(The integral of the limit is not the limit of the integrals.) In this
example we consider a sequence of continuous functions, each of which has the same
integral over the domain. For each n∈IN let f_n be defined on [0, 1] as follows:
f_n(0) =0, f_n(1/(2n)) = 2n, f_n(1/n) = 0,
f_n is linear on [0, 1/(2n)] and on [1/(2n), 1/n], and
f_n = 0 on [1/n, 1].
It is easy to verify that f_n -> 0 on [0, 1]. Now, for each n∈ IN,
∫ from 0 to 1 f_n(x) dx =1 But ∫ from 0 to 1 (lim_n f_n(x)) dx =∫ from 0 to 1 0 dx =0
so that the limit of the integrals is not the integral of the limit.
My question is why
∫ from 0 to 1 f_n(x) dx =1 ?
isn't f_n -> 0 like the above said?
The integral of the limit is not the limit of the integrals.?
2015-12-07 4:57 am
回答 (2)
2015-12-07 6:08 am
✔ 最佳答案
Since f_n is linear on [0, 1/(2n)] and on [1/(2n), 1/n], finding the equations of the lines on these intervals yields f_n(x) = (4n^2)x on [0, 1/(2n)], connecting (0, 0) to (1/(2n), 2n)
...........= 4n - (4n^2)x on [1/(2n), 1/n], connecting (1/(2n), 2n), (1/n, 0)
...........= 0 otherwise.
So, ∫(x = 0 to 1) f_n(x) dx
= ∫(x = 0 to 1/(2n)) 4n^2 x dx + ∫(x = 1/(2n) to 1/n) (4n - (4n^2)x) dx + 0
= 2n^2 x^2 {for x = 0 to 1/(2n)} + (4nx - 2n^2 x^2) {for x = 1/(2n) to 1/n}
= 1/2 + 1/2
= 1.
Yes, f_n --> 0 on [0, 1]. Since the interchange of limit and integral yields different results, there is a good chance that {f_n} is not uniformly convergent to 0 on [0, 1].
if you have more maths/physics question use mathqu homework support app in app store this is the best app to get answers from live tutors who write step by step maths/physics answers search for mathqu in app store
2015-12-07 5:19 am
Since f_n is linear on [0, 1/(2n)] and on [1/(2n), 1/n], finding the equations of the lines on these intervals yields
f_n(x) = (4n^2)x on [0, 1/(2n)], connecting (0, 0) to (1/(2n), 2n)
...........= 4n - (4n^2)x on [1/(2n), 1/n], connecting (1/(2n), 2n), (1/n, 0)
...........= 0 otherwise.
So, ∫(x = 0 to 1) f_n(x) dx
= ∫(x = 0 to 1/(2n)) 4n^2 x dx + ∫(x = 1/(2n) to 1/n) (4n - (4n^2)x) dx + 0
= 2n^2 x^2 {for x = 0 to 1/(2n)} + (4nx - 2n^2 x^2) {for x = 1/(2n) to 1/n}
= 1/2 + 1/2
= 1.
Yes, f_n --> 0 on [0, 1]. Since the interchange of limit and integral yields different results, there is a good chance that {f_n} is not uniformly convergent to 0 on [0, 1].
I hope this helps!
f_n(x) = (4n^2)x on [0, 1/(2n)], connecting (0, 0) to (1/(2n), 2n)
...........= 4n - (4n^2)x on [1/(2n), 1/n], connecting (1/(2n), 2n), (1/n, 0)
...........= 0 otherwise.
So, ∫(x = 0 to 1) f_n(x) dx
= ∫(x = 0 to 1/(2n)) 4n^2 x dx + ∫(x = 1/(2n) to 1/n) (4n - (4n^2)x) dx + 0
= 2n^2 x^2 {for x = 0 to 1/(2n)} + (4nx - 2n^2 x^2) {for x = 1/(2n) to 1/n}
= 1/2 + 1/2
= 1.
Yes, f_n --> 0 on [0, 1]. Since the interchange of limit and integral yields different results, there is a good chance that {f_n} is not uniformly convergent to 0 on [0, 1].
I hope this helps!
收錄日期: 2021-04-18 14:09:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151206205728AAaG5MU