MEANS, g–CONVEX DOMINATED FUNCTIONS & HADAMARD–TYPE INEQUALITIES

S. S. DRAGOMIR, C. E. M. PEARCE, AND J. PEČARIĆ

Abstract. Hadamard–type inequalities are derived for g–convex dominated maps. Applications are given involving two functionals and some common means.

1. Introduction

The Hermite–Hadamard inequality

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

for a convex real–valued function g on a finite interval [a, b] is central to mathematical analysis and is the subject of a huge literature dealing with various generalisations and refinements. In this note we connect together some disparate threads through a Hermite–Hadamard motif. The first of these threads is the unifying concept of a g–convex dominated function (see [6]). Definition 1. Let g : I → ℝ be a given convex function. The real function f : I → ℝ is called g−convex dominated on I if

|λf (x) + (1 − λ) f (y) − f (λx + (1 − λ) y)| (1.1)

≤ λg (x) + (1 − λ) g (y) − g (λx + (1 − λ) y)

for all x, y ∈ I and λ ∈ [0, 1] .

The class of g–convex dominated functions on an interval I is manifestly nonempty. If g is convex on I and e : I → ℝ is defined by

e(x) := x,

then e and g are both g–convex dominated on I. Indeed there are concave functions which are g–convex dominated (for example −g) as well as functions which are neither convex nor concave. The concept of g–convex dominated functions draws together functions with some convex–like properties. We aim to elucidate some of these properties.

A second thread involves several means in common use for a pair x, y of positive numbers, namely the following. For x ≠ y and p ∈ ℝ \ {−1, 0}, we define the p–logarithmic mean (generalised logarithmic mean) L_{p}(x, y) by

L_{p}(x, y) := [\frac{y^{p+1} − x^{p+1}}{(p + 1)(y − x)}]^{1/p} .

alised logarithmic means.

1991 Mathematics Subject Classification. Primary 26D15.

Key words and phrases. Hermite–Hadamard inequality, g–convex dominated functions, gener-

In fact the singularities at p = −1, 0 are removable and L_{p} can be defined for p = −1, 0 so as to make L_{p}(x, y) a continuous function of p. In the limit as p → 0 we obtain the identric mean I(x, y), given by

I(x, y) := \frac{1}{e} (\frac{y^{y}}{x^{x}} )^{1/(y−x)} ,

and in the case p → −1 the logarithmic mean L(x, y), given by

L(x, y) := \frac{y − x}{ln y − ln x} .

In each case we define the mean as x when y = x, which occurs as the limiting value of L_{p}(x, y) for y → x. See [1] Chapter 6, Section 3 for more detail on these means.

In addition we have the arithmetic, geometric and harmonic means, defined respectively by

A(x, y) := \frac{x + y}{2} , G(x, y) := \sqrt{xy} and H(x, y) := \frac{2xy}{x + y} .

The first two arise from L_{p}(x, y) in the respective cases p = 1, p = −2. Remarkably there is no value of p for which L_{p} = H (see [1] p. 347). However H is connected with the generalised logarithmic–mean canon by

(1.2)  H(x, y) = [A (x^{−1}, y^{−1})]^{−1} .

The final thread involves two functionals which interpolate between f((a + b)/2) and ∫_{a}^{b} f(x)dx/(b − a). If f : [a, b] → ℝ with f ∈ L_{1} [a, b], we define the induced mapping H_{f} : [0, 1] → ℝ by

H_{f} (t) := \frac{1}{b − a} ∫_{a}^{b} f (tx + (1 − t) \frac{a + b}{2}) dx.

Similarly for f : [a, b] → ℝ integrable on [a, b] we may define F_{f} : [0, 1] → ℝ by

F_{f} (t) := \frac{1}{(b − a)^{2}} ∫_{a}^{b} ∫_{a}^{b} f (tx + (1 − t) y) dxdy.

For treatments of these functionals see [2]–[6].

In Section 2 we present some general results for g–convex dominated functions. In Section 3 a number of simple particular examples are given in which the means L_{p} and H appear naturally. Finally in Section 4 we show that the two functionals defined above inherit g–convex dominated properties.

2. g–convex dominated maps

We shall make use of the following characterisation of convex–dominated functions established in [6]. Lemma 1. Let g be a convex function on I and f : I → ℝ. Then the following statements are equivalent:

(i) f is g−convex dominated on I; (ii) the mappings g − f and g + f are convex on I; (iii) there exist two convex mappings h, k defined on I such that

f = \frac{1}{2} (h − k) and g = \frac{1}{2} (h + k) .

Proof. “(i) ⟺ (ii) ”. Condition (1.1) is equivalent to

g (λx + (1 − λ) y) − λg (x) − (1 − λ) g (y)
≤ λf (x) + (1 − λ) f (y) − f (λx + (1 − λ) y)
≤ λg (x) + (1 − λ) g (y) − g (λx + (1 − λ) y)

for all x, y ∈ I and λ ∈ [0, 1]. The two inequalities may be rearranged as

λ [f (x) + g (x)] + (1 − λ) [f (y) + g (y))
≥ f (λx + (1 − λ) y) + g (λx + (1 − λ) y)

and

λ [g (x) − f (x)] + (1 − λ) [g (y) − f (y)]
≥ g (λx + (1 − λ) y) − f (λx + (1 − λ) y)

for all x, y ∈ I and λ ∈ [0, 1], which are equivalent to the convexity of g + f and g − f respectively.

The equivalence “(ii) ⟺ (iii) ” is immediate. □

Proposition 1. Suppose f′′, g′′ exist and satisfy

|f′′ (x)| ≤ g′′ (x)

on an interval I. Then f is g–convex dominated on I.

Proof. By the given condition

g′′ (x) , g′′ (x) − f′′ (x) , g′′ (x) + f′′ (x)

are all nonnegative on I, so g, g − f, g + f are all convex on I, whence the stated result follows by Lemma 1. □

Theorem 1. Let g : I → ℝ be a convex function and f : I → ℝ a g–convex dominated mapping. Then for all a, b ∈ I with a < b,

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

and

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

Proof. Since f is g–convex dominated, we have by Lemma 1 that g + f and g − f are convex on [a, b], and so by the classical Hermite–Hadamard inequality

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

and

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

These inequalities are equivalent to those in the enunciation. □

3. Means

We now give several corollaries that provide examples of convex dominated functions and involve generalised logarithmic means. Corollary 1. Suppose [a, b] ⊂ (0, ∞) and p ∈ ℝ\ {−2, −1}. Let f : [a, b] → ℝ be a twice differentiable mapping such that |f′′ (x)| ≤ Mx^{p} (M > 0) for x ∈ [a, b] . Then

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

≤ \frac{M}{(p + 1) (p + 2)} [[L_{p+2} (a, b)]^{p+2} − [A (a, b)]^{p+2}]

and

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

≤ \frac{M}{(p + 1) (p + 2)} [A (a^{p+2}, b^{p+2}) − [L_{p+2} (a, b)]^{p+2}] .

Proof. Define the mapping g : [a, b] → ℝ by

g (x) = \frac{Mx^{p+2}}{(p + 1) (p + 2)} .

Then

g′′ (x) = Mx^{p} ≥ |f′′ (x)|

on [a, b]. The stated results follow from Proposition 1 and Theorem 1. □

In particular we derive the following in the case p = 0. Remark 1. Let f be twice differentiable on [a, b] and suppose that

M := sup_{x∈[a,b]} |f′′ (x)| < ∞.

Then

|f (\frac{a + b}{2}) − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ \frac{M}{24} (b − a)^{2}

and

|\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ \frac{M}{12} (b − a)^{2} .

Remark 2. In the case p = −3, the enunciation yields via (1.2) that

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

and

|\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ \frac{M}{2} · [\frac{L (a, b) − H (a, b)}{H (a, b) L (a, b)}] .

We now examine the two cases p = −2, −1 excluded in the preceding corollary. First we take p = −2. Corollary 2. Suppose [a, b] ⊂ (0, ∞) and let f : [a, b] → ℝ be twice differentiable and such that |f′′ (x)| ≤ M/x^{2} for all x ∈ (a, b). Then

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

and

exp [ |\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ] ≤ [\frac{I (a, b)}{G (a, b)}]^{M} .

Proof. Define g : [a, b] → ℝ by g (x) = −M ln x. Then g′′ (x) = M/x^{2}. Proposition 1 and Theorem 1 provide

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

and

|\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ M [\frac{∫_{a}^{b} ln xdx}{b − a} − \frac{ln a + ln b}{2}] .

The stated inequalities follow from

∫_{a}^{b} ln x dx = b ln b − a ln a − (b − a) = (b − a) ln I (a, b) .
□

For p = −1 we obtain the following. Corollary 3. Suppose [a, b] ⊂ (0, ∞) and let f : [a, b] → ℝ be twice differentiable with |f′′ (x)| ≤ M/x for all x ∈ (a, b) . Then

exp [ |f (\frac{a + b}{2}) − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ] ≤ \begin{matrix}[(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4}(b^{2}−a^{2})}] & \frac{M}{b−a} \\ [(\frac{a+b}{2})^{\frac{a+b}{2}} e^{− \frac{a+b}{2}} ] & M \end{matrix} ,

exp [ |\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx|] ≤ \begin{matrix}[(a^{a}b^{b})^{\frac{1}{2}} e^{− \frac{a+b}{2}} ] & M \\ [(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4} (b^{2}−a^{2})}] & \frac{M}{b−a} \end{matrix} .

Proof. Consider the mapping g (x) = Mx ln x − Mx. Then g′′ (x) = M/x. Proposition 1 and Theorem 1 provide

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

and

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

We have

∫_{a}^{b} (x ln x − x) dx = ln [(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4}(b^{2}−a^{2})}]^{M} ,

g (\frac{a + b}{2}) = ln [(\frac{a + b}{2})^{\frac{a+b}{2}} e^{− \frac{a+b}{2}} ]^{M} ,

\frac{1}{b − a} ∫_{a}^{b} g (x) dx = ln [(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4}(b^{2}−a^{2})}]^{\frac{M}{b−a}}

and

\frac{g (a) + g (b)}{2} = ln [(a^{a}b^{b})^{\frac{1}{2}} e^{− \frac{a+b}{2}} ]^{M} .

These yield

|f (\frac{a + b}{2}) − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ ln \begin{cases}[(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4}(b^{2}−a^{2})}] & \frac{M}{b−a} \\ [(\frac{a+b}{2})^{\frac{a+b}{2}} e^{− \frac{a+b}{2}} ] & M \end{cases} \}

and

|\frac{f (a) + f (b)}{2} − \frac{1}{b − a} ∫_{a}^{b} f (x) dx| ≤ ln \begin{cases}[(a^{a}b^{b})^{\frac{1}{2}} e^{− \frac{a+b}{2}} ] & M \\ [(\frac{b^{\frac{b^{2}}{2}}}{a^{\frac{a^{2}}{2}}} ) e^{− \frac{3}{4} (b^{2}−a^{2})}] & \frac{M}{b−a} \end{cases} \} ,

whence the desired results. □

The remarks in the introduction suggest that the results of Corollaries 2 and 3 may derive in the limit from those of Corollary 1 by letting p → −2 and p → −1 respectively.

We shall have the first inequality in Corollary 2, or rather the version of it obtained by taking natural logarithms of both sides, as a limiting form of the first inequality in Corollary 1 if we can show that

\frac{1}{(p + 1) (p + 2)} [[L_{p+2} (a, b)]^{p+2} − [A (a, b)]^{p+2}] → ln [\frac{A (a, b)}{I (a, b)}]

as p → −2. In extenso this reads as

\frac{1}{(p + 1)(p + 2)} [\frac{b^{p+3} − a^{p+3}}{(p + 3)(b − a)} − (\frac{a + b}{2})^{p+2}] → ln \frac{a + b}{2} − [\frac{b ln b − a ln a}{b − a} − 1] .

Set p = −2 + h. It suffices to show that

(3.1)  \frac{1}{h} [(\frac{a + b}{2})^{h} − \frac{b^{1+h} − a^{1+h}}{(1 + h)(b − a)}] → ln \frac{a + b}{2} − [\frac{b ln b − a ln a}{b − a} − 1]

as h → 0. By L’Hôpital’s rule, the left–hand side has limit equal to

\frac{d}{dx} (\frac{a + b}{2})^{x}|_{x=0} − \frac{d}{dx} \frac{b^{x} − a^{x}}{x(b − a)}|_{x=1} ,

which is readily verified to reduce to the right–hand side of (3.1).

The second inequality of Corollary 2 and the two inequalities of Corollary 3 may be derived similarly.

4. Functionals

We now examine how the two induced maps defined in the introduction inherit g–convex dominated properties. We make use of the following result (see [2] and [3]).

Proposition 2. If g is convex, then

(a) H_{g} is convex;

(b) H_{g} is monotone nondecreasing.

Theorem 2. Let g : [a, b] → ℝ be convex and f : [a, b] → ℝ a g–convex dominated mapping on [a, b]. Then

(i) H_{f} is H_{g}–convex dominated on [0, 1] ; (ii) for all 0 ≤ t_{1} < t_{2} ≤ 1 we have

(4.1)  0 ≤ |H_{f} (t_{2}) − H_{f} (t_{1})| ≤ H_{g} (t_{2}) − H_{g} (t_{1}) ;

(iii) for all t ∈ [0, 1]

0 ≤ |f (\frac{a + b}{2}) − H_{f} (t)| ≤ H_{g} (t) − g (\frac{a + b}{2})

and

0 ≤ |\frac{1}{b − a} ∫_{a}^{b} f (x) dx − H_{f} (t)| ≤ \frac{1}{b − a} ∫_{a}^{b} g (x) dx − H_{g} (t) .

Proof. (i) Since f is g–convex dominated on [a, b] , it follows from Lemma 1 that g − f and g + f are convex on [a, b] and so by Proposition 2 H_{g−f} and H_{g+f} are convex on [0, 1] . By the linearity of the mapping f ↦ H_{f}, we have H_{g−f} = H_{g} − H_{f} and H_{g+f} = H_{g} + H_{f}. Since H_{g} is convex, Lemma 1 yields that H_{f} is H_{g}–dominated on [0, 1] .

(ii) By Proposition 2 H_{g−f} and H_{g+f} are monotone nondecreasing on [0, 1] and thus

H_{g} (t_{1}) − H_{f} (t_{1}) = H_{g−f} (t_{1}) ≤ H_{g−f} (t_{2}) = H_{g} (t_{2}) − H_{f} (t_{2})

and

H_{g} (t_{1}) + H_{f} (t_{1}) = H_{g+f} (t_{1}) ≤ H_{g+f} (t_{2}) = H_{g} (t_{2}) + H_{f} (t_{2}) .

Therefore

H_{f} (t_{2}) − H_{f} (t_{1}) ≤ H_{g} (t_{2}) − H_{g} (t_{1})

and

H_{g} (t_{2}) − H_{g} (t_{1}) ≥ − [H_{f} (t_{2}) − H_{f} (t_{1})] ,

which are equivalent to (4.1) .

(iii) Since

H_{g}(0) = g (\frac{a + b}{2}) and H_{f}(0) = f (\frac{a + b}{2}) ,

the first inequality in (iii) occurs as a special case of (4.1). Likewise the second arises with t = 1.

□

Similarly we have the following result for the second functional, F_{f} (see [3]).

Proposition 3. If g is convex, then (a) F_{g} is convex;

(b) F_{g} is monotone nonincreasing on [0, 1/2] and monotone nondecreasing on [1/2, 1].

This leads to the following result.

Theorem 3. Let g : [a, b] → ℝ be convex and f : [a, b] → ℝ a g–convex dominated function on [a, b] . Then

(i) F_{f} is F_{g}−convex dominated on [0, 1] ; (ii) we have

0 ≤ |F_{f} (t_{2}) − F_{f} (t_{1})| ≤ F_{g} (t_{2}) − F_{g} (t_{1}) for \frac{1}{2} ≤ t_{1} < t_{2} ≤ 1

and

0 ≤ |F_{f} (t_{2}) − F_{f} (t_{1})| ≤ F_{g} (t_{1}) − F_{g} (t_{2}) for 0 ≤ t_{1} < t_{2} ≤ \frac{1}{2};

(iii) for all t ∈ [0, 1] we have

0 ≤ |\frac{1}{b − a} ∫_{a}^{b} f (x) dx − F_{f} (t)| ≤ \frac{1}{b − a} ∫_{a}^{b} g (x) dx − F_{g} (t) ,

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

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

and

0 ≤ |F_{f} (t) − H_{f} (t)| ≤ F_{g} (t) − H_{g} (t) .

References

[1] P. S. Bullen, D. S. Mitrinović & P. M. Vasić, Means and their inequalities, Reidel, Dordrecht (1988).

[2] S. S. Dragomir, A mapping in connection to Hadamard’s inequalities, Ann. Öster. Akad. Wiss. Math.-Natur. Wien, 128 (1991), 17–20.

[3] S. S. Dragomir, Two mappings in connection to Hadamard’s inequalities, J. Math. Anal. Applic. 167 (1992), 49–56.

[4] S. S. Dragomir, On Hadamard’s inequalities for convex functions, Mat. Balkanica, 6 (1992), 215–222.

[5] S. S. Dragomir and R. P. Agarwal, Two new mappings associated with Hadamard’s inequalities for convex functions, Appl. Math. Lett. 11 (3) (1998), 33–38.

[6] S. S. Dragomir and N. M. Ionescu, On some inequalities for convex–dominated functions, Anal. Num. Théor. Approx. 19 (1990), 21–28.

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

Department of Mathematics, University of Adelaide, Adelaide SA 5005, Australia

  E-mail address: cpearce@maths.adelaide.edu.au

Department of Mathematics, University of Adelaide, Adelaide SA 5005, Australia

  E-mail address: pecaric@mahazu.hazu.hr