※ 引述《PttFund (批踢踢基金)》之銘言： : Assume that f has a finite derivative in (a,b) and is continuous : on [a,b]. If |f'(x)| ≦ r < 1 for all x in (a,b). Show that f : has an unique fixed point in [a,b]. 這一題是很經典的題目 @ @" Proof: By Mean-Value Theorem, for any x≠y in [a,b], f(x) - f(y) = f'(ξ)( x - y ), where ξ is between x and y. Since |f'(x)| ≦ r < 1 for all x in (a,b), we have |f(x) - f(y)| ≦ r |x - y|. Take x_0 in [a,b] arbitrarily, and define {x_n} recursively, by setting x_{n+1} = f(x_n) (n = 0, 1, 2, ...). Hence induction gives |x_{n+1} - x_n| ≦ r^n|x_1 - x_0| (n = 0, 1, 2, ...). If n < m, it follows that m |x_n - x_m| ≦ Σ |x_i - x_{i-1}| i=n+1 ≦ (r^n + r^{n+1} + ... + r^{m-1})|x_1 - x_0| ≦ [(1-r)^{-1} |x_1 - x_0|] r^n. Thus {x_n} is a Cauchy sequence on [a,b]. Since [a,b] is complete, lim x_n = x for some x in [a,b]. Since f is continuous on [a,b], f(x) = f( lim x_n) = lim f(x_n) = lim x_{n+1} = x, n→∞ n→∞ n→∞ that is, x is a fixed point. To show the uniqueness, assume that y is another fixed point, then |x - y| = |f(x) - f(y)| ≦ r |x - y|, where r < 1. Thus |x - y| = 0, that is, x = y. --

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