By Erwin Kreyszig
"Provides avenues for making use of sensible research to the sensible learn of usual sciences in addition to arithmetic. comprises labored difficulties on Hilbert house conception and on Banach areas and emphasizes options, rules, tools and significant functions of sensible analysis."
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Def. 3-3. Then for a given E > 0 there is a l) > 0 such that d(x, xo) < Let Xn ------i> implies l) d(Tx, Txo) < E. Xo. Then there is an N such that for all n > N we have Hence for all n > N, By definition this means that TX n ------i> Txo. Conversely, we assume that implies and prove that then T is continuous at Xo. Suppose this is false. ;' E. = lin there is an Xn satisfying 1 d(xm xo)<- n but Clearly Xn ------i> Xo but (TXn) does not converge to Txo. This contradicts TX n ------i> Txo and proves the theorem.
2, that Furthermore, the series in (5) represents [d(xm, x)]P, so that (5) implies that Xm x. Since (xm) was an arbitrary Cauchy sequence in lV, this proves completeness of IV, where 1 ~ P < +00. 5-5 Completeness of C[ a, b]. The function space C[ a, b] is complete; here [a, b] is any given closed interval on R. (Cf. ) Proof. Let (Xm) be any Cauchy sequence in C[a, b]. Then, given any e > 0, there is an N such that for all m, n> N we have (6) where J = [a, b]. Hence for any fixed t = to E J, (m,n>N).
Completeness Proofs The space c is complete. Proof. 4-7. We consider any x = (~i)E c, the closure of c. 4-6(a) there are Xn = (~~n)) E C such that Xn ~ x. Hence, given any E > 0, there is an N such that for n ~ N and all j we have in particular, for n = N and all j. Since XN E C, its terms ~~N) form a convergent sequence. Such a sequence is Cauchy. Hence there is an Nl such that The triangle inequality now yields for all j, k inequality: ~ Nl the following This shows that the sequence x = (~i) is convergent.
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