# Scalar and Asymptotic Scalar Derivatives: Theory and by George Isac

By George Isac

This tremendous worthwhile booklet is dedicated to the learn of scalar and asymptotic scalar derivatives and their purposes to a couple difficulties in nonlinear research, Riemannian geometry and utilized arithmetic. The theoretical effects are constructed specifically with admire to the learn of complementarity difficulties, monotonicity of nonlinear mappings and the non-gradient sort monotonicity on Riemannian manifolds. The textual content is meant for researchers and graduate scholars operating within the fields of nonlinear research, Riemannian geometry and utilized arithmetic.

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Example text

Krasnoselskii’s work and the reader is referred to Krasnoselskii [1964a,b] and Krasnoselskii and Zabreiko [1984]. We note that the main goal of this chapter is to present the notion of the asymptotic scalar derivative and some of its applications. This chapter may be a stimulus for new research in this subject. 1 Asymptotic Differentiability in Banach Spaces Let (E, · ) and (F, · ) be Banach spaces. Let L(E, F ) be the Banach space of linear continuous mappings, where the norm is L = sup x =1 L(x) , for any L ∈ L(E, F ).

We present some results from Hyers–Ulam stability theory related to the asymptotic differentiability. In 1978, a generalized solution to Ulam’s problem for approximately linear mappings was given by Th. M. Rassias (see [Rassias, 1978]). 2 Hyers–Ulam Stability and Asymptotic Derivatives and (F, · ) be Banach spaces. He considered a mapping f : E → F satisfying the condition of continuity of f (tx) in t for each fixed x and such that f (x + y) − f (x) − f (y) ≤ θ( x p + y p ), for any x, y ∈ E and that T : E → F is the unique linear mapping satisfying f (x) − T (x) ≤ 2θ x p.

Proof. 2) for any positive integer n, and for any x ∈ E. 2) follows by induction on n. For n = 1 by ψ-additivity of f we have f (2x) − 2f (x) ≤ 2θψ( x ), 36 2 Asymptotic Derivatives and Asymptotic Scalar Derivatives which implies f (2x) − f (x) ≤ θψ( x ). 2) holds for n and we want to prove it for the case n + 1. 2) we obtain n−1 f (2n 2x) − f (2x) ≤ 2n θ m=0 m ψ(2) 2 ψ(2 x ). Because ψ(2 x ) ≤ ψ(2)ψ( x ) we get n−1 f (2n+1 x) − f (2x) ≤ 2n θ m=0 m ψ(2) 2 ψ(2)ψ( x ). 3) by 1/2 we obtain n f (2n+1 x) f (2x) − ≤ 2n+1 2 ψ(2) 2 θ m=1 m ψ( x ).