Full Text

Turn on search term navigation

http://crossmark.crossref.org/dialog/?doi=10.1007/s40065-015-0135-8&domain=pdf

Web End = Arab. J. Math. (2015) 4:159170DOI 10.1007/s40065-015-0135-8 Arabian Journal of Mathematics

http://crossmark.crossref.org/dialog/?doi=10.1007/s40065-015-0135-8&domain=pdf

Web End = http://crossmark.crossref.org/dialog/?doi=10.1007/s40065-015-0135-8&domain=pdf

Web End = S. S. Dragomir

Some Grss-type results via Pompeius-like inequalities

Received: 9 March 2015 / Accepted: 6 August 2015 / Published online: 27 August 2015 The Author(s) 2015. This article is published with open access at Springerlink.com

Abstract In this paper, some Grss-type results via Pompeius-like inequalities are proved.

Mathematics Subject Classication 26D15 25D10

1 Introduction

In 1946, Pompeiu [18] derived a variant of Lagranges mean value theorem, now known as Pompeius mean value theorem (see also [18, p.83]).

Theorem 1.1 (Pompeiu [18]) For every real valued function f differentiable on an interval [a, b] not containing 0 and for all pairs x1 = x2 in [a, b] , there exists a point between x1 and x2 such thatx1 f (x2) x2 f (x1)

x1 x2 =

f () f () . (1.1)

The following inequality is useful to derive some Ostrowski-type inequalities; see [9].

Corollary 1.2 (Pompeius inequality) With the assumptions of Theorem 1.1 and if f f

= supt(a,b)

f (t) t f (t)

< where (t) = t, t [a, b] , then

|t f (x) x f (t)|

f f

|x t| (1.2)

for any t, x [a, b] .

S. S. Dragomir (B)

Mathematics, College of Engineering and Science, Victoria University, PO Box 14428, Melbourne, MC 8001, AustraliaE-mail: [email protected]://rgmia.org/dragomir

S. S. DragomirSchool of Computational and Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa

123

160 Arab. J. Math. (2015) 4:159170

The inequality (1.2) was obtained by the author in [9].

For other Ostrowski-type inequalities concerning the p-norms f f

p , see [1,2,17,19].

For two Lebesgue integrable functions f, g : [a, b]

R, consider theebyev functional:

C ( f, g) :=

1b a

a f (t)g(t)dt

(b a)2

a f (t)dt

a g(t)dt. (1.3)

Grss [10] showed that

|C ( f, g)|

14 (M m) (N n) , (1.4) provided that there exists the real numbers m, M, n, N such that

m f (t) M and n g (t) N for a.e. t [a, b] . (1.5) The constant 14 is best possible in (1.3) in the sense that it cannot be replaced by a smaller quantity.

Another, however less known, result, though it was obtained byebyev [7], states that

|C ( f, g)|

1 12

(b a)2 , (1.6)

provided that f , g exist and are continuous on [a, b] and f

= supt[a,b]

f (t) . The constant 1

12 cannot

be improved in the general case.Theebyev inequality (1.6) also holds if f, g : [a, b]

R are assumed to be absolutely continuous and

f , g L [a, b] , while

f (t) .

A mixture between Grss result (1.4) andebyevs one (1.6) is the following inequality obtained by Ostrowski [15]:

|C ( f, g)|

18 (b a) (M m)

= ess supt[a,b]

, (1.7) provided that f is Lebesgue integrable and satises (1.5), while g is absolutely continuous and g L [a, b] .

The constant 18 is best possible in (1.7).

The case of Euclidean norms of the derivative was considered by Lupas [12], in which he proved that

|C ( f, g)|

1 2

f 2 g 2 (b a) , (1.8) provided that f, g are absolutely continuous and f , g L2 [a, b] . The constant

2 is the best possible.

Recently, Cerone and Dragomir [3] have proved the following results:

|C ( f, g)| inf R

g q

1b a

f (t)

1b a

a f (s) ds

, (1.9)

where p > 1 and 1p + 1q = 1 or p = 1 and q = , and

|C ( f, g)| inf R

g 1

, (1.10)

provided that f L p [a, b] and g Lq [a, b] (p > 1, 1p + 1q = 1; p = 1, q = or p = , q = 1).

Notice that for q = , p = 1 in (1.9), we obtain

|C ( f, g)| inf R

1b a

ess sup

1b a

a f (s) ds

t[a,b]

f (t)

1b a

f (t)

1b a

a f (s) ds

1b a

f (t)

1b a

a f (s) ds

dt (1.11)

123

Arab. J. Math. (2015) 4:159170 161

and, if g satises (1.5), then

|C ( f, g)| inf R

1b a

f (t)

1b a

a f (s) ds

n + N

1b a

f (t)

1b a

a f (s) ds

dt. (1.12)

The inequality between the rst and the last term in (1.12) has been obtained by Cheng and Sun [8]. However, the sharpness of the constant 12, a generalization for the abstract Lebesgue integral and the discrete version of it have been obtained in [4].

For other recent results on the Grss inequality, see [5,6,11,13,14,16,20] and the references therein. In this paper, some Grss-type results via Pompeius-like inequalities are proved.

2 Some Pompeius-type inequalities

We can generalize the above inequality for the larger class of functions that are absolutely continuous and complex valued as well as for other norms of the difference f f .

Theorem 2.1 Let f : [a, b]

2 (N n)

1b a

f (t)

1b a

a f (s) ds

C be an absolutely continuous function on the interval [a, b] with b > a > 0.

Then for any t, x [a, b] , we have

|t f (x) x f (t)|

f f

|x t| if f f L [a, b] ,

2q1

1/q f f

tq1

xq1

1/q if f f L p [a, b]

p > 1,

1p + 1q = 1,

f f

(2.1)

max{t,x}

min{t,x} ,

or equivalently

f (x)x

1t 1x

if f f L [a, b] ,

f (t) t

f f

2q1

1/q f f

t2q1

x2q1

1/q if f f L p [a, b]

p > 1,

1p + 1q = 1,

f f

(2.2)

min

1 {t2,x2}

Proof If f is absolutely continuous, then f/ is absolutely continuous on the interval [a, b] that does not contain 0 and

f (s) s

ds =

f (x)x

f (t) t

for any t, x [a, b] with x = t.

Since

f (s) s

ds =

f (s) s f (s)s2 ds,

we get the following identity:

t f (x) x f (t) = xt

f (s) s f (s)s2 ds (2.3)

123

162 Arab. J. Math. (2015) 4:159170

for any t, x [a, b] .

We notice that the equality (2.3) was proved for the smaller class of differentiable function and in a different manner in [17].

Taking the modulus in (2.3), we have

|t f (x) x f (t)| =

f (s) s f (s) s2 ds

f (s) s f (s) s2

:= I, (2.4)

and utilizing Hlders integral inequality we deduce

I xt

sups[t,x]([x,t])

f (s) s f (s)

x s2 ds

t f (s) s f (s)

p ds 1/p

x s2q ds

1/q p > 1,

1p + 1q = 1,

t f (s) s f (s)

ds sups[t,x]([x,t])

1 s2

f f

|x t| , 1

2q1

1/q f f

tq1

xq1

1/q p > 1,

1p + 1q = 1,

(2.5)

and the inequality (2.2) is proved. Remark 2.2 The rst inequality in (2.1) also holds in the same form for 0 > b > a.

3 Some Grss-type inequalities

We have the following result of Grss type.

Theorem 3.1 Let f, g : [a, b]

f f

max{t,x}

min{t,x} ,

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

If f , g L [a, b] , then

b3 a3 3

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

112 (b a)4

f f

g g

. (3.1)

The constant 1

12 is best possible.

Proof From the rst inequality in (2.1), we have

a (t f (x) x f (t)) (tg (x) xg (t)) dtdx

|(t f (x) x f (t)) (tg (x) xg (t))| dtdx

f f

g g

a (x t)2 dtdx. (3.2)

123

Arab. J. Math. (2015) 4:159170 163

Observe that

a (t f (x) x f (t)) (tg (x) xg (t)) dtdx =

t2 f (x) g (x) + x2 f (t) g (t) tg (t) x f (x) f (t) txg (x)

dtdx

= 2

a t2dt

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

and

a (x t)2 dtdx =

1 3

(b x)3 + (x a)3

dx =

16 (b a)4 .

Utilizing the inequality (3.2), we deduce the desired result (3.1).Now, assume that the inequality (3.1) holds with a constant B > 0 instead of 1

12 , i.e.,

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

b3 a3 3

B (b a)4

f f

g g

. (3.3)

If we take f (t) = g (t) = 1, t [a, b] , then b3 a3

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

b3 a3

3 (b a)

b2 a2

112 (b a)4

and f f

g g

= 1

12 , which proves the sharpness of the constant.

The following result for the complementary (p, q)-norms, with p, q > 1 and 1p + 1q = 1, holds.

Theorem 3.2 Let f, g : [a, b]

and by (3.3) we get B

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

If f L p [a, b] , g Lq [a, b] with p, q > 1, p, q = 2 and 1p + 1q = 1, then

b3 a3 3

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

2 (2q 1)1/q (2p 1)1/p

f f

p g g

q M1/qq (a, b) M1/pp (a, b) , (3.4)

where

Mq (a, b) :=

xq tq1

tq xq1

dtdx.

We have the bounds

Mq (a, b) (b a) N1/2q (a, b)

and

Mp (a, b) (b a) N1/2p (a, b)

123

164 Arab. J. Math. (2015) 4:159170

where, for r > 1,

Nr (a, b) :=

, r = 32

2 b2r+1a2r+1

2r+1 b2r+3a2r+32r+3

b2a2

b2 a2

b2+a2

2 ln ba b2a22

, r = 32.

Proof From the second inequality in (2.1), we have

|t f (x) x f (t)|

(2q 1)1/q

f f

xq tq1

1/q

tq xq1

and

g g

x pt p1

t px p1

1/p

|tg (x) xg (t)|

(2p 1)1/p

for any t, x [a, b] .

If we multiply these inequalities and integrate, then we get

a (t f (x) x f (t)) (tg (x) xg (t)) dtdx

|(t f (x) x f (t)) (tg (x) xg (t))| dtdx

(2q 1)1/q (2p 1)1/p

f f

p g g

xq tq1

tq xq1

1/q

x pt p1

t px p1

1/p

dtdx. (3.5)

Utilizing Hlders integral inequality for double integrals, we have

xq tq1

tq xq1

1/q

x pt p1

1/p

dtdx

t px p1

xq tq1

tq xq1

dtdx

1/q b

x pt p1

t px p1

dtdx

1/p

= M1/qq (a, b) M1/pp (a, b) (3.6) for p, q > 1 and 1p + 1q = 1.

Utilizing CauchyBunyakowskySchwarz integral inequality for double integrals, we have

Mq (a, b) =

xq tq1

tq xq1

dtdx

a dtdx

1/2 b

xq tq1

tq xq1

2 dtdx

1/2

= (b a)

xq tq1

tq xq1

2 dtdx

1/2

123

Arab. J. Math. (2015) 4:159170 165

Observe that

Nq (a, b) :=

xq tq1

tq xq1

2 dtdx

x2q

t2(q1)

dtdx 2

xq tq1

tq xq1

dtdx +

t2q

x2(q1)

dtdx

= 2

a x2qdx

a t2(q1)dt 2

a xdx

= 2

b2q+1 a2q+1 2q + 1

b2q+3 a2q+3

2q + 3

b2 a22

provided q = 32. If q = 32, then

Nq (a, b) =

b2 a2

b2 + a2

2 ln

b2 a2 2

Therefore,

Mq (a, b) (b a) N1/2q (a, b)

and, similarly,

Mp (a, b) (b a) N1/2p (a, b) .

Remark 3.3 The double integral

Mq (a, b) :=

xq tq1

tq xq1

dtdx

can be computed exactly by iterating the integrals. However, the nal form is too complicated to be stated here.

The Euclidian norms case is as follows:

Theorem 3.4 Let f, g : [a, b]

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

If f , g L2 [a, b] , then

b3 a3 3

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

1 9

f f

2 g g

b3 + a3

ln b

b3 a3

. (3.7)

Proof From the second inequality in (2.1), we have

|t f (x) x f (t)|

1 3

f f

x2t

1/2

t2 x

and

g g

x2t

t2 x

1/2

|tg (x) xg (t)|

1 3

for any t, x [a, b] .

123

166 Arab. J. Math. (2015) 4:159170

If we multiply these inequalities and integrate, then we get

a (t f (x) x f (t)) (tg (x) xg (t)) dtdx

|(t f (x) x f (t)) (tg (x) xg (t))| dtdx

1 3

f f

2 g g

x2t

t2 x

dtdx. (3.8)

Since

x2t

t2 x

dtdx

x2t

t2 x

dt +

t2x

x2 t

x2 (2 ln x ln a ln b) +

b3 + a3 2x3 3x

and

a x2 (2 ln x ln a ln b) dx

= a 2x2 ln xdx ln (ab)

a x2dx

b3 + a3

ln ba

2 9

b3 a3

while

b3 + a3 2x3

3x dx =

b3 + a3

ln ba

2 9

b3 a3

then we conclude that

x2t

t2 x

dtdx =

b3 + a3

ln b

b3 a3

Making use of the inequality (3.8), we deduce the desired result (3.7). Remark 3.5 It is an open question to the author if 19 is best possible in (3.7).

Theorem 3.6 Let f, g : [a, b]

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

Then,

a f (t) g (t) dt

a t f (t) dt

a tg (t) dt

b3 a3 3

f f

1 g g

2b3 + a3 3ab2

6a . (3.9)

Proof From the third inequality in (2.1), we have

a (t f (x) x f (t)) (tg (x) xg (t)) dtdx

|(t f (x) x f (t)) (tg (x) xg (t))| dtdx

f f

1 g g

max {t, x}

min {t, x}

2 dtdx. (3.10)

123

Arab. J. Math. (2015) 4:159170 167

Observe that

max {t, x}

min {t, x}

2 dtdx

max {t, x}

min {t, x}

max {t, x}

x min {t, x}

dt +

x t

t x

dt +

2b3 + a3 3ab2 6a ,

which together with (3.10) produces the desired inequality (3.9).

4 Some related results

The following result holds.

Theorem 4.1 Let f, g : [a, b]

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

If f , g L [a, b] , then

(b a)

f (t) g (t)t2 dt

f (t)t dt

g (t)t dt

(b a)2

L2 (a, b) G2 (a, b) L2 (a, b) G2 (a, b)

f f

g g

, (4.1)

where G (a, b) := ab is the geometric mean and

L (a, b) :=

b a

ln b ln a

is the Logarithmic mean.

The inequality (4.1) is sharp.

Proof From the rst inequality in (2.2), we have

f (x)x

f (t) t

g (x)x

g (t) t

f f

1 x

f f

(4.2)

for any t, x [a, b] .

Integrating this inequality on [a, b]2 , we get

f (x)x

f (t) t

g (x)x

g (t) t

dtdx

f (x)x

f (t) t

g (x)x

g (t) t

dtdx

f f

1 x

2 dtdx. (4.3)

123

168 Arab. J. Math. (2015) 4:159170

We have

f (x)x

f (t) t

g (x)x

g (t) t

dtdx

= 2

(b a)

f (t) g (t)t2 dt

f (t)t dt

g (t)t dt

and

1 x

2 dtdx = 2 (b a)2

L2 (a, b) G2 (a, b) L2 (a, b) G2 (a, b) .

Making use of (4.3), we get the desired result (4.1).

If we take f (t) = g (t) = 1, then we have

(b a)

f (t) g (t)t2 dt

f (t)t dt

g (t)t dt

= (b a)2

L2 (a, b) G2 (a, b) L2 (a, b) G2 (a, b)

and

f f

g g

= 1

and we obtain in both sides of (4.1) the same quantity

(b a)2

L2 (a, b) G2 (a, b) L2 (a, b) G2 (a, b) .

The case of Euclidian norms is as follows:

Theorem 4.2 Let f, g : [a, b]

C be absolutely continuous functions on the interval [a, b] with b > a > 0.

If f , g L2 [a, b] , then (b a)

f (t) g (t)t2 dt

f (t)t dt

g (t)t dt

(b a)3

a2b2 . (4.4)

Proof From the second inequality in (2.2) for p = q = 2, we have

f (x)x

1 6

f f

2 g g

f (t) t

1 3

f f

1t3

1 x3

1/2

(4.5)

and

g (x)x

g (t) t

1 3

g g

1t3

1 x3

1/2

(4.6)

for any t, x [a, b] .

On multiplying (4.5) with (4.6), we derive

f (x)x

f (t) t

g (x)x

g (t) t

1 3

f f

2 g g

1t3

1 x3

(4.7)

for any t, x [a, b] .

123

Arab. J. Math. (2015) 4:159170 169

Integrating this inequality on [a, b]2 , we get

f (x)x

f (t) t

g (x)x

g (t) t

dtdx

f (x)x

f (t) t

g (x)x

g (t) t

dtdx

1 3

f f

2 g g

1t3

1 x3

dtdx. (4.8)

We have

1t3

1 x3

dtdx =

1t3

1 x3

dt +

1x3

1 t3

1t3

1 x3

dt +

(b a)3

a2b2 .

From (4.8), we then obtain the desired result (4.4). Remark 4.3 It is an open question to the author if 16 is the best possible constant in (4.4).

The interested reader may obtain other similar results in terms of the norms f f

p g g

1x3

1 t3

dx =

q with

p, q > 1, p, q = 2 and 1p + 1q = 1. However, the details are omitted.

Acknowledgments The author would like to thank the anonymous referees for their valuable comments that have been implemented in the nal version of the paper.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/

Web End =http:// http://creativecommons.org/licenses/by/4.0/

Web End =creativecommons.org/licenses/by/4.0/ ), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

References

1. Acu, A.M.; Sofonea, F.D.: On an inequality of Ostrowski type. J. Sci. Arts 3(16), 281287 (2011)2. Acu, A.M.; Babos, A.; Sofonea, F.D.: The mean value theorems and inequalities of Ostrowski type. Sci. Stud. Res. Ser. Math. Inform. 21(1), 516 (2011)

3. Cerone, P.; Dragomir, S.S.: New bounds for theebyev functional. Appl. Math. Lett. 18, 603611 (2005)4. Cerone, P.; Dragomir, S.S.: A renement of the Grss inequality and applications. Tamkang J. Math. 38(1), 3749 (2007)5. Cerone, P.; Dragomir, S.S.: Some bounds in terms of -seminorms for Ostrowski-Grss type inequalities. Soochow J. Math. 27(4), 423434 (2001)

6. Cerone, P.; Dragomir, S.S.; Roumeliotis, J. Grss inequality in terms of -seminorms and applications. Integr. Transforms Spec. Funct. 14(3), 205216 (2003)

7. Chebyshev, P.L.: Sur les expressions approximatives des intgrals dnis par les outres prises entre les mme limites. Proc. Math. Soc. Charkov 2, 9398 (1882)

8. Cheng, X-L.; Sun, J.: Note on the perturbed trapezoid inequality. J. Ineq. Pure Appl. Math. 3(2), (2002) (art. 29, 7 pp)9. Dragomir, S.S.: An inequality of Ostrowski type via Pompeius mean value theorem. J. Inequal. Pure Appl. Math. 6(3), (2005) (article 83, 9 pp)

10. Grss, G.: ber das Maximum des absoluten Betrages von 1

a f (x)g(x)dx

1 (ba)2

a f (x)dx

ba g(x)dx. Math. Z.

39, 215226 (1935)11. Li, X.; Mohapatra, R.N.; Rodriguez, R.S.: Grss-type inequalities. J. Math. Anal. Appl. 267(2), 434443 (2002)12. Lupas, A.: The best constant in an integral inequality. Mathematica (Cluj, Romania) 15(38)(2), 219222 (1973)13. Mercer, A.M.: An improvement of the Grss inequality. J. Inequal. Pure Appl. Math. 6(4), (2005) (article 93, 4 pp)14. Mitrinovi, D.S.; Peari, J.E.; Fink, A.M.: Classical and New Inequalities in Analysis. Kluwer Academic Publishers, Dordrecht/Boston/London (1993)

15. Ostrowski, A.M.: On an integral inequality. Aequat. Math. 4, 358373 (1970)16. Pachpatte, B.G.: On Grss like integral inequalities via Pompeius mean value theorem. J. Inequal. Pure Appl. Math. 6(3), (2005) (article 82, 5 pp)

17. Peari, J.; Ungar, .: On an inequality of Ostrowski type. J. Inequal. Pure Appl. Math. 7(4), (2006) (art. 151, 5 pp)18. Pompeiu, D.: Sur une proposition analogue au thorme des accroissements nis. Mathematica (Cluj, Romania) 22, 143 146 (1946)

123

170 Arab. J. Math. (2015) 4:159170

19. Popa, E.C.: An inequality of Ostrowski type via a mean value theorem. Gen. Math. 15(1), 93100 (2007)20. Sahoo, P.K.; Riedel, T.: Mean Value Theorems and Functional Equations. World Scientic, Singapore, New Jersey, London, Hong Kong (2000)

123

Word count: 4039

Show less

The Author(s) 2015

Abstract

Translate

In this paper, Some Grüss-type results via Pompeiu's-like inequalities are proved.

Details

Title

Some Grüss-type results via Pompeiu's-like inequalities

Author

Dragomir, S S

Pages

159-170

Publication year

2015

Publication date

Aug 2015

Publisher

Springer Nature B.V.

ISSN

21935351

e-ISSN

21935343

Source type

Scholarly Journal

Language of publication

English

DOI

https://doi.org/10.1007/s40065-015-0135-8

ProQuest document ID

1713730294

The Author(s) 2015

Some Grüss-type results via Pompeiu's-like inequalities

Jump to:

Full Text

Abstract

Details

Suggested sources