Results 1 to 8 of 8

Math Help - Help with limsup proof?

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    168

    Help with limsup proof?

    Let {a_n} be a sequence and let S_n={a_k:k\geq n}. Prove:

    limsup a_n = inf{sup Sn | n in N}

    Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?
    So then limsup a_n = sup Sn which implies Sn = T, where T is the set of all subsequential limits?

    Is that the correct way to interpret this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by paupsers View Post
    Let {a_n} be a sequence and let S_n=\{a_k:k\geq n\}. Prove:

    limsup a_n = inf{sup Sn | n in N}

    Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?
    What do you mean? \sup S_n=\sup_{k\geq n} a_kdepends of course on n.

    If you want us to help you, you should give your definition of \limsup_n a_n (because usually, it is \limsup_n a_n = \inf_n \sup_{k\geq n} a_k, which is your question...)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by paupsers View Post
    Let {a_n} be a sequence and let S_n={a_k:k\geq n}. Prove:

    limsup a_n = inf{sup Sn | n in N}

    Now, since sup Sn is a single point, isn't inf{sup Sn} the same point?
    So then limsup a_n = sup Sn which implies Sn = T, where T is the set of all subsequential limits?

    Is that the correct way to interpret this problem?

    I think there's a mistake in the notation: it should be S_n=\{a_k\}_{k\geq n}=\{a_k, a_{k+1},...\},

    Now you loot at \inf (\sup_{\substack{n\in \mathbb{N}}}\,S_n) , and of course that, in case it exists and is finite, the supremum is a single point, but you're taking the infimum over a possibly infinite set of these numbers.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by tonio View Post
    I think there's a mistake in the notation: it should be S_n=\{a_k\}_{k\geq n}=\{a_k, a_{k+1},...\},

    Now you loot at \inf (\sup_{\substack{n\in \mathbb{N}}}\,S_n) , and of course that, in case it exists and is finite, the supremum is a single point, but you're taking the infimum over a possibly infinite set of these numbers.

    Tonio
    Sorry to correct you: it is \inf_{\substack{n\in \mathbb{N}}}\,(\sup S_n), i.e. we take the supremum of the set S_n, and then the infimum of these values when n varies. Maybe it was good to clarify this again, since this was the first cause of problem in the original post, I think.

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Laurent View Post
    Sorry to correct you: it is \inf_{\substack{n\in \mathbb{N}}}\,(\sup S_n), i.e. we take the supremum of the set S_n, and then the infimum of these values when n varies. Maybe it was good to clarify this again, since this was the first cause of problem in the original post, I think.

    Of course: that \substack thing did whatever it wanted (in fact, what I told it to but that was not what I meant)
    It also should be S_n=\{a_k\}_{k\geq n}=\{a_n, a_{n+1},...\}.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2009
    Posts
    168
    Quote Originally Posted by Laurent View Post
    What do you mean? \sup S_n=\sup_{k\geq n} a_kdepends of course on n.

    If you want us to help you, you should give your definition of \limsup_n a_n (because usually, it is \limsup_n a_n = \inf_n \sup_{k\geq n} a_k, which is your question...)
    Sorry for the confusion. I can't figure out how to get a bracket ({ }) to show in latex.

    I'm just trying to prove that limsup a_n=inf{ sup S_n|n \in N} where S_n={ a_k|k \geq n} and a_n is a bounded sequence using only the fact that limsup a_n=sup(T), where T is the set of all subsequential limits of a_n.

    I realize that what I'm asking is a "definition," but I'm just trying to find a way to derive it, basically. Any help on this?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by paupsers View Post
    I'm just trying to prove that limsup a_n=inf{ sup S_n|n \in N} where S_n={ a_k|k \geq n} and a_n is a bounded sequence using only the fact that limsup a_n=sup(T), where T is the set of all subsequential limits of a_n.
    Let L=\inf_n \sup S_n. You should note first that the sequence (\sup S_n)_n is decreasing hence L=\lim_n \sup S_n.

    Since \limsup_n a_n = \sup T, you must prove:

    a) that L\geq\lim_n a_{\varphi(n)} for any convergent subsequence (a_{\varphi(n)})_n
    b) that, for any \varepsilon>0, there is a convergent subsequence (a_{\varphi(n)})_n such that \lim_n a_{\varphi(n)}>L-\varepsilon.

    I let you try a) (compare a_{\varphi(n)} with terms from the sequence (\sup S_n)_n)

    As for b), you have to "construct" a convergent subsequence. In fact, instead of using \varepsilon like above, you can directly construct a subsequence that converges toward \sup T (which means that in fact \sup T=\max T). To do so, intuitively, choose a_{\varphi(1)} close to \sup S_1 (it is a sup, hence it is a limit of values a_k in S_1), then a_{\varphi(2)} close(r) to \sup S_{\varphi(n)+1}, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)


    NB: you'll probably have to distinguish cases depending whether \sup T is finite or not.


    --
    About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2009
    Posts
    168
    Quote Originally Posted by Laurent View Post
    Let L=\inf_n \sup S_n. You should note first that the sequence (\sup S_n)_n is decreasing hence L=\lim_n \sup S_n.

    Since \limsup_n a_n = \sup T, you must prove:

    a) that L\geq\lim_n a_{\varphi(n)} for any convergent subsequence (a_{\varphi(n)})_n
    b) that, for any \varepsilon>0, there is a convergent subsequence (a_{\varphi(n)})_n such that \lim_n a_{\varphi(n)}>L-\varepsilon.

    I let you try a) (compare a_{\varphi(n)} with terms from the sequence (\sup S_n)_n)

    As for b), you have to "construct" a convergent subsequence. In fact, instead of using \varepsilon like above, you can directly construct a subsequence that converges toward \sup T (which means that in fact \sup T=\max T). To do so, intuitively, choose a_{\varphi(1)} close to \sup S_1 (it is a sup, hence it is a limit of values a_k in S_1), then a_{\varphi(2)} close(r) to \sup S_{\varphi(n)+1}, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)


    NB: you'll probably have to distinguish cases depending whether \sup T is finite or not.


    --
    About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.
    Sigh... all of that seems way above my head. Any way you could elaborate a bit more?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. manipulating limsup
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 20th 2011, 04:28 AM
  2. limsup calculation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 25th 2010, 08:43 AM
  3. limsup
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 4th 2009, 04:27 PM
  4. Liminf to limsup
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 20th 2009, 10:49 AM
  5. limsup.
    Posted in the Differential Geometry Forum
    Replies: 14
    Last Post: November 15th 2009, 11:10 AM

Search Tags


/mathhelpforum @mathhelpforum