Soe rears o the paper Soe eleetary iequalities of G. Beett Dag Ah Tua ad Luu Quag Bay Vieta Natioal Uiversity - Haoi Uiversity of Sciece Abstract We give soe couterexaples ad soe rears of soe of the corollaries of Beett s paper, ivolvig the weighted ea atrices o l p. We also discuss Littlewood s proble. 1 Itroductio Let p be fixed ad 1 p. The l p spaces are the spaces of real-valued sequeces x x ) 1 with ors ) 1/p x p x p < with 1 p <, 1 x sup x <. N For 1 < p <, let p p/p 1) deote the cougate of p. A weighted ea atrix A a ),1 is defied by { a /A whe 1, a 0 otherwise, 1) where A a, a 0, a 1 > 0. 1 The atrix A fro l p ito l q has the or A p,q sup Ax q. x p 1 I [1], G. Beett gave soe corollaries that ivolve the boudedess of weighted ea atrices o l p. To show that the coverses of soe of the corollaries do ot hold, we prove soe couterexaples i sectio. We also preset soe propositios which give sufficiet coditios for a weighted ea atrix A ot to be bouded o l p. I [1], whe cosiderig Littlewood s proble, G. Beett gave a exaple to show that there does ot exist a positive costat K for reverse of Littlewood s iequality to be true. I sectio 3, by odifyig the exaple of G. Beett we show that the reverse of Littlewood s iequality is ot true i geeral. 1
Soe rears about weighted ea atrices I paper [1], G. Beett obtaied sufficiet coditios for a weighted ea atrix to be a bouded liear operator fro l p ito l q as show by followig theore. Theore 1 [1]). Let A be the weighted ea atrix give by 1). a) If 1 < p q < the followig coditios are equivalet : i) A aps l p ito l q. ii) For soe costat K 1 ad all 1,,... such that 1 1 ) q K 1 1 iii) For soe costat K ad all 1,,... such that A q ) 1/q 1 ) 1/p iv) For soe costat K 3 ad all 1,,... such that a A q K 3 b) If 1 q < p the A does ot ap l p ito l q. ) q/p. ) K. 3) A q /q. 4) I additio to applyig above theore, G. Beett also gave soe corollaries. Corollary 1 [1]). Let p be fixed ad 1 < p <. If the A is bouded o l p. Corollary [1]). Let p be fixed ad 1 < p <. If 1 + + O ) a 1/p 1) A, 5) the A is bouded o l p. a OA 1 p ), 6) I the followig exaple we show that the coditio 5) is ot ecessary; i.e., A is bouded ad li sup a 1/p 1) A. 7)
Exaple 1. Let a { 1 if, if. The For < +1, we have I : A r0 a a 1/p 1) A r + 1 +, r0 [ rp + p 1 p 1 > 1 +1)p ] +. /p 1) + ) ) /p 1). + As, we get I, ad 7) is true. Next, we chec the boudedess of A. By Corollary we eed oly show that We have So Thus, A p 1 a OA 1 p ). A 1 + + log. a p 1. p p. p p p 1 p 1 p p, ad A is bouded. I the followig propositio, we state a sufficiet coditio for A ot to be bouded. Propositio 1. Let p be fixed ad 1 < p <. If the ad A is ot bouded. li if sup N 1 1 + + a 1/p 1) A +, 8) 1 1 + 9) 3
Proof. Let K > 0. To show that 9) is true we have to fid a N such that ) p Case 1: 1 I : 1 M < +. We have 1 1 > K. A 1 a 1/p 1 ) 1/p. The, Thus, I : 1 1 1 1 1 1 > 1 ) 1/p > ap /p 1. 1 1 I order to have I > K we eed to fid a such that i.e., 1 l M > K; > e KM/ap 1. 1 > ap 1 l M. Hece, we ca choose 0 [e KM/ap 1 ] + 1. The I > K. We have show that 9) is true. I additio, by part ii) of Theore 1, we deduce that A is ot bouded. Case : +. By 8), there exists 0 such that, for all > 0 1 1 > K)1/p a 1 p. Thus, 1 > K. 4
For > 0 we have I : > > 1 0 1 0 1 K 1 1 0 +1 1 1 1. 0 +1 + 0 +1 0 + K 1 1 I order to have I > K we eed to fid a such that K 1 0 +1 > K; i.e., Sice 1 +, the 0 +1 0 +1 > 0 1 as.. 10) Thus we ca always fid a large eough such that 10) holds. Therefore we obtai 9) ad A is ot bouded. We ow provide a exaple i which A is bouded ad li sup a A 1 p +. 11) 5
Exaple. Taig For < +1, we have a { 1 if, if. 1) A a + +1 a r + ) + ) r0 + 1) +, ad r p + p +1 +. r0 Thus + 1) A + log 1) log + log, + 1) A + log log + 1) +. It the follows that A p 1 a ) p 1 1 p 1 p 1) 1 p), p 1 ad 11) is true. To chec that A is bouded, usig Corollary 1, it is oly ecessary to show that Oa 1/p 1) A ). Fro the defiitio of a, Sice a 1/p 1) A p +1 + + 1)/ + log +1 + / log +1 log li +1 0, + 1/. 6
there exists a C > 0 such that Therefore ad A is bouded. If +, the 1 log +1 < C. a 1/p 1) A C + 1, sup N ) 1/p 1 ) 1/p 1 ) 1/p a 1 +. Fro part iii) of Theore 1, A is ot bouded. Now, we have the propositio i the case < +. 1 Propositio. Let p be fixed ad 1 < p <. If the ad A is ot bouded. sup N li if a A 1 p a 1 < + ad +, 13) + 14) Proof. Let K > 0. Fro 13), there exists a 0 such that, for all > 0, Let > 0. The or 0 +1 a a 0 +1 a 0 +1 > K 1/p A 1 p. 7 > K 0 +1 > K.,
This above iequality is equivalet to 14) ad, fro part iv) of the Theore 1, A is ot bouded. Next, we cosider aother exaple to show that the coverse of two above Corollaries is false. Exaple 3. Let For < +1, we get For, 1 if, a if, is odd, if, is eve. A a +. A A 1 i0 1 r0 a i+1 + a r + a i + i1 i + 1) + i0 Siilarly, for + 1, we have A i + i1 1) + + 1 1 1 ) + 3 4 + + 1 1 1 ). 3 4 i + 1) + i0 i + +1 1 i1 + 1) + + 1 1 1 ) + +1 1 3 4 + +1 + 1 1 1 ) 1. 3 4 Thus A { ) + 1 3 1 1 4 + if, ) + 1 3 1 1 4 1 + if +1. 15) 8
It is clear that If is eve A p 1 a > A A > >. 1 1 1 p) p [ 1 ], p 1 ad the left had side teds to ifiity as teds to ifiity. If is odd, the 1 +1 a 1/p 1) +1 A +1 +1 +1)/p 1) + +1 + 1 3 ) +1)p + +1 + 1 3 ) 1 1 The left had side of the above iequality also teds to ifiity as to ifiity. To show the boudedess of A, fro part ii) of the Theore 1, it is sufficiet to show that l l K 1 l Z + for soe costat K 1. For s l < s+1, O other had, The l 1 +1 1 l 1 s 1 s s. 16) +1 r p + +1 + 1 +1 + +1. r0 s+1 1 1 s 0 < +1 s 0 9 + 1) p +1 + +1.
Sice + 1 +1 li 0, the there exists a K > 0 such that, for all, Hece l 1 Fro 16) ad 17), we get l 1 l 1 + 1 +1 < K. s K + < K + s+1. 17) 0 K + )p s+1 s s 4K +, ad hece A is bouded. We have show that the coverses of Corollaries 1 ad are ot true. We ow cosider aother Corollary i the paper of G. Beett [1]. Corollary 3 [1]). Let p be fixed, 1 < p <. If A is bouded o l p, the O ). 18) Exaple 4. I this exaple we show that the coverse of Corollary 3 does ot hold by taig 1 whe 1, a whe + 1 for soe Z + {0}, 0 otherwise. For < +1, the 19) A a 1 + l +1, l0 1 + l0 lp +1)p + p. 0) p 1 10
The so that +1 + +1 +1)p + +1 +1 l+1 +1)p +1 + 1 ) + +1)p +1 + 1 ) + l < l+1 l+1)p l+1 l < l+1 l+1 l l+1)p +1)p +1 + 1 ) + p +1)1 p) 1 1 p +1)p +1 + 1 ) + +1)1 p) [ p +1 + 1 ) p ) + +1] +1)p, 1) p +1 + 1 ) p ) + +1 p ) < p p. The a defied by 19) also satisfy 18). We shall ow show that A is ot bouded. It is sufficiet to show that ) 1/p ) 1/p sup. By 0) ad 1), we have ) 1/p ) 1/p [ +1 + 1 ) p ) + +1] ) 1/p +1)p +1)p + p p p 1 ) 1/p > [ +1 + 1 ) p ) + +1] 1/p +1) > 1/p 8, +1 11
ad A is ot bouded. Returig to Propositio 1 ad Propositio, with a as chose i Exaple 4, we have Hece I ap 1 + + a 1/p 1) A +1)p 1 p 1) /p 1) p 1 I p 1 as. O other had, for a as i Exaple 4, li if 3 Littlewood s proble a A 1 p 0. 1 p 1). p +1 +1 +1)p We cosider the Littlewood s proble that is discussed i [1]; i.e., does there exists a costat K such that a A K a 4 A. ) a 3 1 1 By applyig Holder s iequality ad Hardy s iequality G. Beett showed that, for decreasig sequece a, K. He also showed that there does ot exist a positive costat K such that a 4 A K a A. 3) by taig 1 1 a 3 1 1 0 if > N, a 1 if r for r 1,,.., N, ɛ otherwise, where ɛ N / N. The geeral case of Littlewood s proble is preseted by the followig theore. Theore [1]). Let p, q, r 1. If a ) 1 is a sequece of o-egative ubers with partial su A a 1 + + a, the ) r ) r pq + r) q a p A q a 1+p/q a p p A q ) 1+r/q. 5) 1 We ow cosider the reverse iequality of 5) for p, q, r 1; i.e., does there exist a positive costat K such that ) r a p A q ) 1+r/p K a p A q a 1+p/q? 6) 1 1 1 1 4)
To aswer this questio we cosider the followig cases. If r 1, p 1, q 1 the the left-had side of 6) becoes LHS a A q ) 1+1/q a 1+1/q A q+1, 1 ad the right-had side of 6) becoes RHS a A q 1 Thus, we eed oly to copare A q+1 1 a 1+1/q 1 1 a 1+1/q a A q. 1 ad a A q. For ay q 1, 1 a A q A q+1 q a A q. Thus 6) holds with K q, for p 1, r 1, q 1 ad the iverse of 6) is true for this case. I the other cases we shall show that there does ot exist a positive costat K such that 6) holds by taig a as i 4), ad with ɛ N α / N, for soe α > 1 to be chose later. For r < r+1, r A a s + a r + r)ɛ. s1 Estiatig the left had side of 6, 1 < s 1 LHS N 1 a p A q ) 1+r/q N a p saq + s s1 a 1 < N s p A q ) 1+r/q. LHS > > N [s + s s)ɛ] q+r s1 N [ 1 ɛ) q+r s q+r + sq+r) ɛ q+r] s1 N 1 ɛ) q+r s q+r + ɛ q+r s1 N sq+r) s1 > 1 ɛ) q+r N q+r + ɛ q+r Nq+r) > N αq+r) Sice ɛ N α N ). 13
Cosiderig the right had side of 6), RHS I 1 N 1 N s1 a p A q a p saq s I 1 + ɛ p I. < < N a 1+p/q N a 1+p/q s r r + 1 < N s p aa q N N [s + s s)ɛ] q [ N + 1 + N s )ɛ 1+p/q] r s1 N s + s ɛ) q N + 1 + N ɛ 1+p/q) r s1 N q s q + sq ɛ q ) [ r N + 1) r + Nr ɛ r1+p/q)] s1 < [ q+r N) r + Nr ɛ r1+p/q)] N s q + sq ɛ q ) s1 < [ N q+r N) r + Nr ɛ r1+p/q)] N q + ɛ q s1 N s1 sq ) a 1+p/q < [ q+r N) r + Nr ɛ r1+p/q)] ) N q+1 + ɛ q N+1)q q q 1 < q+r [ N) r + Nr ɛ r1+p/q)] N q+1 + ɛ q N+1)q q 1 ). r [ ) ] N I 1 < q+r r N r + Nr α r1+p/q) [ ] N q+1 + q N αq N [ ) ] N < q+r r N r + Nr α r1+p/q) N q+1 N αq q 1. q 1 Sice ɛ N α N ) Sice N αr1+p/q) li 0, N Np/q N r there exists a N 1 N 1 α, r, p, q) such that, for all N > N 1 ) N Nr α r1+p/q) < r N r. 7) N 14
Therefore We also have I < < < 1 < N s N 1 q A I 1 < q+r+ q 1 N αq+r for N > N 1. 8) N A q s0 s << s+1 N 1 s0 N 1 s0 s A q s+1 N a 1+p/q 1 N a 1+p/q r a 1+p/q r s [ s + 1 + s+1 ɛ ] q [ N + N ɛ 1+p/q] r N 1 s [ q s + 1) q + qs+1) ɛ q] [ r N r + Nr ɛ r1+p/q)] s0 < [ [ ] q+r N r + Nr ɛ r1+p/q)] N 1 N 1 s s + 1) q + ɛ q s qs+1) s0 r < [ q+r N r + Nr ɛ r1+p/q)] N N q + ɛ q Nq+1)) [ ) ] N < q+r N r + Nr α r1+p/q) N N q + N N αq) Sice ɛ N α N ). N Sice 7) ad α > 1, If p > 1, r 1, the I < q+r 1 + r )N r N+1 N αq for N > N 1. 9) li N q+r+1 1 + r ) N αp+αq+r 0, Np 1) so there exists a N N α, r, p, q) > N 1 such that, for all N > N, s0 We ay choose α. If p 1, r > 1, the RHS I 1 + ɛ p I < q+r+3 q 1 N αq+r. ɛ p I < q+r+1 1 + r )N α+αq+r for N > N 1, ad LHS > N αq+r), 15
RHS I 1 + ɛ p I < q+r+ q 1 N αq+r + q+r+1 1 + r )N α+αq+r for N > N 1. Thus, we choose α > r r 1, for istace α r + 1 r 1. Suary, for p r 1, q 1, we obtai the Littlewood s iequality ad reverse Littlewood s iequality. For other cases, we oly get the Littlewood s iequality ad we does ot fid the positive costat K such that the reverse Littlewood s iequality holds. Acowledgets We would lie to express y sicere tha to Haoi Uiversity of Sciece. Moreover, the first author give thafuless to the fud of VIASM. Last, we especially wish to tha Professor B. E. Rhoades. Refereces [1] G. Beett, Soe eleetary iequalities, Quart. J. Math. Oxford Ser. ) 38 1987), 401-45. 16