Erratum to: Cesàro summable difference sequence space

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.

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

$$\begin{array}{rcl}{\parallel x-\{x_{1}e-\ell \overline{e}+\sum _{k=1}^n(x_k-x_1+(k-1)\ell )e_k\}\parallel }_{\mathrm{\Delta }}& =& \underset{k\ge n}{sup}|\frac{1}{k}(x_1-x_{k+1})-\ell |\\ \to & 0\text{as }n\to \mathrm{\infty }\end{array}$$

so that $x=x_{1}e-\ell \overline{e}+{\sum }_k(x_k-x_1+(k-1)\ell )e_k$. If also we had $x=ae+b\overline{e}+{\sum }_k{a}_k{e}_k$, then

$$s_n=(x_1-a)e-(\ell +b)\overline{e}+\sum _{k=1}^n(x_k-x_1+(k-1)\ell -a_k)e_k\to 0\text{as }n\to \mathrm{\infty }.$$

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

$$\begin{array}{rcl}{\parallel x-x^{[n]}\parallel }_{\mathrm{\Delta }}& =& {\parallel (0,0,0,\dots ,n+1,n+2,\dots )\parallel }_{\mathrm{\Delta }}\\ =& \underset{k\ge n}{sup}\left|\frac{0-(k+1)}{k}\right|\\ =& 1+\frac{1}{n}\end{array}$$

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