# A Sharp Bound for the Čebyšev Functional of Convex or Concave Functions

Authors: Mohammad W. Alomari
Journal: Chinese Journal of Mathematics
 (2013)

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## Abstract

A sharp bound for the Čebyšev functional of convex or concave functions is proved.


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## Body

## 1. Introduction

For two Lebesgue functionsf,g:[a,b]→ℝ, the Čebyšev functionaldefined by
(1)𝒯(f,g)=1b-a∫ab‍f(t)g(t)dt-1b-a∫ab‍f(t)dt·1b-a∫ab‍g(t)dt,
has interesting applications in various integral approximations as pointed out in the references below.The problem of bounding the Čebyšev functional has a long history, starting with Grüss [1] inequality in 1935, where he had proved that, for two integrable functions f, g such that ϕ≤f(x)≤Φ and γ≤g(x)≤Γ for any x∈[a,b], the inequality
(2)|C(f,g)|≤14(Φ-ϕ)(Γ-γ)
holds, and the constant 1/4 is the best possible.After that, many authors have studied the functional (1), and therefore, several bounds under various assumptions for the functions involved have been obtained.In 1971, Atkinson [2] proved that if f, g are twice differentiable and convex on [a,b] and
(3)∫ab‍(t-a+b2)g(t)dt=0,
then 𝒯(f,g)≥0.This result is implied by that of Lupaş [3] who proved the following: if f,g are convex functions on the interval [a,b], then
(4)𝒯(f,g)≥12(b-a)3∫ab‍(t-a+b2)f(t)dt·∫ab‍(t-a+b2)g(t)dt,
with equality when at least one of the functions f and g is a linear function on [a,b] (see also [4]).In recent years, several bounds for the Čebyšev functional in various cases including convexity assumptions for the functions involved are proved. For other results for convex integrands, see [5–9] and [10, page 256], [10, page 262] where further references are given.In this work, a sharp bound for the Čebyšev functional of convex or concave functions is proved.


## 2. The Result

We may begin with the following lemma.Lemma 1.
IfF,G:[a,b]→ℝ are both nonnegative, decreasing (increasing), and convex on [a,b], then H(t):=F(t)G(t) is nonnegative, decreasing (increasing), and convex function on [a,b].Proof.
The nonnegativity and monotonicity ofH follow directly by assumptions. To prove that H is convex, by the monotonicity of F, G and for x<y, we have
(5)(F(x)-F(y))(G(y)-G(x))≤0,
which implies that
(6)F(x)G(y)+F(y)G(x)≤F(x)G(x)+F(y)G(y).
Now, by (6) if α,β>0 and α+β=1, then
(7)H(αx+βy)=F(αx+βy)G(αx+βy)≤[αF(x)+βF(y)][αG(x)+βG(y)]=α2F(x)G(x)+αβ[F(x)G(y)+F(y)G(x)]+β2F(y)G(y)≤α2F(x)G(x)+αβ[F(x)G(x)+F(y)G(y)]+β2F(y)G(y)=αF(x)G(x)+βF(y)G(y)=αH(x)+βH(y),
which proves that H is convex and the proof is completely finished.Throughout this paper, letf,g:[a,b]→ℝ be any functions. Define the functions F,G:[a,b]→ℝ, such that
(8)F(t):=f(t)-f(a)+f(b)2,G(t):=g(t)-1b-a∫ab‍g(s)ds,∀t∈[a,b].Now, we may state our first result as follows.Theorem 2.
Under the assumptions of Lemma1, one has
(9)𝒯(f,g)≤112(f(b)-f(a))(g(b)-g(a)).
The constant 1/12 is the best possible in the sense that it cannot be replaced by smaller one.Proof.
Firstly, we note that, for any convex functionh defined on [a,b], we have
(10)h(t)≤t-ab-ah(b)+b-tb-ah(a).
Using the identity [10, page 246],
(11)𝒯(f,g)=1b-a∫ab‍F(t)G(t)dt=1b-a∫ab‍[f(t)-f(a)+f(b)2]×[g(t)-1b-a∫ab‍g(s)ds]dt,
and since F and G are two nonnegative monotonic convex functions on [a,b], then by Lemma 1, we have
(12)𝒯(f,g)=1b-a∫ab‍[f(t)-f(a)+f(b)2]×[g(t)-1b-a∫ab‍g(s)ds]dt≤1b-a∫ab‍[t-ab-af(b)+b-tb-af(a)-f(a)+f(b)2]×[{∫ab‍g}t-ab-ag(b)+b-tb-ag(a)-1b-a∫ab‍g(s)ds]dt=1b-a{112af(a)g(b)-112af(a)g(a)-112af(b)g(b)+112af(b)g(a)-112bf(a)g(b)+112bf(a)g(a)+112bf(b)g(b)-112bf(b)g(a)}=112(f(b)-f(a))(g(b)-g(a)),
which gives the desired inequality (6). To prove the sharpness, assume that (6) holds with constant C>0; that is,
(13)𝒯(f,g)≤C(f(b)-f(a))(g(b)-g(a)).
Let [a,b]=[0,1]; consider that f(x)=g(x)=x, x∈[0,1], so that we have ∫01‍f(x)g(x)dx=1/3 and ∫01‍f(x)dx=∫01‍g(x)dx=1/2. Making use of (9), we get
(14)𝒯(f,g)=13-14=112≤C,
which shows that the constant 1/12 is the best possible in both cases, and thus, the proof is completely finished.A parallel result for concave functions may be considered as follows.Lemma 3.
IfF:[a,b]→ℝ is nonnegative, decreasing (increasing), and concave on [a,b], while G:[a,b]→ℝ is nonnegative, increasing (decreasing), and concave on [a,b], then H(t):=F(t)G(t) is nonnegative, decreasing (increasing), and concave function on [a,b].Proof.
The nonnegativity and monotonicity ofH follow directly by assumptions. To prove that H is concave, by the monotonicity of F, G and for x<y, we have
(15)(F(x)-F(y))(G(y)-G(x))≥0,
which implies that
(16)F(x)G(y)+F(y)G(x)≥F(x)G(x)+F(y)G(y).
Now, by (16) if α,β>0 and α+β=1, then
(17)H(αx+βy)=F(αx+βy)G(αx+βy)≥[αF(x)+βF(y)][αG(x)+βG(y)]=α2F(x)G(x)+αβ[F(x)G(y)+F(y)G(x)]+β2F(y)G(y)≥α2F(x)G(x)+αβ[F(x)G(x)+F(y)G(y)]+β2F(y)G(y)=αF(x)G(x)+βF(y)G(y)=αH(x)+βH(y),
which proves that H is concave and the proof is completely finished.Now, we may state the reverse of (9) as follows.Theorem 4.
Under the assumptions of Lemma3, one has
(18)𝒯(f,g)≥112(f(b)-f(a))(g(b)-g(a)).
The constant 1/12 is the best possible in the sense that it cannot be replaced by smaller one.



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*Source: 295146-2013-09-23.xml*