This is what I tried:Let and be three sequences of real numbers.
Suppose that for every n, and let . Assuming that limsup a_n and limsup b_n are finite, prove the following inequality:
Assume
Therefore
This is only true if eventually or eventually. Is there a way to adapt this proof for sequences which constantly oscillate between or ? (ie.
I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.What can you say about the value of as compared to and ?
Justify your answer.
Does anyone have any ideas?
Let , let and let . By the definition of limsup, the set N is infinite. Let and let . Then , so at least one of the sets must be infinite. Say is infinite. That says that there are infinitely many values of n for which , and so . Since this holds for all , it follows that .
WOW!! Well firstly I got this problem from an analysis book. I just wondered what the answer was since someone pointed out my proof wasn't completely correct.
For the last part of the question I don't actually think anything definite can be stated, but I can't justify it. It would be better if it asked me to prove something else! =D
Also:
Where does this part come from?
So is this a typo? Surely the maths you posted implies what Opalg originally put? (or am I wrong? :s)
I'll also try and find a way to justify the fact that limsup has no other special qualities apart from what has just been proved. It's a little like proving the impossible! 8-p
To bkarpuz, about your 1st post: I feel like you only prove in a complicated way, or maybe I missed something? Then you can deduce .
It is a true fact that , but not a trivial one. Actually this is a particular case of the initial problem, and the answer Opalg gave to it gives also a proof of this fact (it proves the inequality " ", and " " is almost trivial).
Here's a variant of Opalg's proof, maybe more complicated, but maybe more intuitive: there exists a subsequence which converges toward . For any , either or , so that one case happens for infinitely many . For instance, there are infinitely many such that . Then we can pick a subsequence of which converges toward a limit (for instance, you can take ). From , we deduce (taking the limsups)
.
(we have because the limsup is the maximum of the limits of subsequences, a fact I already used before. )
Well, you could say
,
but you're right, you can't give any lower bound with limsups.
I now think that proving
was silly (because of the convexity), okay.
I am nowadays working on and values of , and I think I considered that term in the form at first sight, and gave a proof similar (but complicated) to those in my works.
But I think that
is trivial.
Thanks for the comments.