Abstract :
In this work we prove a theorem which shows that a Cesáro matrix of order α>−1α>−1 is a bounded operator on AkAk, defined below by (2); i.e., (C,α)∈B(Ak)(C,α)∈B(Ak).
Keywords
Absolute summability;
Bounded operator;
Cesáro matrix;
Conservative matrix
Let ∑an∑an be an infinite series with partial sums (sn)(sn), (C,α)(C,α) the Cesáro matrix of order αα. The concept of absolute summability of order kk was introduced and studied by Flett [2]. A series ∑an∑an is summable |C,α|k|C,α|k, k≥1k≥1, α>−1α>−1, if
equation(1)
View the MathML source∑n=1∞nk−1|σn−1α−σnα|k<∞,
Turn MathJax on
where View the MathML sourceσnα denotes the nnth term of the (C,α)(C,α) transform of (sn)(sn).
He also proved the following inclusion theorem. If series ∑an∑an is summable |C,α|k|C,α|k, it is summable |C,β|r|C,β|r for each r≥k≥1r≥k≥1, α>−1α>−1, β>α+1/k−1/rβ>α+1/k−1/r. It then follows that, if one chooses r=kr=k, then a series ∑an∑an which is |C,α|k|C,α|k summable is also |C,β|k|C,β|k summable for k≥1k≥1, β>α>−1β>α>−1.
Let ∑an∑an be a series with partial sums snsn. Define
equation(2)
View the MathML sourceAk≔{(sn)n=0∞:∑n=1∞nk−1|an|k<∞;an=sn−sn−1}.
Turn MathJax on
A matrix TT is said to be a bounded linear operator on AkAk, written T∈B(Ak)T∈B(Ak), if T:Ak→AkT:Ak→Ak.
If one sets α=0α=0 in the inclusion statement involving (C,α)(C,α) and (C,β)(C,β), then one obtains the fact that (C,β)∈B(Ak)(C,β)∈B(Ak) for each β>0β>0.
Let TT be a sequence-to-sequence transformation transforming the sequence (sn)(sn) into (tn)(tn). If, whenever (sn)(sn) converges absolutely, (tn)(tn) converges absolutely, TT is called absolutely conservative. If the absolute convergence of (sn)(sn) implies absolute convergence of (tn)(tn) to the same limit, TT is called absolutely regular.
Das [1] defined a matrix T=(tnv)T=(tnv) to be absolutely kkth-power conservative for k≥1k≥1, if T∈B(Ak)T∈B(Ak); i.e., if (sn)(sn) is a sequence satisfying
View the MathML source∑n=1∞nk−1|sn−sn−1|k<∞,
Turn MathJax on
then
View the MathML source∑n=1∞nk−1|tn−tn−1|k<∞,
Turn MathJax on
where
View the MathML sourcetn=∑v=0∞tnvsv.
Turn MathJax on
He also showed that every conservative Hausdorff matrix H∈B(Ak)H∈B(Ak). We know that if β≥0β≥0, then (C,β)(C,β) is regular and if β<0β<0, then (C,β)(C,β) is neither conservative nor regular.
In this work we extend the result of Flett to the case β>−1β>−1, thus demonstrating that being a conservative matrix is not a necessary condition for a matrix to belong to B(Ak)B(Ak).
