Anonymous

高微 均勻連續証明題

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Update:

| sin x - sin y | < | x- y |

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• L
Lv 7

Let f be a continuous real function with period T.

Since [nT, (n+1)T] is a compact set for each integer n, f is uniformly continuous on [nT, (n+1)T].

Thus fix e>0, there exists d>0 such that |f(x) - f(y)| < e/2 whenever x and y are both in [nT, (n+1)T] and |x-y| < d.

Now we show f is uniformly continuous on |R. For the same e, we will have that

|f(x) - f(y)| < e whenever |x-y| < d. This is since (*)

case1. x and y are both in [nT, (n+1)T] for some integer n.

(*) holds obviously

case2. W.l.o.g. we assume x < y.

x in [nT, (n+1)T] and y in [(n+1)T, (n+2)T] for some integer n. In this case we have

|f(x) - f(y)| <= |f(x) - f((n+1)T)| + |f((n+1)T) - f(y)| < e/2 + e/2 = e

(*) holds.

Since e is arbitrary, f is uniformly continuous on |R.

• Anonymous
6 years ago

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• 蜉蝣
Lv 6

Let f be a continuous real function with period T.

Since [nT, (n+1)T] is a compact set for each integer n, f is uniformly continuous on [nT, (n+1)T].

我覺得這好像要證一下比較好,應該不完全是因為連續於compact set,所以就均勻連續

跟f(nT)=f(nT+T)應該要扯得上

sorry,純個人想法,所以也可能是錯的

2008-11-09 03:44:15 補充：

sorry again,我剛查了,的確有此性質:若f在一compact set 連續,則f為均勻連續

2008-11-09 15:42:03 補充：

Thus fix e>0, there exists d>0 such that |f(x) - f(y)| < e/2 whenever x and y are both in [nT, (n+1)T] and |x-y| < d

最好加註0

2008-11-09 15:44:21 補充：

d小於T(比原先小的都一定能成立),這樣最後才能假設x,y必落在相鄰兩個周期內

2008-11-09 15:58:27 補充：

這也正是週期函數的T可派上場的地方

像(0,1)雖可寫成無窮多個compact set的聯集,但它做不到讓這些set的寬度都大於等於一個固定正數

(0,1)為[1/(n+1),1/n](n=2,3.........)[1-(1/n),1-(1/(n+1)](n=2,3.....)這些緊緻集的聯集

但寬度會趨近於0

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這只是額外提醒

天大大的證明已很完整周延了