math problem~急>
a. let fn(t), n=1,2,....be continuous functions on an interval [a,b] and {fn(t)} converge uniformly to f(t) on [a,b] show that lim n-> ∞∫上b下a fa(t)dt= ∫ 上b下af(t)dt
b. construct {fn(t)} on [0,1] such that the above equality does not hold true.
Answers & Comments
Verified answer
f_n->f uniformly, f_n is continuous on [a,b] for n=1,2,....
=>f is continuous on [a,b]
且 f_n is uniformly bounded 故存在M,N >0 使得 |f_n-f|<=M+N
由bounded convergent theorem
lim(n->∞)∫_[a,b]f_n(t)d=∫_[a,b]f(t)dt
第二部分 考慮這樣的函數:
f_n(t)=n/(1+n^2t^2) on [0,1]
f_n(t)-->∞ at t=0
f_n(t)->0 on (0,1]
故 f_n(t)->f(t) 這裡f(t)=∞ for t=0, f(t)=0 on (0,1]
∫_[0,1]f(t)=0 但是 lim(n->∞)∫_[0,1]ndt/(1+n^2t^2)
=lim(n->∞)tan^(-1)(n)
=π/2
兩者並不相等