Scharze space asked in 科學數學 · 1 decade ago

實變數函數論14

(1) Use definition to show that the Cantor-Lebesgue function f(x)

: [0,1]->[0,1] is not a Lipschitz continuous

(2)Show that the Cantor- Lebesgue function f(x):[0,1]->[0,1] satisfies the following:

|f(x)-f(y)|<=2|x-y|^α, for all x,y€[0,1]

where α€(0,1) is a constant given by α=(ln2/ln3)

(Hint:Use the fact that if x,y€[0,1] with |x-y|<=3^(-k) for some k€|N,then the difference |f(x)-f(y)| is at most 2^(-k).

For arbitrary x,y €[0,1] one can choose an unique k€|N such that 3^(-k-1)<|x-y|<=3^(-k)

which implies |f(x)-f(y)|<=2^(-k).Rewrite the estimate without involving k)

兩題只要證一題也可以

兩題證明具有關聯性

我的證明不太能夠說服我

Update:

prime大大:

我另外還有一題要請教

如果可以的話..............

Update 2:

第一題的證明我是這樣寫的:

Suppose f is Lipschite continuous

then there exist a real number M>0 such that

|f(x)-f(y)|<=M|x-y| for all x,y€[0,1]

For arbitrary x,y€[0,1],choose k€N such that 1/(3^(k+1))<|x-y|<=1/3^k

=>|f(x)-f(y)|<=2^(-k)<=M|x-y|

But if 1/3^(k+1)<|x-y|<=1/3^k

Update 3:

M/3^(k+1)<M|x-y|<=M/3^k

so M/3^(k+1)<1/2^k<=M|x-y|<=M/3^k

=>(3/2)^k<=M<(3/2)^k*3

M->∞ as k->∞

Update 4:

這樣證邏輯上有無問題?

Update 5:

您真內行!

提示的確是給(2)用

1 Answer

Rating
  • prime
    Lv 4
    1 decade ago
    Favorite Answer

    1的證明不是非常明顯

    就是找到一組( Xn, Yn)使得

    | Xn - Yn | = 3^(-k)

    | f(Xn) - f(Yn) | = 2^(-k)

    如果 Lip. condition 成立

    必然要有正數L, 使得 2^(-k) < = L * 3^(-k)

    => L >= (2/3) ^ k -> 0 as k->oo矛盾...

    我們取 Xn = 0, Yn = 3^(-n)

    很明顯 f(Xn)=0, f(Yn)= 2^(-n).

    ( Xn, Yn)即為所求

    第二個證明也非常簡單

    假設 (1/3)^(k+1) < | x-y | <= (1/3) ^(k)

    不失ㄧ般性,假設 x<y

    考慮 f_(k+1)(趨近C-L-function那個函數序列的第k+1項)

    Cantor set構成的第k+1次,相鄰2塊cantor set的距離是 (1/3)^(k+1)

    跟0最近那塊距離也是(1/3)^(k+1)

    跟1最近那塊距離也是(1/3)^(k+1)

    每塊cantor set最小的長度是 (1/3)^(k+1)

    所以可以找到p1, p2, p3

    p1<= x <= p2 <= y <= p3

    其中 p1 p2,p3 落在連續3個的第k+1次cantor集或是{0},{1}

    若 p1 落在第k+1次cantor 集或是{0},{1}, 我們知道

    f_(k+1) (p1) = f_(k+2) (p_1) = ....

    故 f(p1) = f_(k+1) (p1)

    同理, f(p2) = f_(k+1) (p2), f(p3) = f_(k+1) (p3),

    因為在f_(k+1)中,相鄰的cantor set(第k+1次)的賦值差為(1/2)^k+1

    故 f(y) - f(x) <= f_(k+1)(p_3) - f_(k+1) = 2*( 1/ 2 )^(k+1) = ( 1/ 2 )^(k)

    所以

    log_2 ( 2 |x-y|^α)

    = log_2 ( 2 ) + α* log_2 ( | x - y |)

    >= 1 + log_3 ( | x - y | )

    >= - k

    log_2 ( | f(x) - f(y) | )

    <= - k

    故 |f(x)-f(y)|<=2|x-y|^α, for all x,y€[0,1]

    2007-10-06 03:14:56 補充:

    2^(-k) < = L * 3^(-k)

    => L >= (3/2) ^ k -> oo as k->oo矛盾...

    2007-10-06 13:37:25 補充:

    你就把問題post出來..

    2007-10-07 00:35:07 補充:

    so M/3^(k+1)<1/2^k<=M|x-y|<=M/3^k

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

    這步有問題

    你不知道 | f(x) - f(y) | 的下界

    怎麼知道它比 M/3^(k+1) 大

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

    我猜提示是給(2)用的...

Still have questions? Get your answers by asking now.