PROPERTIES OF SOME SEQUENCES OF MAPPINGS ASSOCIATED TO THE HERMITE-HADAMARD INEQUALITY

S.S. DRAGOMIR

Abstract. The properties of some sequences of functions defined by multiple integrals associated with the Hermite-Hadamard integral inequality for convex functions are studied.

1. Introduction

The following integral inequality

(1.1)  f (\frac{a + b}{2}) ≤ \frac{1}{b − a} ∫_{a}^{b} f (t) dt ≤ \frac{f (a) + f (b)}{2},

which holds for any convex function f : [a, b] → ℝ, is well known in the literature as the Hermite-Hadamard inequality.

There is an extensive amount of literature devoted to this simple and nice result which has many applications in the Theory of Special Means and in Information Theory for divergence measures, from which we would like to refer the reader to the book [5].

The main aim of this paper is to consider some natural sequences of functions defined by multiple integrals and study their properties in relation to the Hermite-Hadamard inequality.

2. Some Sequences of Multiple Integrals

For an integrable mapping f : [a, b] → ℝ, let us define the sequences of functionals defined by the following multiple integrals:

L_{1} (f) : = \frac{f (a) + f (b)}{2},
L_{n} (f) : = \frac{1}{2 (b − a)^{n−1}} ∫_{a}^{b} ... ∫_{a}^{b} [f (\frac{x_{1} + ... + x_{n−1} + b}{n})

+f (\frac{x_{1} + ... + x_{n−1} + a}{n})] dx_{1}...dx_{n−1}

for n ≥ 2 and

A_{n} (f) := \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n} for n ≥ 1.

In [4], the authors proved the following result which connects the functional A_{n} (f) to the Hermite-Hadamard inequality (1.1).

1991 Mathematics Subject Classification. Primary 26D15, 26D10; Secondary 26D99.

Key words and phrases. Hermite-Hadamard Inequality.

Theorem 1. Let f : [a, b] → ℝ be a convex function on [a, b] . Then

(2.1)  f (\frac{a + b}{2}) ≤ A_{n+1} (f) ≤ A_{n} (f) ≤ ⋯ ≤ A_{2} (f) ≤ \frac{1}{b − a} ∫_{a}^{b} f (t) dt

for any n ∈ ℕ, n ≥ 1.

The sequence L_{n} (f) may be also connected to the Hermite-Hadamard inequality through the following result. Theorem 2. Assume that f : [a, b] → ℝ is convex on [a, b] . Then for all n ≥ 2 one has the inequalities:

(2.2) f (\frac{a + b}{2}) ≤ \frac{1}{2} \{f [\frac{(n − 1) \frac{a+b}{2} + b}{n}] + f [\frac{(n − 1) \frac{a+b}{2} + a}{n}]\}
≤ L_{n} (f)
≤ \frac{n − 1}{n} · \frac{1}{b − a} ∫_{a}^{b} f (x) dx + \frac{1}{n} · \frac{f (a) + f (b)}{2}
≤ \frac{f (a) + f (b)}{2}.

Proof. Using Jensen’s inequality for multiple integrals, we have

\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} f (\frac{x_{1} + ⋯ + x_{n−1} + b}{n}) dx_{1} … dx_{n−1}

≥ f [\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} [\frac{x_{1} + ⋯ + x_{n−1} + b}{n}] dx_{1} … dx_{n−1}]
= f [\frac{(n − 1) \frac{a+b}{2} + b}{n}]

and

\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} f (\frac{x_{1} + ⋯ + x_{n−1} + a}{n}) dx_{1} … dx_{n−1}

≥ f [\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} [\frac{x_{1} + ⋯ + x_{n−1} + a}{n}] dx_{1} … dx_{n−1}]
= f [\frac{(n − 1) \frac{a+b}{2} + a}{n}]

which gives, for n ≥ 2, that

L_{n} (f) ≥ \frac{1}{2} [f (\frac{(n − 1) \frac{a+b}{2} + b}{n}) + f (\frac{(n − 1) \frac{a+b}{2} + a}{n})]
≥ f (\frac{a + b}{2}) .

By the convexity of f we also have (n ≥ 2) that

f (\frac{x_{1} + ⋯ x_{n−1} + b}{n}) ≤ \frac{f (x_{1}) + ⋯ + f (x_{n−1}) + f (b)}{n}

HERMITE-HADAMARD INEQUALITIES 3

and

f (\frac{x_{1} + ⋯ + x_{n−1} + a}{n}) ≤ \frac{f (x_{1}) + ⋯ + f (x_{n−1}) + f (a)}{n}.

Integrating these inequalities on [a, b]^{n−1} , we deduce

\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} f (\frac{x_{1} + ⋯ + x_{n−1} + b}{n}) dx_{1} … dx_{n−1}

≤ (\frac{n − 1}{n}) · \frac{1}{b − a} ∫_{a}^{b} f (t) dt + \frac{1}{n}f (b)

and

\frac{1}{(b − a)^{n−1}} ∫_{a}^{b} ⋯ ∫_{a}^{b} f (\frac{x_{1} + ⋯ + x_{n−1} + a}{n}) dx_{1} … dx_{n−1}

≤ (\frac{n − 1}{n}) · \frac{1}{b − a} ∫_{a}^{b} f (t) dt + \frac{1}{n}f (a)

giving

L_{n} (f) ≤ \frac{1}{2} [\frac{2 (n − 1)}{n} \frac{1}{b − a} ∫_{a}^{b} f (t) dt + \frac{1}{n} (f (a) + f (b))] .

Since

\frac{1}{b − a} ∫_{a}^{b} f (x) dx ≤ \frac{f (a) + f (b)}{2}.

The last part of (2.2) is also proved.

The following lemma holds. Lemma 1. Let f : I ⊆ ℝ be a differentiable function on I̊ (I̊ is the interior of I) and a, b ∈I̊ with a < b. If f′ is integrable on [a, b] , then we have the equality: L_{n} (f) − A_{n} (f) (2.3)

= \frac{1}{n} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (x_{n} − \frac{a + b}{2}) dx_{1}...dx_{n}

for all n ≥ 1. Proof. For n = 1, we must prove that

\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx = \frac{1}{b − a} ∫_{a}^{b} f′ (x) (x − \frac{a + b}{2}) dx.

Indeed, by an integration by parts, we have that:

∫_{a}^{b} f′ (x) (x − \frac{a + b}{2}) dx = f (x) (x − \frac{a + b}{2})|_{a}^{b} − ∫_{a}^{b} f (x) dx
= \frac{(b − a) (f (a) + f (b))}{2} − ∫_{a}^{b} f (x) dx

and the required identity is proved.

Let us prove the equality (2.3) for n ≥ 2.

By an integration by parts, we have:

∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (x_{n} − \frac{a + b}{2}) dx_{n}

= n f (\frac{x_{1} + ... + x_{n}}{n} ) (x_{n} − \frac{a + b}{2})|_{a}^{b} − n ∫_{a}^{b} f (\frac{x_{1} + ... + x_{n}}{n} ) dx_{n}
= n \{\frac{b − a}{2} [f (\frac{x_{1} + ... + x_{n−1} + b}{n}) + f (\frac{x_{1} + ... + x_{n−1} + a}{n})]

− ∫_{a}^{b} f (\frac{x_{1} + ... + x_{n}}{n} ) dx_{n}\} .

If we integrate this equality on [a, b]^{n−1} , we have that:

\frac{1}{n (b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (x_{n} − \frac{a + b}{2}) dx_{1}...dx_{n}

= \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} \frac{b − a}{2} [f (\frac{x_{1} + ... + x_{n−1} + b}{n})

+f (\frac{x_{1} + ... + x_{n−1} + a}{n})] dx_{1}...dx_{n−1}

−\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}..dx_{n}

= \frac{1}{2 (b − a)^{n−1}} ∫_{a}^{b} ... ∫_{a}^{b} [f (\frac{x_{1} + ... + x_{n−1} + b}{n})

+f (\frac{x_{1} + ... + x_{n−1} + a}{n})] dx_{1}...dx_{n−1}

−\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}..dx_{n}

= L_{n} (f) − A_{n} (f)

and the identity (2.3) is proved.

3. Counterpart Inequalities for L_{n} (f) and A_{n} (f)

By the use of the lemma in the above seection, we can point out the following estimation results for the sequences defined above. Theorem 3. Let f : I ⊆ ℝ → ℝ be a convex function defined on the interval I and a, b ∈I̊ with a < b. Then we have the inequality:

(3.1)  0 ≤ A_{n} (f) − f (\frac{a + b}{2}) ≤ n [L_{n} (f) − A_{n} (f)]

for all n ≥ 1.

Proof. The first inequality in (3.1) follows from Theorem 1.

For n = 1 we have

\frac{1}{b − a} ∫_{a}^{b} f (x) dx − f (\frac{a + b}{2}) ≤ \frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx,

which is Bullen’s inequality (see for example [5, p. 140]).

Since any continuous convex function on [a, b] is the uniform limit of a sequence of differentiable convex functions on (a, b) , we can assume, without loss of generality, that f is differentiable convex on I. Thus, we have the inequality

f (\frac{a + b}{2}) − f (\frac{x_{1} + ... + x_{n}}{n} ) ≥ (\frac{a + b}{2} − \frac{x_{1} + ... + x_{n}}{n} ) f′ (\frac{x_{1} + ... + x_{n}}{n} )

for all x_{1}, ..., x_{n} ∈ [a, b] .

Integrating on [a, b]^{n} , we get that

f (\frac{a + b}{2}) − A_{n} (f)

≥ \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{a + b}{2} − \frac{x_{1} + ... + x_{n}}{n} ) f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}
= \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{a + b}{2} − x_{n}) f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}

as

∫_{a}^{b} ... ∫_{a}^{b} x_{1}f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}

= ... = ∫_{a}^{b} ... ∫_{a}^{b} x_{n}f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}.

Using Lemma 1, we have that

\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{a + b}{2} − x_{n}) f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}

= n (A_{n} (f) − L_{n} (f)) ,

and from the above inequality we get (3.1) .

Another result of this type is embodied in the following theorem.

Theorem 4. Let f : I ⊆ ℝ → ℝ be a convex function defined on the interval I and a, b ∈I̊ with a < b. Then we have the inequality:

(3.2)  0 ≤ A_{n} (f) − A_{n+1} (f) ≤ \frac{n}{n + 1} [L_{n} (f) − A_{n} (f)]

for all n ≥ 1.

Proof. The first inequality in (3.2) follows from Thorem 1.

For n = 1 we have to prove that

0 ≤ \frac{1}{b − a} ∫_{a}^{b} f (x) dx − \frac{1}{(b − a)^{2}} ∫_{a}^{b} ∫_{a}^{b} f (\frac{x + y}{2} ) dxdy

≤ \frac{1}{2} [\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx] ,

which is known (see for example [3]).

We can assume, without loss of generality, that f is differentiable and convex on I. Thus, we have the inequality

f (\frac{x_{1} + ... + x_{n+1}}{n + 1} ) − f (\frac{x_{1} + ... + x_{n}}{n} )

≥ (\frac{x_{1} + ... + x_{n+1}}{n + 1} − \frac{x_{1} + ... + x_{n}}{n} ) f′ (\frac{x_{1} + ... + x_{n}}{n} )
= [\frac{nx_{n+1} − (x_{1} + ... + x_{n})}{n (n + 1)}] f′ (\frac{x_{1} + ... + x_{n}}{n} ) ,

for all x_{1}, ..., x_{n+1} ∈ [a, b] .

Integrating on [a, b]^{n+1} , we get:

A_{n+1} (f) − A_{n} (f)

≥ \frac{1}{(b − a)^{n+1} n (n + 1)} [∫_{a}^{b} ... ∫_{a}^{b} nx_{n+1}f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n+1}

− ∫_{a}^{b} ... ∫_{a}^{b} (x_{1} + ... + x_{n}) f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n+1}]

= \frac{1}{(b − a)^{n+1} n (n + 1)} [n · \frac{b^{2} − a^{2}}{2} ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}

−n (b − a) ∫_{a}^{b} ... ∫_{a}^{b} x_{n}f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}]

= \frac{1}{(b − a)^{n} (n + 1)} [∫_{a}^{b} ... ∫_{a}^{b} (\frac{a + b}{2} − x_{n}) f′ (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}]

(by Lemma 1)

= \frac{n}{n + 1} [A_{n} (f) − L_{n} (f)] ,

and the inequality (3.2) is proved.

Next, we shall point out some estimations for the difference L_{n} (f) − A_{n} (f).

Theorem 5. Let f : I ⊆ ℝ → ℝ be a differentiable function on I̊ and a, b ∈I̊ with a < b. If |f′|^{2} is integrable on [a, b], then we have the inequality:

|L_{n} (f) − A_{n} (f)| (3.3)

≤ \frac{\sqrt{3} (b − a)}{6n\sqrt{n}} [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{x_{1} + ... + x_{n}}{n} )|^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

for all n ≥ 1.

Proof. If |f′|^{2} is integrable on [a, b] , then f′ is integrable on [a, b] and we have the equality (see Lemma 1):

L_{n} (f) − A_{n} (f)
= \frac{1}{n} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (x_{n} − \frac{a + b}{2}) dx_{1}...dx_{n}.

HERMITE-HADAMARD INEQUALITIES 7

On the other hand, it is clear that

∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) x_{n} dx_{1}...dx_{n}

= ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n},

and thus

L_{n} (f) − A_{n} (f)
= \frac{1}{n (b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f′ (\frac{x_{1} + ... + x_{n}}{n} ) (\frac{x_{1} + ... + x_{n}}{n} − \frac{a + b}{2}) dx_{1}...dx_{n}

for all n ≥ 1.

If we apply the Cauchy-Buniakowsky-Schwartz integral inequality, we have |L_{n} (f) − A_{n} (f)| (3.4)

≤ \frac{1}{n} (\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{x_{1} + ... + x_{n}}{n} )|^{2} dx_{1}...dx_{n})^{\frac{1}{2}}

× (\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |\frac{x_{1} + ... + x_{n}}{n} − \frac{a + b}{2}|^{2} dx_{1}...dx_{n})^{\frac{1}{2}} .

Let us compute

U := \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{x_{1} + ... + x_{n}}{n} − \frac{a + b}{2})^{2} dx_{1}...dx_{n}.

We have

U = \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} [\frac{1}{n^{2}} (x_{1}^{2} + ... + x_{n}^{2} + 2 ∑_{1≤i<j≤n} x_{i}x_{j})] dx_{1}...dx_{n}

−2 · \frac{a + b}{2} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n} + (\frac{a + b}{2})^{2} .

However, a simple calculation shows us that

\frac{1}{n^{2}} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (x_{1}^{2} + ... + x_{n}^{2} + 2 ∑_{1≤i<j≤n} x_{i}x_{j}) dx_{1}...dx_{n}

= \frac{1}{n^{2}} [n · \frac{1}{(b − a)} ∫_{a}^{b} x^{2}dx + 2 · \frac{n (n − 1)}{2} (\frac{1}{b − a} ∫_{a}^{b} xdx)^{2}]
= \frac{1}{n} [\frac{b^{3} − a^{3}}{3 (b − a)} + \frac{2 (n − 1)}{2} · (\frac{a + b}{2})^{2}]

and

\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{x_{1} + ... + x_{n}}{n} ) dx_{1}...dx_{n}

= \frac{1}{(b − a)} ∫_{a}^{b} xdx = \frac{a + b}{2}.

Thus,

U = \frac{1}{n} [\frac{a^{2} + ab + b^{2}}{3} + (n − 1) (\frac{a + b}{2})^{2}] − (\frac{a + b}{2})^{2}

= \frac{1}{n} [\frac{a^{2} + ab + b^{2}}{3} − (\frac{a + b}{2})^{2}]
= \frac{1}{12n} (b − a)^{2} .

Using inequality (3.4) with U as above, we easily obtain inequality (3.3) . We shall omit the details.

Corollary 1. Let f : I ⊆ ℝ → ℝ be a differentiable function on I̊ and a, b ∈I̊ with a < b. If M := sup_{x∈[a,b]} |f′ (x)| < ∞, then we have the inequality:

(3.5)  |L_{n} (f) − A_{n} (f)| ≤ \frac{\sqrt{3} (b − a) M}{6n\sqrt{n}}

for all n ≥ 1.

The above corollary allows us to state the following estimation result for convex mappings. Theorem 6. Let f : I ⊆ ℝ → ℝ be a differentiable convex function on I and a, b ∈I̊ with a < b. If M := sup_{x∈[a,b]} |f′ (x)| < ∞, then we have the inequalities:

(3.6)  0 ≤ A_{n} (f) − f (\frac{a + b}{2}) ≤ \frac{\sqrt{3} (b − a) M}{6\sqrt{n}} , n ≥ 1

and

(3.7)  0 ≤ A_{n} (f) − A_{n+1} (f) ≤ \frac{\sqrt{3} (b − a) M}{6 (n + 1) \sqrt{n}} , n ≥ 1.

Moreover, we have that:

lim_{n→∞} n^{p} [A_{n} (f) − f (\frac{a + b}{2})] = 0 for 0 ≤ p < \frac{1}{2}

and

lim_{n→∞} n^{q} [A_{n} (f) − A_{n+1} (f)] = 0 for 0 ≤ q < \frac{3}{2}.

4. Counterpart Inequalities for A_{n} (f)

Next, we shall point out some estimation results for the weighted sequence

A_{n} (f, q) = \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} f (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) dx_{1}...dx_{n}

for all n ≥ 1, where q_{i} > 0, i = \overline{1, n} and Q_{n} := ∑_{i=1}^{n} q_{i}.

This sequence is connected to the Hermite-Hadamard inequality through the following result obtained in [1]. Theorem 7. Let f : [a, b] → ℝ be a convex function on [a, b] . Then for any q_{i} > 0 (i = \overline{1, n}) one has the inequalities:

(4.1)  f (\frac{a + b}{2}) ≤ A_{n} (f) ≤ A_{n} (f, q) ≤ \frac{f (a) + f (b)}{2}.

In what follows we point out other results for this sequence.

Theorem 8. Let f : I ⊆ ℝ → ℝ be a differentiable convex function on I and a, b ∈I̊ with a < b. If |f′|^{2} is integrable on [a, b], then we have the inequality

(4.2) 0 ≤ A_{n} (f, q) − f (\frac{a + b}{2})

≤ \frac{\sqrt{3} (∑_{j=1}^{n} q_{j}^{2})^{\frac{1}{2}} (b − a)}{6Q_{n}}

× [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )|^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

for n ≥ 1.

Proof. The first inequality in (4.2) is obvious by (4.1).

By the convexity of f, we can write that

f (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) − f (\frac{a + b}{2})

≤ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{a + b}{2}) f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )

for all x_{1}, ..., x_{n} ∈ [a, b] .

If we integrate over [a, b]^{n} , we obtain

A_{n} (f, q) − f (\frac{a + b}{2}) (4.3)

≤ \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{a + b}{2})

×f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) dx_{1}...dx_{n}

≤ [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{a + b}{2})^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

× [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )|^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

on using the Cauchy-Buniakowsky-Schwartz inequality for the last inequality. Now, denote

B := \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{a + b}{2})^{2} dx_{1}...dx_{n}, n ≥ 1.

10 S.S. DRAGOMIR

Then we have

B = \frac{1}{Q_{n}^{2}} · \frac{1}{(b − a)^{n}}

× ∫_{a}^{b} ... ∫_{a}^{b} [q_{1}^{2}x_{1}^{2} + ... + q_{n}^{2}x_{n}^{2} + 2 ∑_{1≤i<j≤n} q_{i}x_{i}q_{j}x_{j}] dx_{1}...dx_{n}

−2 · \frac{a + b}{2} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) dx_{1}...dx_{n}

+ (\frac{a + b}{2})^{2} .

However,

\frac{1}{Q_{n}^{2}} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} [∑_{j=1}^{n} q_{j}^{2}x_{j}^{2} + 2 ∑_{1≤i<j≤n} q_{i}x_{i}q_{j}x_{j}] dx_{1}...dx_{n}

= \frac{1}{Q_{n}^{2}} [(∑_{j=1}^{n} q_{j}^{2}) · \frac{1}{b − a} ∫_{a}^{b} x^{2}dx + 2 ∑_{1≤i<j≤n} q_{i}q_{j} (\frac{1}{b − a} ∫_{a}^{b} xdx)^{2}]
= \frac{1}{Q_{n}^{2}} [\frac{b^{2} + ab + a^{2}}{3} ∑_{j=1}^{n} q_{j}^{2} + 2 ∑_{1≤i<j≤n} q_{i}q_{j} (\frac{a + b}{2})^{2}]

and

\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} [\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ] dx_{1}...dx_{n}

= \frac{1}{b − a} ∫_{a}^{b} xdx = \frac{a + b}{2}.

Then we have

B = \frac{1}{Q_{n}^{2}} [∑_{j=1}^{n} q_{j}^{2} (\frac{b^{2} + ab + a^{2}}{3} ) + 2 ∑_{1≤i<j≤n} q_{i}q_{j} (\frac{a + b}{2})^{2}] − (\frac{a + b}{2})^{2}
= \frac{1}{Q_{n}^{2}} [∑_{j=1}^{n} q_{j}^{2} (\frac{b^{2} + ab + a^{2}}{3} ) + 2 ∑_{1≤i<j≤n} q_{i}q_{j} (\frac{a + b}{2})^{2} − Q_{n}^{2} (\frac{a + b}{2})^{2}] .

As

Q_{n}^{2} = ∑_{j=1}^{n} q_{j}^{2} + 2 ∑_{1≤i<j≤n} q_{i}q_{j},

then

B = \frac{1}{Q_{n}^{2}} · ∑_{j=1}^{n} q_{j}^{2} [\frac{b^{2} + ab + a^{2}}{3} − (\frac{a + b}{2})^{2}]
= \frac{(b − a)^{2} ∑_{j=1}^{n} q_{j}^{2}}{12Q_{n}^{2}} .

Using inequality (4.3) , we deduce the desired inequality (4.2) .

Corollary 2. With the above assumptions, given that M := sup_{x∈[a,b]} |f′ (x)| < ∞, we have the inequality:

(4.4)  0 ≤ A_{n} (f, q) − f (\frac{a + b}{2}) ≤ \frac{\sqrt{3} (b − a) (∑_{j=1}^{n} q_{j}^{2})^{\frac{1}{2}}}{6Q_{n}} M

for n ≥ 1.

Remark 1. Note that if lim_{n→∞} \frac{∑_{j=1}^{n} q_{j}^{2}}{Q_{n}^{2}} = 0, then, from (4.4) , we recapture the result from [1].

The following result also holds: Theorem 9. Let f : I ⊆ ℝ → ℝ be a differentiable convex function on I and a, b ∈I̊ with a < b. If |f′|^{2} is integrable on [a, b] and q_{i} > 0 (i ≥ 1) , then one has the estimation:

(4.5) 0 ≤ A_{n} (f, q) − A_{n} (f)

≤ \frac{\sqrt{3} (b − a)}{6} [∑_{j=1}^{n} (\frac{q_{j}}{Q_{n}} − \frac{1}{n})^{2}]^{\frac{1}{2}}

× [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )|^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

for all n ≥ 1, where Q_{n} := ∑_{i=1}^{n} q_{i}.

Proof. The first inequality follows by Theorem 7.

Using the convexity of f, we have that

f (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) − f (\frac{x_{1} + ... + x_{n}}{n} )

≤ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{x_{1} + ... + x_{n}}{n} ) f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )

for all x_{1}, ..., x_{n} ∈ [a, b] .

Integrating on [a, b]^{n} , we obtain

A_{n} (f, q) − A_{n} (f) (4.6)

≤ \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{x_{1} + ... + x_{n}}{n} )

×f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} ) dx_{1}...dx_{n}

≤ [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{x_{1} + ... + x_{n}}{n} )^{2} dx_{1}...dx_{n}]^{\frac{1}{2}}

× [\frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} |f′ (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} )|^{2} dx_{1}...dx_{n}]^{\frac{1}{2}} ,

by applying the Cauchy-Buniakowsky-Schwartz integral inequality for the last inequality.

12 S.S. DRAGOMIR

Let us define

C := \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} (\frac{q_{1}x_{1} + ... + q_{n}x_{n}}{Q_{n}} − \frac{x_{1} + ... + x_{n}}{n} )^{2} dx_{1}...dx_{n}.

Then we have:

C = \frac{1}{Q_{n}^{2}n^{2}} · \frac{1}{(b − a)^{n}}

× ∫_{a}^{b} ... ∫_{a}^{b} [(nq_{1} − Q_{n}) x_{1} + ... + (nq_{n} − Q_{n}) x_{n}]^{2} dx_{1}...dx_{n}

= \frac{1}{n^{2}Q_{n}^{2}} · \frac{1}{(b − a)^{n}} ∫_{a}^{b} ... ∫_{a}^{b} [∑_{j=1}^{n} (nq_{j} − Q_{n})^{2} x_{j}^{2}

+ 2 ∑_{1≤i<j≤n} (nq_{i} − Q_{n}) (nq_{j} − Q_{n}) x_{i}x_{j}] dx_{1}...dx_{n}

= \frac{1}{n^{2}Q_{n}^{2}} [∑_{j=1}^{n} (nq_{j} − Q_{n})^{2} \frac{1}{b − a} ∫_{a}^{b} x^{2}dx

+ 2 ∑_{1≤i<j≤n} (nq_{i} − Q_{n}) (nq_{j} − Q_{n}) (\frac{1}{b − a} ∫_{a}^{b} xdx)^{2}]

= \frac{1}{n^{2}Q_{n}^{2}} [∑_{j=1}^{n} (nq_{j} − Q_{n})^{2} \frac{a^{2} + ab + b^{2}}{3}

+ 2 ∑_{1≤i<j≤n} (nq_{i} − Q_{n}) (nq_{j} − Q_{n}) (\frac{a + b}{2})^{2}]

= \frac{1}{n^{2}Q_{n}^{2}} [∑_{j=1}^{n} (nq_{j} − Q_{n})^{2} [\frac{a^{2} + ab + b^{2}}{3} − (\frac{a + b}{2})^{2}]

+ (\frac{a + b}{2})^{2} [∑_{j=1}^{n} (nq_{i} − Q_{n})^{2} + 2 ∑_{1≤i<j≤n} (nq_{i} − Q_{n}) (nq_{j} − Q_{n})]] .

However, it is easy to see that

∑_{j=1}^{n} (nq_{j} − Q_{n})^{2} + 2 ∑_{1≤i<j≤n} (nq_{i} − Q_{n}) (nq_{j} − Q_{n})

= [∑_{j=1}^{n} (nq_{j} − Q_{n})]^{2} = 0.

Hence,

C = \frac{∑_{j=1}^{n} (Q_{n} − nq_{j})^{2}}{Q_{n}^{2}n^{2}} · \frac{(b − a)^{2}}{12} .

Finally, by using inequality (4.6), we deduce the desired inequality (4.5) .

Corollary 3. With the above assumptions, given that M := sup_{x∈[a,b]} |f′ (x)| < ∞, we have the inequality:

(4.7)  0 ≤ A_{n} (f, q) − A_{n} (f) ≤ \frac{\sqrt{3} (b − a) M}{6} [∑_{j=1}^{n} (\frac{q_{j}}{Q_{n}} − \frac{1}{n})^{2}]^{\frac{1}{2}}

for all n ≥ 1. Remark 2. If we assume that q_{i} > 0 (i ≥ 1) are such that

lim_{n→∞} \frac{∑_{j=1}^{n} (Q_{n} − nq_{j})^{2}}{Q_{n}^{2}n^{2}} = 0,

then we have

lim_{n→∞} [A_{n} (f, q) − A_{n} (f)] = 0.

References

[1] C. BUŞE, S.S. DRAGOMIR and D. BARBU, The convergence of some sequences connected to Hadamard’s inequality, Demostratio Math. (Poland), 29 (1) (1996), 53-59.

[2] S.S. DRAGOMIR and C. BUŞE, Refinements of Hadamard’s inequality for multiple integrals, Utilitas Math (Canada), 47 (1995), 193-195.

[3] S. S. DRAGOMIR and N. M. IONESCU, Some integral inequalities for differentiable convex functions, Coll. Pap. of the Fac. of Sci. Kragujevac (Yugoslavia), 13 (1992), 11-16, ZBL No. 770.

[4] S.S. DRAGOMIR, J.E. PEČARIĆ and J. SÁNDOR, A note on the Jensen-Hadamard inequality, Anal. Num. Theor. Approx. (Romania), 19 (1990), 21-28. MR 93b : 260 14.ZBL No. 733:26010.

[5] J. PEČARIC, F. PROSCHAN and Y. L. TONG, Convex Functions, Partial Orderings and Statistical Applications, Academic Press, Inc., 1992.

School of Communications and Informatics, Victoria University of Technology, PO Box 14428, Melbourne City MC, 8001, Victoria, Australia.

E-mail address: sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html