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

Abstract

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

1. Introduction

For two Lebesgue functions 𝑓,𝑔∶[𝑎,𝑏]→ℝ, the Čebyšev functional defined by 𝒯=1𝑓,𝑔𝑏−𝑎^𝑏_𝑎𝑓(𝑡)𝑔(𝑡)−1𝑑𝑡𝑏−𝑎^𝑏_𝑎𝑓(𝑡)1𝑑𝑡⋅𝑏−𝑎^𝑏_𝑎𝑔(𝑡)𝑑𝑡, (1) 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 𝐶≤1𝑓,𝑔4Φ−𝜙Γ−𝛾 (2) 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 ^𝑏_𝑎𝑡−𝑎+𝑏2𝑔(𝑡)𝑑𝑡=0, (3) 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 𝒯𝑓,𝑔≥12(𝑏−𝑎)^3𝑏𝑎𝑡−𝑎+𝑏2𝑓(𝑡)⋅𝑑𝑡_𝑏𝑎𝑡−𝑎+𝑏2𝑔(𝑡)𝑑𝑡, (4) 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. If 𝐹,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 of 𝐻 follow directly by assumptions. To prove that 𝐻 is convex, by the monotonicity of 𝐹, 𝐺 and for 𝑥<𝑦, we have 𝐹(𝑥)−𝐹𝑦𝐺𝑦−𝐺(𝑥)≤0, (5) which implies that 𝐹(𝑥)𝐺𝑦𝑦𝐺(𝑥)(𝑥)𝐺(𝑥)𝑦𝐺𝑦.+𝐹≤𝐹+𝐹 (6) Now, by (6) if 𝛼,𝛽>0 and 𝛼+𝛽=1, then 𝐻𝐺≤(𝑥)𝑦(𝑥)𝑦𝛼𝑥+𝛽𝑦=𝐹𝛼𝑥+𝛽𝑦𝛼𝑥+𝛽𝑦𝛼𝐹+𝛽𝐹𝛼𝐺+𝛽𝐺=𝛼²𝐹(𝑥)𝐺(𝑥)+𝛼𝛽𝐹(𝑥)𝐺𝑦+𝐹𝑦𝐺(𝑥)+𝛽²𝐹𝑦𝐺𝑦≤𝛼²𝐹(𝑥)𝐺(𝑥)+𝛼𝛽𝐹(𝑥)𝐺(𝑥)+𝐹𝑦𝐺𝑦+𝛽²𝐹𝑦𝐺𝑦(𝑥)𝐺(𝑥)𝑦𝐺𝑦(𝑥)𝑦,=𝛼𝐹+𝛽𝐹=𝛼𝐻+𝛽𝐻 (7) which proves that 𝐻 is convex and the proof is completely finished.

Throughout this paper, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be any functions. Define the functions 𝐹,𝐺∶[𝑎,𝑏]→ℝ, such that 𝐹(𝑡)(𝑡)−𝑓(𝑎)(𝑏)∶=𝑓+𝑓2,𝐺(𝑡)(𝑡)−1∶=𝑔𝑏−𝑎^𝑏_𝑎𝑔(𝑠)[].𝑑𝑠,∀𝑡∈𝑎,𝑏 (8)

Now, we may state our first result as follows.

Theorem 2. Under the assumptions of Lemma 1, one has 𝒯≤1𝑓,𝑔𝑓(𝑏)(𝑎𝑔(𝑏)(𝑎.12−𝑓)−𝑔) (9) 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 function ℎ defined on [𝑎,𝑏], we have ℎ(𝑡)≤𝑡−𝑎ℎ(𝑏)+𝑏−𝑎𝑏−𝑡ℎ(𝑎).𝑏−𝑎 (10) Using the identity [10, page 246], 𝒯=1𝑓,𝑔𝑏−𝑎^𝑏_𝑎𝐹(𝑡)𝐺(𝑡)=1𝑑𝑡𝑏−𝑎^𝑏_𝑎𝑓(𝑡)−𝑓(𝑎)+𝑓(𝑏)2×𝑔(𝑡)−1𝑏−𝑎^𝑏_𝑎𝑔(𝑠)𝑑𝑠𝑑𝑡, (11) and since 𝐹 and 𝐺 are two nonnegative monotonic convex functions on [𝑎,𝑏], then by Lemma 1, we have 𝒯=1𝑓,𝑔𝑏−𝑎^𝑏_𝑎𝑓(𝑡)−𝑓(𝑎)+𝑓(𝑏)2×𝑔(𝑡)−1𝑏−𝑎^𝑏_𝑎𝑔(𝑠)≤1𝑑𝑠𝑑𝑡𝑏−𝑎^𝑏_𝑎𝑡−𝑎𝑓(𝑏)+𝑏−𝑎𝑏−𝑡𝑓(𝑎)−𝑓(𝑎)(𝑏)𝑏−𝑎+𝑓2×𝑡−𝑎𝑔(𝑏)+𝑏−𝑎𝑏−𝑡𝑔(𝑎)−1𝑏−𝑎𝑏−𝑎^𝑏_𝑎𝑔(𝑠)=1𝑑𝑠𝑑𝑡1𝑏−𝑎(𝑎)𝑔(𝑏)−112𝑎𝑓(𝑎)𝑔(𝑎)−112𝑎𝑓(𝑏)𝑔(𝑏)+112𝑎𝑓(𝑏)𝑔(𝑎)−112𝑎𝑓(𝑎)𝑔(𝑏)+112𝑏𝑓(𝑎)𝑔(𝑎)+112𝑏𝑓(𝑏)𝑔(𝑏)−112𝑏𝑓(𝑏)𝑔(𝑎=112𝑏𝑓)𝑓(𝑏)(𝑎)𝑔(𝑏)(𝑎),12−𝑓−𝑔 (12) which gives the desired inequality (6). To prove the sharpness, assume that (6) holds with constant 𝐶>0; that is, 𝒯𝑓,𝑔≤𝐶𝑓(𝑏)−𝑓(𝑎)𝑔(𝑏)−𝑔(𝑎). (13) Let [𝑎,𝑏]=[0,1]; consider that 𝑓(𝑥)=𝑔(𝑥)=𝑥, 𝑥∈[0,1], so that we have ∫¹₀𝑓(𝑥)𝑔(𝑥)𝑑𝑥=1/3 and ∫¹₀𝑓(𝑥)𝑑𝑥=∫¹₀𝑔(𝑥)𝑑𝑥=1/2. Making use of (9), we get 𝒯=1𝑓,𝑔3−14=112≤𝐶, (14) 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. If 𝐹∶[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 of 𝐻 follow directly by assumptions. To prove that 𝐻 is concave, by the monotonicity of 𝐹, 𝐺 and for 𝑥<𝑦, we have 𝐹(𝑥)−𝐹𝑦𝐺𝑦−𝐺(𝑥)≥0, (15) which implies that 𝐹(𝑥)𝐺𝑦+𝐹𝑦𝐺(𝑥)≥𝐹(𝑥)𝐺(𝑥)+𝐹𝑦𝐺𝑦. (16) Now, by (16) if 𝛼,𝛽>0 and 𝛼+𝛽=1, then 𝐻𝐺≥(𝑥)𝑦(𝑥)𝑦𝛼𝑥+𝛽𝑦=𝐹𝛼𝑥+𝛽𝑦𝛼𝑥+𝛽𝑦𝛼𝐹+𝛽𝐹𝛼𝐺+𝛽𝐺=𝛼²𝐹(𝑥)𝐺(𝑥)+𝛼𝛽𝐹(𝑥)𝐺𝑦+𝐹𝑦𝐺(𝑥)+𝛽²𝐹𝑦𝐺𝑦≥𝛼²𝐹(𝑥)𝐺(𝑥)+𝛼𝛽𝐹(𝑥)𝐺(𝑥)+𝐹𝑦𝐺𝑦+𝛽²𝐹𝑦𝐺𝑦(𝑥)𝐺(𝑥)𝑦𝐺𝑦(𝑥)𝑦,=𝛼𝐹+𝛽𝐹=𝛼𝐻+𝛽𝐻 (17) 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 Lemma 3, one has 𝒯≥1𝑓,𝑔𝑓(𝑏)(𝑎𝑔(𝑏)(𝑎.12−𝑓)−𝑔) (18) The constant 1/12 is the best possible in the sense that it cannot be replaced by smaller one.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.