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Erratum to: Cesàro summable difference sequence space

  • 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.

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 +(k1)) e k . If also we had x=ae+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 0as n.

But for all nN, | 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 1kn1 and | a 1 + k ( + b ) k | s n Δ for all kn. Letting n, we see that x 1 =a, b=, a 1 =0 and a k + 1 = x k + 1 kb x 1 + a 1 =k x 1 + x k + 1 , for k1, so that the representation x= x 1 e e ¯ + k ( x k x 1 +(k1)) 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. □

References

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    Bhardwaj VK, Gupta S: Cesàro summable difference sequence space. J. Inequal. Appl. 2013., 2013: Article ID 315

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Correspondence to Vinod K Bhardwaj.

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The online version of the original article can be found at 10.1186/1029-242X-2013-315

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Bhardwaj, V.K., Gupta, S. Erratum to: Cesàro summable difference sequence space. J Inequal Appl 2014, 11 (2014). https://doi.org/10.1186/1029-242X-2014-11

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Keywords

  • sequence space
  • AK property
  • Schauder basis