## Trending News

# 高微 均勻連續証明題

証明：周期連續函數必均勻連續

--

我的想法是假設 f 具有周期 L ，在 [L, L]上均勻連續

但是在 [L, L] 外面的就不知道怎麼証？

請教一下

再請教一小題

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

是不是會成立？謝謝

### 4 Answers

- LLv 71 decade agoFavorite Answer
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.

- Login to reply the answers

- Anonymous6 years ago
【亞洲36588合法彩券公司直營 官網: A36588.NET 】

【 最新活動→迎接新會員，首存狂送20% 】

【運動→電子→對戰→現場→彩球 】

【免費服務 →電影區、討論區、KTV歡唱、運動轉播、即時比分、24H客服 】

【亞洲36588合法彩券公司直營 官網: A36588.NET 】

- Login to reply the answers

- 蜉蝣Lv 61 decade ago
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

----

這只是額外提醒

天大大的證明已很完整周延了

- Login to reply the answers