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Math Help - limsup.

  1. #1
    Super Member Showcase_22's Avatar
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    limsup.

    Let (a_n; n \geq 1), \ (b_n;n \geq 1) and (\lambda_n; n \geq 1) be three sequences of real numbers.

    Suppose that \lambda_n \in [0,1] for every n, and let c_n=\lambda_n a_n+(1-\lambda_n)b_n. Assuming that limsup a_n and limsup b_n are finite, prove the following inequality:

    limsup \  c_n \leq max (limsup\ a_n, limsup\ b_n)
    This is what I tried:

    (1-\lambda_n)b_n > (1-\lambda_n)a_n
    \Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_na_n+(1-\lambda_n)a_n=a_n

    \Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ a_n

    Assume b_n>a_n

    \lambda_nb_n> \lambda_na_n

    \Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_nb_n+(1-\lambda_n)b_n=b_n

    \Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ b_n

    Therefore limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq max(limsup \ a_n,limsup \ b_n)

    This is only true if a_n>b_n eventually or b_n>a_n eventually. Is there a way to adapt this proof for sequences which constantly oscillate between a_n>b_n or b_n>a_n? (ie. a_n=5 \ (n \ even), \ 2 \ (n \ odd).

    What can you say about the value of limsup_{n \rightarrow \infty} c_n as compared to limsup_{n \rightarrow \infty} a_n and limsup_{n \rightarrow \infty}b_n?

    Justify your answer.
    I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

    Does anyone have any ideas?
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    This is what I tried:

    (1-\lambda_n)b_n > (1-\lambda_n)a_n
    \Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_na_n+(1-\lambda_n)a_n=a_n

    \Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ a_n

    Assume b_n>a_n

    \lambda_nb_n> \lambda_na_n

    \Rightarrow \lambda_na_n+(1-\lambda_n)b_n \leq \lambda_nb_n+(1-\lambda_n)b_n=b_n

    \Rightarrow limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq limsup \ b_n

    Therefore limsup \ (\lambda_na_n+(1-\lambda_n)b_n) \leq max(limsup \ a_n,limsup \ b_n)

    This is only true if a_n>b_n eventually or b_n>a_n eventually. Is there a way to adapt this proof for sequences which constantly oscillate between a_n>b_n or b_n>a_n? (ie. a_n=5 \ (n \ even), \ 2 \ (n \ odd).



    I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

    Does anyone have any ideas?
    I hope this helps.
    Set f_{n}(\lambda):=\lambda a_{n}+(1-\lambda)b_{n} for \lambda\in[0,1], then f_{n} is nondecreasing if a_{n}-b_{n}\geq 0 and decreasing if a_{n}-b_{n}<0 (check the derivative of f_{n}).
    Now, considering the nondecreasing/decreasing nature of f_{n}, define
    c_{n}^{\ast}:=\begin{cases} f_{n}(1)=a_{n},& a_{n}\geq b_{n}\\ f_{n}(0)=b_{n},& a_{n}<b_{n}\end{cases}
    for \mathbb{N}.
    Then, we have c_{n}=f_{n}(\lambda_{n})\leq c_{n}^{\ast}=\max\{a_{n},b_{n}\} for all n\in\mathbb{N}.
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  3. #3
    Super Member Showcase_22's Avatar
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    Does what you have imply:

    limsup \ c_n=limsup f_n(\lambda) \leq limsup \c*_n=max \{limsup \ a_n, limsup \  b_n \}?

    Would the addition of this step then make it a complete proof?
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    Does what you have imply:

    \limsup c_n=\limsup f_n(\lambda) \leq \limsup c^{*}_n=\max \{\limsup a_n, \limsup b_n \}?

    Would the addition of this step then make it a complete proof?
    Sure!
    I wonder from where you got this problem?
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  5. #5
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    Let C = \limsup c_n, let \varepsilon>0 and let N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}. By the definition of limsup, the set N is infinite. Let N_1 = \{n\in N: a_n\geqslant c_n\} and let N_2 = \{n\in N: b_n\geqslant c_n\}. Then N\subseteq N_1\cup N_2, so at least one of the sets N_1,\,N_2 must be infinite. Say N_1 is infinite. That says that there are infinitely many values of n for which a_n>C-\varepsilon, and so \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon. Since this holds for all \varepsilon>0, it follows that \max\{ \limsup a_n,\,\limsup b_n\}\geqslant C.
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  6. #6
    Super Member Showcase_22's Avatar
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    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:

    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon<br />

    Where does this part come from?
    Last edited by Showcase_22; March 11th 2009 at 12:15 PM.
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  7. #7
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Opalg View Post
    Let C = \limsup c_n, let \varepsilon>0 and let N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}. By the definition of limsup, the set N is infinite. Let N_1 = \{n\in N: a_n\geqslant c_n\} and let N_2 = \{n\in N: b_n\geqslant c_n\}. Then N\subseteq N_1\cup N_2, so at least one of the sets N_1,\,N_2 must be infinite. Say N_1 is infinite. That says that there are infinitely many values of n for which a_n>C-\varepsilon, and so \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon. Since this holds for all \varepsilon>0, it follows that \max\{ \limsup a_n,\,\limsup b_n\}\geqslant C.
    Hi Opalg, having your answer read carefully, I cant see where the requirement \lambda_{n}\in[0,1] is used.
    Do you think it is extra?

    Showcase_22,
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon<br />
    this is just a typo, it should be
    <br />
\max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon<br />
    Last edited by bkarpuz; March 11th 2009 at 12:25 PM. Reason: Pointed out the typo
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  8. #8
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    Quote Originally Posted by Showcase_22 View Post
    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
    I agree with that. I don't see that there is anything else that can be said about limsup(c_n)

    Quote Originally Posted by Showcase_22 View Post
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon<br />

    Where does this part come from?
    It follows from the definition of limsup that if there are infinitely many values of n for which a_n>C-\varepsilon, then \limsup a_n>C-\varepsilon.
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  9. #9
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Showcase_22,
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon<br />
    this is just a typo, it should be
    <br />
\max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon<br />
    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 c_n has no other special qualities apart from what has just been proved. It's a little like proving the impossible! 8-p
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  10. #10
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    Quote Originally Posted by bkarpuz View Post
    Hi Opalg, having your answer read carefully, I cant see where the requirement \lambda_{n}\in[0,1] is used.
    The condition \lambda_{n}\in[0,1] says that c_n lies between a_n and b_n. That implies that at least one of a_n and b_n must be \geqslant c_n. In my previous comment, that is the reason that N\subseteq N_1\cup N_2.

    Quote Originally Posted by bkarpuz View Post
    Showcase_22,
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon<br />
    this is just a typo, it should be
    <br />
\max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon<br />
    No it's not a typo, it needs to be what I stated.
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  11. #11
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Opalg View Post
    The condition \lambda_{n}\in[0,1] says that c_n lies between a_n and b_n. That implies that at least one of a_n and b_n must be \geqslant c_n. In my previous comment, that is the reason that N\subseteq N_1\cup N_2.


    No it's not a typo, it needs to be what I stated.
    If its not a typo, then you should have written
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon.<br />
    In any case, there is a typo.
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  12. #12
    MHF Contributor
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    Quote Originally Posted by bkarpuz View Post
    If its not a typo, then you should have written
    <br />
\max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon.<br />
    In any case, there is a typo.
    Humph! Okay, I'll admit that.
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  13. #13
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    To bkarpuz, about your 1st post: I feel like you only prove c_n\leq \max(a_n,b_n) in a complicated way, or maybe I missed something? Then you can deduce \limsup_n c_n\leq \limsup_n \max(a_n,b_n).

    It is a true fact that \limsup_n \max(a_n,b_n)=\max(\limsup_n a_n,\limsup_n b_n), 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 " \leq", and " \geq" is almost trivial).

    Here's a variant of Opalg's proof, maybe more complicated, but maybe more intuitive: there exists a subsequence (c_{\phi(n)})_{n\geq 0} which converges toward \limsup_n c_n. For any n, either c_{\phi(n)}\leq a_{\phi(n)} or c_{\phi(n)}\leq b_{\phi(n)}, so that one case happens for infinitely many n. For instance, there are infinitely many n such that c_{\phi(n)}\leq a_{\phi(n)}. Then we can pick a subsequence (a_{\phi(\psi(n))})_{n\geq 0} of (a_{\phi(n)})_{n\geq 0} which converges toward a limit \ell\in\mathbb{R}\cup\{\pm\infty\} (for instance, you can take \ell=\limsup_n a_{\phi(n)}). From c_{\phi(\psi(n))}\leq a_{\phi(\psi(n))}, we deduce (taking the limsups)

    \limsup_n c_n\leq \ell\leq \limsup_n a_n\leq \max(\limsup_n a_n,\limsup_n b_n).

    (we have \ell\leq \limsup_n a_n because the limsup is the maximum of the limits of subsequences, a fact I already used before. )


    Quote Originally Posted by Showcase_22 View Post

    I'm not really sure what to put for this part of the question. It seems like nothing conclusive can be written.

    Does anyone have any ideas?
    Well, you could say

    \min(\liminf_n a_n,\liminf_n b_n)\leq \liminf_n c_n\leq\limsup_n c_n \leq \max(\limsup_n a_n, \limsup_n b_n),

    but you're right, you can't give any lower bound with limsups.
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  14. #14
    Senior Member bkarpuz's Avatar
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    Arrow

    Quote Originally Posted by Laurent View Post
    To bkarpuz, about your 1st post: I feel like you only prove c_n\leq \max(a_n,b_n) in a complicated way, or maybe I missed something? Then you can deduce \limsup_n c_n\leq \limsup_n \max(a_n,b_n).

    It is a true fact that \limsup_n \max(a_n,b_n)=\max(\limsup_n a_n,\limsup_n b_n), 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 " \leq", and " \geq" is almost trivial).
    I now think that proving
    c_{n}\leq\max\{a_{n},b_{n}\}
    was silly (because of the convexity), okay.
    I am nowadays working on \max and \min values of \lambda(\lambda-1)\cdots(\lambda-k), and I think I considered that term in the form a\lambda(\lambda-b) at first sight, and gave a proof similar (but complicated) to those in my works.
    But I think that
    \limsup_{n\to\infty}\max\{a_{n},b_{n}\}=\max\bigg\  {\limsup_{n\to\infty}a_{n},\limsup_{n\to\infty}b_{  n}\bigg\}
    is trivial.

    Thanks for the comments.
    Last edited by bkarpuz; March 16th 2009 at 04:18 AM.
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