By P.N. Natarajan

Ultrametric research has emerged as a huge department of arithmetic lately. This publication provides, for the 1st time, a short survey of the examine to this point in ultrametric summability idea, that is a fusion of a classical department of arithmetic (summability thought) with a latest department of study (ultrametric analysis). numerous mathematicians have contributed to summability thought in addition to sensible research. The publication will attract either younger researchers and more matured mathematicians who're seeking to discover new parts in analysis.

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Extra resources for An Introduction to Ultrametric Summability Theory

Example text

Since x diverges, p→∞ |xk+1 − xk | → 0, k → ∞ so that there exists ε {k( j)} of positive integers such that > 0 and an increasing sequence |xk( j)+1 − xk( j) | > ε , j = 1, 2, . . 24) We may assume that k( j + 1) − k( j) > 1, j = 1, 2, . . 25) where M = sup |xk |. Before proving the claim, we show that if A is not a Schur k≥0 matrix, then x is necessarily bounded under the hypothesis of the theorem. Suppose x is unbounded. We consider two cases: Case (i). If A is such that ank = 0 for some n and k = k(i), i = 1, 2, .

The lack of ordering on K makes it difficult to find an analogue for classical “convexity”. Classical convexity is replaced, in the ultrametric setting, by a notion called “K convexity” (see [3]), which is defined as follows. 1 A set S of vectors is said to be “absolutely K -convex” if ax + by ∈ S whenever |a|, |b| ≤ 1 and x, y ∈ S; translates w + S of such sets S are called “K convex”. A topological linear space X (which is defined as in the classical case) is said to be “locally K -convex” if its topology has a base of K -convex sets at 0.

We claim that the set of spheres {C x (Ax)/x ∈ M} is a nest. , ||Ax1 − Ax2 || ≤ max{ x1 , x2 }. It now follows that Ax1 ∈ C 2 (Ax2 ) or Ax2 ∈ C 1 (Ax1 ). Consequently, the spheres {C x (Ax)/x ∈ M} form a nest. Since Y is spherically complete, there exists z 0 ∈ ¯ x∈M C x (Ax). Define an extension A of A to [M, x 0 ] by ¯ + λx0 ) = Ax + λz 0 . A(x It is clear that A¯ is a linear extension of A. Now, ¯ + λx0 )|| = ||Ax + λz 0 || || A(x = |λ|||A(λ−1 x) + z 0 || = |λ|||A(−λ−1 x) − z 0 || ≤ |λ| −λ−1 x = |λ|||A|||| − λ−1 x − x0 || = ||A||||x + λx0 ||.

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