Jensen type inequalities for twice differentiable functions

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Research Article

Jensen type inequalities for twice differentiable functions

Abdallah El Frissi, Benharrat Bela¨ıdi^∗, Zinelaˆabidine Lareuch

Department of Mathematics, Laboratory of Pure and Applied Mathematics, University of Mostaganem (UMAB), B. P. 227 Mostaganem-(Algeria)

Dedicated to George A Anastassiou on the occasion of his sixtieth birthday Communicated by Professor C. Park

Abstract

In this paper, we give some Jensen-type inequalities forϕ:I →R, I = [α, β]⊂R,whereϕis a continuous function on I, twice differentiable on ˚I = (α, β) and there exists m = inf

x∈˚I

ϕ⁰⁰(x) or M = sup

x∈˚I

Keywords: Jensen inequality, Convex functions, Twice differentiable functions.

2010 MSC: 26D15.

1. Introduction and main results

Throughout this note, we writeI and ˚I for the intervals [α, β] and (α, β) respectively−∞ ≤α < β≤+∞.

A function ϕ is said to be convex on I if λϕ(x) + (1−λ)ϕ(y) ≥ ϕ(λx+ (1−λ)y) for all x, y ∈ I and 0 ≤ λ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A functionϕthat is continuous function onI and twice differentiable on ˚I is convex on I ifϕ⁰⁰(x)≥0 for all x∈˚I (concave if the inequality is flipped).

The famous inequality of Jensen states that:

∗Corresponding author

Email addresses: [email protected](Abdallah El Frissi),[email protected](Benharrat Bela¨ıdi ), [email protected](Zinelaˆabidine Lareuch)

Received 2009-4-22

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Theorem 1.1. ([1], [3]) Let ϕ be a convex function on the interval I ⊂ R, x = (x₁, x₂,· · ·, x_n) ∈ Iⁿ (n≥2), let pi ≥0,i= 1,2,· · ·, n and Pn=Pn

i=1pi. Then

ϕ 1 P_n

n

X

i=1

pixi

!

≤ 1 P_n

n

X

i=1

piϕ(xi). (1.1)

If ϕ is strictly convex, then inequality in (1.1) is strict except when x₁ =x₂ =· · · =x_n.If ϕ is a concave function, then inequality in (1.1) is reverse.

Theorem 1.2. [3] Letϕbe a convex function on I ⊂R,and letf : [0,1]−→I be a continuous function on [0,1]. Then

ϕ Z 1

0

f(x)dx

≤ Z 1

0

ϕ(f(x))dx. (1.2)

If ϕ is strictly convex, then inequality in (1.2) is strict. If ϕis a concave function, then inequality in (1.2) is reverse.

In [2], Malamud gave some complements to the Jensen and Chebyshev inequalities and in [4], Saluja gave some necessary and sufficient conditions for three-step iterative sequence with errors for asymptotically quasi-nonexpansive type mapping converging to a fixed point in convex metric spaces. In this paper, we give some inequalities of the above type forϕ:I →Rsuch thatϕis a continuous on I,twice differentiable on ˚I and there existsm= inf

x∈˚I

ϕ⁰⁰(x) or M = sup

x∈˚I

ϕ⁰⁰(x).We obtain the following results:

Theorem 1.3. Letϕ:I −→Rbe a continuous function onI,twice differentiable on˚I, x= (x₁, x₂,· · ·, x_n)∈ Iⁿ (n≥2), let pi ≥0, i= 1,2,· · ·, n and Pn=Pn

i=1pi. (i) If there exists m= inf

x∈˚I

ϕ⁰⁰(x), then 1 P_n

n

X

i=1

piϕ(xi)−ϕ 1 P_n

n

X

i=1

pixi

!

≥ m 2



 1 Pn

n

X

i=1

p_ix²_i − 1 Pn

n

X

i=1

p_ix_i

!2

. (1.3)

(ii) If there exists M = sup

x∈˚I

ϕ⁰⁰(x), then 1 P_n

n

X

i=1

p_iϕ(x_i)−ϕ 1 P_n

n

X

i=1

p_ix_i

!

≤ M 2



 1 Pn

n

X

i=1

pix²_i − 1 Pn

n

X

i=1

pixi

!2

. (1.4)

Equality in (1.3) and (1.4) hold if x₁ =x₂ =· · ·=x_n or if ϕ(x) =αx²+βx+γ, α, β, γ∈R.

Theorem 1.4. Let ϕ : I −→ R be a continuous function on I, twice differentiable on ˚I. Suppose that f : [a, b]−→I and p: [a, b]−→R⁺ are continuous functions on [a, b].

(i) If there exists m= inf

x∈˚I

ϕ⁰⁰(x), then Rb

ap(x)ϕ(f(x))dx Rb

ap(x)dx

−ϕ Rb

ap(x)f(x)dx Rb

a p(x)dx

!

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≥ m 2



 Rb

ap(x) (f(x))²dx Rb

ap(x)dx

− Rb

ap(x)f(x)dx Rb

ap(x)dx

!2

. (1.5)

x∈˚I

ϕ⁰⁰(x), then Rb

ap(x)ϕ(f(x))dx R_b

ap(x)dx

−ϕ Rb

ap(x)f(x)dx R_b

a p(x)dx

!

≤ M 2



 Rb

a p(x) (f(x))²dx Rb

ap(x)dx

− Rb

ap(x)f(x)dx Rb

ap(x)dx

!2

. (1.6)

Equality in (1.5) and (1.6) hold if ϕ(x) =αx²+βx+γ, α, β, γ ∈R.

Corollary 1.5. Let ϕ:I −→Rbe a continuous function onI,twice differentiable on˚I and letf : [a, b]−→

I be a continuous function on [a, b]. (i) If there exists m= inf

x∈˚I

ϕ⁰⁰(x), then 1 b−a

Z _b

a

ϕ(f(x))dx−ϕ 1

b−a Z _b

a

f(x)dx

≥ m 2

1 b−a

Z b a

(f(x))²dx− 1

b−a Z b

a

f(x)dx ²!

. (1.7)

x∈˚I

ϕ⁰⁰(x), then 1 b−a

Z b a

ϕ(f(x))dx−ϕ 1

b−a Z b

a

f(x)dx

≤ M 2

1 b−a

Z b a

(f(x))²dx− 1

b−a Z b

a

f(x)dx ²!

. (1.8)

Equality in (1.7) and (1.8) hold if ϕ(x) =αx²+βx+γ, α, β, γ ∈R.

Corollary 1.6. Letϕ:I −→Rbe a continuous function onI,twice differentiable on˚I,x= (x₁, x₂,· · ·, x_n)∈ Iⁿ (n≥2), let pi ≥0, i= 1,2,· · ·, n and Pn=Pn

i=1pi. If there existm= inf

x∈˚I

ϕ⁰⁰(x) and M = sup

x∈˚I

ϕ⁰⁰(x), then we have

m 2



 1 P_n

n

X

i=1

p_ix²_i − 1 P_n

n

X

i=1

p_ix_i

!2



≤ 1 Pn

n

X

i=1

p_iϕ(x_i)−ϕ 1 Pn

n

X

i=1

p_ix_i

!

≤ M 2



 1 Pn

n

X

i=1

p_ix²_i − 1 Pn

n

X

i=1

p_ix_i

!2

. (1.9)

Equality in (1.9) occurs, if ϕ(x) =αx²+βx+γ, α, β, γ∈R.

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Corollary 1.7. Letϕ:I −→Rbe a continuous function on I,twice differentiable on˚I, and letf : [0,1]−→

I be a continuous function on [0,1].If there exist m= inf

x∈˚I

ϕ⁰⁰(x) and M = sup

x∈˚I

ϕ⁰⁰(x), then we have

m 2

Z 1 0

(f(x))²dx− Z 1

0

f(x)dx 2!

≤ Z 1

0

ϕ(f(x))dx−ϕ Z 1

0

f(x)dx

≤ M 2

Z 1 0

(f(x))²dx− Z 1

0

f(x)dx 2!

. (1.10)

Equality in (1.10) holds if ϕ(x) =αx²+βx+γ α, β, γ∈R.

Corollary 1.8. Let ϕ:I −→R be a convex function on I, twice differentiable on˚I, and let f : [a, b]−→I be a continuous function on[a, b].If there exists m= inf

x∈˚I

ϕ⁰⁰(x), then 1

b−a Z b

a

ϕ(f(x))dx−ϕ 1

b−a Z b

a

f(x)dx

≥ m 2

1 b−a

Z b a

(f(x))²dx− 1

b−a Z b

a

f(x)dx 2!

≥0 (1.11)

and

1 P_n

n

X

i=1

piϕ(xi)−ϕ 1 P_n

n

X

i=1

pixi

!

≥ m 2



 1 P_n

n

X

i=1

pix²_i − 1 P_n

n

X

i=1

pixi

!2

≥0. (1.12)

Corollary 1.9. Let ϕ:I −→Rbe a concave function onI, twice differentiable on˚I, and letf : [a, b]−→I be a continuous function on[a, b]. If there exists M = sup

x∈˚I

ϕ⁰⁰(x),then 1

b−a Z b

a

ϕ(f(x))dx−ϕ 1

b−a Z b

a

f(x)dx

≤ M 2

1 b−a

Z b a

(f(x))²dx− 1

b−a Z b

a

f(x)dx 2!

≤0 (1.13)

and

1 P_n

n

X

i=1

piϕ(xi)−ϕ 1 P_n

n

X

i=1

pixi

!

≤ M 2



 1 P_n

n

X

i=1

pix²_i − 1 P_n

n

X

i=1

pixi

!2

≤0. (1.14)

Remark 1.10. In the above ifϕ∈C²([α, β]),then we can replace inf and sup by min and max respectively.

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2. Lemma

Our proofs depend mainly upon the following lemma.

Lemma 2.1. Let ϕbe a convex function on I ⊂Rand differentiable on ˚I. Suppose thatf : [a, b]−→I and p: [a, b]−→R⁺ are continuous functions on [a, b]. Then

ϕ Rb

ap(x)f(x)dx Rb

ap(x)dx

!

≤ R_b

ap(x)ϕ(f(x))dx Rb

ap(x)dx . (2.1)

If ϕ is strictly convex, then inequality in (2.1) is strict. If ϕis a concave function, then inequality in (2.1) is reverse.

Proof. Suppose that ϕ is a convex function on I ⊂ R and differentiable on ˚I. Then for each x, y ∈˚I, we have

ϕ(x)−ϕ(y)≥(x−y)ϕ⁰(y). (2.2)

Replacex byf(x) and sety=

Rb

ap(x)f(x)dx Rb

ap(x)dx in (2.2), we obtain ϕ(f(x))−ϕ

Rb

ap(x)f(x)dx Rb

ap(x)dx

!

≥ f(x)− Rb

ap(x)f(x)dx Rb

ap(x)dx

! ϕ⁰

Rb

ap(x)f(x)dx Rb

ap(x)dx

!

. (2.3)

Multiplying both sides of inequality (2.3) byp(x) we obtain p(x)ϕ(f(x))−p(x)ϕ

Rb

ap(x)f(x)dx Rb

ap(x)dx

!

≥ p(x)f(x)−p(x) Rb

ap(x)f(x)dx R_b

a p(x)dx

! ϕ⁰

Rb

a p(x)f(x)dx R_b

ap(x)dx

!

. (2.4)

By integration in (2.4) we obtain (2.1).

3. Proof of the Theorems

Proof of Theorem 1.3. Suppose thatϕ:I →Ris a continuous function onI and twice differentiable on

˚I. Setg(x) =ϕ(x)−^m₂x².Differentiating twice times both sides ofg we getg⁰⁰(x) =ϕ⁰⁰(x)−m≥0.Then g is a convex function onI. By formula (1.1), we have

g 1 Pn

n

X

i=1

p_ix_i

!

≤ 1 Pn

n

X

i=1

p_ig(x_i) (3.1)

which implies that

ϕ 1 Pn

n

X

i=1

p_ix_i

!

−m 2

1 Pn

n

X

i=1

p_ix_i

!2

≤ 1 P_n

n

X

i=1

p_iϕ(x_i)−m 2

1 P_n

n

X

i=1

p_ix²_i. (3.2)

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Then, by (3.2) we can write

1 Pn

n

X

i=1

p_iϕ(x_i)−ϕ 1 Pn

n

X

i=1

p_ix_i

!

≥ m 2



 1 Pn

n

X

i=1

pix²_i − 1 Pn

n

X

i=1

pixi

!2

. (3.3)

If we putg(x) = −ϕ(x) +^M₂ x²,then by differentiating both sides of g we get g⁰⁰(x) =−ϕ⁰⁰(x) +M ≥0.

Henceg is a convex function on I and by similar proof as above, we obtain 1

Pn n

X

i=1

piϕ(xi)−ϕ 1 Pn

n

X

i=1

pixi

!

≤ M 2



 1 Pn

n

X

i=1

pix²_i − 1 Pn

n

X

i=1

pixi

!2

. (3.4)

Proof of Theorem 1.4. Suppose thatϕ:I →Ris a continuous function onI and twice differentiable on

˚I. Set g(x) = ϕ(x)− ^m₂x². Differentiating both sides of g we get g⁰⁰(x) = ϕ⁰⁰(x)−m ≥ 0. Hence g is a convex function onI and by formula (2.1) we have

g Rb

ap(x)f(x)dx Rb

ap(x)dx

!

≤ Rb

a p(x)g(f(x))dx Rb

ap(x)dx (3.5)

which implies that

ϕ Rb

ap(x)f(x)dx Rb

ap(x)dx

!

−m 2

Rb

a p(x)f(x)dx Rb

ap(x)dx

!2

≤ Rb

ap(x)ϕ(f(x))dx Rb

ap(x)dx

−m 2

Rb

ap(x) (f(x))²dx Rb

ap(x)dx . (3.6)

Then by (3.6), we can write Rb

ap(x)ϕ(f(x))dx Rb

ap(x)dx

−ϕ Rb

ap(x)f(x)dx Rb

a p(x)dx

!

≥ m 2



 Rb

ap(x) (f(x))²dx Rb

ap(x)dx

− Rb

ap(x)f(x)dx Rb

ap(x)dx

!2

. (3.7)

If we putg(x) =ϕ(x)−^M₂ x²,then by differentiating both sides ofg, we getg⁰⁰(x) =ϕ⁰⁰(x)−M ≤0.Thus, g is a concave function onI and by a similar proof as above, we obtain

Rb

ap(x)ϕ(f(x))dx R_b

ap(x)dx

−ϕ Rb

ap(x)f(x)dx R_b

a p(x)dx

!

≤ M 2



 Rb

a p(x) (f(x))²dx Rb

ap(x)dx − Rb

ap(x)f(x)dx Rb

ap(x)dx

!2

. (3.8)

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References

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[2] S. M. Malamud,Some complements to the Jensen and Chebyshev inequalities and a problem of W. Walter, Proc.

Amer. Math. Soc.129(2001), no. 9, 2671–2678. 1

[3] D. S. Mitrinovi´c, J. E. Peˇcari´c and A. M. Fink,Classical and new inequalities in analysis. Mathematics and its Applications(East European Series), 61. Kluwer Academic Publishers Group, Dordrecht, 1993. 1.1, 1.2

[4] G. S. Saluja, Convergence of fixed point of asymptoticaly quasi-nonexpansive type mappings in convex metric spaces, J. Nonlinear Sci. Appl.1(2008), no. 3, 132-144. 1