Open Access

Erratum to: Cesàro summable difference sequence space

Journal of Inequalities and Applications20142014:11

https://doi.org/10.1186/1029-242X-2014-11

Received: 18 December 2013

Accepted: 18 December 2013

Published: 9 January 2014

The original article was published in Journal of Inequalities and Applications 2013 2013:315

Abstract

Theorem 3.7 of Bhardwaj and Gupta, Cesàro summable difference sequence space, J. Inequal. Appl. 2013:315, 2013, is incorrect as it stands. The corrected version of this theorem is given here.

MSC:40C05, 40A05, 46A45.

Keywords

sequence space AK property Schauder basis

In [1], Bhardwaj and Gupta have introduced the Cesàro summable difference sequence space C 1 ( Δ ) as the set of all complex sequences x = ( x k ) with ( x k x k + 1 ) C 1 , where C 1 is the linear space of all ( C , 1 ) summable sequences.

Unfortunately, Theorem 3.7 of [1] is incorrect, as it stands. Consequently the assertions of Corollaries 3.8 and 3.9 of [1] remain actually open. The corrected version of Theorem 3.7 of [1] is obtained here as Corollary 2 to Theorem 1, which is itself a negation of Corollary 3.8 of [1]. Finally Corollary 3.9 of [1] is proved as Theorem 3.

It is well known that C 1 is separable (see, for example, Theorem 4 of [2]). In view of the fact [[3], Theorem 3] that ‘if a normed space X is separable, then so is X ( Δ ) ’, it follows that Theorem 3.7 of [1] is untrue. The mistake lies in the third line of the proof where it is claimed that A is uncountable. In fact, A is countable.

The following theorem provides a Schauder basis for C 1 ( Δ ) and hence negates Corollary 3.8 of [1].

Theorem 1 C 1 ( Δ ) has Schauder basis namely { e ¯ , e , e 1 , e 2 , } , where e ¯ = ( 0 , 1 , 2 , 3 , ) , e = ( 1 , 1 , 1 , ) and e k = ( 0 , 0 , 0 , , 1 , 0 , 0 , ) , 1 is in the kth place and 0 elsewhere for k = 1 , 2 ,  .

Proof Let x = ( x k ) C 1 ( Δ ) with 1 k i = 1 k Δ x i , i.e., lim k 1 k ( x 1 x k + 1 ) = . We have
x { x 1 e e ¯ + k = 1 n ( x k x 1 + ( k 1 ) ) e k } Δ = sup k n | 1 k ( x 1 x k + 1 ) | 0 as  n
so that x = x 1 e e ¯ + k ( x k x 1 + ( k 1 ) ) e k . If also we had x = a e + b e ¯ + k a k e k , then
s n = ( x 1 a ) e ( + b ) e ¯ + k = 1 n ( x k x 1 + ( k 1 ) a k ) e k 0 as  n .

But for all n N , | x 1 a a 1 | s n Δ , | k b x k + 1 + x 1 + a k + 1 a 1 k | s n Δ for 1 k n 1 and | a 1 + k ( + b ) k | s n Δ for all k n . Letting n , we see that x 1 = a , b = , a 1 = 0 and a k + 1 = x k + 1 k b x 1 + a 1 = k x 1 + x k + 1 , for k 1 , so that the representation x = x 1 e e ¯ + k ( x k x 1 + ( k 1 ) ) e k is unique. □

The following is a correction of Theorem 3.7 of [1].

Corollary 2 C 1 ( Δ ) is separable.

The result follows from the fact that if a normed space has a Schauder basis, then it is separable.

Finally, we prove a theorem which is in fact Corollary 3.9 of [1].

Theorem 3 C 1 ( Δ ) does not have the AK property.

Proof Let x = ( x k ) = ( 1 , 2 , 3 , ) C 1 ( Δ ) . Consider the n th section of the sequence ( x k ) written as x [ n ] = ( 1 , 2 , 3 , , n , 0 , 0 , ) . Then
x x [ n ] Δ = ( 0 , 0 , 0 , , n + 1 , n + 2 , ) Δ = sup k n | 0 ( k + 1 ) k | = 1 + 1 n

which does not tend to 0 as n . □

Notes

Authors’ Affiliations

(1)
Department of Mathematics, Kurukshetra University
(2)
Department of Mathematics, Arya P. G. College

References

  1. Bhardwaj VK, Gupta S: Cesàro summable difference sequence space. J. Inequal. Appl. 2013., 2013: Article ID 315Google Scholar
  2. Bennet G: A representation theorem for summability domains. Proc. Lond. Math. Soc. 1972, 24: 193-203.View ArticleGoogle Scholar
  3. Çolak R: On some generalized sequence spaces. Commun. Fac. Sci. Univ. Ank. Ser. 1989, 38: 35-46.Google Scholar

Copyright

© Bhardwaj and Gupta; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.