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Some integral inequalities of H

¨

older and

Minkowski type

Ondrej Hutn´ık

1

Abstract. A number of integral inequalities of H¨older and Minkowski

type involving a class of generalized weighted quasi-arithmetic means

in integral form is established. Some well known inequalities and their

generalizations as consequences of our results are derived.

1 Introduction

The celebrated H¨older and Minkowski inequalities belong to the fundamen-

tal and classical inequalities in mathematics. They can be found in many

books on real functions, analysis, functional analysis or L

p

-spaces. Their

integral analogues are as follows, cf. [6].

Proposition 1.1 Let γ and δ be conjugate exponents, i.e. γ

−1

+ δ

−1

= 1,

with 1 < γ < ∞. Let (X, M, µ) be a measurable space and f, g : X → [0, ∞]

be meas urable functions. Then

Z

X

fg dµ ≤

Z

X

f

γ

dµ

1

γ

Z

X

g

δ

dµ

1

δ

, (H¨older)

and

Z

X

(f + g)

γ

dµ

1

γ

≤

Z

X

f

γ

dµ

1

γ

+

Z

X

g

γ

dµ

1

γ

. (Minkowski)

Because of their usefulness in analysis and its applications, these in-

equalities have received a considerable attention in the past decades and

a number of papers have appeared which deal with their various gener-

alizations, extensions and a pplications. In connection with the theory of

special means we can ﬁnd some extensions and applications of the H¨older

and Minkowski inequality e.g. in [8 ], [9], [11], [12], and [17].

The main purpose of this paper is to establish some integral inequalities

for a class of generalized weighted quasi-arithmetic means in integral form,

mainly connected with the classical H¨older and Minkowski inequalities.

1

Mathematics Subject Classiﬁcation (2000): 26D15

Key words and phrases: Integral inequality, H¨older inequality, Minkowski inequality,

generalized weighted quasi-arithmetic mean, weight function, convex function.

Acknowledgement. This paper was supported by Grants VEGA 2/5065/05 and APVT-

51-006904.

1

The structure o f this article is as follows. In Section 2 we recall the def-

inition of the generalized weighted quasi-arithmetic mean M

[a,b],g

(p, f) and

state some preliminary results. In Section 3 we give a number of weighted

integral inequalities of H¨older type involving M

[a,b],g

(p, f) and state a few

suﬃcient conditions for their validity. In the fourth section we give anal-

ogous results for Minkowski type inequalities. These results are natural

generalizations of results from [1]. Some applications and generalizations of

well known inequalities are given in last section.

2 Preliminaries

Let L

1

([a, b]) be the vector space of all real Lebesgue integrable functions

deﬁned on the interval [a, b] ⊂ R, a < b, with resp ect to the usual Lebesgue

measure. Denote by L

+

1

([a, b]) the positive cone of L

1

([a, b]), consisting

of non-negative functions. In what follows kpk

[a,b]

denotes the L

1

-norm of

p ∈ L

+

1

([a, b]). For the deﬁnition below, cf. [4].

Deﬁnition 2.1 Let p ∈ L

+

1

([a, b]), f : [a, b] → [α, β] be measurable and

g : [α, β] → R be continuous and strictly monotone, where −infty < α <

β < ∞. The generalized weighted quasi-arithmetic mean of f with respect

to the weight function p is the real number M

[a,b], g

(p, f) given by

M

[a,b], g

(p, f) = g

−1

1

kpk

[a,b]

Z

b

a

p(x)g(f(x)) dx

, (1)

where g

−1

denotes the inverse function to g.

The means M

[a,b], g

(p, f) include many commonly used two-variable inte-

gral means as particular cases, cf. [5]. In particular, for g(x) = x we obta in

the classical weighted arithmetic means A

[a,b]

(p, f).

Note that a further possible extension of M

[a,b], g

(p, f) could be con-

sidered in the case of analytic functions. Indeed, let f be of the form

f(θ) = |h(re

ıθ

)|, where 0 < r < 1 and h is an analytic function in the open

unit disk D = {z : |z| < 1} of the complex plane. In that case choosing

a = 0, b = 2 π, g(x) = x

q

for 0 < q < ∞ and p(x) ≡ 1 on [0, 2π] yields the

integral mean of order q,

M

q

(r, h) =

1

2π

Z

2π

0

|h(re

ıθ

)|

q

dθ

1/q

.

Much research has been devoted to the dependence of the op erator

of means on the behavior of the input functions p, f and g. The following

lemma gives a generalization of the well known Jensen inequality to the

2

class of means of Deﬁnition 2.1. This enables us to derive va rious inequali-

ties for the means M

[a,b],g

(p, f) depending on the convexity properties of f

and g.

Lemma 2.2 (Jensen Inequality) Let p ∈ L

+

1

([a, b]) and f : [a, b] →

[α, β] be measurable, where −∞ < α < β < ∞. If g : [α, β] → R is

convex (resp. concave), then

g

A

[a,b]

(p, f)

≤ (resp. ≥) A

[a,b]

(p, g ◦ f).

An elementary proof of Lemma 2.2 is given in [7]. Some basic properties

of M

[a,b], g

(p, f) derived using the weighted integral analogue of the Jensen

inequality can be found in [4] and [5]. As an easy consequence of the Jensen

inequality we get the following useful result.

Corollary 2.3 Let p ∈ L

+

1

([a, b]) and f : [a, b] → [α, β] be measurable ,

where −∞ < α < β < ∞. If g : [α, β] → R is convex increasing or con cave

decreasing (resp. convex decreasing or concave increasing) on (α, β), then

A

[a,b]

(p, f) ≤ (resp. ≥) M

[a,b], g

(p, f).

In the following lemma we summarize results which will be useful in the

rest of this paper.

Lemma 2.4 Let P ∈ L

1

([a, b]) and F : [a, b] → R be measurable. Then the

inequality

Z

b

a

P (t)F (t) dt ≤ 0

holds in each of the following cases:

(a) F is non-negative and non-increas i ng and

Z

x

a

P (t) dt ≤ 0, x ∈ [a, b];

(b) F is non-negative and non-decreasing and

Z

b

x

P (t) dt ≤ 0, x ∈ [a, b];

(c) F ∈ L

+

1

([a, b]) is symmetrical on [a, b], non-increasing on [

a+b

2

, b] and

Z

b−x

a+x

P (t) dt ≤ 0, x ∈

0,

b − a

2

;

3

(d) F is non-neg ative an d non-increasin g on [

a+b

2

, b] such that F (a + x) ≥

F (b − x) for all x ∈ [0,

b−a

2

],

P (x) ≤ 0, x ∈

a,

a + b

2

,

and

Z

b−x

a+x

P (t) dt ≤ 0, x ∈

0,

b − a

2

;

(e) F is n on-negative and non- decreasing on [a,

a+b

2

] such that F (a + x) ≤

F (b − x) for x ∈ [0,

b−a

2

],

P (x) ≤ 0, x ∈

a + b

2

, b

,

and

Z

b−x

a+x

P (t) dt ≤ 0, x ∈

0,

b − a

2

.

Remark 2.5 Recall that F is symmetrical on [a, b] if

F (a + x) = F (b − x), for all x ∈

0,

b − a

2

.

The statement of Lemma 2.4 in case (a) was proved in [6] for the interval

[0, 1]. For the proof of the other cases, cf. [1].

3 H¨older-type inequalities

In what follows we always consider weight functions p

i

∈ L

+

1

([a, b]) for

i = 1, 2, . . . , n + 1, where n ∈ N (the set of all natural numbers). Put

P

i

(x) =

1

kp

i

k

[a,b]

Z

x

a

p

i

(t) dt, x ∈ [a, b],

for i = 1, 2, . . . , n + 1. We establish a few integral inequalities of H¨o lder and

Minkowski type for the means M

[a,b], g

(p, f) involving P

i

, i = 1, 2, . . . , n + 1,

and give some suﬃcient conditions for their validity.

Theorem 3.1 Let p

i

∈ L

+

1

([a, b]) for i = 1, 2, . . . , n + 1 a nd f : [a, b] →

[α, β] be a non-negative measurable function, where −∞ < α < β < ∞.

Let γ

i

, i = 1, 2, . . . , n be positive real numbers such that

P

n

i=1

1

γ

i

= 1 an d

g : [α, β] → R be continuous.

4

(a) If f is non-increasing, g is either convex increasing or concave de-

creasing, and

P

n+1

(x) ≤

n

Y

i=1

P

i

(x)

1/γ

i

, x ∈ [a, b], (2)

then

A

[a,b]

(p

n+1

, f) ≤

n

Y

i=1

M

[a,b], g

(p

i

, f)

1/γ

i

. (3)

(b) If f is non-decreasing, g is either convex decreasing or concave in-

creasing, and (2) is reversed, then (3) is reversed.

Proof. We will prove (a). From Corollary 2.3 we have M

[a,b], g

(p

i

, f) ≥

A

[a,b]

(p

i

, f) f or all i = 1, 2, . . . , n. Then

n

Y

i=1

M

[a,b], g

(p

i

, f)

1/γ

i

≥

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

.

Using integration by parts, we have

n

Y

i=1

1

kp

i

k

[a,b]

Z

b

a

p

i

(x)f(x) dx

1/γ

i

=

n

Y

i=1

f(b) +

Z

b

a

P

i

(x) d

f(x)

1/γ

i

,

where

f(x) = −f(x). Fro m the discrete and integral H¨older inequalities,

we obta in

n

Y

i=1

f(b) +

Z

b

a

P

i

(x) d

f(x)

1/γ

i

≥ f(b) +

n

Y

i=1

Z

b

a

P

i

(x) d

f(x)

1/γ

i

≥ f(b) +

Z

b

a

n

Y

i=1

P

i

(x)

1/γ

i

d

f(x),

and by the use of inequality (2), we get

n

Y

i=1

M

[a,b], g

(p

i

, f)

1/γ

i

≥ f(b) +

Z

b

a

P

n+1

(x) d

f(x) =

Z

b

a

P

′

n+1

(x)f(x) dx

=

1

kp

n+1

k

[a,b]

Z

b

a

p

n+1

(x)f(x) dx = A

[a,b]

(p

n+1

, f).

The proof of (b) is similar, with the so called Popoviciu inequality

from [10] used instead of the discrete H¨older’s inequality.

5

Remark 3.2 Observe that the term

Q

n

i=1

P

i

(x)

1/γ

i

in condition (2) is the

weighted (discrete) geometric mean G

(n)

(P

1

(x), . . . , P

n

(x)) of non-negative

terms P

i

(x) with weights γ

i

, i = 1 , 2, . . . , n. Therefore, (2) may be rewritten

as

A

(n)

(P

n+1

(x), . . . , P

n+1

(x)) ≤ G

(n)

(P

1

(x), . . . , P

n

(x)),

where A

(n)

(P

1

(x), . . . , P

n

(x)) stands for the (discrete) arithmetic mean.

As a kind of duality to Theorem 3.1 we directly have

Theorem 3.3 Let p

i

, γ

i

and f, g be as in Theorem 3.1.

(a) If f is no n-increa sing, g is either convex d ecreasing or concave in-

creasing, and the inequality (2) is valid, then

M

[a,b], g

(p

n+1

, f) ≤

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

. (4)

(b) If f is non-decreasing, g is either convex increasing or concave de-

creasing, and (2) is reversed, then (4) is reversed.

Proof. From the proof of Theorem 3.1, we have

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

≥ A

[a,b]

(p

n+1

, f).

If g is either convex decreasing or concave increasing, then A

[a,b]

(p

n+1

, f) ≥

M

[a,b], g

(p

n+1

, f), which completes the proof.

Note that Theorems 3.1 and 3.3 seem to b e closely related to the com-

parison problem b etween means (cf. [5], Theorem 3.1).

Our purpose now is to weaken the assumption (2) using Lemma 2.4.

Therefore, the following theorem involves the derivatives of t he weight func-

tions P

i

, i = 1, 2, . . . , n + 1.

Theorem 3.4 Let p

i

, γ

i

, and f, g be as in Theorem 3.1.

(a) If f is non-increasing, g is either convex increasing or concave de-

creasing,

P

′

n+1

(x) ≤

n

Y

i=1

P

1/γ

i

i

!

′

(x), x ∈

a,

a + b

2

,

and

P

n+1

(b − x) − P

n+1

(a + x) ≤

n

Y

i=1

P

1/γ

i

i

(b − x) −

n

Y

i=1

P

1/γ

i

i

(a + x), (5)

for x ∈ [0,

b−a

2

], then the inequality (3) holds.

6

(b) If f is non-decreasing, g is either convex decreasing or concave in-

creasing,

P

′

n+1

(x) ≥

n

Y

i=1

P

1/γ

i

i

!

′

(x), x ∈

a + b

2

, b

,

and (5) is re versed, then (3) is reversed.

Proof. (a) Setting

F = f, P = P

′

n+1

−

n

Y

i=1

P

1/γ

i

i

!

′

,

and applying Lemma 2.4 (d), we get

Z

b

a

n

Y

i=1

P

1/γ

i

i

!

′

(x)f(x) dx ≥

Z

b

a

P

′

n+1

(x)f(x) dx = A

[a,b]

(p

n+1

, f).

Since g is either convex increasing or concave decreasing, using the proof of The-

orem 3.1 we have

n

Y

i=1

M

[a,b], g

(p

i

, f)

1/γ

i

≥

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

≥ f(b) +

Z

b

a

n

Y

i=1

P

i

(x)

1/γ

i

d

f(x)

=

Z

b

a

n

Y

i=1

P

1/γ

i

i

!

′

(x)f(x) dx ≥ A

[a,b]

(p

n+1

, f).

Item (b) may be proved similarly, by applying Lemma 2.4 (e) to F = f,

and P =

Q

n

i=1

P

1/γ

i

i

′

− P

′

n+1

.

Obviously, the integral and diﬀerential calculus plays a fundamental role

when establishing conditions for the inequality (3) to be valid. Thus, it is

natural to give the following suﬃcient conditions.

Theorem 3.5 Let p

i

, γ

i

, f and g be as in Theorem 3.1 and f be diﬀeren-

tiable. Then the inequality (3) holds in each of the following cases:

(a) f

′

(x) ≤ 0, f is co nvex (or f

′

(x) ≥ 0, f is co ncave), g is ei ther convex

increasing or concave decreasing, and

Z

x

a

P

n+1

(t) dt ≤

Z

x

a

n

Y

i=1

P

i

(t)

1/γ

i

dt, x ∈ [a, b]; (6)

7

(b) f

′

(x) ≤ 0, f is co ncave (or f

′

(x) ≥ 0, f is co nvex), g is either convex

increasing or concave decreasing, and

Z

b

x

P

n+1

(t) dt ≤

Z

b

x

n

Y

i=1

P

i

(t)

1/γ

i

dt, x ∈ [a, b]; (7)

(c) f

′

is non-positive and symmetrical on [a, b], non-decreasing on [

a+b

2

, b]

(or f

′

is non-negative and symmetrical on [a, b], non-increasing o n

[

a+b

2

, b]), g is either convex increasing or co ncave decreasing, and

Z

b−x

a+x

P

n+1

(t) dt ≤

Z

b−x

a+x

n

Y

i=1

P

i

(t)

1/γ

i

dt, (8)

for all x ∈ [0 ,

b−a

2

].

Proof. Let us prove (a). Suppose that f

′

≤ 0. Put

F = −f

′

and P = P

n+1

−

n

Y

i=1

P

1/γ

i

i

.

Since f is convex, then f

′

is non-decreasing and since f

′

≤ 0, it follows that

F is a non-negative and non-increasing function on [a, b]. Then

Z

x

a

P (t) dt =

Z

x

a

P

n+1

(t) dt −

Z

x

a

n

Y

i=1

P

i

(t)

1/γ

i

dt ≤ 0,

for all x ∈ [a, b] and using Lemma 2.4 (a) we have

Z

b

a

P

n+1

(x)F (x) dx ≤

Z

b

a

n

Y

i=1

P

i

(x)

1/γ

i

F (x) dx. (9)

Replacing −f(x) by

f(x) a nd adding f(b) to both sides of (9), we get

f(b) +

Z

b

a

P

n+1

(x) d

f(x) ≤ f(b) +

Z

b

a

n

Y

i=1

P

i

(x)

1/γ

i

d

f(x). (10)

For the left-hand side of (10) we have

f(b) +

Z

b

a

P

n+1

(x) d

f(x) =

Z

b

a

P

′

n+1

(x)f(x) dx = A

[a,b]

(p

n+1

, f). (11)

For the right-hand side of (10) we use the H¨older inequality to get

f(b) +

Z

b

a

n

Y

i=1

P

i

(x)

1/γ

i

d

f(x) ≤ f(b) +

n

Y

i=1

Z

b

a

P

i

(x) d

f(x)

1/γ

i

≤

n

Y

i=1

f(b) +

Z

b

a

P

i

(x) d

f(x)

1/γ

i

=

n

Y

i=1

Z

b

a

P

′

i

(x)f(x) dt

1/γ

i

=

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

.

8

Since g is either convex increasing or concave decreasing, we have A

[a,b]

(p

i

, f) ≤

M

[a,b]

(p

i

, f), for i = 1, 2, . . . , n, and therefore

n

Y

i=1

A

[a,b]

(p

i

, f)

1/γ

i

≤

n

Y

i=1

M

[a,b], g

(p

i

, f)

1/γ

i

. (12)

Substituting (11 ) and (12) into (10) we obtain the desired inequality. If

f

′

≥ 0, we replace f

′

by F and use the same method.

Similarly we may prove items (b) and (c) by the use of items (b) and (c)

of Lemma 2.4, respectively.

The same metho d yields a kind of dual to Theorem 3.5:

Theorem 3.6 Let p

i

, γ

i

and f, g be as in Theorem 3.5. Then the inequal-

ity (4) holds in each of the f ollowing cases:

(a) f

′

(x) ≤ 0, f is co nvex (or f

′

(x) ≥ 0, f is co ncave), g is ei ther convex

decreasing or concave increasing, and (6) is valid;

(b) f

′

(x) ≤ 0, f is co ncave (or f

′

(x) ≥ 0, f is co nvex), g is either convex

decreasing or concave increasing on [α, β], and (7) is valid;

(c) f

′

is non-positive and symmetrical on [a, b], non-decreasing on [

a+b

2

, b]

(or f

′

is non-negative and symmetrical on [a, b], non-increasing o n

[

a+b

2

, b]), g is either convex decreasing or concave increasing, and (8)

is valid.

4 Minkowski-type inequalities

In this section we establish some analogous inequalities of Minkowski type.

Theorem 4.1 Let p

i

, f, g be as in Theorem 3.1.

(a) Let q > 1 or q < 0. If f is non-increasing, g is either convex increasing

or concave decreasing, and

P

n+1

(x) ≤

n

X

i=1

δ

i

P

i

(x)

1/q

!

q

, x ∈ [a, b], (13)

where δ

i

, i = 1, 2, . . . , n, are positive numbers such that

P

n

i=1

δ

i

= 1,

then

A

[a,b]

(p

n+1

, f) ≤

n

X

i=1

δ

i

M

[a,b], g

(p

i

, f)

1/q

!

q

. (14)

If f is non-decreasing, g is either convex decreas i ng or co ncave in -

creasing, and (13) is valid, then (14) is re versed.

9

(b) Let 0 < q < 1. If f is non-decreasing, g is either con vex increasing or

concave decreasing, and (13) is rev ersed, then (14) holds.

If f is non-increasing, g is either convex decreasi ng or concave in-

creasing, and (13) is reversed, then (14) is reversed.

Proof. Suppose that q > 1, f is non-increasing, and (13) is valid. Since g

is either convex increasing or concave decreasing, according to Corollary 2.3

we have M

[a,b], g

(p

i

, f) ≥ A

[a,b]

(p

i

, f) for all i = 1, 2, . . . , n, and therefore

n

X

i=1

δ

i

M

[a,b], g

(p

i

, f)

1/q

≥

n

X

i=1

δ

i

A

[a,b]

(p

i

, f)

1/q

.

Using integration by parts, we get

n

X

i=1

δ

i

A

[a,b]

(p

i

, f)

1/q

=

n

X

i=1

δ

i

f(b) +

Z

b

a

P

i

(t) d

f(t)

1/q

,

where

f(t) = −f(t). Applying the discrete and integral versions of the

Minkowski inequality, we obtain

n

X

i=1

δ

i

f (b) +

Z

b

a

P

i

(t) d

f (t)

1

q

≥

n

X

i=1

δ

i

f (b)

1

q

!

q

+

n

X

i=1

δ

i

Z

b

a

P

i

(t) d

f (t)

1

q

q

1

q

≥

f (b) +

Z

b

a

n

X

i=1

δ

i

P

1/q

i

!

q

(t) d

f (t)

!

1

q

.

According to (13), we have

f(b) +

Z

b

a

n

X

i=1

δ

i

P

1/q

i

!

q

(t) d

f(t)

!

1

q

≥

f(b) +

Z

b

a

P

n+1

(t) d

f(t)

1

q

=

Z

b

a

P

′

n+1

(t)f(t) dt

1

q

=

A

[a,b]

(p

n+1

, f)

1

q

.

In the case q < 0 the Bellman inequality (cf. [10]) is used instead of the

discrete Minkowski inequality.

Remark 4.2 Note that the term

P

n

i=1

δ

i

P

1/q

i

q

in the previous theorem

is, in fact, the weighted (discrete) power mean P

[1/q]

(n)

(P

1

(x), . . . , P

n

(x)) of o r -

der 1/q for the n-tuple (P

1

(x), . . . , P

n

(x)) with weights (δ

1

, . . . , δ

n

). Thus,

the condition (13) may be equivalently rewritten as

A

(n)

(P

n+1

(x), . . . , P

n+1

(x)) ≤ P

[1/q]

(n)

(P

1

(x), . . . , P

n

(x)).

10

From the proof of Theorem 4.1(a) and Corollary 2.3 we immediately

have the following result.

Theorem 4.3 Let p

i

, f, g be as in Theorem 3.1.

(a) Let q > 1 or q < 0. If f is non-increasing, g is either convex decreasing

or concave increasing, and inequality (13) is valid, then

M

[a,b], g

(p

n+1

, f) ≤

n

X

i=1

δ

i

A

[a,b]

(p

i

, f)

1/q

!

q

. (15)

If f is non-decreasing, g is either convex increasing or concave de-

creasing, and (13) is valid, then (15) is re versed.

(b) Let 0 < q < 1. If f is non-decreasing, g is either convex decreasing or

concave increa sing, and (13) is reversed, then (15) holds.

If f is non-increasing, g is either conv ex increasing or concave de-

creasing, and (13) is reversed, then (15) is reversed.

As in the theorems stated in t he previous section, the requirement (13)

could be given in a weaker form. In what follows we will consider only the

case when q > 1 or q < 0 . The similar results hold for 0 < q < 1.

Theorem 4.4 Let p

i

, δ

i

, f and g be as in Theorem 4.1 and f be diﬀeren-

tiable.

(a) If f is non-increasing, g is either convex increasing or concave de-

creasing,

P

′

n+1

(x) ≤

n

X

i=1

δ

i

P

i

(x)

1/q

!

q

!

′

, for x ∈

a,

a + b

2

,

and

P

n+1

(b−x)−P

n+1

(a+x) ≤

n

X

i=1

δ

i

P

1/q

i

!

q

(b−x)−

n

X

i=1

δ

i

P

1/q

i

!

q

(a+x),

(16)

for x ∈ [0,

b−a

2

], then (14) holds.

(b) If f is non-decreasing, g is either convex decreasing or concave in-

creasing,

P

′

n+1

(x) ≥

n

X

i=1

δ

i

P

i

(x)

1/q

!

q

!

′

, for x ∈

a + b

2

, b

,

and (16) is reversed, then (14) is reversed.

11

The proof is analogous to the proof of Theorem 3.4. The following result

may be proved similarly to Theorem 3.5.

Theorem 4.5 Let p

i

, δ

i

, f and g be as in Theorem 4.1 and f be diﬀeren-

tiable. Then the inequality (14) holds in each of the following cases :

(a) f

′

(x) ≤ 0, f is co nvex (or f

′

(x) ≥ 0, f is co ncave), g is ei ther convex

increasing or concave decreasing, and

Z

x

a

P

n+1

(t) dt ≤

Z

x

a

n

X

i=1

δ

i

P

i

(t)

1/q

!

q

dt, x ∈ [a, b]; (17)

(b) f

′

(x) ≤ 0, f is co ncave (or f

′

(x) ≥ 0, f is co nvex), g is either convex

increasing or concave decreasing, and

Z

b

x

P

n+1

(t) dt ≤

Z

b

x

n

X

i=1

δ

i

P

i

(t)

1/q

!

q

dt, x ∈ [a, b]; (18)

(c) f

′

is non-positive and symmetrical on [a, b], non-decreasing on [

a+b

2

, b]

(or f

′

is non-negative and symmetrical on [a, b], non-increasing o n

[

a+b

2

, b]), g is either convex increasing or co ncave decreasing, and

Z

b−x

a+x

P

n+1

(t) dt ≤

Z

b−x

a+x

n

X

i=1

δ

i

P

i

(t)

1/q

!

q

dt, (19)

for all x ∈ [0 ,

b−a

2

].

5 Applications

In this section we deduce some inequalities from int egra l inequalities stated

in Section 3 and 4. Since the means M

[a,b], g

(p, f) cover many known two-

variable integral means, the inequalities obtained are generalizations of some

well known ones.

Corollary 5.1 Let f : [a, b] → R be non-negative and non-increasing ,

h

i

: [a, b] → R, i = 1, 2, . . . , n, be non-negative a nd non-decreasing with

continuous ﬁrst de rivative and h

i

(a) = 0 for all i = 1, 2, . . . , n. If γ

i

,

i = 1, 2, . . . , n are positive numbers such that

P

n

i=1

1/γ

i

= 1, then

Z

b

a

n

Y

i=1

h

i

(t)

1/γ

i

!

′

f(t) dt ≤

n

Y

i=1

Z

b

a

h

′

i

(t)f(t) dt

1/γ

i

. (20)

12

Proof. Put

p

i

= h

′

i

, i = 1, . . . , n, and p

n+1

=

n

Y

i=1

h

1/γ

i

i

!

′

.

Then

P

i

(x) =

h

i

(x)

h

i

(b)

, i = 1, 2, . . . , n, and P

n+1

(x) =

Q

n

i=1

h

i

(x)

1/γ

i

Q

n

i=1

h

i

(b)

1/γ

i

.

Thus P

n+1

(x) =

Q

n

i=1

P

1/γ

i

i

(x), for x ∈ [a, b]. If we now choose g(x) = x,

then the result follows from Theorem 3.1(a).

Remark 5.2 Note that the inequality (20) (cf. [15]) is a generalization

of the so-called Gauss-P´olya ineq uali ty (cf. [16]). Namely, for n = 2, a = 0,

γ

1

= γ

2

= 2, h

1

(t) = t

2u+1

, h

2

(t) = t

2v+1

for u, v > −1/2 and f a non-

negative non-increasing function, we have

Z

b

0

t

u+v

f(t) dt

2

≤

1 −

u − v

u + v + 1

2

!

Z

b

0

t

2u

f(t) dt

Z

b

0

t

2v

f(t) dt,

whenever the integrals exist. Putting u = 0 , v = 2 and letting b → ∞ we

obtain the result of C.F. Gauss between the second and the fourth order

moments (cf. [6]):

Z

∞

0

t

2

f(t) dt

2

≤

5

9

Z

∞

0

f(t) dt ·

Z

∞

0

t

4

f(t) dt.

Remark 5.3 Alzer [2] derived the inequality

Z

b

a

G

(2)

(h

1

(t), h

2

(t))

′

f(t) dt ≤ G

(2)

Z

b

a

h

′

1

(t)f(t) dt,

Z

b

a

h

′

2

(t)f(t) dt

,

which holds for non-negative decreasing functions h

1

, h

2

, f : [a, b] → R such

that h

1

, h

2

and

√

h

1

h

2

are continuously diﬀerentiable with h

1

(a) = h

2

(a)

and h

1

(b) = h

2

(b). Obviously, this is a special case of (20) for n = 2.

An analogous result connected with the weighted p ower mean P

[r]

(2)

is as

follows (cf. [14]):

Corollary 5.4 Let h

1

, h

2

: [a, b] → R be non-negative non-decreasing func-

tions with continuous ﬁrst derivatives and h

1

(a) = h

2

(a), h

1

(b) = h

2

(b). Let

γ

1

, γ

2

> with γ

1

+ γ

2

= 1.

(a) If f : [a, b] → R is non-negative and non-decrea sing, then

P

[r]

(2)

Z

b

a

h

′

1

(t)f(t) dt,

Z

b

a

h

′

2

(t)f(t) dt

≤

Z

b

a

P

[s]

(2)

(h

1

(t), h

2

(t))

′

f(t) dt

(21)

for r, s < 1, and for r, s > 1 the inequality is rev ersed.

13

(b) If f : [a, b] → R is non-nega tive and non-increa s i ng, then for r < 1 < s

the inequality (21) hold s and for r > 1 > s the inequality is reve rsed.

For some analogous results related to the Gauss-P´olya inequality in-

volving quasi-arithmetic means and logarithmic means, see [14] and [15].

A generalization of P´olya inequality for Stolarsky and Gini means is given

in [13].

Remark 5.5 Similarly, if f : [0, 1] → R is non-negative and non-decreasing,

then

Z

1

0

t

u+v

f(t) dt

2

≥

1 −

u − v

u + v + 1

2

!

Z

1

0

t

2u

f(t) dt

Z

1

0

t

2v

f(t) dt.

In the following corollary we give a generalization of the above inequality.

Corollary 5.6 Let f : [0, 1] → R be non-negative and non-decreasing, and

γ

i

, i = 1, 2 , . . . , n be posi tive numbers s uch that

P

n

i=1

1/γ

i

= 1. If λ

i

>

−1/γ

i

for i = 1, . . . , n, then

Z

1

0

t

P

n

i=1

λ

i

f(t) dt ≥

Q

n

i=1

(1 + λ

i

γ

i

)

1/γ

i

1 +

P

n

i=1

λ

i

n

Y

i=1

Z

1

0

t

λ

i

γ

i

f(t) dt

1/γ

i

.

Proof. Since λ

i

> −1/γ

i

for i = 1, . . . , n, we put

p

i

(t) =

t

1+λ

i

γ

i

′

, and p

n+1

(t) =

n

Y

i=1

t

λ

i

+1/γ

i

!

′

.

Then for a = 0, b = 1, we have

P

i

(x) = x

1+λ

i

γ

i

, i = 1, . . . , n, and P

n+1

(x) = x

1+

P

n

i=1

λ

i

,

and therefore

P

n+1

(x) =

n

Y

i=1

P

1/γ

i

i

(x).

Choosing g to be the identity and applying Theorem 3.1(b), we obtain

A

[0,1]

(p

n+1

, f) =

1 +

n

X

i=1

λ

i

!

Z

1

0

t

P

n

i=1

λ

i

f(t) dt,

and

n

Y

i=1

M

[0,1],g

(p

i

, f)

1/γ

i

=

n

Y

i=1

(1 + λ

i

γ

i

)

1/γ

i

·

n

Y

i=1

Z

1

0

t

λ

i

γ

i

f(t) dt

1/γ

i

.

Hence the result.

The following corollary is a consequence of Theorem 4.1 and may be

found in [17].

14

Corollary 5.7 Let f : [a, b] → R be non-negative and non-increasing, h

i

:

[a, b] → R, i = 1, . . . , n, be non-negative non-decreasing functions with

continuous ﬁrst derivatives. If q > 1, then

Z

b

a

n

X

i=1

h

i

(t)

q

!

′

f(t) dt

1/q

≤

n

X

i=1

Z

b

a

h

i

(t)

q

′

f(t) dt

1/q

. (22)

Proof. Put

δ

i

=

h

i

(b)

P

n

i=1

h

i

(b)

and

p

i

(x) =

h

i

(x)

1/q

′

, i = 1, . . . , n, p

n+1

(x) =

P

n

i=1

h

i

(x)

P

n

i=1

h

i

(b)

q

′

.

If g is the identity, then the functions f, g, p

i

, and numbers δ

i

, i =

1, . . . , n, satisfy the assumptions of Theorem 4.1(a), which yields (22).

References

[1] Abramovich, S.—Peˇcari´c, J.—Varoˇsanec, S., New generalization o f

Gauss-P´olya’s inequality, Math. Inequal. Appl. 1(3) (1998), 331– 342.

[2] Alzer, H., An extension of an inequality of G. P´olya, Buletin Instittutilui

Polytechnic Din, Iasi. Tomal 36(40) Fasc. 14 (1990), 17-18.

[3] Gauss, C.F., Theoria combinationis o bservationum 1821, German

transl. in Abhandlugen zur Methode der kleines Quadrate, Neudruck,

W¨urzburg 1964.

[4] Haluˇska, J.— Hutn´ık, O.: On generalized weighted quasi-arithmetic

means in integral form. Jour. Electrical Engineering 56, 12/s ( 2005),

3–6.

[5] Haluˇska, J.—Hutn´ık, O., Some inequalities involving integral means,

Tatra Mt. Math. Publ. (to appear).

[6] Hardy, G.H.—Littlewood, J.E.—P´olya, G., Inequalities, Cambridge Uni-

versity Press, 1967.

[7] Hutn´ık, O.: On Hadamard type inequalities for generalized weighted

quasi-arithmetic means. J. Inequal. Pure Appl. Math. 7(3 ) (2006), Art.

96.

15

[8] Losonczi, L .—P´ales, Zs., Minkowski’s inequality for two variable Gini

means, Acta Sci. Math. (Szeged) 62 (1996), no. 3–4, 413–425.

[9] Losonczi, L.—P´ales, Zs., Minkowski’s inequality for two variable diﬀer-

ence means, Proc. Amer. Math. Soc. 126, (1998), 779–789.

[10] Mitrinovi´c, D.S.—Peˇcari´c, J.E.—Fink, A.M., Classical and new inequal-

ities in analysis, Dordrecht, Kluwer Acad. Publishers, 1993.

[11] P´ales, Zs., H¨older-type inequalities for quasiarithmetic means, Acta

Math. Hungar. 47 (1986), no. 3–4, 395–399.

[12] P´ales, Zs., Strong H¨older and Minkowski inequalities for quasiarith-

metic means, Acta Sci. Math. (Szeged) 65 (1999), no. 3–4, 493–503.

[13] Pearce, C.E.M—Peˇcari´c, J.—

ˇ

Sunde, J., A generalization of P´olya’s in-

equality to Stolar sky and Gini means, Math. Inequal. Appl. 1 (2)

(1998), 211–222.

[14] Peˇcari´c, J.—

ˇ

Sunde, J.—Varoˇsanec, S., On Gauss-P´olya’s Inequality,

Sitzungsber. Abt. II, 207 (1998), 71–82.

[15] Peˇcari´c, J.—Varoˇsanec, S., A generalization of P´olya’s Inequality, In-

equalities and Applications. World Scientiﬁc Publishing Company, Sin-

gapore, (1 994), 501–504.

[16] P´olya, G.—Szeg¨o, G., Aufgaben und Lehrs¨atze aus der Analysism I, II,

Berlin, Springer Verlag, 1956.

[17] Varoˇsanec, S., Inequalities of Minkowski’s type, Real Anal. Ex-

change 20, No. 1, (1994–95), 250–255.

Ondrej Hutn´ık, Institute of Mathematics, Faculty of Science, P. J.

ˇ

Saf´arik

University, Current address: Jesenn´a 5, 041 54 Koˇsice, Slovakia

E-mail address: ondrej.hutnik@upjs.sk

16