Advnces in Dynmicl Systems nd Alictions ISSN 0973-5321, Volume 10, Number 1,. 1 9 (2015 htt://cmus.mst.edu/ds Some Hrdy Tye Inequlities with Weighted Functions vi Oil Tye Inequlities Rvi P. Agrwl Tes A&M University Kingsville Dertment of Mthemtics Kingsville, Tes, 78363, USA Donl O Regn Ntionl University of Irelnd School of Mthemtics, Sttistics nd Alied Mthemtics Glwy, Irelnd Smir H. Ser Mnsour University Dertment of Mthemtics, Fculty of Science Mnsour 35516, Egyt shser@mns.edu.eg Abstrct In this er, we will rove severl new inequlities of Hrdy tye with elicit constnts. The min results will be roved using generliztions of Oil s inequlity. AMS Subject Clssifictions: 26A15, 26D10, 26D15, 39A13, 34A40. Keywords: Hrdy s inequlity, Oil s inequlity. 1 Introduction The clssicl Hrdy inequlity (see [10] sttes tht for f 0 integrble over ny finite intervl (0, nd f integrble nd convergent over (0, nd > 1, then ( 1 ( f(tdt d f (d. (1.1 1 0 0 Received Februry 20, 2014; Acceted August 10, 2014 Communicted by Mrtin Bohner 0
2 R. Agrwl, D. O Regn nd S. Ser The constnt (/ ( 1 is the best ossible. Some etensions of Hrdy s inequlity were considered in Beesc [5]. Our im in this er is to rove some inequlities with weighted functions of Hrdy tye using Oil tye inequlities. 2 Min Results Throughout the er, ll functions re ssumed to be ositive nd mesurble nd ll the integrls which er in the inequlities re ssumed to eist nd be finite. To obtin inequlities of Hrdy tye we loo t inequlities for R(, b = r(t dt nd F ( = R(, bf (F (d, f(tdt. Ech Oil tye inequlity will give Hrdy tye inequlity. We will use number of Oil tye inequlities to illustrte this oint. Boyd nd Wong [8] roved if > 0 nd if y is n bsolutely continuous function on [, b] with y( = 0 (or y(b = 0, then q(t y(t y (t dt 1 λ 0 ( + 1 w(t y (t +1 dt, (2.1 q nd w re nonnegtive functions in C 1 [, b], nd such tht the boundry vlue roblem hs solution (q(t (u (t = λw (tu (t, with u( = 0 nd q(b [u (b] = λ w(b u (b, for which u > 0 in [, b] (let λ 0 be the smllest eigenvlue of the boundry vlue roblem. Alying the inequlity (2.1 on the term (+1 ( + 1 R(, bf (F (d, we hve R(, bf (F (d 1 s(t(f ( +1 d, (2.2 λ 0 r nd s re nonnegtive functions, r C[, b], s C 1 [, b], nd such tht the boundry vlue roblem hs solution (R(, b = (R(, b (u ( = λs (u (, (2.3 r(tdt nd note R C 1 [, b] since r C[, b] with u( = 0 nd u(b = 0, for which u > 0 in [, b] (let λ 0 be the smllest eigenvlue of the boundry vlue roblem.
Hrdy Tye Inequlities 3 Theorem 2.1. Assume tht r, s re nonnegtive functions with r C[, b], s C 1 [, b] nd > 0. Then +1 r( f(tdt d 1 λ 0 s( (f( +1 d, for ll integrble functions f 0 λ 0 is the smllest eigenvlue of the boundry vlue roblem (2.3. Proof. Let F ( = f(tdt. Since f is integrble on [, b] then F is bsolutely continuous on [, b]. Note F ( = 0, F ( = f( > 0 nd Integrtion by rts gives R(, b = +1 r( f(tdt d = r(f +1 (d. +1 r( f(tdt d = R(, bf +1 ( b + ( + 1 R(, bf (F (d r(tdt. Using R(b, b = 0 nd F ( = 0, we hve +1 r( f(tdt d = ( + 1 Now (2.2 estblishes the result. R(, bf (F (d. (2.4 Boyd in [7] etended the results of [8]. In [7, Theorem 2.1] the uthor estblished inequlities (best ossible constnts of the form s(t y(t y (t q dt ( b +q r(t y (t dt, λ 0 ( + q > 0, > 1, 0 q with r, s C 1 (, b nd r > 0, s > 0.e. on (, b; here λ 0 is the smllest eigenvlue of n rorite boundry vlue roblem (ssuming certin conditions re stisfied; see [7]. With these conditions (with q = 1 nd > 1 we obtin using the rocedure before nd in Theorem 2.1 +1 r( f(tdt d ( +1 r(t (f(t dt, λ 0
4 R. Agrwl, D. O Regn nd S. Ser λ 0 is the smllest eigenvlue of n rorite boundry vlue roblem. Insted of this inequlity (nd resenting the conditions to gurntee the eistence of λ 0 we will consider two secil cses of this result, one found in [7] nd the other in [6]. In the following, we ly n inequlity due to Boyd [7] nd the Hölder inequlity. The Boyd inequlity sttes tht: If y is bsolutely continuous on [, b] with y( = 0 (or y(b = 0, then y(t ν y (t η dt N(ν, η, s(b ν ( ν > 0, s > 1, 0 η < s, nd N(ν, η, s := I(ν, η, s := (s η ν ν σ ν+η s (s 1(ν + η (I(ν, η, s ν, σ := 1 0 { 1 + ν+η s y (t s dt, (2.5 { } 1 ν(s 1 + (s η s, (2.6 (s 1(ν + η } (ν+η+sν/sν s(η 1 s η t [1 + (η 1t]t 1/ν 1 dt. Aly the Hölder inequlity nd inequlity (2.5 to obtin R(, bf (F (d ( ( R b (, bd ( N 1 q (q, q, s(b ( (F ( s d +1 s F q ( (F ( q q d R (, bd, (2.7 > 1, 1/ + 1/q = 1, s > 1 nd 1 < q < s; here N(q, q, s is determined from (2.6 by utting ν = q nd η = q. Theorem 2.2. Assume tht r is nonnegtive mesurble function on (, b, > 1, s > 1, 1 < q < s nd 1/ + 1/q = 1. Then +1 ( r( f(tdt +1 d C (f( s s d, for ll integrble functions f 0; here ( C = ( + 1 N 1 q (q, q, s(b R (, bd.
Hrdy Tye Inequlities 5 Proof. The result follows from (2.4 nd (2.7. As in the roof of Theorem 2.1, by utting F ( = f(tdt, we hve the following result. Theorem 2.3. Assume tht r is nonnegtive mesurble function on (, b, > 1, 1 < q < s nd 1/ + 1/q = 1. Then ( +1 ( r( f(tdt +1 d C (f( s s d, for ll integrble functions f 0; here ( C = ( + 1 N 1 q (q, q, s(b When η = s eqution (2.5 becomes R (, d y(t ν y (t η dt L(ν, η(b ν ( L(ν, η := ( ν ηνη ν η ν + η ν + η Γ ( Γ ( η+1 + 1 η ν η+1 η nd R(, = r(tdt. ν+η η y (t η dt, (2.8 Γ ( 1 ν nd Γ is the Gmm function. Aly inequlity (2.8 to obtin F q ( (F ( q d L(q, q(b q ( Using (2.10, we see tht ( L(q, q = (qq + 1 + 1 Γ R(, bf (F (d ( ( Γ ( ν, (2.9 q+q (F ( q q d, (2.10 q+1 + 1 q q q+1 q Γ ( R b (, bd ( 1 q ( L 1 q (q, q(b R (, bd ( (F ( q d +1 q > 1 nd 1/ + 1/q = 1. This gives us the following results.,. (2.11 F q ( (F ( q q d
6 R. Agrwl, D. O Regn nd S. Ser Theorem 2.4. Assume tht r is nonnegtive mesurble function on (, b, > 1, q > 1 nd 1/ + 1/q = 1. Then +1 ( r( f(tdt +1 d C (f( q q d, for ll integrble functions f 0; here nd L(q, q is defined s in (2.11. ( C = ( + 1 L 1 q (q, q(b R (, bd Theorem 2.5. Assume tht r is nonnegtive mesurble function on (, b, > 1, q > 1 nd 1/ + 1/q = 1. Then ( +1 ( r( f(tdt +1 d C (f( q q d, for ll integrble functions f 0; here nd L(q, q is defined s in (2.11. ( C = ( + 1 L 1 q (q, q(b R (, d Finlly we ly n Oil tye inequlity due to Beesc [6] to rove inequlities of Hrdy tye. The inequlity due to Beesc is given in the following theorem. Theorem 2.6. Let r, s be nonnegtive, mesurble functions on (α, τ. Further ssume tht > 1, > 0, 0 < q <, nd let y be bsolutely continuous in [α, τ] such tht y(α = 0. Then τ α [ τ (+q/ r(t y(t y (t q dt K 1 (, q, s(t y (t dt], (2.12 α K 1 (, q, = q ( q q + ( τ (r(y q (s(y α q q ( y s 1 1 (tdt ( 1/( q dy q.
Hrdy Tye Inequlities 7 If insted [α, τ] is relced by [τ, β] nd y(α = 0 is relced by y(β = 0, then β τ [ β (+q/ r(t y(t y (t q dt K 2 (, q, s(t y (t dt], (2.13 τ K 2 (, q, = q ( q q + ( β (r(y q (s(y τ q q ( β y s 1 1 (tdt ( 1/( q dy q. Now, we ly inequlity (2.12 nd (2.13. For comleteness we ly (2.12 with > 1 to obtin [ (+1/ R(, b F ( F (d K 1 (, 1, s((f ( d], (2.14 ( 1 K 1 (, 1, = 1 + ( (R(, b 1 (s( 1 1 1 s 1 1 (tdt d. (2.15 Theorem 2.7. Let > 0, > 1 nd let r, s be nonnegtive mesurble functions on (, b. Then +1 [ (+1/ r( f(tdt d ( + 1 K 1 (, 1, s((f( d], for ll integrble functions f 0; here K 1 (, 1, is defined s in (2.15. Proof. The result follows from (2.4 nd (2.14. The roof of the following theorem cn be obtined by lying inequlity (2.13 nd hence is omitted. Theorem 2.8. Let > 0, > 1 nd let r, s be nonnegtive mesurble functions on (, b. Then ( +1 [ (+1/ r( f(tdt d ( + 1 K 2 (, 1, s((f( d],
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