數學問題:高微

Question

1. Let f:[0,+無窮)-->R be continuous
and let f be differentiable on (0,+無窮).
Assume f(0)= 0 and f(x) --> 0 as x--> +無窮 .
Show that there is a c 屬於 (0,+無窮) such that f\'(c) = 0 .

2. Given an example of a continuous map f:( 0, 1 )-->R
whose graph is not closed .
Can this happen for f : A包含於R --> R where A is closed ?

Answer 1

1.先證:limx->0f(x)=0,f 是有界的函數,要證f(x)是有界函數,我們必須找到一正數K,使得|f(x)|≦K因為limx->∞f(x)=0,給定一正數ε>0,會存在一正數M>0,使得當x≧M,我們有:|f(x)|<ε當x屬於[0,M),取[0,M)的閉包Cl([0,M))=[0,M]因為[0,M]為一緊致集,且f(x)為連續函數,根據最大-最小值定理,存在y屬於[0,M],使得|f(x)|≦f(y)=a取K=Max{a,ε},則|f(x)|≦K,對於所有x屬於[0,+∞)=>f(x)是有界函數再回到本題,利用反證法,假設不存在c屬於(0,+∞),使得f'(c)=0考慮f'(x)>0,對於所有x屬於(0,+∞)則f(x)在開區間(0,+∞)是嚴格遞增函數,也就是說f(x)>0,for all x屬於(0,+∞)=>f(x)不是有界函數,此與命題矛盾,故存在c屬於(0,+∞),使得f'(c)=02.取f(x)=x,f(x)在開區間(0,1)是連續函數,f((0,1))=(0,1)(0,1)不是閉集不可能發生,A是閉集,則A必定是有界,故A是有界的閉集=>A是緊致集=>f(A)也是緊緻集(Compact set)