Multivariable Advanced Calculus

Sequences4.1 Vector Valued Sequences And Their LimitsFunctions defined on the set of integers larger than a given integer which have valuesin a vector space are called vector valued sequences. I will always assume the vectorspace is a normed vector space. Actually, it will specialized even more to F n , althougheverything can be done for an arbitrary vector space and when it creates no difficulties,I will state certain definitions and easy theorems in the more general context and usethe symbol ||·|| to refer to the norm. Other than this, the notation is almost the same asit was when the sequences had values in C. The main difference is that certain variablesare placed in bold face to indicate they are vectors. Even this is not really necessarybut it is conventional to do it.The concept of subsequence is also the same as it was forsequences of numbers. To review,Definition 4.1.1 Let {a n } be a sequence and let n 1 < n 2 < n 3 , · · · be anystrictly increasing list of integers such that n 1 is at least as large as the first number inthe domain of the function. Then if b k ≡ a nk , {b k } is called a subsequence of {a n } .Example 4.1.2 Let a n = ( n + 1, sin ( 1n)). Then {an } ∞ n=1is a vector valued sequence.The definition of a limit of a vector valued sequence is given next. It is just like thedefinition given for sequences of scalars. However, here the symbol |·| refers to the usualnorm in F n . In a general normed vector space, it will be denoted by ||·|| .Definition 4.1.3 A vector valued sequence {a n } ∞ n=1converges to a in a normedvector space V, written aslim a n = a or a n → an→∞if and only if for every ε > 0 there exists n ε such that whenever n ≥ n ε ,||a n − a|| < ε.In words the definition says that given any measure of closeness ε, the terms of thesequence are eventually this close to a. Here, the word “eventually” refers to n beingsufficiently large.Theorem 4.1.4 If lim n→∞ a n = a and lim n→∞ a n = a 1 then a 1 = a.Proof: Suppose a 1 ≠ a. Then let 0 < ε < ||a 1 − a|| /2 in the definition of the limit.It follows there exists n ε such that if n ≥ n ε , then ||a n − a|| < ε and |a n − a 1 | < ε.Therefore, for such n,a contradiction.||a 1 − a|| ≤ ||a 1 − a n || + ||a n − a||< ε + ε < ||a 1 − a|| /2 + ||a 1 − a|| /2 = ||a 1 − a|| ,71