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2011 DIFFERENTIATION ON VILENKIN GROUPS USING A MATRIX
Medo Pepić
Taiwanese J. Math. 15(6): 2387-2402 (2011). DOI: 10.11650/twjm/1500406477

Abstract

Given a Vilenkin group G, a scalar matrix $\Lambda = [\lambda_{ij}]_{i \in \mathbb{N}, j \in \mathbb{N}_0}$, a function $f \in L^1(G)$, and a point $x \in G$ we introduce, for each $\alpha \in \mathbb{R}$, the $(\Lambda,\alpha)$ − derivative $f$ at $x$ denoted by $f^{(\Lambda,\alpha)}(x)$. We also introduce the sets: $$ M_\alpha = M(G,\Lambda,\alpha,x) := \big\{ f \in L^1(G):\exists f^{(\Lambda,\alpha)}(x) \big\}, $$ $$ M = M(G,\Lambda,x) := \{f \in L^1(G):\exists f^\Lambda(x)\}; $$ where $f^\Lambda(x)$ derivative in [8], which is a generalization of Onneweer’s derivative $f^{[1]}(x)$ in [6]. We proved:

(a) Five theorems which express essential characteristics of $(\Lambda,\alpha)$− derivative,

(b) $M = M_0$,

(c) $(\forall \alpha,\beta \in \mathbb{R}) \wedge (\alpha \lt \beta) \Rightarrow (M_\alpha \subseteq M_\beta) \wedge (M_\beta \setminus M_\alpha \not= \theta)$.

Statement b) states that the method $(\Lambda,\alpha)$ − differentiation, for $\alpha = 0$, is equal to $\Lambda$ − differentiation and statement c) says that $(\Lambda,\alpha)$− differentiation increases with increasing $\alpha \in \mathbb{R}$.

Citation

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Medo Pepić. "DIFFERENTIATION ON VILENKIN GROUPS USING A MATRIX." Taiwanese J. Math. 15 (6) 2387 - 2402, 2011. https://doi.org/10.11650/twjm/1500406477

Information

Published: 2011
First available in Project Euclid: 18 July 2017

zbMATH: 1253.43005
MathSciNet: MR2896124
Digital Object Identifier: 10.11650/twjm/1500406477

Subjects:
Primary: 43A75

Keywords: differentiation of functions , Vilenkin group

Rights: Copyright © 2011 The Mathematical Society of the Republic of China

Vol.15 • No. 6 • 2011
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