Proving limsup Inequality for Bounded Sequences

In summary: I just don't see it. I can see that s_n*t_n is bounded, but I don't see how any subsequences of that have to converge. I was looking through my notes and I have the statement "If a_n is bounded, then every subsequence of a_n is bounded" written down. I don't really know what to do with that. I don't really know how to handle the problem. I'm sorry to be wasting your time. Thanks for your help.Ok, thanks for all your help. I'll try to think about it some more.
  • #1
steelphantom
159
0

Homework Statement


Let (s_n) and (t_n) be bounded sequences of nonnegative numbers. Prove that limsup(s_n*t_n) <= (limsup(s_n))*(limsup(t_n)).


Homework Equations



The Attempt at a Solution


I know that (s_n) and (t_n) have convergent subsequences, but that's about it. Where could I go from here? Thanks!
 
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  • #2
What you could do is suppose the contrary; i.e. suppose limsup(s_n*t_n) > (limsup(s_n))*(limsup(t_n)), and then find a counterexemple to this by choosing the sequences s_n and t_n adequately.
 
  • #3
quasar987 said:
What you could do is suppose the contrary; i.e. suppose limsup(s_n*t_n) > (limsup(s_n))*(limsup(t_n)), and then find a counterexemple to this by choosing the sequences s_n and t_n adequately.

Ok, I can do that. So, basically I'm saying: there exists x such that not A fails => A is true for all x. Is this the idea? Thanks! :smile:
 
  • #4
Yep.
 
  • #5
When you negate a for every you get a there exists.

Not A should say:

There exists two bounded non-negative sequences s_n, t_n such that
limsup(s_n*t_n) > limsup(s_n)*limsup(t_n).


Finding a sequence that doesn't fit that doesn't prove a damn.
 
  • #6
Oh yes, Vid is right. What I suggested doesn't work.
 
  • #7
Ok, this just doesn't seem right to me. Isn't this just equivalent to coming up with an example of what I want to prove, and not really proving it?

For example, what if I tried to "prove" x + 1 > 0 for all x in R. If I assume the converse, i.e. x + 1 <= 0 for all x in R, and I take x to be 1, I get x + 1 = 2 > 0. So the converse is false, but the original statement is obviously not true.

Edit: I was posting this while Vid posted. That's what I was suspecting. How can I try to prove this then?
 
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  • #8
You could still assume not and try and derive a contradiction.
 
  • #9
Ok, well intuitively, I see why this is true, because limsup(s_n) is the biggest number in S = {limits of all subsequences of s_n} and limsup(t_n) is the biggest number in T = {limits of all subsequences of t_n}. So multiplied together, limsup(s_n)*limsup(t_n) has to be >= limsup(s_n*t_n) because s_n*t_n can't ever be bigger than sup(T)*sup(S).

I think I know what's going on. But I'm really not sure on how to prove it, either by deriving a contradiction or otherwise. Thanks for your help.
 
  • #10
Use the fact that they are bounded sequences. If |s_n|<=M and |t_n| <=N. Since everything is positive |s_n||t_n| <= MN. If limsup(s_n*t_n) > limsup(s_n)*limsup(t_n)...

Can you get it from here?
 
  • #11
Vid said:
Use the fact that they are bounded sequences. If |s_n|<=M and |t_n| <=N. Since everything is positive |s_n||t_n| <= MN. If limsup(s_n*t_n) > limsup(s_n)*limsup(t_n)...

Can you get it from here?

Well, can I say that limsup(s_n) <= s_n <= M, and also that limsup(t_n) <= t_n <= N? So limsup(s_n)limsup(t_n) <= MN. And limsup(s_n*t_n) <= (s_n*t_n) <= MN. Then we have MN >= limsup(s_n*t_n) > limsup(s_n)limsup(t_n). But this is a contradiction, because we could have limsup(s_n)limsup(t_n) = MN.

Is this right?
 
  • #12
The idea is sound but limsup(s_n) <= s_n is not right.

What you want to invoke is the fact that s_n is bounded by M, and thus so is every subsequence of s_n, and thus so is the limit of any convergent subsequence of s_n, and thus so is limsup(s_n) (as the sup over all these limits of convegent subsequences)
 
  • #13
I'm talking too fast again.

limsup(s_n)limsup(t_n) < MN

is not a contradiction with

limsup(s_n)limsup(t_n) <= MN.
 
  • #14
Ok, so I can say that since (s_n) <= M, then any subsequence (s_n_k) <= M. Since s_n is bounded, it has a convergent subsequence. If lim(s_n_k) = s, then s <= M. limsup(s_n) = sup(S), where S = {limits of all subsequences of s_n}, so limsup(s_n) <= M. Similarly, limsup(t_n) <= N.

Then since u_n = (s_n*t_n) <= MN, any subsequence of u_n <= MN. Since u_n is bounded, it has a convergent subsequence. If lim(u_n_k) = u, then u <=MN. limsup(u_n) = sup(U), where U = {limits of all subsequences of u_n}, so limsup(u_n) <=MN. Then we have MN >= limsup(s_n*t_n) > limsup(s_n)limsup(t_n).

Edit: I just saw your other post, so this isn't right. This is pretty frustrating. It's a pretty easy proof, I know, but I'm just not seeing it. Thanks again for your assistance.
 
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  • #15
Since I don't know what Vid was hinting at, I can try to guide you through a solution I found if you'd like?

Suppose that for some sequences {s_n}, {t_n}, we have

limsup(s_n*t_n) > (limsup(s_n))*(limsup(t_n))

Let's try to derive a contradiction. Because both sequences are non negative and bounded, then so is s_n*t_n. Set limsup(s_n*t_n) = L (with 0<L<+infinity). This means there exists a subsequence {s_n_k*t_n_k} such that s_n_k*t_n_k-->L.

Now use the fact that {s_n_k} is bounded to deduce that there exists a convergent subsequence s_n_k_l-->s. In particular, s_n_k_l*t_n_k_l-->L , and so

[tex]\frac{1}{t_n_k_l}=\frac{s_n_k_l}{s_n_k_lt_n_k_l}\rightarrow s/L[/tex]

(note that because L>0, then s_n_k_l*t_n_k_l >0 for l large enough)

This means that t_n_k_l -->L/s. So we can write limsup(s_n)>=s and limsup(t_n)>=L/s, and hence limsup(s_n)*limsup(t_n)>=L, a contradiction. (!)

Surely there is simpler.
 
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FAQ: Proving limsup Inequality for Bounded Sequences

What is a limsup in real analysis?

A limsup, short for limit superior, is a concept in real analysis that represents the largest limit point of a sequence of numbers. It is the supremum or the least upper bound of the set of all limit points of the sequence.

How is limsup different from limit?

The limit of a sequence represents the actual value that the sequence approaches, while the limsup represents the largest value that the sequence can approach infinitely. In other words, the limit is a concrete value, while the limsup is a theoretical bound.

How is limsup calculated?

The limsup of a sequence can be calculated by taking the limit of the supremum of the sequence. In other words, the limsup is the limit of the maximum values of the sequence as the number of terms approaches infinity.

What is limsup used for in real analysis?

Limsup is a useful tool in real analysis for understanding the behavior and properties of sequences of numbers. It is often used in proofs and theorems to establish the existence of certain limits and to prove convergence or divergence of sequences.

Can limsup be infinite?

Yes, the limsup of a sequence can be infinite if the sequence has no upper bound. In this case, the limsup represents the largest possible value that the sequence can approach infinitely.

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