問一題均勻收斂

這其實是不是均勻收斂的反例!!!!

f_n 不可能均勻收斂在[0,1] ，也不可能均勻收斂在 ]0,1] 。

因為當x > 0 ，有
lim_n f_n(x) →0 ...(*)
但當x=0, lim_n f_n(0) =1 !!
如此應可以看出 {f_n}不可能均勻收斂在[0,1] 。
更嚴格地看，由(*)，可假設f=0。
取 0 < ε < 1/2 ，取 x=1/n 。當 n > 1 ，我們有
| f_n(x) - f(x) | = | f_n(1/n) - 0| = 1/2 > ε
所以 f_n 不可能均勻收斂在[0,1] 。

2009-02-02 20:37:03 補充：
對不起，不知為甚麼看錯題目答錯題。
我會改正。
如改不了我會自行刪除答案。

2009-02-02 20:56:07 補充：
可以了，雖然真係解錯題，但神奇地無需改任何步驟。答案依舊。

2009-02-02 21:07:05 補充：
這題是不是收斂到

f(x) = 1, if x=0
f(x) = 0, otherwise

Ans: Converge pointwise

f_n(x)為連續函數 in [0, 1]
收歛結果之函數在[0,1]內不連續,怎可能uniform?

2009-02-02 21:20:30 補充：
及時兄004的意見是對的!
只有點收歛,沒有均勻收歛!

I think your question is wrong

F_n(x) = x/(1+(n^2)(x^2))

This is a sequence of functions: for each value of n = 1, 2 , 3..., we get a function:

For example:

F_1(x) = x/(1+x^2)

F_2(x) = x/(1+4x^2)

...

F_10(x) = 1/(1+100x^2)

Now we want to show that F_n(x) converges uniformly on [0,1]. What does that mean? Well, a sequence of functions F_n(x) converges uniformly on a set X if there exists a function F(x) such that, given ε > 0, there exists N so that for n ≥ N:

sup|F_n(x) - F(x)| < ε

where the supremum is taken over X. So this is kind of like saying that, for a fixed value x_0, the sequence of values F_n(x_0) converges to the value F(x_0), except all of these sequences have to converge "at the same rate".

So first we have to cook up our function F(x). To do this, note that for a fixed value x_0:

lim_{n->inf} x_0/(1+(n^2)(x_0^2)) = 0

Indeed, if x0 ≠ 0, the numerator is fixed and the denominator goes off to infinity, and if x0 = 0, this is just the constant sequence 0. So we let F(x) = 0, and we claim F_n(x) converges to F(x). Now we need to show this convergence happens UNIFORMLY, or at the same rate.

So let ε > 0. We want to find a value N so that for n ≥ N and ALL x:

|x/(1+(n^2)(x^2))| = |x|/(1+(n^2)(x^2)) < ε

Solving for n, this inequality gives:

|x|/ε < 1+(n^2)(x^2)

[(1/ε)|x| - 1]/x^2 < n^2

(|x|-ε)/εx^2 < n^2

I.e., we want to find a value of N so that this inequality is true for ALL values of x and all n ≥ N. Since the quantity on the right is increasing, it's enough to show that it holds for n = N. Well, if |x| < ε, this is less than 0, so any value of N works. If |x| ≥ ε, then:

(|x|-ε)/εx^2 < |x|/εx^2 = 1/ε|x| ≤ 1/ε^2

So let N ≥ 1/ε. Then we get the desired inequality, and Fn(x) converges to F(x) uniformly

2009-02-02 20:38:33 補充：
F_10(x) = x/(1+100x^2)

2009-02-02 20:41:09 補充：
1/(1+ n^2 * x^2) 找不到對應的n

2009-02-02 20:50:38 補充：
GOOD JOB