An Introduction to Ultrametric Summability Theory by P.N. Natarajan

By P.N. Natarajan

This is the second one, thoroughly revised and extended variation of the author’s first e-book, masking quite a few new issues and up to date advancements in ultrametric summability idea. Ultrametric research has emerged as a big department of arithmetic lately. This publication provides a quick survey of the study up to now in ultrametric summability concept, that's a fusion of a classical department of arithmetic (summability idea) with a latest department of study (ultrametric analysis). numerous mathematicians have contributed to summability idea in addition to useful research. The ebook will attract either younger researchers and more matured mathematicians who're trying to discover new parts in research. The ebook is additionally priceless as a textual content in case you desire to specialise in ultrametric summability theory.

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Extra info for An Introduction to Ultrametric Summability Theory

Example text

Though not relevant to the present monograph, it is worth noting that the concept of derivative and its properties have been studied in ultrametric analysis (see [2]). With regard to derivatives, we need the following definition. 1 If U is any subset of an ultrametric field K without isolated points and f : U → K , we say that f is differentiable at x ∈ U if lim y→0 f (x + y) − f (x) exists. 5) exists, it is called the derivative of f at x, denoted by f (x). , any set which is both open and closed) is differentiable everywhere with χU = 0 everywhere.

And so we can find a positive integer N such that |ank − δk | < M , n > N , k = 0, 1, 2, . . , k0 , M = sup |xk − s|. Thus k≥0 k0 (ank − δk )(xk − s) < k=0 M . M = , n > N. 32 4 Ultrametric Summability Theory Also ∞ (ank − δk )(xk − s) ≤ sup |ank − δk | |xk − s| k>k0 k=k0 +1 N , k=0 from which it follows that ∞ (ank − δk )(xk − s) = 0. 3). 3) are clear by considering the sequences {0, 0, . . , 0, 1, 0, . . }, 1 occurring in the kth place, k = 0, 1, 2, .

N≥1 k∈S n,k≥1 Proof Necessity. Let ϕ be ultrametric full. Let (ank ) be an infinite matrix for which sup ank < ∞ for every S ∈ ϕ. Let S ∈ ϕ and k0 ∈ S. Since ϕ is hereditary, n≥1 k∈S S = S\{k0 } ∈ ϕ. e. sup |ank0 | < ∞, n≥1 for every k0 ∈ S and so sup sup |ank | < ∞ for every S ∈ ϕ. 5 that sup |ank | < ∞. n,k≥1 Sufficiency. Let (ank ) be an infinite matrix such that sup n≥1 every S ∈ ϕ. 2 Steinhaus-Type Theorems 41 ank ≤ sup sup |ank | sup n≥1 k∈S n≥1 k∈S <∞ for every S ∈ ϕ. 5, completing the proof.

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