# Erratum to: Cesàro summable difference sequence space

- Vinod K Bhardwaj
^{1}Email author and - Sandeep Gupta
^{2}

**2014**:11

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

© Bhardwaj and Gupta; licensee Springer. 2014

**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

In [1], Bhardwaj and Gupta have introduced the Cesàro summable difference sequence space ${C}_{1}(\mathrm{\Delta})$ as the set of all complex sequences $x=({x}_{k})$ with $({x}_{k}-{x}_{k+1})\in {C}_{1}$, 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 $X(\mathrm{\Delta})$’, 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 ${C}_{1}(\mathrm{\Delta})$ and hence negates Corollary 3.8 of [1].

**Theorem 1** ${C}_{1}(\mathrm{\Delta})$ *has Schauder basis namely* $\{\overline{e},e,{e}_{1},{e}_{2},\dots \}$, *where* $\overline{e}=(0,1,2,3,\dots )$, $e=(1,1,1,\dots )$ *and* ${e}_{k}=(0,0,0,\dots ,1,0,0,\dots )$, 1 *is in the* *kth place and* 0 *elsewhere for* $k=1,2,\dots $ .

*Proof*Let $x=({x}_{k})\in {C}_{1}(\mathrm{\Delta})$ with $\frac{1}{k}{\sum}_{i=1}^{k}\mathrm{\Delta}{x}_{i}\to \ell $,

*i.e.*, ${lim}_{k}\frac{1}{k}({x}_{1}-{x}_{k+1})=\ell $. We have

But for all $n\in \mathbb{N}$, $|{x}_{1}-a-{a}_{1}|\le {\parallel {s}_{n}\parallel}_{\mathrm{\Delta}}$, $|\frac{kb-{x}_{k+1}+{x}_{1}+{a}_{k+1}-{a}_{1}}{k}|\le {\parallel {s}_{n}\parallel}_{\mathrm{\Delta}}$ for $1\le k\le n-1$ and $|\frac{-{a}_{1}+k(\ell +b)}{k}|\le {\parallel {s}_{n}\parallel}_{\mathrm{\Delta}}$ for all $k\ge n$. Letting $n\to \mathrm{\infty}$, we see that ${x}_{1}=a$, $b=-\ell $, ${a}_{1}=0$ and ${a}_{k+1}={x}_{k+1}-kb-{x}_{1}+{a}_{1}=k\ell -{x}_{1}+{x}_{k+1}$, for $k\ge 1$, so that the representation $x={x}_{1}e-\ell \overline{e}+{\sum}_{k}({x}_{k}-{x}_{1}+(k-1)\ell ){e}_{k}$ is unique. □

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

**Corollary 2** ${C}_{1}(\mathrm{\Delta})$ *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** ${C}_{1}(\mathrm{\Delta})$ *does not have the AK property*.

*Proof*Let $x=({x}_{k})=(1,2,3,\dots )\in {C}_{1}(\mathrm{\Delta})$. Consider the

*n*th section of the sequence $({x}_{k})$ written as ${x}^{[n]}=(1,2,3,\dots ,n,0,0,\dots )$. Then

which does not tend to 0 as $n\to \mathrm{\infty}$. □

## Notes

## Authors’ Affiliations

## References

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

## Copyright

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