Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem: Let $(f_n)$ be a uniformly bounded sequence of real valued continuous functions on $[0,1]$. Prove that there is ONE subsequence $(f_{n_k})$ such that for every $0\le a < b \le 1$, we have $$\lim_{k\to\infty} \int_a^b \! f_{n_k}(t) dt $$ exists.

Context: Advanced Undergraduate Analysis. Familiar with Real Analysis by Carothers and Principles of Analysis by Rudin

I think it would be obvious to show this for all rationals inbetween a and b but I do not know how to start showing that there is a single subsequence. Any help would be appreciated, Thank you.

share|improve this question
2  
I doubt there would be only one. It was most probably meant 'at least one'. Any subsequence of such a sequence will be suitable, too. –  Berci Nov 26 '12 at 2:01
    
Hint: To prove it for rationals $a,b$ use a diagonal argument. –  Jose27 Nov 26 '12 at 2:25
    
Thank for the replies, I was of the impression it was 'at least one' but 'ONE' was emphasized so I'm still confused as to what it meant. Also, Jose could you expand a bit on the application of a diagonal argument in this particular case? –  jimmywho Nov 26 '12 at 2:43
add comment

1 Answer 1

up vote 4 down vote accepted

Hint: Apply Arzelà–Ascoli theorem to the sequence $\int_0^xf_n(t)dt$, $x\in[0,1]$.

share|improve this answer
    
Thanks for your reply. I was able to prove that there exist a subsequence of $F_n = \int_0^x f_n(t)dt$ that converges uniformly for all $x \in [0,1]$. –  jimmywho Nov 27 '12 at 0:43
1  
My question is that since this is true then for $ a < b \in [0,1] \int_0^a f_n(t)dt$ and $\int_0^b f_n(t)dt$ have uniformly converging subsequences then we have that by subtraction $\int_a^b f_n(t)dt$ has a uniformly converging subsequence? –  jimmywho Nov 27 '12 at 0:50
1  
@jimmywho: Since you have proved that there exists $n_k\to\infty$, such that $F_{n_k}(x)=\int_0^xf_{n_k}(t)d t$ uniformly converges to some $F$ on $[0,1]$, then by the definition of uniformly convergence, $\int_a^b f_{n_k}(t)d t=F_{n_k}(b)-F_{n_k}(a)$ uniformly converges to $F(b)-F(a)$. –  23rd Nov 27 '12 at 3:27
    
Thanks, I believe that was what I was attempting to state in my comment. I believe that this is the result I was looking for but am I missing anything? –  jimmywho Nov 27 '12 at 3:51
    
@jimmywho: Sorry, I cannot understand what you mean. Is there anything still unclear to you? –  23rd Nov 27 '12 at 3:58
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.