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Real Analysis - limsups

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data
    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)).


    2. Relevant equations

    3. 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!
     
  2. jcsd
  3. Apr 15, 2008 #2

    quasar987

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    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.
     
  4. Apr 15, 2008 #3
    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:
     
  5. Apr 15, 2008 #4

    quasar987

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  6. Apr 15, 2008 #5

    Vid

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    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.
     
  7. Apr 15, 2008 #6

    quasar987

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    Oh yes, Vid is right. What I suggested doesn't work.
     
  8. Apr 15, 2008 #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?
     
    Last edited: Apr 15, 2008
  9. Apr 15, 2008 #8

    Vid

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    You could still assume not and try and derive a contradiction.
     
  10. Apr 15, 2008 #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.
     
  11. Apr 15, 2008 #10

    Vid

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    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?
     
  12. Apr 15, 2008 #11
    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?
     
  13. Apr 15, 2008 #12

    quasar987

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    The idea is sound but limsup(s_n) <= s_n is not right.

    What you wanna 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)
     
  14. Apr 15, 2008 #13

    quasar987

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    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.
     
  15. Apr 15, 2008 #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.
     
    Last edited: Apr 15, 2008
  16. Apr 15, 2008 #15

    quasar987

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    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.
     
    Last edited: Apr 15, 2008
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