Some inequalities for weighted and integral means of convex functions on linear spaces

Some inequalities for weighted and integral means of convex functions on linear spaces

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Dragomir, Sever S (2020) Some inequalities for weighted and integral means of convex functions on linear spaces. Proceedings of the Institute of

Mathematics and Mechanics, 46 (2). pp. 197-209. ISSN 2409-4986

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National Academy of Sciences of Azerbaijan Volume 46, Number 2, 2020, Pages 197–209 https://doi.org/10.29228/proc.29

SOME INEQUALITIES FOR WEIGHTED AND INTEGRAL MEANS OF CONVEX FUNCTIONS ON LINEAR SPACES

SILVESTRU SEVER DRAGOMIR

Abstract. Letf be a convex function on a convex subsetC of a linear space andx, y∈C,withx6=y.Ifp: [0,1]→Ris a Lebesgue integrable and symmetric function, namely p(1−t) = p(t) for all t ∈ [0,1] and such that the condition

0≤ Z τ

0

p(s)ds≤ Z 1

0

p(s)dsfor allτ∈[0,1]

holds, then we have

1 R1

0 p(τ)dτ Z 1

0

p(τ)f((1−τ)x+τ y)dτ− Z 1

0

f((1−τ)x+τ y)dτ

≤ 1

R1 0 p(τ)dτ

Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ[∇−f_y(y−x)− ∇+f_x(y−x)]

≤ 1

2[∇₋fy(y−x)− ∇+fx(y−x)].

Some applications for norms and semi-inner products are also pro- vided.

1. Introduction Let X be a real linear space,x, y ∈X,x6=y and let

[x, y] :={(1−λ)x+λy, λ∈[0,1]}

be thesegment generated byxandy. We consider the functionf : [x, y]→Rand the attached functionϕ_(x,y) : [0,1]→R,ϕ_(x,y)(t) :=f[(1−t)x+ty],t∈[0,1].

It is well known that f is convex on [x, y] iff ϕ(x, y) is convex on [0,1], and the following lateral derivatives exist and satisfy

(i) ϕ⁰_±(x,y)(s) =∇_±f(1−s)x+sy(y−x),s∈[0,1), (ii) ϕ⁰_+(x,y)(0) =∇₊f_x(y−x),

(iii) ϕ⁰_−(x,y)(1) =∇₋f_y(y−x),

2010Mathematics Subject Classification. 26D15; 46B05.

Key words and phrases. Convex functions, Linear spaces, Integral inequalities, Hermite- Hadamard inequality, F´ejer’s inequalities, Norms and semi-inner products.

197

where∇_±fx(y) are theGˆateaux lateral derivatives, we recall that

∇₊f_x(y) : = lim

h→0+

f(x+hy)−f(x)

h ,

∇₋fx(y) : = lim

k→0−

f(x+ky)−f(x)

k , x, y∈X.

The following inequality is the well-known Hermite-Hadamard integral inequal- ity for convex functions defined on a segment [x, y]⊂X:

f

x+y 2

≤ Z 1

0

f[(1−t)x+ty]dt≤ f(x) +f(y)

2 , (HH)

which easily follows by the classical Hermite-Hadamard inequality for the convex functionϕ(x, y) : [0,1]→R

ϕ_(x,y) 1

2

≤ Z 1

0

ϕ_(x,y)(t)dt≤ ϕ_(x,y)(0) +ϕ_(x,y)(1)

2 .

For other related results see the monograph on line [8]. For some recent results in linear spaces see [1], [2] and [9]-[12].

In the recent paper [7] we established the following refinements and reverses of F´ejer’s inequality for functions defined on linear spaces:

Theorem 1.1. Let f be an convex function on C and x, y ∈ C with x 6=y. If p: [0,1]→[0,∞) is Lebesgue integrable and symmetric, namely p(1−t) =p(t) for all t∈[0,1],then

0≤ 1 2

h∇₊f^x+y

2

(y−x)− ∇₋f^x+y

2

(y−x)iZ 1 0

t−1 2

p(t)dt (1.1)

≤ Z 1

0

f((1−t)x+ty)p(t)dt−f

x+y 2

Z 1 0

p(t)dt

≤ 1

2[∇−fy(y−x)− ∇₊fx(y−x)]

Z 1 0

t−1 2

p(t)dt

and 0≤ 1

2 h

∇₊f^x+y

2 (y−x)− ∇₋f^x+y

2 (y−x) iZ 1

0

1 2 −

t−1 2

p(t)dt (1.2)

≤ f(x) +f(y) 2

Z 1 0

p(t)dt− Z 1

0

f((1−t)x+ty)p(t)dt

≤ 1

2[∇₋f_y(y−x)− ∇₊f_x(y−x)]

Z 1 0

1 2 −

t−1 2

p(t)dt.

If we take p≡1 in (1.1), then we get 0≤ 1

8

h∇₊f^x+y

2

(y−x)− ∇₋f^x+y

2

(y−x)i

(1.3)

≤ Z 1

0

f[(1−t)x+ty]dt−f

x+y 2

≤ 1

8[∇₋f_y(y−x)− ∇₊f_x(y−x)]

that was firstly obtained in [4], while from (1.2) we recapture the result obtained in [5]

0≤ 1 8

h∇₊f^x+y

2

(y−x)− ∇₋f^x+y

2

(y−x)i

(1.4)

≤ f(x) +f(y)

2 −

Z ₁

0

f[(1−t)x+ty]dt

≤ 1

8[∇₋fy(y−x)− ∇₊fx(y−x)].

Motivated by the above results, we establish in this paper some upper and lower bounds for the difference

Z 1 0

p(τ)f((1−τ)x+τ y)dτ− Z 1

0

p(τ)dτ Z 1

0

f((1−τ)x+τ y)dτ where f is a convex function onC and x, y∈C,with x6=y while p: [0,1]→R is a Lebesgue integrable function such that

0≤ Z τ

0

p(s)ds≤ Z 1

0

p(s)dsfor all τ ∈[0,1].

Some applications for norms and semi-inner products are also provided.

2. Main Results

We start to the following identity that is of interest in itself as well:

Lemma 2.1. Let f be a convex function on C and x, y ∈ C, with x 6= y. If g: [0,1]→Cis a Lebesgue integrable function, then we have the equality

Z 1 0

g(τ)ϕ_(x,y)(τ)dτ− Z 1

0

g(τ)dτ Z 1

0

ϕ_(x,y)(τ)dτ (2.1)

= Z 1

0

Z 1 τ

g(s)ds

τ ϕ⁰_(x,y)(τ)dτ +

Z ₁

0

Z τ

0

g(s)ds

(τ −1)ϕ⁰_(x,y)(τ)dτ.

Proof. Integrating by parts in the Lebesgue integral, we have Z τ

0

tϕ⁰_(x,y)(t)dt+ Z 1

τ

(t−1)ϕ⁰_(x,y)(t)dt

=τ ϕ_(x,y)(τ)− Z τ

0

ϕ_(x,y)(t)dt−(τ −1)ϕ_(x,y)(τ)− Z 1

τ

ϕ_(x,y)(t)dt

=ϕ_(x,y)(τ)− Z 1

0

ϕ_(x,y)(t)dt that holds for allτ ∈[0,1].

If we multiply this identity by g(τ) and integrate overτ in [0,1],then we get Z 1

0

g(τ)ϕ_(x,y)(τ)dτ− Z 1

0

g(τ)dτ Z 1

0

ϕ_(x,y)(t)dt (2.2)

= Z 1

0

g(τ) Z τ

0

tϕ⁰_(x,y)(t)dt

dτ+ Z 1

0

g(τ) Z 1

τ

(t−1)ϕ⁰_(x,y)(t)dt

dτ.

Using integration by parts, we get Z 1

0

g(τ) Z τ

0

tϕ⁰_(x,y)(t)dt

dτ (2.3)

= Z 1

0

Z τ

0

tϕ⁰_(x,y)(t)dt

d Z τ

0

g(s)ds

= Z τ

0

g(s)ds Z τ

0

tϕ⁰_(x,y)(t)dt

1 0

− Z 1

0

Z τ

0

g(s)ds

τ ϕ⁰_(x,y)(τ)dτ

= Z 1

0

g(s)ds Z 1

0

tϕ⁰_(x,y)(t)dt

− Z 1

0

Z τ

0

g(s)ds

τ ϕ⁰_(x,y)(τ)dτ

= Z 1

0

Z 1 0

g(s)ds− Z τ

0

g(s)ds

τ ϕ⁰_(x,y)(τ)dτ

= Z ₁

0

Z 1 τ

g(s)ds

τ ϕ⁰_(x,y)(τ)dτ and

Z 1 0

g(τ) Z 1

τ

(t−1)ϕ⁰_(x,y)(t)dt

dτ (2.4)

= Z ₁

0

Z 1 τ

(t−1)ϕ⁰_(x,y)(t)dt

d Z τ

0

g(s)ds

= Z 1

τ

(t−1)ϕ⁰_(x,y)(t)dt Z τ

0

g(s)ds

1 0

+ Z ₁

0

Z τ

0

g(s)ds

(τ −1)ϕ⁰_(x,y)(τ)dτ

= Z 1

0

Z τ

0

g(s)ds

(τ −1)ϕ⁰_(x,y)(τ)dτ,

which proves the identity in (2.1).

Theorem 2.1. Let f be an operator convex function on C and x, y ∈ C, with x6=y. If p: [0,1]→R is a Lebesgue integrable function such that

0≤ Z τ

0

p(s)ds≤ Z 1

0

p(s)ds for all τ ∈[0,1], (2.5)

then we have the inequalities Z 1

0

Z 1 τ

p(s)ds

τ dτ∇₊f_x(y−x) (2.6)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇−fy(y−x)

≤ Z 1

0

p(τ)f((1−τ)x+τ y)dτ − Z 1

0

p(τ)dτ Z 1

0

f((1−τ)x+τ y)dτ

≤ Z 1

0

Z 1 τ

p(s)ds

τ dτ∇₋f_y(y−x)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₊f_x(y−x) or, equivalently,

Z 1 0

(1−τ) Z τ

0

[p(1−s)∇₋f_y(y−x)−p(s)∇₊f_x(y−x)]ds

dτ (2.7)

≤ Z 1

0

p(τ)f((1−τ)x+τ y)dτ− Z 1

0

p(τ)dτ Z 1

0

f((1−τ)x+τ y)dτ

≤ Z 1

0

(1−τ) Z τ

0

[p(1−s)∇₊fx(y−x)−p(s)∇₋fy(y−x)]ds

dτ.

Proof. We have forϕ_(x,y) and p: [0,1]→Ra Lebesgue integrable function that Z 1

0

p(τ)ϕ_(x,y)(τ)dτ− Z 1

0

p(τ)dτ Z 1

0

ϕ_(x,y)(τ)dτ (2.8)

= Z 1

0

Z 1 τ

p(s)ds

(τ)ϕ⁰_(x,y)(τ)dτ

− Z 1

0

Z τ

0

p(s)ds

(1−τ)ϕ⁰_(x,y)(τ)dτ.

By the gradient inequalities forϕ_(x,y) we have

τ∇₋fy(y−x)≥τ ϕ⁰_(x,y)(τ)≥τ∇₊fx(y−x) (2.9) and

(1−τ)∇₋f_y(y−x)≥(1−τ)ϕ⁰_(x,y)(τ)≥(1−τ)∇₊f_x(y−x) (2.10) for all τ ∈(0,1).

From

Z τ

0

p(s)ds≤ Z 1

0

p(s)ds= Z τ

0

p(s)ds+ Z 1

τ

p(s)ds, we get that R1

τ p(s)ds≥0 for allτ ∈(0,1). From (2.9) we derive that

Z 1 τ

p(s)ds

τ∇₋fy(y−x)≥ Z 1

τ

p(s)ds

τ ϕ⁰_(x,y)(τ)

≥ Z 1

τ

p(s)ds

τ∇₊f_x(y−x)

and from (2.10) that

− Z τ

0

p(s)ds

(1−τ)∇₊fx(y−x)≤ − Z τ

0

p(s)ds

(1−τ)ϕ⁰_(x,y)(τ)

≤ − Z τ

0

p(s)ds

(1−τ)∇−fy(y−x) all τ ∈(0,1).

If we integrate these inequalities over τ ∈[0,1] and add the obtained results, then we get

Z 1 0

Z 1 τ

p(s)ds

τ dτ∇₋f_y(y−x)− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₊f_x(y−x)

≥ Z 1

0

Z 1 τ

p(s)ds

τ ϕ⁰_(x,y)(τ)dτ− Z 1

0

Z τ

0

p(s)ds

(1−τ)ϕ⁰_(x,y)(τ)dτ

≥ Z 1

0

Z 1 τ

p(s)ds

τ dτ∇₊fx(y−x)− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇−fy(y−x). By using the equality (2.1) we obtain

Z 1 0

Z 1 τ

p(s)ds

τ dτ∇₊fx(y−x) (2.11)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇−fy(y−x)

≤ Z 1

0

p(τ)ϕ_(x,y)(τ)dτ− Z 1

0

p(τ)dτ Z 1

0

ϕ_(x,y)(τ)dτ

≤ Z 1

0

Z 1 τ

p(s)ds

τ dτ∇₋f_y(y−x)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₊f_x(y−x), namely (2.6).

If we change the variabley= 1−τ,then we have Z 1

0

Z 1 τ

p(s)ds

τ dτ = Z 1

0

Z 1 1−y

p(s)ds

(1−y)dy.

Also by the change of variable u= 1−s, we get Z 1

1−y

p(s)ds= Z y

0

p(1−u)du, which implies that

Z ₁

0

Z 1 τ

p(s)ds

τ dτ = Z ₁

0

Z τ

0

p(1−s)ds

(1−τ)dτ.

Therefore Z 1

0

Z 1 τ

p(s)ds

τ dτ∇−fy(y−x)− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₊fx(y−x)

= Z 1

0

Z τ

0

p(1−s)ds

(1−τ)dτ∇₋fy(y−x)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₊fx(y−x)

= Z ₁

0

(1−τ) Z τ

0

[p(1−s)∇₋f_y(y−x)−p(s)∇₊f_x(y−x)]ds

dτ and

Z 1 0

Z 1 τ

p(s)ds

τ dτ∇₊f_x(y−x)− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇−f_y(y−x)

= Z 1

0

Z τ

0

p(1−s)ds

(1−τ)dτ∇₊fx(y−x)

− Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ∇₋fy(y−x)

= Z 1

0

(1−τ) Z τ

0

[p(1−s)∇₊fx(y−x)−p(s)∇₋fy(y−x)]ds

dτ,

and by (2.11) we get (2.7).

We say that the functionp: [0,1]→R is symmetric on [0,1] if p(1−t) =p(t) for allt∈[0,1].

Corollary 2.1. Let f be a convex function on C and x, y ∈ C, with x 6= y.

If p : [0,1] → R is a Lebesgue integrable and symmetric function such that the condition (2.5) holds, then we have

1 R1

0 p(τ)dτ Z 1

0

p(τ)f((1−τ)x+τ y)dτ− Z 1

0

f((1−τ)x+τ y)dτ

(2.12)

≤ 1

R1

0 p(τ)dτ Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ[∇₋f_y(y−x)− ∇₊f_x(y−x)]

≤ 1

2[∇₋f_y(y−x)− ∇₊f_x(y−x)].

Proof. Since p is symmetric, then p(1−s) = p(s) for all s∈[0,1] and by (2.7) we get

Z 1 0

Z τ

0

p(s)ds

(1−τ)dτ[∇₊fx(y−x)− ∇−fy(y−x)]

≤ Z 1

0

p(τ)ϕ_(x,y)(τ)dτ − Z 1

0

p(τ)dτ Z 1

0

ϕ_(x,y)(τ)dτ

≤[∇₋f_y(y−x)− ∇₊f_x(y−x)]

Z 1 0

Z τ

0

p(s)ds

(1−τ)dτ, which is equivalent to the first inequality in (2.12).

Since 0≤R_τ

0 p(s)ds≤R₁

0 p(τ)dτ, hence Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ ≤ Z 1

0

p(τ)dτ Z 1

0

(1−τ)dτ = 1 2

Z 1 0

p(τ)dτ

and the last part of (2.12) is proved.

Remark 2.1. If the function p is nonnegative and symmetric then the inequality (2.12) holds true.

If we consider the weightp: [0,1]→[0,∞), p(s) = s−¹₂

,then Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ

= Z ₁

0

Z τ

0

s−1 2

ds

(1−τ)dτ

= Z ¹₂

0

Z τ

0

s−1 2

ds

(1−τ)dτ +

Z 1

1 2

Z τ

0

s−1 2

ds

(1−τ)dτ

= Z ¹

2

0

Z τ

0

1 2−s

ds

(1−τ)dτ +

Z 1

1 2

Z ¹

2

0

1 2 −s

ds+

Z τ

1 2

s−1

2 !

(1−τ)dτ

= Z ¹

2

0

1 2τ −τ²

2

(1−τ)dτ +

Z 1

1 2

Z ¹

2

0

1 2 −s

ds+

Z τ

1 2

s−1

2

ds

!

(1−τ)dτ.

We have

Z ¹

2

0

1 2τ −τ²

2

(1−τ)dτ = 1 2

Z ¹

2

0

(1−τ)τ(1−τ)dτ

= 1 2

Z ¹

2

0

(1−τ)²τ dτ = 11 384 and

Z 1

1 2

Z ¹₂

0

1 2 −s

ds+

Z τ

1 2

s−1

2

ds

!

(1−τ)dτ

= Z 1

1 2

1 8 +1

2

τ −1 2

2!

(1−τ)dτ

= 1 8

Z 1

1 2

(1−τ)dτ+1 2

Z 1

1 2

τ −1

2 2

(1−τ)dτ = 7 384.

Therefore

Z 1 0

Z τ

0

p(s)ds

(1−τ)dτ = 3 64. Since R1

0

τ −¹₂

dτ = ¹₄,hence 1 R1

0 p(τ)dτ Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ = 3 16. Utilising (2.12) for symmetric weightp: [0,1]→[0,∞), p(s) =

s−¹₂

,we get

4

Z 1 0

τ −1 2

f((1−τ)x+τ y)dτ− Z 1

0

f((1−τ)x+τ y)dτ

(2.13)

≤ 3

16[∇₋fy(y−x)− ∇₊fx(y−x)]

where f is a convex function onC and x, y∈C,with x6=y.

Consider now the symmetric function p(s) = (1−s)s, x∈[0,1].Then Z τ

0

p(s)ds= Z τ

a

(1−s)sds=−1

6τ²(2τ −3), τ ∈[0,1]

and Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ =−1 6

Z 1 0

τ²(2τ −3) (1−τ)dτ = 1 40. Also

Z 1 0

p(τ)dτ = Z 1

0

(1−τ)τ dτ = 1 6 and

1 R₁

0 p(τ)dτ Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ = 3 20 and by (2.12) we obtain

6

Z 1 0

(1−τ)τ f((1−τ)x+τ y)dτ− Z 1

0

f((1−τ)x+τ y)dτ

(2.14)

≤ 3

20[∇−fy(y−x)− ∇₊fx(y−x)],

where f is a convex function onC and x, y∈C,with x6=y.

3. Examples for Norms

Now, assume that (X,k·k) is a normed linear space. The function f₀(s) =

1

2kxk²,x∈X is convex and thus the following limits exist (iv) hx, yi_s:=∇₊f_0,y(x) = lim

t→0+

ky+txk²−kyk²

2t ;

(v) hx, yi_i:=∇₋f_0,y(x) = lim

s→0−

ky+sxk²−kyk²

2s ;

for any x, y ∈ X. They are called the lower and upper semi-inner products associated to the norm k·k.

For the sake of completeness we list here some of the main properties of these mappings that will be used in the sequel (see for example [2] or [6]), assuming thatp, q∈ {s, i} and p6=q:

(a) hx, xi_p =kxk² for all x∈X;

(aa) hαx, βyi_p =αβhx, yi_p ifα, β≥0 andx, y∈X;

(aaa) hx, yi_p

≤ kxk kykfor all x, y ∈X;

(av) hαx+y, xi_p =αhx, xi_p+hy, xi_p ifx, y ∈X and α∈R;

(v) h−x, yi_p =− hx, yi_q for all x, y∈X;

(va) hx+y, zi_p ≤ kxk kzk+hy, zi_p for all x, y, z∈X;

(vaa) The mapping h·,·i_p is continuous and subadditive (superadditive) in the first variable forp=s (or p=i);

(vaaa) The normed linear space (X,k·k) is smooth at the point x0 ∈ X\ {0} if and only ifhy, x₀i_s=hy, x₀i_i for ally∈X; in general hy, xi_i ≤ hy, xi_sfor allx, y ∈X;

(ax) If the normk·kis induced by an inner producth·,·i,thenhy, xi_i =hy, xi= hy, xi_s for allx, y ∈X.

The function f_r(x) = kxk^r (x ∈ X and 1 ≤ r < ∞) is also convex. There- fore, the following limits, which are related to the superior (inferior) semi-inner products,

∇±fr,y(x) := lim

t→0±

ky+txk^r− kyk^r t

=rkyk^r−1 lim

t→0±

ky+txk − kyk

t =rkyk^r−2hx, yi_s(i)

exist for all x, y ∈X whenever r ≥2; otherwise, they exist for any x ∈ X and nonzeroy∈X. In particular, if r= 1, then the following limits

∇_±f_1,y(x) := lim

t→0±

ky+txk − kyk

t = hx, yi_s(i) kyk exist for x, y∈X and y6= 0.

Ifp: [0,1]→Ris a Lebesgue integrable and symmetric function such that the condition

0≤ Z τ

0

p(s)ds≤ Z 1

0

p(s)ds for all τ ∈[0,1], is valid, then by (2.12) we get

1 R1

0 p(τ)dτ Z 1

0

p(τ)k(1−τ)x+τ yk^rdτ− Z 1

0

k(1−τ)x+τ yk^rdτ

(3.1)

≤ r

R₁

0 p(τ)dτ Z 1

0

Z τ

0

p(s)ds

(1−τ)dτ

×h

kyk^r−2hy−x, yi_i− kxk^r−2hy−x, xi_si .

Ifr ≥2,then the inequality (3.1) holds for allx, y ∈X. If r∈[1,2), then the inequality (3.1) holds for all x, y∈X withx, y 6= 0.

For r= 2 we get

1 R1

0 p(τ)dτ Z 1

0

p(τ)k(1−τ)x+τ yk²dτ− Z 1

0

k(1−τ)x+τ yk²dτ

(3.2)

≤ 2

R1

0 p(τ)dτ Z ₁

0

Z τ

0

p(s)ds

(1−τ)dτ[hy−x, yi_i− hy−x, xi_s] for all x, y ∈X.

If we take p(τ) = τ −¹₂

, τ ∈[0,1] in (3.1), then we obtain

4

Z 1 0

τ −1 2

k(1−τ)x+τ yk^rdτ− Z 1

0

k(1−τ)x+τ yk^rdτ

(3.3)

≤ 3r 16

hkyk^r−2hy−x, yi_i− kxk^r−2hy−x, xi_si . If X=H a real inner product space, then from (3.2) we get

1 R1

0 p(τ)dτ Z 1

0

p(τ)k(1−τ)x+τ yk²dτ− Z 1

0

k(1−τ)x+τ yk²dτ

(3.4)

≤ 2

R1

0 p(τ)dτ Z 1

0

Z τ

0

p(s)ds

(1−τ)dτky−xk² for all x, y ∈H.

4. Examples for Functions of Several Variables

Now, let Ω⊂Rⁿbe an open convex set inRⁿ. IfF : Ω→Ris a differentiable convex function on Ω, then, obviously, for any ¯c∈Ω we have

∇F_¯c(¯y) =

n

X

i=1

∂F (¯c)

∂xi

·y_i, y¯= (y₁, ..., y_n)∈Rⁿ, where _∂x^∂F

i are the partial derivatives of F with respect to the variable xi (i = 1, ..., n).

Ifp: [0,1]→Ris a Lebesgue integrable and symmetric function such that the condition (2.5) holds, then we have for all ¯a, ¯b∈Ω that

1 R1

0 p(τ)dτ Z 1

0

p(τ)F (1−τ) ¯a+τ¯b dτ−

Z 1 0

f (1−τ) ¯a+τ¯b dτ

(4.1)

≤ 1

R1

0 p(τ)dτ Z ₁

0

Z τ

0

p(s)ds

(1−τ)dτ

×

n

X

i=1

∂F ¯b

∂x_i −∂F(¯a)

∂x_i

!

(b_i−a_i)

≤ 1 2

n

X

i=1

∂F ¯b

∂xi

−∂F(¯a)

∂xi

!

(b_i−a_i).

If we take p(τ) = τ− ¹₂

, τ ∈[0,1] in (4.1), then we get

4

Z 1 0

τ− 1 2

F (1−τ) ¯a+τ¯b dτ−

Z 1 0

f (1−τ) ¯a+τ¯b dτ

(4.2)

≤ 3 16

n

X

i=1

∂F ¯b

∂xi

−∂F(¯a)

∂xi

!

(b_i−a_i) for all ¯a,¯b∈Ω.

References

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[12] E. Kikianty, S. S. Dragomir and P. Cerone, Sharp inequalities of Ostrowski type for convex functions defined on linear spaces and application.Comput. Math. Appl.56 (2008), no. 9, 2235–2246.

Silvestru Sever Dragomir

Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia

DST-NRF Centre of Excellence in the Mathematical, and Statistical Sciences, School of Computer Science, & Applied Mathematics, University of the Witwa- tersrand,, Private Bag 3, Johannesburg 2050, South Africa

E-mail address: sever.dragomir@vu.edu.au

Received: December 11, 2019; Accepted: July 27, 2020