Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 143-151 , 2010 Applied Mathematics
A New Type of Hardy-Hilbert’s Integral Inequalities Aziz Sa¼glam1, Hüseyin Y¬ld¬r¬m2, Mehmet Zeki Sar¬kaya3
Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University, Afyon, Türkiye
e-mail:1azizsaglam @ aku.edu.tr,2hyildir@ aku.edu.tr,3sarikaya@ aku.edu.tr
Received Date: December 12, 2009 Accepted Date: March 3, 2010
Abstract. In this paper, a new type of Hardy-Hilbert’s integral inequality in which the weight function is homogeneous fuction are given. Other results are also obtained.
Key words: Hardy-Hilbert-type integral inequality; weight function; B- func-tion; Hölder’s inequality:
2000 Mathematics Subject Classi…cation: 26D15 1. Introduction
First, let us recall the well-known Hilbert’s integral inequality: If f (x) ; g (x) 0; 0 <1R 0 f2(x)dx < 1 and 0 <R1 0 g2(x)dx < 1; then (see [1]) (1.1) 1 Z 0 1 Z 0 f (x)g(y) x + y dxdy < 0 @ 1 Z 0 f2(x)dx 1 A 1 20 @ 1 Z 0 g2(x)dx 1 A 1 2 ; (1.2) 1 Z 0 1 Z 0 f (x)g(y) max fx; ygdxdy < 4 0 @ 1 Z 0 f2(x)dx 1 A 1 20 @ 1 Z 0 g2(x)dx 1 A 1 2 ;
where the constant factors and 4 are the best possible in (1.1) and (1.2), respectively. Inequality (1.1) is called Hilbert’s integral inequality and (1.2) is called Hilbert’s type which have been extended by Hardy (see [2]) as follows:
If p > 1; 1p + 1q = 1; f (x) ; g (x) 0; such that 0 < 1R 0 fp(x)dx < 1 and 0 <R1 0 gq(x)dx < 1; then (1.3) 1 Z 0 1 Z 0 f (x)g(y) x + y dxdy < sin(p) 0 @ 1 Z 0 fp(x)dt 1 A 1 p0 @ 1 Z 0 gq(x)dt 1 A 1 q ; (1.4) 1 Z 0 1 Z 0 f (x)g(y) max fx; ygdxdy < pq 0 @ 1 Z 0 fp(x)dt 1 A 1 p0 @ 1 Z 0 gq(x)dt 1 A 1 q ;
where the constant factors sin(
p) and pq are the best possible in (1.3) and (1.4),
respectively. The inequality (1.3) is important in in analysis and its applications (cf. Mitrinovic et. al.[3]). In tecent years, Yang [5], [8] gave two di¤erent nice generalizations of (1.3) by introducing a parameter > 0 and in ([4], [6], [7], [10], [11], [15]) authors gave some strengthened versions and extension of the inequality (1.3). In particular, in [5], B Yang gave the following extension of (1.3) as follows: If > 2 min fp; qg ; f; g 0 satisfy
0 < 1 Z 0 t1 fp(t)dt < 1 and 0 < 1 Z 0 t1 gq(t)dt < 1; then (1.5) 1R 0 1 R 0 f (x)g(y) (x+y) dxdy < k (p) 1 R 0 t1 fp(t)dt 1 p 1R 0 t1 gq(t)dt 1 q ;
where the constant factor k (p) = B p+p 2;q+q 2 is the best possible and B is the beta function.
Another result of the same type inequality (1.4) is given by Li et. al., (1.6) R1 0 1 R 0 f (x)g(y)
A minfx;yg+B maxfx;ygdxdy < D (A; B) 1 R 0 f2(x)dx 1 2 R1 0 g2(x)dx 1 2 ; where the constant factor D (A; B) (see [12, Lemma 2.1]) is the best possible in inequality (1.6). For more information related to this subject see, for example, [9, 13].
By introducing a parameter > 0; in [14], Azar gave a result of similar type inequality to (1.6): (1.7) 1 R 0 1 R 0 f (x)g(y)
A minfx ;y g+B maxfx ;y gdxdy
< C (A; B) R1 0 xp(1 2) 1fp(x)dx 1 p R1 0 xq(1 2) 1gq(x)dx 1 q ;
where the constant factor C (A; B) (see [14, Theorem 2.3]) is the best possible in inequality (1.7).
In [16], Sulaiman gave the following extension of ()1.5 as follows:
Let f; g 0, K (u; v) be positive, increasing, homogeneous function of degree ; 0 < < min f(1 b) q=p; (1 a) q=pg ; a; b > 0; p > 1; 1 p + 1 q = 1: Set F (u) = u Z 0 f (t)dt; G (v) = v Z 0 g(t)dt: Then (1.8) T R 0 T R 0 F (u)G(v) K(u;v) dudv T p p pK1pqqK2 T R 0 (T t) Fp 1(t)f (t)dt !1=p T R 0 (T t) Gq 1(t)g(t)dt !1=q ; where K1= 1 Z 0 ta 1 K (1; t)dt and K2= 1 Z 0 tb 1 K (t; 1)dt:
The object of this paper is to give some new inequalities similar to that of Hardy-Hilbert’s integral inequality for homogeneous functions.
2. Main Result
Lemma 1. Let K (t; 1) ; K (1; t) be positive increasing functions, 0 < +1 : Set for s 1; A1; A2 0; f (s) = s s Z 0 t K (1; t) + A1min f1; tg + A2max f1; tg dt; and g (s) = s s Z 0 t K (t; 1) + A1min f1; tg + A2max f1; tg dt: Then f (s) f (1) ; g (s) g (1) :
Proof: We have
f0(s) = s K(1;s)+A s
1minf1;sg+A2maxf1;sg
s 1Rs 0
t
K(1;t)+A1minf1;tg+A2maxf1;tgdt
s
K(1;s)+A1minf1;sg+A2maxf1;sg
s 1
K(1;s)+A1minf1;sg+A2maxf1;sg
s R 0
t dt
=K(1;s)+A1minsf1;sg+A2maxf1;sg 1 +1 0:
This shows that f is nonincreasing and hence f (s) f (1) : The other part has a similar proof
The following is our main result.
Theorem 1. Let f; g 0; K (u; v) be positive, increasing, homogeneous func-tion of degree ; 0 < min f(1 b) q=p; (1 a) q=pg ; a; b > 0; p > 1;
1 p + 1 q = 1: Set F (u) = u Z 0 f (t) dt; G (v) = v Z 0 g (t) dt: Then T R 0 T R 0 F (u)G(v)
K(u;v)+A1minfu ;v g+A2maxfu ;v gdudv
T pppK 1pqqK2 T R 0 (T t) Fp 1(t) f (t) dt !1=p T R 0 (T t) Gq 1(t) g (t) dt !1=q ; where K1= 1 Z 0 ta 1 K (1; t) + A1min f1; t g + A2max f1; t g dt; and K2= 1 Z 0 tb 1 K (t; 1) + A1min f1; t g + A2max f1; t g dt:
Proof: T R 0 T R 0 F (u)G(v)
K(u;v)+A1minfu ;v g+A2maxfu ;v gdudv
= T R 0 T R 0 v a 1 p F (u) u b 1
q [K(u;v)+A1minfu ;v g+A2maxfu ;v g]1=p
u
b 1 q G(v)
v
a 1
p [K(u;v)+A1minfu ;v g+A2maxfu ;v g]1=qdudv
T R 0 T R 0 va 1Fp(u)
u(b 1)p=q[K(u;v)+A1minfu ;v g+A2maxfu ;v g]dudv
!1=p T R 0 T R 0 ub 1Gq(v)
v(a 1)q=p[K(u;v)+A1minfu ;v g+A2maxfu ;v g]dudv
!1=q = M1=pN1=q: We …rst consider M = T Z 0 u(1 b)p=qFp(u) du T Z 0 va 1
K (u; v) + A1min fu ; v g + A2max fu ; v g dv:
Observe that on putting v = uy; dv = udy; 0 y t=u; we have, in view of Lemma 1, by writting = a + (1 b) p=q ;
T R 0
va 1
K(u;v)+A1minfu ;v g+A2maxfu ;v gdv
= T =uR
0
(uy)a 1u
K(u;uy)+A1minfu ;(uy) g+A2maxfu ;(uy) gdy
= ua T =uR
0
ya 1
K(1;y)+A1minf1;y g+A2maxf1;y gdy
= ua Tu Tu
T =uR 0
ya 1
K(1;y)+A1minf1;y g+A2maxf1;y gdy
T ua R1 0
ya 1
K(1;y)+A1minf1;y g+A2maxf1;y gdy
By above, we obtain M T K1 T R 0 ua+(1 b)p=q Fp(u) du = T K1 T R 0 Fp(u) du: Since, Fp(u) = u Z 0 (Fp(t))0dt = p u Z 0 Fp 1(t) f (t) dt;
and after an interchange of the order of integration we obtain
M pT K1 T R 0 u R 0 Fp 1(t) f (t) dtdu = pT K1 T R 0 (T t) Fp 1(t) f (t) dt:
Similarly, the other part follows by using Lemma 1, replacing by where = b + (1 a) q=p to obtain N qT K2 T Z 0 (T t) Gq 1(t) g (t) dt:
This completes the proof of the theorem.
Corollary 1. By an application of Theorem 1, for the special case a = b = =2, we have T R 0 T R 0 F (u)G(v)
K(u;v)+A1minfu ;v g+A2maxfu ;v gdudv
T pppK 1pqqK3 T R 0 (T t) Fp 1(t) f (t) dt !1=p T R 0 (T t) Gq 1(t) g (t) dt !1=q ; where K1= 1 Z 0 t =2 1 K (1; t) + A1min f1; t g + A2max f1; t g dt;
and K3= 1 Z 1 t =2 1 K (1; t) + A1min f1; t g + A2max f1; t gdt < 1:
Furthermore, when for B1; B2 0; K (u; v) = (B1u + B2v) and B1; B2; A1 and A2 are not zero meanwhile, we have
T R 0 T R 0 F (u)G(v)
(B1u+B2v) +A1minfu ;v g+A2maxfu ;v gdudv
T K6ppppqq T R 0 (T t) Fp 1(t) f (t) dt !1=p T R 0 (T t) Gq 1(t) g (t) dt !1=q ; where K6= 1 Z 1 t =2 1 K (B1; B2t) + A1min f1; t g + A2max f1; t gdt < 1: Proof: Let K4= 1 Z 0 t =2 1 K (B1; B2t) + A1min f1; t g + A2max f1; t g dt; K5= 1 Z 0 t =2 1 K (B1t; B2) + A1min f1; t g + A2max f1; t g dt: For a = b = =2; we have K5= K6 1 R 0 t =2 1
K(B1t;B2)+A1minf1;t g+A2maxf1;t gdt
= 1 R 0
t =2 1
K(B1t;B2tt 1)+A1minft t ;t g+A2maxft t ;t gdt
= 1 R 0
t =2 1
K(B1;B2t 1)+A1minf1;t g+A2maxf1;t gdt
=1R 1
t=2 1
The other part follows from the fact that for K (B1t; B2) = (B1+ B2t) for B1; B2 0 and K4= K5= K6; T R 0 T R 0 F (u)G(v)
(B1u+B2v) +A1minfu ;v g+A2maxfu ;v gdudv
= T K6ppppqq T R 0 (T t) Fp 1(t) f (t) dt !1=p T R 0 (T t) Gq 1(t) g (t) dt !1=q :
Corollary 2. By an application of Theorem 1 with K (u; v) = B1u + B2v ; we have T R 0 T R 0 F (u)G(v)
B1u +B2v +A1minfu ;v g+A2maxfu ;v gdudv
= T pppK apqqKb T R 0 (T t) Fp 1(t) f (t) dt !1=p T R 0 (T t) Gq 1(t) g (t) dt !1=q ; where Kr= 1 Z 0 tr 1 B1+ B2t + A1min f1; t g + A2max f1; t g dt (r 2 fa; bg) ;
and B1; B2; A1 and A2 are not zero meanwhile. References
1. G. H. Hardy, J. E. Littlewood and G. Polya, Inequalities, Cambridge University Press, Cambridge, 1952.
2. G. H. Hardy, Note on a theorem of Hilbert concerning series of positive terms, Proc. Math. Soc. 23(2) (1925), Records of Proc. XLV-XLVI.
3. D. S Mitrinovic, J. E. Pecaric and A.M. Fink, Inequalities involving functions and their integrals and derivatives, Bostan: Kluwer Academic Publishers, 1991.
4. M. Krni´c and J. Peµcari´c, General Hilbert’s and Hardy’s inequalities, Math. Inequal. Appl., 8(1) 2005, 29-51.
5. B. Yang, On Hardy-Hilbert’s integral inequality,J. Math. Anal. Appl., 261 (2001), 295-306.
6. B. Yang, On best extensions of Hardy-Hilbert’s integral inequality with two para-meters, J. Inequal. Pure Appl. Math., 6(3) (2005), Art. 81.
7. B. Yang, A new Hilbert-type inequality, Bull. Belg. Math. Soc., 13 (2006), 479-487. 8. B. Yang, On the extended Hilbert’s integral inequality, J. Inequal. Pure Appl. Math., 5(4) (2004), Art. 96.
9. B. Yang, On a Hilbert-type operator with a synnetric homogeneous Kernel of 1-order and applications, Journal of Inequalities and Applications, vol. 2007, Article ID 47812, 9 pages, 2007.
10. A. Shamandy, On new extensions of integral inequalities similar to Hilbert’s in-equality (I), Proc. Pakistan Acad. Sci., 42(2) 2005, 89-98.
11. A. Shamandy, On new extensions of integral inequalities similar to Hilbert’s in-equality (II), Proc. Pakistan Acad. Sci., 42(2) 2005, 99-109.
12. Y. Li, Y. Qian, and B. He, On further analogs of Hilbert’s Inequality, International Journal of Mathematics and Mathematical Scienses, vol. 2007, Article ID 76329, 6 pages, 2007.
13. G. Xi, A reverse Hardy-Hilbert-type inequality, J. Inequal. Appl., 2007 (2007), Art. 79758.
14. L. E. Azar, On Some Extensions of Hardy-Hilbert’s Inequality and Applications, Journal of Inequalities and Applications, vol. 2008, Article ID 546829, 14 pages, 2008. 15. H. Y¬ld¬r¬m, U. M. Özkan and M. Z. Sar¬kaya, On the Hilbert Type Integral Inequalities with some parameters and Its Reverse, Kyunkgpook Mathematical Jour-nal(In Press).
16. Waad T. Sulaiman, A study on some new types of Hardy-Hilbert’s integral in-equalities, Banach J. Math. Anal., 2 (2008), no. 1, 16-20.
