Scharze space asked in 科學數學 · 1 decade ago

高微[固定點]

Let f :|R->|R be monotonically increasing (may be discontinuous),

Suppose 0<f(0),f(50)<50

試證 f has a fixed point

2 Answers

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  • 蜉蝣
    Lv 6
    1 decade ago
    Favorite Answer

    consider a= sup{x|x=

    2008-11-05 03:58:47 補充:

    Let f :|R->|R be monotonically increasing (may be discontinuous),

    Suppose 0<f(0),f(50)<50 ,試證 f has a fixed point

    ------------------------------

    幾何圖形是很直覺的,但還是用代數及定義推論它

    Let A={x|x=<f(x),x∈[0,50]}. Since 0<f(0),hence A is not an empty set.

    And A has an upper bound 50, then supA does exist and supA=<50.

    Let a be supA. We will show that f(a)=a.

    (1)If f(a)<a,then by the definition of a, ∃b s.t. f(a)<b<a and b∈ A(i.e.b=<f(b))

    f is monotonically increasing, b<a =>f(b)=<f(a)<b ,a contradiction.

    (2)If f(a)>a. By the definition of a, we have0=< a=<50.

    f is monotonically increasing ,then 0<f(0)=<f(a)=<f(50)<50,i.e. f(a)∈[0,50]

    Consider b=[a+f(a)]/2, then a<b<f(a)

    f is monotonically increasing, a<b =>f(a)=<f(b)<f(f(a)),i.e.f(a)∈ A

    since f(a)>a and f(a)∈ A ,it contradicts the definition of a.

    By (1)(2),we obtain f(a)=a. Hence f has a fixed point.

    2008-11-05 04:15:21 補充:

    (2)Consider b這行多此一舉,可以刪掉(但也沒錯啦)

    1號大大,好像漏證了某一方,雖然很明顯(其實我就是為了想這個,一開始繞錯方向,繞半天才繞出來)

    2008-11-05 04:24:22 補充:

    由定義應尚不足以判定u(我的a)屬於X(我的A),仍是需加上sup與increasing條件來說明

    雖然圖形很直觀,但若可以直觀,那整題不就用直觀解釋就好了

    拍謝,純是個人看法

    2008-11-05 04:29:56 補充:

    且一開始的0<50也下得太快(雖然此題是對的情形),

    即使50不屬於X(A),但也不表示上界不可能為50

    同理0雖屬於X(A),但也不表示,上界一定大於0

    2008-11-05 05:18:23 補充:

    且一開始的0<50也下得太快(雖然此題是對的情形),

    即使50不屬於X(A),但也不表示最小上界不可能為50

    同理0雖屬於X(A),但也不表示,最小上界一定大於0

    Source(s): 國中數學老師的腦袋
  • 1 decade ago

    設X={x : x in [0, 50], f(x)>=x}, 則

    X不是空集合, X 有上界<50, 設u=sup{X} 則f(u)=u ??

    (反證)

    設f(u)>u (即f(u)≠u)

    (1) f monotone, 0<u< 50 => 0<f(u)<50

    (2) f( f(u) ) >= f(u) (因 f mono.)

    又 f(u)>u (已知, 假設)

    => f( f(u) ) > u => f(u) in X => f(u) <= sup{X}=u

    => f(u) <=u 與 f(u)>u矛盾

    由反證法, 故 f(u)=u

    即u為 f 之 fixed point

    2008-11-05 05:13:58 補充:

    (1) 0<50 => 0<50 改為0<=u<=50=> 0<=f(u)<=f(50)<50

    2008-11-05 05:28:54 補充:

    (1)0<50=>0<50改為 0<=u<=50=>0<=f(u)<=f(50)<50

    Source(s): me
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