Get An Introduction to Ultrametric Summability Theory PDF

By P.N. Natarajan

Ultrametric research has emerged as a tremendous department of arithmetic lately. This ebook provides, for the 1st time, a quick survey of the examine up to now in ultrametric summability thought, that is a fusion of a classical department of arithmetic (summability concept) with a contemporary department of study (ultrametric analysis). a number of mathematicians have contributed to summability thought in addition to sensible research. The publication will attract either younger researchers and more matured mathematicians who're trying to discover new parts in analysis.

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127) obtained the same theorem using the Banach-Steinhaus theorem. Later to Monna, we have only the papers by Rangachari and Srinivasan [4], Srinivasan [5] and Somasundaram [6, 7] till Natarajan took up a detailed study of matrix transformations, special methods of summability and other aspects of summability theory in the ultrametric set up. In the sequel, K is a complete, non-trivially valued, ultrametric field unless otherwise stated. We shall suppose that the entries of sequences, series and infinite matrices are in K .

If y ∈ K (A(x)), for any z ∈ K , |y − z| ≤ lim |z − (Ax)n |. , y ∈ C α lim |z − xk | (z) for any z ∈ K , k→∞ which implies that K (A(x)) ⊆ K (α) (x). Conversely, let K (A(x)) ⊆ K (α) (x). Then it is clear that A is regular by considering convergent sequences for which K (α) (x) = lim xn . 22). Let, if possible, ∞ |ank | > α. lim n→∞ Then k=0 ∞ |ank | = α + h, for some h > 0. lim n→∞ k=0 46 4 Ultrametric Summability Theory Using the hypothesis and the fact that A is regular, we can now choose two strictly increasing sequences {n(i)} and {k(n(i))} of positive integers such that k(n(i−1)) h , 8 |an(i),k | < k=0 k(n(i)) |an(i),k | > α + k=k(n(i−1))+1 and ∞ |an(i),k | < k=k(n(i))+1 h 4 h .

N + r − 1 , n = 0, 1, 2, . . , = 0, otherwise r being a fixed positive integer. (xn+1 + xn+2 + · · · + xn+r ) − (xn + xn+1 + · · · + xn+r −1 ) r (xn+r − xn ) → 0, n → ∞, = r (Ax)n+1 − (Ax)n = if x = {xk } ∈ r . Thus, A sums all sequences in r . It is clear that A is regular. Consequently (c, c; P) ∩ ( r , c) = φ, proving our claim. We note that (c, c; P) ∩ ( r , c) = φ when K = R or C. Since ( ∞ , c) ⊆ ( r , c), it follows that (c, c; P) ∩ ( ∞ , c) = φ, which is Steinhaus theorem. We call such results Steinhaus type theorems.

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