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Thread: limsup.

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

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

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

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

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

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

    Assume $\displaystyle b_n>a_n$

    $\displaystyle \lambda_nb_n> \lambda_na_n$

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

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

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

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

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

    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:

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

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

    Assume $\displaystyle b_n>a_n$

    $\displaystyle \lambda_nb_n> \lambda_na_n$

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

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

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

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



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

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

    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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    Exclamation

    Quote Originally Posted by Showcase_22 View Post
    Does what you have imply:

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

    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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    Let $\displaystyle C = \limsup c_n$, let $\displaystyle \varepsilon>0$ and let $\displaystyle N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}$. By the definition of limsup, the set N is infinite. Let $\displaystyle N_1 = \{n\in N: a_n\geqslant c_n\}$ and let $\displaystyle N_2 = \{n\in N: b_n\geqslant c_n\}$. Then $\displaystyle N\subseteq N_1\cup N_2$, so at least one of the sets $\displaystyle N_1,\,N_2$ must be infinite. Say $\displaystyle N_1$ is infinite. That says that there are infinitely many values of n for which $\displaystyle a_n>C-\varepsilon$, and so $\displaystyle \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon$. Since this holds for all $\displaystyle \varepsilon>0$, it follows that $\displaystyle \max\{ \limsup a_n,\,\limsup b_n\}\geqslant C$.
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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:

    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon
    $

    Where does this part come from?
    Last edited by Showcase_22; Mar 11th 2009 at 12:15 PM.
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  7. #7
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Opalg View Post
    Let $\displaystyle C = \limsup c_n$, let $\displaystyle \varepsilon>0$ and let $\displaystyle N = \{n\in\mathbb{N}:c_n>C-\varepsilon\}$. By the definition of limsup, the set N is infinite. Let $\displaystyle N_1 = \{n\in N: a_n\geqslant c_n\}$ and let $\displaystyle N_2 = \{n\in N: b_n\geqslant c_n\}$. Then $\displaystyle N\subseteq N_1\cup N_2$, so at least one of the sets $\displaystyle N_1,\,N_2$ must be infinite. Say $\displaystyle N_1$ is infinite. That says that there are infinitely many values of n for which $\displaystyle a_n>C-\varepsilon$, and so $\displaystyle \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon$. Since this holds for all $\displaystyle \varepsilon>0$, it follows that $\displaystyle \max\{ \limsup a_n,\,\limsup b_n\}\geqslant C$.
    Hi Opalg, having your answer read carefully, I cant see where the requirement $\displaystyle \lambda_{n}\in[0,1]$ is used.
    Do you think it is extra?

    Showcase_22,
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon
    $
    this is just a typo, it should be
    $\displaystyle
    \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon
    $
    Last edited by bkarpuz; Mar 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
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon
    $

    Where does this part come from?
    It follows from the definition of limsup that if there are infinitely many values of n for which $\displaystyle a_n>C-\varepsilon$, then $\displaystyle \limsup a_n>C-\varepsilon$.
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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,
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon
    $
    this is just a typo, it should be
    $\displaystyle
    \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon
    $
    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 $\displaystyle c_n$ 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 $\displaystyle \lambda_{n}\in[0,1]$ is used.
    The condition $\displaystyle \lambda_{n}\in[0,1]$ says that $\displaystyle c_n$ lies between $\displaystyle a_n$ and $\displaystyle b_n$. That implies that at least one of $\displaystyle a_n$ and $\displaystyle b_n$ must be $\displaystyle \geqslant c_n$. In my previous comment, that is the reason that $\displaystyle N\subseteq N_1\cup N_2$.

    Quote Originally Posted by bkarpuz View Post
    Showcase_22,
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n>C-\varepsilon
    $
    this is just a typo, it should be
    $\displaystyle
    \max\{ a_n,\,b_n\}\geqslant a_n>C-\varepsilon
    $
    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 $\displaystyle \lambda_{n}\in[0,1]$ says that $\displaystyle c_n$ lies between $\displaystyle a_n$ and $\displaystyle b_n$. That implies that at least one of $\displaystyle a_n$ and $\displaystyle b_n$ must be $\displaystyle \geqslant c_n$. In my previous comment, that is the reason that $\displaystyle N\subseteq N_1\cup N_2$.


    No it's not a typo, it needs to be what I stated.
    If its not a typo, then you should have written
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon.
    $
    In any case, there is a typo.
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  12. #12
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    Quote Originally Posted by bkarpuz View Post
    If its not a typo, then you should have written
    $\displaystyle
    \max\{ \limsup a_n,\,\limsup b_n\}\geqslant \limsup a_n {\color{red}{\geqslant}} C-\varepsilon.
    $
    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 $\displaystyle c_n\leq \max(a_n,b_n)$ in a complicated way, or maybe I missed something? Then you can deduce $\displaystyle \limsup_n c_n\leq \limsup_n \max(a_n,b_n)$.

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

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

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

    (we have $\displaystyle \ell\leq \limsup_n a_n$ 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

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

    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 $\displaystyle c_n\leq \max(a_n,b_n)$ in a complicated way, or maybe I missed something? Then you can deduce $\displaystyle \limsup_n c_n\leq \limsup_n \max(a_n,b_n)$.

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

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