Research Inventy: International Journal Of Engineering And ScienceVol.2, Issue 9 (April 2013), Pp 19-23Issn(e): 2278-4721, Issn(p):2319-6483, Www.Researchinventy.Com19ON SOME ASPECTS OF u -IDEALS DETERMINED BY 1AND 0cJohn Wanyonyi Matuya1, Achiles Simiyu2and Shem aywa21Department of Mathematics, Maasai Mara University, P.O Box 861-20500, Narok , Kenya2Department of Mathematics, Masinde Muliro University of science and Technology,P.O BOX 190-50100, Kakamega, Kenya.Abstract: The M -ideals defined on a real Banach space are called u -ideals. The u -ideals containingisomorphic copies of 1 are not strict u -ideals. In this paper we show that u -ideals with unconditionalbasis nx which is shrinking has no isomorphic copy of 1 and thus a strict u -ideal. Finally we show thatu -ideals with unconditional basis nx which is boundedly complete is not homeomorphic to copies of 0cimplying that they are weakclosed in their biduals X.Key Words: u -ideals, strict u -ideals, Shrinking, boundedly complete2010 Mathematics Subject Classification: 47B10; 46B10; 46A25 This work was supported by NationalCouncil of Science and Technology (NCST), Kenya.I. INTRODUCTIONA sequence 1n ixis called a basis of a normed space X if for every x X there exists a uniqueseries1i iia x that converges to x . The basis 1n ixfor a Banach space X is an unconditional basis if foreach x X there exists a unique expansion of the form1n nnx a x where the sum convergesunconditionally. The basis nx is said to be boundedly complete whenever given a sequence na of scalarsfor which 1: 1nk kka x n is bounded, then 1limnn k Kka x exists. If nx X is not boundedlycomplete then X contains an isomorphic copy of 0c .Let y X which belongs to X . We say that nx is shrinking,1 1, , . , , .n ni i i ii in nu y u x y u x u converges to ,y u for every u X .The notion of u -ideals was introduced and studied thoroughly by Godfrey et al . They generalized M -ideals defined on a real Banach space. In their paper on unconditional Ideals they established that u -idealscontaining copies of 1 are not strict u -ideals. A Banach space X is said to be a strict u -ideal in its bidualwhen the canonical decomposition XXX is unconditional. In other words for X to be a strictu -ideal the u -complement of Xmust be norming, that is, the range V of the induced projection on Xis a norming subspace of X. Vegard and Asvald 3 characterized Banach spaces which are strict u -ideals in their biduals and showed that X is a strict u -ideal in a Banach space Y if it contains 0c . Matuya etal  using the approximation properties, hermitian conditions, isometry studied properties of u -ideals and
On Some Aspects Of u -Ideals Determined….20their characterization. They showed that u -ideals containing no copies of sequence space 1 are strict u -ideals. In this paper we show that Banach spaces with unconditional basis nx that is shrinking is nothomeomorphic to copies of 1 and so it a strict u -ideal. We also show that the u -ideals with unconditionalbasis which are boundedly complete are not bicontinuous to 0c meaning their u -complement is weakclosed.II. RESULTS ON STRICT u -IDEALS2.1 PropositionLet X be u -ideal with unconditional basis nx . The following statements are equivalent: The sequence nx is shrinking X contains no copy of 1 X is a strict u -idealProof: i ii it is clear from the definition that if a sequence nx is shrinking then X contains noisomorphic copy of 1 . In fact proving by contradiction it suffices to show that X contains a copy of 1 . Let nc be a sequence of coefficient functional associated with nx . Our hypothesis applies that, for somec X , the series1, .n nnc x c does not converge in , ,X X X implying that1, .kn nn kc x c cannot be Cauchy in , ,X X X . We therefore find a bounded set 1L andfor each k we define maps , :P Q L Xby 1, .n nnP c x c and 1, .kn nn kQ c x c which implies together with 2L P Q that 2L Q L P holds. P is continuous since nx is bounded, so that P maps Lhomeomorphically into X . Now Lis dense in 1 . Since 1 is weakly sequentially complete the same is truefor the subspace 1P of X . In particular nc has a weak limit in X and so does nx because of, lim , ,k n knx c x c k , this limit has to be x and this is the desired contradiction.ii iii Considering the map P it is clear from the definition that if x X and P x x then x Uwhere U p x . Now for each X we consider the set |F x X P x x thenthere is a net d Xx B converging in the weak-topology of Xto u . However, u x X so thatdxconverges to x. Thus lim ddu x x so that u x x . Now P x ux so we conclude that x F . Hence Fis norming.iii i Suppose X is strict u -ideal. We proceed to show that nx is shrinking. Let y X whichbelongs to X . Since nx is shrinking,1 1, , . , , .n ni i i ii in nu y u x y u x u
On Some Aspects Of u -Ideals Determined….21converges to ,y u for every u X . We conclude from this that1, .ni iiy u x is bounded in X so ithas a limit sayx .1, , . , ,i iix u y u x u y u , for all u X ,we get y x X .2.2 Proposition Let X be u -ideal with unconditional basis nx . Show that if X contains no copy of 0cthen nx is boundedly complete nxis weakclosed in XSuppose nx is not boundedly complete. Then there exists iT with1ni ii nx bounded but notCauchy in X . Let now e and let andi im n be increasing sequences in such that1i i im n m andii j j ema x u , that is, 1,ie a i . Here we set , ,.....,i i i k iM m m n . Choose0i eu U such that , 1i iu a , i and let f besuch that ,, , andMe e x f x x X M .By construction of the sets iM , it follows thatii i i gi i j Mg a g x e ,Whence , i giv a e0, gg v U .There exists a constant 0c such that, .i giv a c e0, gg v U .Consider now T as a subspace of 0c and define: T X by n nna .Then is linear and continuous, because of n nng g a 0lub , . .gn n gUnv a c e , ,g T . is also open , because given T and j such that j , then we can choose such that
On Some Aspects Of u -Ideals Determined….22, ,i i j i i j iu a u a for all i .Therefore ,j n j nnu a ,j n n n n n nn nv a e a f .Thus is bicontinuous between T and X . This implies that T is dense in 0c and since X is sequenciallycomplete, extends a linear homeomorphic embedding 0c into X which is a contradiction.We shall show that when an dn nx X x X satisfy1 n nnx x and 10n nnx s x for ,s F X X ,then 10n nnx x .We may assume that0nx and1nM x .Let 0 and choose N such that4n Nx . Since X has the weak approximating sequence,for ˆˆ ,LXK X Z , there exists a net ˆ ˆ,s F X X such that lub 1s and L s x k L x k L k x .Therefore s x x weakfor all x L k . In particular s x x for all x F . Since1,s we also have 1,s L if 1x then x L k for some k K . Thus 1s x , so s x is weakconvergent. Hence there is some ssuch that2n nx s xM , 1,....,n NNow we have 1 1n n n nn nx x x s x 1Nn n nnx x s x n n nn Nx L k s L k 2n nn Nx L s L k
On Some Aspects Of u -Ideals Determined….232 2 Hence 10n nnx x III. CONCLUSIONWe have shown that u -ideals can be characterized using sequence spaces. In particular we consideredthe sequence spaces 1 and 0c . The u -ideals containing no isomorphic copies of 1 are strict u -idealswhereas those that are not homeomorphic to the copies of 0c their u -complement is weak-closedIV. ACKNOWLEDGEMENTThe authors convey their appreciation to the National Council of Science and Technology (NCST),Kenya for funding this research work and Masinde Muliro University of science and Technology for theirtechnical support.REFERENCES Diestel. J. “ Sequences and Series in Banach spaces’’. Springer-Verlag Newyork , 1984. Godefrey, G. Kalton, J. and Saphar,P. “ Unconditional ideals in Banach spaces ’’ .studiaMathematica.104 No.1,(1993) 13-59. Lima, V. Lima, A. “ Strict u -ideals in Banach spaces ’’.studiaMath.3 No.195 (2009) 275-285. Matuya . J. W. et al “On characterization of u -ideals determined by sequences’’ A.J.Curr.Eng and Math.1( 2012) 247-250.