x=2ab/a+bg(x)=g1/1/a+1/b-1/x,∀x∈[a,b]. For some other inequalities in connection with Fejér inequalities see [3–8] and the references therein.In [9], Hwang found out the Fejér trapezoidal inequality related to convex functions as follows:Theorem 1.
Letf:Io⊂R→R be a differentiable mapping, where a,b∈I with a<b, and let g:[a,b]→[0,∞) be a continuous positive mapping symmetric to a+b/2. If the mapping f′ is convex on [a,b], then the following inequality holds: (2)fa+fb2∫abgxdx-∫abfxgxdx≤b-a4f′a+f′b∫1+t/2a+1-t/2b1-t/2a+1+t/2bgxdxdt.In [8], Sarıkaya revealed new lemma and the difference between the right and middle part of (1) using Hölder’s inequality for convex function as follows:Lemma 2 (see [8]).
Letf:Io⊂R→R be a differentiable mapping on Io, a,b∈Io with a<b, and let w:[a,b]→[0,∞) be a differentiable mapping. If f′∈L[a,b], then the following equality holds:(3)1b-a∫abfxwxdx-1b-afa+b2∫abwxdx=b-a∫01ktf′ta+1-tbdtfor each t∈[0,1], where(4)kt=∫0twas+1-sbds,0≤t<12-∫t1was+1-sbds,12≤t<1Theorem 3 (see [8]).
Letf:Io⊆R→R be a differentiable mapping, a,b∈Io with a<b, and let w:[a,b]→R+ be a differentiable mapping symmetric to a+b/2. If f′q is convex on [a,b], q>1, then the following inequality holds: (5)1b-afa+fb2∫abwxdx-1b-a∫abfxwxdx≤12∫01gxpdt1/pf′aq+f′bq21/q,where 1/p+1/q=1 and (6)gx=∫a+b-atb-b-atwxdx.Theorem 4 ([1] Theorem 2.1).
Letf:I→R be a mapping that is differentiable on Io, let a,b∈Io be points with a<b, and let w:[a,b]→R be a nonnegative integrable mapping that is differentiable on (a,b). If w is symmetric to a+b/2 and if f′ is convex on [a,b], then (7)fa+fb2∫abwxdx-∫abfxwxdx≤f′a+f′b∫a+b/2bwxb-xdx,In [6], İşcan identified the harmonically convex function and proved Hermite Hadamard type inequality connected with harmonically convex function as follows:Definition 5 (see [6]).
LetI⊂R∖{0} be a real interval. A function f:I→R is said to be harmonically convex if(8)fxytx+1-ty≤tfy+1-tfxfor ∀x,y∈I and t∈[0,1]. If the inequalities in (1) are reversed then f is said to be harmonically concave.Theorem 6 (see [6]).
Letf:I⊂R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L[a,b] then the following inequalities hold(9)f2aba+b≤abb-a∫abfxx2dx≤fa+fb2Theorem 7 (see [6]).
Letf:I⊂(0,∞)→R be differentiable function on Io, a,b∈I, with a<b and f′∈L[a,b]. If f′q is harmonically convex on [a,b]b for q≥1, then(10)fa+fb2-abb-a∫abfxx2dx≤abb-a2λ1-1/qλ2f′aq+λ3f′bq1/q.where(11)λ1=1ab-2b-a2ln⁡a+b24abλ2=-1bb-a+3a+bb-a3ln⁡a+b24abλ3=1ab-a+3b+ab-a3ln⁡a+b24ab=λ1-λ2Definition 8 (see [10]).
A functionf:[a,b]→R is said to satisfy Lipschitz condition on [a,b] if there is a constant K so that for any two points x,y∈[a,b],(12)fx-fy≤Kx-y.In this paper, I obtain a new trapezoidal form of Fejér inequality via the absolute value of the derivative of the considered function is the harmonically convex function. In addition, I get features ofM(t) mapping as lemma and the new theorems and new corollaries. Furthermore, some applications in connection with special means are given.


## 2. Main Results

In the section, we have obtained the new theorem and corollary about Hermite Hadamard Fejér type inequality for the both harmonically convex functions.We use this lemma for harmonically convex function and motivated by above works and results we consider a mappingM(t) and obtain some introductory properties related to it. Also a new trapezoidal form of Fejér inequality is proved in the case that the absolute value of considered function is harmonically convex.Related to a functiong:[a,b]→R consider the mapping M:[0,1]→R as the following:(13)Mt=∫t1gabsb+1-sads-∫0tgabsb+1-sads.There exist some properties for the mapping M(t), compiled in the following lemma which are used to obtain our main results.Lemma 9.
Suppose thatI⊆R∖{0} is an interval, a,b∈Io with a<b and g:[a,b]⊆R∖{0}→R is an integrable function on [a,b].(1)
Ifg is symmetric to 2ab/a+b, then(14)Mt=2∫t1/2gabsb+1-sads,0<t<12-2∫1/2tgabsb+1-sads,12<t<1(2)
For anyt∈[0,1], (15)Mt+M1-t=0.(3)
Ifg is a nonnegative function, then(16)Mt≥0,0<t<12,Mt≤0,12<t<1.(4)
The following inequalities hold.(17)∫01Mtdt≤12g∞,and(18)∫01Mtdt≤2gq∫01t-121/pdt(5)
LetI⊆R∖{0} be an interval. f:Io→R is a differentiable mapping on Io, Io is an interior of I, and g is a differentiable nonnegative mapping. If f′∈L[a,b], then we get the following equality:(19)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx=abb-a2∫01Mtf′ab/tb+1-tatb+1-ta2dtProof.
(1) By taking the change of variablex=ab/sb+(1-s)a to (13), for 0≤t≤1/2, we obtain(20)Mt=abb-a∫aab/tb+1-tagxx2dx-∫ab/tb+1-tabgxx2dx,where 2ab/a+b≤ab/tb+(1-t)a≤b. Because of g being symmetric to 2ab/a+b,(21)∫2ab/a+bbgxx2dx=∫a2ab/a+bgxx2dx,and so(22)∫aab/tb+1-tagxx2dx=∫a2ab/a+bgxx2dx+∫2ab/a+bab/tb+1-tagxx2dx=∫2ab/a+bbgxx2dx+∫2ab/a+bab/tb+1-tagxx2dx.However(23)∫2ab/a+bbgxx2dx=∫2ab/a+bab/tb+1-tagxx2dx+∫ab/tb+1-tabgxx2dx.If we use (22) and (23) to (20), we get(24)Mt=2∫t1/2gabsb+1-sads,where 0≤t≤1/2.
By using the same argument as above, we can prove that(25)Mt=-2∫1/2tgabsb+1-sads,where 1/2≤t≤1.
(2) It is easy consequence of assertion of (1)
(3) By the assertion (3), we can get the following relations:(26)∫01Mtdt=∫01/2Mt-∫1/21Mtdt=2∫01/2∫t1/2gabsb+1-sadt+2∫1/21∫1/2tgabsb+1-sadsdt≤2∫01/2∫t1/2sups∈t,1/2⁡gabsb+1-sadsdt+2∫1/2tsups∈1/2,t⁡gabsb+1-sadsdt2∫01/212-tg∞dt+2∫1/21t-12g∞dt=2g∞∫01t-12dt=12g∞For the second part of (4), we conceive the following assertion which is not hard to prove:(27)∫01Mtdt=2∫01∫t1/2gabsb+1-sadsdt.By using Hölder inequality to the last inequality we get(28)∫t1/2gabsb+1-sads≤∫t1/2ds1/p∫t1/2gabsb+1-saqds1/q≤gqt-121/p.Now using (29) in (27) inequality, we get(29)∫01Mtdt≤2gq∫01t-121/pdt.(4) Firstly we calculate the equality as follows: (30)∫01Mtdfabtb+1-ta=abb-a∫01∫aab/bt+1-tagxx2dx-∫ab/tb+1-tabgxx2dxdfabtb+1-ta=abb-a∫aab/tb+1-tagxx2dxfabbt+1-ta01+b-aab∫01fabtb+1-tagabtb+1-tadt-∫ab/tb+1-tabggxx2dxfabtb+1-ta01+b-aab∫01fabtb+1-tagabtb+1-tadt=abb-a-∫abgxx2dxfa+fb+2∫abfxgxx2dx-abb-a∫01Mtf′ab/tb+1-tadttb+1-ta2=abb-a-∫abgxx2dxfa+fb+2∫abfxgxx2dxIf we pulse both of the last equal with -1/2, then the proof is completed.Theorem 10.
Suppose thatf:I⊆R∖{0} is a differentiable mapping on Io, a,b∈Io with a<b and g:[a,b]→R is a differentiable mapping. Assume that f′ is an integrable on [a,b] and there exist constants m<M such that(31)-∞<m≤x2f′x≤M<∞for all x∈[1/b,1/a]. Then (32)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-b-am+M4ab∫01Mtdt≤b-aM-m4ab∫01Mtdt.Proof.
From (5) of Lemma9, we get (33)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx=abb-a2∫01Mtf′ab/tb+1-tatb+1-ta2-m+M2a2b2+m+M2a2b2dt=b-am+M4ab∫01Mtdt+abb-a2∫01Mtf′ab/tb+1-tatb+1-ta2-m+M2a2b2dtas a result of (34)J=abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-b-am+M4ab∫01Mtdt=abb-a2∫01Mtf′ab/tb+1-tatb+1-ta2-m+M2a2b2dt.If it is used the absolute value to both sides of the last equality, we have (35)J≤abb-a2∫01Mtf′ab/tb+1-tatb+1-ta2-m+M2a2b2dt,since from the inequality m≤a2b2/tb+(1-t)a2f′ab/tb+(1-t)a≤M, we have (36)ma2b2-m+M2a2b2≤f′ab/tb+1-tatb+1-ta2-m+M2≤Ma2b2-m+M2a2b2which implies that(37)f′ab/tb+1-tatb+1-ta2-m+M2≤M-m2a2b2Remark 11.
If we takeg is symmetric to 2ab/a+b, then from Lemma 9, we get (38)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-b-am+M4ab∫01Mtdt≤b-aM-m8abg∞,and (39)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-b-am+M4ab∫01Mtdt≤b-aM-m2abgq∫01t-121/pdt.Theorem 12.
Suppose thatf:I⊂R∖{0}→R is a differentiable mapping on Io, a,b∈Io with a<b and g:[a,b]→R is a differentiable mapping symmetric to 2ab/a+b. If f′ is a harmonically convex mapping on [a,b], then(40)fa+fb2∫abgxx2dx-∫abfxgxx2dx≤b-aab∫a2ab/b-agxx2A1xf′a+A2x,tf′bdxwhere(41)A1x=∫01/a-1/x/1/a-1/bttb+1-ta2dt+∫01/a-1/x/1/a-1/b1-t1-tb+ta2dtand(42)A2x=∫01/a-1/x/1/a-1/b1-ttb+1-ta2dt+∫01/a-1/x/1/a-1/bt1-tb+ta2dt.Proof.
From the definition ofM(t), claim of Lemma 9, and f′ being a harmonically convex functions, we have(43)fa+fb2∫abgxx2dx-∫abfxgxx2dx=b-a22∫01Mtf′ab/tb+1-tatb-1-ta2≤b-a22∫01Mtf′abtb+1-tadttb-1-ta2≤b-a2∫01/2∫t1/2gabsb+1-sadstf′a+1-tf′btb+1-ta2dt+∫1/21∫1/2tgabsb+1-sadstf′a+1-tf′btb+1-ta2dt.If we change the order of integration, we get(44)fa+fb2∫abgxxdx-∫abfxgxxdx≤b-a2∫01/2∫0sgabsb+1-satf′a+1-tf′btb+1-ta2dtds+∫1/21∫s1gabsb+1-satf′a+1-tf′btb+1-ta2dtds=b-aab∫2ab/a+bb∫01/x-1/b/1/x-1/bgxx2tf′a+1-tf′btb+1-ta2dtdx+∫a2ab/a+b∫1/x-1/b/1/x-1/b1gxx2tf′a+1-tf′btb+1-ta2dtdx.Since g(x) function is symmetric to 2ab/a+b, then(45)∫2ab/a+bb∫01/x-1/b/1/x-1/bgxx2tf′a+1-tf′btb+1-ta2dtdx=∫a2ab/a+b∫01/a-1/x/1/a-1/bgxx2tf′a+1-tf′btb+1-ta2dtdx.Also it is not hard to see that(46)∫1/x-1/b/1/a-1/b1tdttb+1-ta2=∫01/a-1/x/1/a-1/b1-tdt1-tb+ta2,and(47)∫1/x-1/b/1/a-1/b11-tdttb+1-ta2=∫01/a-1/x/1/a-1/btdt1-tb+ta2.By using (45), (46), and (47) to (44), then we complete this proof.Theorem 13.
Letf:I⊂R∖{0}→R be a mapping differentiable on Io, let a,b∈Io be points w,th a<b, and let g:[a,b]→R be a nonnegative integrable mapping that is differentiable on (a,b). Assume that f′ is integrable on [a,b] and satisfies a Lipschitz condition for some K>0. Then(48)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-abb-a2∫01Mtf′2ab/a+btb+1-ta2dt≤Ka2b2b-a22a+b∫01Mt1-2ttb+1-ta3.Proof.
By using (5) of Lemma9, we have (49)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx=abb-a2∫01Mttb+1-ta2f′abtb+1-ta-f′2aba+b+f′2aba+bdt=abb-a2∫01Mttb+1-ta2f′abtb+1-ta-f′2aba+bdt+abb-a2∫01Mttb+1-ta2f′2aba+bdt.From the last equality, we get (50)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-abb-a2∫01Mttb+1-ta2f′2aba+bdt=abb-a2∫01Mttb+1-ta2f′abtb+1-ta-f′2aba+bdt.If f′ satisfies a Lipschitz condition as (12) for some K>0, then (51)f′abtb+1-ta-f′2aba+b≤Kabb-aa+b1-2ttb+1-ta.Because of this inequality, the proof is completed.Remark 14.
In Theorem13, suppose that g is symmetric to 2ab/a+b. If we use (1) of Lemma 9, we obtain(52)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx-abb-a2∫01Mttb+1-ta2f′2aba+bdt≤Ka2b2b-a2a+b∫01∫t1/2gabsb+1-sa1-2ttb+1-ta3dsdt≤Ka2b2b-a2a+b∫011/2-tsups∈t,1/2⁡gab/sb+1-sa1-2tinft∈0,1⁡tb+1-ta3dt≤2Kb2b-a2g∞aa+b∫0112-t2dt=Kb2b-a2g∞6aa+bwhich implies that(53)abb-afa+fb2∫abgxx2dx-∫abfxgxx2dx≤Kb2b-a26aa+b+bb-a4af′2aba+bg∞.Corollary 15.
In Theorem13, if we take g(x)=1, for ∀x∈[a,b], then(54)fa+fb2-abb-a∫abfxx2dx≤Kb2b-a26aa+b+bb-a4af′2aba+b


## 3. Application

Recall the following means which could be considered extensions of arithmetic, geometric, harmonic, and generalized logarithmic from positive to real numbers.(1)
The arithmetic mean:(55)A=Aa,b=a+b2;a,b∈R(2)
The geometric mean:(56)G=Ga,b=ab;a,b∈0,∞(3)
The harmonic mean:(57)H=Ha,b=2aba+b;a,b∈R(4)
The generalized logarithmic mean:(58)Lna,b=bn+1-an+1b-an+11/n;n∈Z∖-1,0,a,b∈R,a≠b.Now, we use these means in Corollary 15, Theorem 12 and Remark 11.Proposition 16.
Leta,b∈(0,∞), a<b, and r>2. Then(59)Aar,br-G2a,bLr-2r-2a,b≤bb-a2aKbA-1a,b6+r2Hr-1a,bProof.
The claim follows from Corollary15 for f(x)=xr,x∈(0,∞),r>2. Since f is a convex and nondecreasing function, f is a harmonically convex function.Proposition 17.
Leta,b∈(0,∞),a<b, and r>2. Then(60)Aar,br-G2a,bLr-2r-2a,b≤G4a,b∫a2ab/b-a1x2A1xf′a+A2x,tf′bdxwhere(61)A1x=∫01/a-1/x/1/a-1/bttb+1-ta2dt+∫01/a-1/x/1/a-1/b1-t1-tb+ta2dtand(62)A2x=∫01/a-1/x/1/a-1/b1-ttb+1-ta2dt+∫01/a-1/x/1/a-1/bt1-tb+ta2dt.Proof.
The claim follows from Theorem12 for f(x)=xr and g(x)=1,x∈(0,∞),r>2. Since f is a convex and nondecreasing function, f is a harmonically convex function.Proposition 18.
Leta,b∈(0,∞),a<b, and r>2. Then(63)Aar,br-G2a,bLr-2r-2a,b≤b-aM-m8abProof.
The claim follows from Remark11 for f(x)=xr and g(x)=1,x∈(0,∞),r>2. Since f is a convex and nondecreasing function, f is a harmonically convex function.



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