# A Generalized Hermite-Hadamard Inequality for Coordinated Convex Function and Some Associated Mappings

Authors: Atiq Ur Rehman; Gulam Farid; Sidra Malik
Journal: Journal of Mathematics
 (2016)

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## Abstract

We have discussed the generalization of Hermite-Hadamard inequality introduced by Lupaş for convex functions on coordinates defined in a rectangle from the plane. Also we define that mappings are related to it and their properties are discussed.


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## Body

## 1. Introduction

Convex functions have great importance in many areas of Mathematics. A real valued functionf:I→R, where I is interval in R, is said to be convex if for α,β≥0 with α+β=1(1)fαx+βy≤αfx+βfy.In particular, if α=β=1/2 we have that(2)fx+y2≤fx+fy2.One of the most important inequalities that has attracted many mathematician in this field in the last few decades is the famous Hermite-Hadamard inequality, which establishes a refinement of (2) and it involves the notions of convexity.Letf:I→R be a convex mapping defined on the interval I of real numbers and a,b∈I with a<b. The double inequality(3)fa+b2≤1b-a∫abfxdx≤fa+fb2is known as the Hermite-Hadamard inequality. This inequality was published by Hermite in 1883 and was independently proved by Hadamard in 1893 (see Mitrinović and Lacković [1] for the whole history). It gives us an estimation of the mean value of a convex function f and it is important to note that (3) provides a refinement to the Jensen inequality. Fink [2] has worked on a best possible Hermite-Hadamard inequality and has deduced this inequality with least restrictions. Dragomir in [3, 4] worked on Hermite-Hadamard inequality and a mapping in connection to it. Dragomir and Milośević [5] gave some refinements of Hadamard's inequalities and its applications (see also [6–8] for more results). Dragomir and Pearce [9] wrote a monograph on selected topics on Hermite-Hadamard inequalities.The aim of this paper is to discuss an analogue of the generalization of Hermite-Hadamard inequality introduced by Lupaş for convex functions defined in a rectangle from the plane. Also some related mappings are defined and their properties are discussed.In [10] Dragomir has given the concept of convex functions on the coordinates in a rectangle from the plane and established the Hermite-Hadamard inequality for it.Definition 1.
LetΔ2=a,b×c,d⊂R with a<b and c<d. A function f:Δ2→R is called convex on coordinates if the partial mappings fy:[a,b]→R, fy(u)≔f(u,y) and fx:[c,d]→R, fx(v)≔f(x,v) are convex, defined for all y∈[c,d] and x∈[a,b].Note that every convex mappingf:Δ2→R is convex on the coordinates but the converse is not generally true.Theorem 2.
Suppose thatf:Δ2→R is convex on the coordinates on Δ2. Then we have (4)fa+b2,c+d2≤121b-a∫abfx,c+d2dx+1d-c∫cdfa+b2,ydy≤1b-ad-c∫ab∫cdfx,ydydx≤141b-a∫abfx,cdx+1b-a∫abfx,ddx+1d-c∫cdfa,ydy+1d-c∫cdfb,ydy≤fa,c+fa,d+fb,c+fb,d4.In [10], Dragomir for f:Δ2→R, defined a mapping H:[0,1]2→R as (5)Ht,s=1b-ad-c∫ab∫cdftx+1-ta+b2,sy+1-sc+d2dydxand proved the properties of this mapping in following theorem.Theorem 3.
Suppose thatf:Δ2→R is convex on the coordinates on Δ2.(i)
The mappingH is convex on the coordinates on [0,1]2.(ii)
We have the bounds(6)sups,t∈0,12Hs,t=1b-ad-c∫ab∫cdfx,ydydx=H1,1,infs,t∈0,12Hs,t=fa+b2,c+d2=H0,0.(iii)
The mappingH is monotonic nondecreasing on the coordinates.Also, in [10] Dragomir gave the following mapping, which is closely connected with Hadamard's inequality, H~:0,12→R defined as (7)H~t,s≔1b-a2d-c2∫ab∫ab∫cd∫cdftx+1-ty,sz+1-sudxdydzdu.and proved the following properties of this mapping.Theorem 4.
Suppose thatf:Δ2→R is convex on the coordinates on Δ2. (i)
We have the equalities(8)H~t+12,s=H~12-t,s∀t∈0,12,s∈0,1,H~t,s+12=H~t,12-s∀t∈0,1,s∈0,12,H~1-t,s=H~t,s,H~t,1-s=H~t,s∀t,s∈Δ2.(ii)
H~ is convex on the coordinates.(iii)
We have the bounds(9)inft,s∈0,12H~t,s=H~12,12=1b-a2d-c2∫ab∫ab∫cd∫cdfx+y2,z+u2dxdydzdu,supt,s∈0,12H~t,s=H~0,0=H~1,1=1b-ad-c∫ab∫cdfx,zdzdx.(iv)
The mappingH~·,s is monotonic nondecreasing on [0,1/2) and nondecreasing on [1/2,1] for all s∈[0,1]. A similar property has the mapping H~t,· for all t∈[0,1].(v)
We have the inequality(10)H~t,s≥max⁡Ht,s,H1-t,s,Ht,1-s,H1-t,1-s.
A generalized form of Hermite-Hadamard inequality is given by Lupaş in [11] (see also [12, page 143]).Theorem 5.
Let p, q be given positive numbers anda<b. Then the inequality,(11)fpa+qbp+q≤12y∫A-yA+yftdt≤pfa+qfbp+q,holds for A=pa+qb/p+q, y>0 and all continuous convex functions f:[a,b]→R iff y≤b-a/p+qmin⁡p,q.


## 2. Main Results

The following results comprise generalization of Hermite-Hadamard inequality introduced by Lupaş on coordinates in a rectangle from the plane.Theorem 6.
LetΔ2=a1,b1×a2,b2⊂R2 and p1, p2, q1, and q2 be positive real numbers and (12)Ai=piai+qibipi+qi,yi>0,whereyi≤bi-aipi+qimin⁡pi,qifori=1,2.Also let f:Δ2→R be convex on the coordinates on Δ2.(13)fA1,A2≤1212y1∫A1-y1A1+y1fs1,A2ds1+12y2∫A2-y2A2+y2fA1,s2ds2≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2≤1p1+q1p12y1∫A1-y1A1+y1fs1,a2ds1+q12y1∫A1-y1A1+y1fs1,b2ds1+1p2+q2p22y2∫A2-y2A2+y2fa1,s2ds2+q22y2∫A2-y2A2+y2fb1,s2ds2≤p1p2fa1,a2+p1q2fa1,b2+q1p2fb1,a2+q1q2fb1,b2×1p1+q1+1p2+q2.Proof.
Sincef:Δ2→R is convex on coordinates, it follows that the mapping fs1:[a2,b2]→R, fs1(s2)≔f(s1,s2), is convex on [a2,b2] for all s1∈[a1,b1], so by inequality (11) one has (14)fs1A2≤12y2∫A2-y2A2+y2fs1s2ds2≤p2fs1a2+q2fs1b2p2+q2,that is (15)fs1,A2≤12y2∫A2-y2A2+y2fs1,s2ds2≤p2fs1,a2+q2fs1,b2p2+q2.Observe that for 0<y1≤b1-a1/p1+q1min⁡p1,q1, by considering two cases 0<p1≤q1 and 0<q1<p1 we can easily verify that a1≤A1-y1<A1+y1≤b1, so that f is defined on A1-y1,A1+y1. Integrating the above inequality on A1-y1,A1+y1, we have(16)12y1∫A1-y1A1+y1fs1,A2ds1≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2≤1p2+q2p22y1∫A1-y1A1+y1fs1,a2ds1+q22y1∫A1-y1A1+y1fs1,b2ds1.By a similar argument applied for the mapping fs2:[a1,b1]→R, fs2(s1)≔f(s1,s2), we get(17)12y2∫A2-y2A2+y2fA1,s2ds2≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2≤1p1+q1p12y2∫A2-y2A2+y2fa1,s2ds2+q12y2∫A2-y2A2+y2fb1,s2ds2.Summing the inequalities (16) and (17), we get the second and third inequality in (13). Since f is convex on coordinates, using convexity of f on first coordinate and inequality (11), we have(18)fA1,A2≤12y1∫A1-y1A1+y1fs1,A2ds1and, using convexity of f on second coordinate, we have(19)fA1,A2≤12y2∫A2-y2A2+y2fA1,s2ds2.Adding the inequalities (18) and (19) we get the first inequality in (13). Finally, by the same Hermite-Hadamard inequality introduced by Lupaş we can also state(20)12y1∫A1-y1A1+y1fs1,a2ds1≤p1fa1,a2+q1fb1,a2p1+q1,(21)12y1∫A1-y1A1+y1fs1,b2ds1≤p1fa1,b2+q1fb1,b2p1+q1,(22)12y2∫A2-y2A2+y2fa1,s2ds2≤p2fa1,a2+q2fa1,b2p2+q2,(23)12y2∫A2-y2A2+y2fb1,s2ds2≤p2fb1,a2+q2fb1,b2p2+q2.Multiply (20) and (21) by p2 and multiply (22) and (23) by q2 then on addition we get the last inequality in (13).


## 3. Some Associated Mappings

In this section we will discuss some mappings associated with generalized Hermite-Hadamard inequality introduced by Lupaş for convex mappings on coordinates.Letp1, p2, q1, and q2 be positive real numbers and Ai=piai+qibi/pi+qi and yi>0 where yi≤bi-ai/pi+qimin⁡pi,qi for i=1,2. For mapping f:Δ2=a1,b1×a2,b2→R, we can define a mapping H:[0,1]2→R, (24)Hu,v≔14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fus1+1-uA1,vs2+1-vA2ds1ds2,where a1≤A1-y1<A1+y1≤b1 and a2≤A2-y2<A2+y2≤b2, so that f is defined on [A1-y1,A1+y1] and [A2-y2,A2+y2]. The properties of this mapping are embodied in the following theorem.Theorem 7.
Suppose thatp1,p2,q1, and q2 are positive real numbers and (25)Ai=piai+qibipi+qi,yi>0,whereyi≤bi-aipi+qimin⁡pi,qifori=1,2.Also let f:Δ2→R be convex on the coordinates on Δ2. (i)
The mappingH is convex on the coordinates on [0,1]2.(ii)
We have the bounds(26)supu,v∈0,12Hu,v=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2=H1,1,infu,v∈0,12Hu,v=fA1,A2=H0,0.(iii)
The mappingH is monotonic nondecreasing on the coordinates.Proof.
(i) Fixv∈[0,1], for all u1,u2∈[0,1] and α,β≥0 with α+β=1; we have (27)Hαu1+βu2,v=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fαu1s1+1-u1A1+βu2s1+1-u2A1,vs2+1-vA2ds1ds2≤α4y1y2∫A1-y1A1+y1∫A2-y2A2+y2fu1s1+1-u1A1,vs2+1-vA2ds1ds2+β4y1y2∫A1-y1A1+y1∫A2-y2A2+y2fu2s1+1-u2A1,vs2+1-vA2ds1ds2=αHu1,v+βHu2,v.If u∈[0,1] is fixed then, for all v1, v2∈[0,1] and α,β≥0 with α+β=1, we also have H(u,αv1+βv2)≤αH(u,v1)+βH(u,v2) and the statement is proved.
(ii) Sincef is convex on coordinates, we have by Jensen's inequality for integrals that (28)
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. By the convexity of H on the coordinates, we have (29)Hu,v≤12y1∫A1-y1A1+y1v12y2∫A2-y2A2+y2fus1+1-uA1,s2ds2+1-v12y2∫A2-y2A2+y2fus1+1-uA1,A2ds2ds1≤v12y2∫A2-y2A2+y2u12y1∫A1-y1A1+y1fs1,s2ds1+1-u12y1∫A1-y1A1+y1fA1,s2ds1ds2+1-v12y2∫A2-y2A2+y2u12y1∫A1-y1A1+y1fs1,A2ds1+1-u12y1∫A1-y1A1+y1fA1,A2ds1ds2=uv14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2+1-uv14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fA1,s2ds1ds2+u1-v14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,A2ds1ds2+1-u1-v14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fA1,A2ds1ds2=uv14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2+1-uv12y2∫A2-y2A2+y2fA1,s2ds2+u1-v12y1∫A1-y1A1+y1fs1,A2ds1+1-u1-vfA1,A2.
By Hadamard’s inequality introduced by Lupas (11), we also have (30)fA1,s2≤12y1∫A1-y1A1+y1fs1,s2ds1,fs1,A2≤12y2∫A2-y2A2+y2fs1,s2ds2. Thus by integration, we get that (31)12y2∫A2-y2A2+y2fA1,s2ds2≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2,12y1∫A1-y1A1+y1fs1,A2ds1≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2.Also using the result (32)fA1,A2≤14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2we deduce the inequality (33)Hu,v≤uv+1-uv+u1-v+1-u1-v14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2,for all (u,v)∈[0,1]2 and the second bound in (ii) is proved.
(iii) Firstly, we show thatH(u,v)≥H(0,v) for all [u,v]∈[0,1]2. By Jensen’s inequality for integrals, we have (34)Hu,v≥12y2∫A2-y2A2+y2f12y1∫A1-y1A1+y1us1+1-uA1ds1,vs2+1-vA2ds2=12y2∫A2-y2A2+y2fA1,vs2+1-vA2ds2=H0,vfor all [u,v]∈[0,1]2.
Now let0≤u1≤u2≤1. By the convexity of mapping H(·,v) for all [u,v]∈[0,1]2, we have (35)Hu2,v-Hu1,vu2-u1≥Hu1,v-H0,vu1≥0.The following theorem also holds.Theorem 8.
Suppose thatp1,p2,q1, and q2 are positive real numbers and (36)Ai=piai+qibipi+qi,yi>0,whereyi≤bi-aipi+qimin⁡pi,qifori=1,2.Also let f:Δ2→R be convex on the coordinates on Δ2. (i)
The mappingH is convex on [0,1]2.(ii)
Define the mappingh:[0,1]→R, h(u)≔H(u,u). Then h is convex, monotonic nondecreasing on [0,1] and one has the bounds (37)supu∈0,1hu=h1=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2,infu∈0,1hu=h0=fA1,A2.Proof.
(i) Let(u1,u2), (v1,v2)∈[0,1]2 and α,β≥0 with α+β=1. Since f:Δ2→R is convex on Δ2, we have (38)Hαu1,u2+βv1,v2=Hαu1+βv1,αu2+βv2=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fαu1s1+1-u1A1,u2s2+1-u2A2+βv1s1+1-v1A1,v2s2+1-v2A2ds1ds2≤α·14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fu1s1+1-u1A1,u2s2+1-u2A2ds1ds2+β·14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fv1s1+1-v1A1,v2s2+1-v2A2ds1ds2=αHu1,u2+βHv1,v2,which shows H is convex on [0,1]2.
(ii) Letu1,u2∈[0,1] and α,β≥0 with α+β=1. (39)hαu1+βu2=Hαu1+βu2,αu1+βu2=Hαu1,u1+βu2,u2≤αHu1,u1+βHu2,u2=αhu1+βhu2,which shows the convexity of h on [0,1].
We have, by the above theorem, that(40)hu=Hu,u≥H0,0=fA1,A2,u∈0,1,hu=Hu,u≤H1,1=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds1ds2,u∈0,1,which proves the required bounds.
Now let0≤u1≤u2≤1. By the convexity of mapping h, we have that (41)hu2-hu1u2-u1≥hu1-h0u1≥0, and the theorem is proved.Next, for positive real numbersp1,p2,q1,q2 and Ai=piai+qibi/pi+qi and yi>0 where yi≤bi-ai/pi+qimin⁡pi,qi for i=1,2, then for mapping f:Δ2=a1,b1×a2,b2→R we shall consider the mapping H~:[0,1]2→[0,∞), which is given as (42)H~u,v≔116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fus1+1-ut1,vs2+1-vt2dt2ds2dt1ds1,where a1≤A1-y1<A1+y1≤b1 and a2≤A2-y2<A2+y2≤b2, so that f is defined on [A1-y1,A1+y1] and [A2-y2,A2+y2].The next theorem contains the main properties of this mapping.Theorem 9.
Suppose thatp1,p2,q1, and q2 are positive real numbers and (43)Ai=piai+qibipi+qi,yi>0,whereyi≤bi-aipi+qimin⁡pi,qifori=1,2.Also let f:Δ2→R be convex on the coordinates on Δ2.(i)
We have the equalities(44)H~u+12,v=H~12-u,v,∀u∈0,12,v∈0,1,H~u,v+12=H~u,12-v,∀u∈0,12,v∈0,1,H~1-u,v=H~u,v,H~u,1-v=H~u,v,∀u,v∈Δ2.(ii)
H~ is convex on the coordinates.(iii)
We have the bounds(45)infu,v∈0,12H~u,v=H~12,12=116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,s2+t22dt2ds2dt1ds1,supu,v∈0,12H~u,v=H~0,0=H~1,1=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds2ds1.(iv)
The mappingH~(·,v) is monotonic nonincreasing on 0,1/2 and nondecreasing on 1/2,1 for all v∈[0,1]. A similar property has the mapping H~(u,·) for all u∈[0,1].(v)
We have the inequality(46)H~u,v≥max⁡Hs1,s2,H1-s1,s2,Hs1,1-s2,H1-s1,1-s2.Proof.
(i) and (ii) are obvious.
(iii) By the convexity off in the first variable, we get that (47)12fus1+1-ut1,vs2+1-vt2+f1-us1+ut1,vs2+1-vt2≥fs1+t12,vs2+1-vt2for all (s1,t1)∈[a1,b1]2, (s2,t2)∈[a2,b2]2, and (u,v)∈[0,1]2. Integrating on A1-y1,A1+y12, we get (48)14y12∫A1-y1A1+y1∫A1-y1A1+y1fus1+1-ut1,vs2+1-vt2ds1dt1≥14y12∫A1-y1A1+y1∫A1-y1A1+y1fs1+t12,vs2+1-vt2ds1dt1.Similarly,(49)14y22∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,vs2+1-vt2ds2dt2≥14y22∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,s2+t22ds2dt2.Now integrating this inequality on [A1-y1,A1+y1]2 and taking into account the above inequality, we deduce (50)H~u,v≥116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,s2+t22dt2ds2dt1ds1for (u,v)∈[0,1]2. The first bound in (iii) is proved and the second bound goes on likewise.
(iv) The monotonicity ofH~·,s follows by a similar argument from (iii).
(v) By Jensen’s integral inequality, we have successively for all(u,v)∈[0,1]2 that (51)H~u,v≥18y1y22∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2f12y1∫A1-y1A1+y1us1+1-ut1dt1,vs2+1-vt2dt2ds2ds1=18y1y22∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fus1+1-uA1,vs2+1-vt2dt2ds2ds1≥14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fus1+1-uA1,vs2+1-vA2ds2ds1=Hs1,s2.Similarly, we can easily show H~(1-u,v)≥H(1-s1,s2), H~(u,1-v)≥H(s1,1-s2), and H~(1-u,1-v)≥H(1-s1,1-s2).
In addition, asH~(u,v)=H~(1-u,v)=H~(u,1-v)=H~(1-u,1-v) for all (u,v)∈[0,1]2, so we deduce H~(u,v)≥max⁡H(s1,s2),H(1-s1,s2),H(s1,1-s2),H(1-s1,1-s2).
The theorem is thus proved.Theorem 10.
Suppose thatp1, p2, q1, and q2 are positive real numbers and (52)Ai=piai+qibipi+qi,yi>0,whereyi≤bi-aipi+qimin⁡pi,qifori=1,2.Also let f:Δ2→R be convex on the coordinates on Δ2. (i)
The mappingH~ is convex on Δ2.(ii)
Define the mappingh~:[0,1]→R, h~u≔H~(u,u). Then h~ is convex, monotonic nonincreasing on 0,1/2 and nondecreasing on 1/2,1 and one has the bounds(53)infu∈0,1h~u=h~12=116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,s2+t22dt2ds2dt1ds1,supu∈0,1h~u=h~1=h~0=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds2ds1.(iii)
One has the inequality(54)h~u≥max⁡hu,h1-u.Proof.
(i) Let(u1,u2), (v1,v2)∈[0,1] and α,β≥0 with α+β=1. Since f is convex on Δ2, we have (55)H~αu1,u2+βv1,v2=H~αu1+βv1,αu2+βv2=116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fαu1s1+1-u1t1,u2s2+1-u2t2+βv1s1+1-v1t1,v2s2+1-v2t2dt2ds2dt1ds1≤α·116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fu1s1+1-u1t1,u2s2+1-u2t2dt2ds2dt1ds1+β·116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fv1s1+1-v1t1,v2s2+1-v2t2dt2ds2dt1ds1=αH~u1,u2+βH~v1,v2,which shows H~ is convex on Δ2.
(ii) Letu1,u2, ∈[0,1] and α,β≥0 with α+β=1. (56)h~αu1+βu2=H~αu1+βu2,αu1+βu2=H~αu1,u1+βu2,u2≤αH~u1,u1+βH~u2,u2=αh~u1+βh~u2,which shows the convexity of h~ on [0,1].
As(57)h~u=H~u,u≥H~12,12=h~12=116y12y22∫A1-y1A1+y1∫A1-y1A1+y1∫A2-y2A2+y2∫A2-y2A2+y2fs1+t12,s2+t22dt2ds2dt1ds1,h~u=H~u,u≤H~0,0=H~0,0=14y1y2∫A1-y1A1+y1∫A2-y2A2+y2fs1,s2ds2ds1,which proves the required bounds.
It is obvious thath~ is monotonic nonincreasing on 0,1/2 and nondecreasing on 1/2,1.
(iii) As we know(58)h~u=H~u,u≥max⁡Hu,u,H1-u,1-u=max⁡hu,h1-uand the theorem is proved.



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*Source: 1631269-2016-12-13.xml*