By Yehudah Dagan
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Additional info for Khirbet Qeiyafa in the Judean Shephelah: Some Considerations
Prove this inequality. Given that α ∈ ]0, 1[, at T xn = xn+1 , and xn → x. Choose to any ε ∈ R+ an N , such that we for all p ≥ N have d(x, xp ) < ε. If p ≥ N and p ≥ n + 1, then d(x, xn ) ≤ d(x, xp ) + d(xp , xn ) < ε + d(xp , xn ) ≤ ε + d(xp , xp−1 ) + d(xp−1 , xp−2 ) + · · · + d(xn+1 , xn ) = ε + d(T xp−1 , T xp−2 ) + d(T xp−2 , T xp−3 ) + · · · + d(T xn , T xn−1 ) 1 − αp−n · d(xn−1 , xn ) ≤ ε+α· 1−α α · d(xn−1 , xn ). ≤ ε+ 1−α This is true for every ε > 0, thus d(x, xn ) ≤ α · d(xn−1 , xn ).
Since T m is a contraction, the ﬁxpoint is unique, so T x = x, and we have proved that x is a ﬁxpoint for T . Conversely, if x is a ﬁxpoint for T , then x is also a ﬁxpoint for T m , because T x = x implies that T m x = T m−1 (T x) = T m−1 = · · · = T x = x. We have assumed that T m is a contraction, hence the ﬁxpoint for T m is unique. This is true for every ﬁxpoint x for T , hence it must be unique. com 32 1. Topological and metric spaces Topological and Metric Spaces... 23 We consider the metric space Rk with the metric k |xi − yi | d1 (x, y) = i=1 and a mapping T : Rk → Rk given by T x = Cx + b, where C = (cij ) is a k × k matrix and b ∈ Rk .
In particular, f has both a maximum value and a minimum value. Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME Are you about to graduate as an engineer or geoscientist? Or have you already graduated? P. Moller - Maersk. com 64 2. Banach spaces Topological and Metric Spaces... 19 Show that any ﬁnite dimensional subspace of a normed vector space is a Banach space. Let (V, · ) be the normed space, and let U be a ﬁnite dimensional subspace of V .