By Mohamed A. Khamsi

Provides up to date Banach area results.

* beneficial properties an in depth bibliography for out of doors reading.

* presents targeted workouts that elucidate extra introductory fabric.

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**Additional info for An Introduction to Metric Spaces and Fixed Point Theory**

**Example text**

Imip(f(xa)) a Let ε > 0. Then there exists ao 6 / such that a > ao implies r < tp(f(xa)) a, then m a x { d ( i a , i ß ) , c d ( / ( i 0 ) , / ( i ß ) ) } < (/(*))· This proves that x > xa for each a £ I.

Suppose T has Lipschitz constant k e (0,1). Then for each x € M, d(T(x),T2(x)) Adding d(x,T(x)) < kd(x,T(x)). to both sides of the above gives + d(T(x),T2(x)) d(x,T(x)) < d(x,T(x)) + kd(x,T(x)), which can be rewritten as d(x, T(x)) - k d(x, T(x)) < d(x, T(x)) - d(T(x), T2(x)). This, in turn, is equivalent to d{x,T(x)) < (1 - k)-l[d{x,T{x)) - d(T(x),T2(x))}. Now define the function φ : M —♦ R + by setting ψ{χ) = (1 — for x e M. Thus d(x, T{x)) < φ(χ) - φ(Τ(χ)), k)~1d(x,T(x)), χ£Μ. Therefore if x G M and m, n G N with n < m, m d(T"(x),T m + 1 (x)) < Y^d(Ti(x),Ti+\x)) < φ{Τη{χ))-φ{Τ™+\χ)).

But by assumption, x < g(x)—a contradiction. Those familiar with the logical foundations of mathematics might note that the proof of implication (C) => (E) uses the Axiom of Choice, whereas the proof that (E) => (C) does not. In fact, these two theorems are equivalent only if one (*) 56 CHAPTER 3. METRIC CONTRACTION PRINCIPLES assumes (as we do) the Axiom of Choice. (A proof of (C) can be given within the basic axioms of Zermelo-Praenkel set theory [116]. ) We first prove Caristi's theorem using the principle of transfinite induction, and then we prove a more general version of Caristi's theorem using a Zorn's lemma argument.