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Math Help - Limsups

  1. #1
    Super Member Showcase_22's Avatar
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    Limsups

    If \{a_n\}_{n=1}^{\infty} and \{b_n\}_{n=1}^{\infty} are bounded and positive sequences, prove that:

    i). \limsup_{n \rightarrow \infty} a_nb_n \leq \limsup_{n \rightarrow \infty} a_n \limsup_{n \rightarrow \infty}b_n.

    ii). there is equality if either \{a_n\}_{n=1}^{\infty} or \{b_n\}_{n=1}^{\infty} converges.

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    For part i). I tried rewriting \limsup_{n \rightarrow \infty}{a_n} as \lim_{k \rightarrow \infty} sup \{a_n|n \geq k\}. Similarly, \limsup_{ n \rightarrow \infty} b_n=\lim_{k \rightarrow \infty} sup \{b_n|n \geq k\}.

    I have no idea where to go from here!

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    If \{a_n\}_{n=1}^{\infty} and \{b_n\}_{n=1}^{\infty} are bounded and positive sequences, prove that:

    i). \limsup_{n \rightarrow \infty} a_nb_n \leq \limsup_{n \rightarrow \infty} a_n \limsup_{n \rightarrow \infty}b_n.
    Notice that if a^{(k)} is the supremum for \{a_k,a_{k+1},...\} and b^{(k)} is supremum for \{b_k,b_{k+1},...\} then it means a^{(k)}\geq a_j for j\geq k and b^{(k)}\geq b_j for j\geq k. Thus, a^{(k)}b^{(k)}\geq a_jb_j for j\geq k. Thus, \sup\{ a_jb_j|j\geq k\} \leq a^{(k)}b^{(k)} = \sup\{ a_j|j\geq k\} \cdot \sup \{b_j |j\geq k\}. Take the limit, \limsup (a_nb_n) \leq \limsup (a_n) \limsup (b_n).

    ii). there is equality if either \{a_n\}_{n=1}^{\infty} or \{b_n\}_{n=1}^{\infty} converges.
    Say that a_n converge so \limsup a_n = \lim a_n = a for some a\geq 0.
    Thus, from above, we know that \limsup (a_nb_n) \leq a\limsup(b_n).

    Let b=\limsup(b_n), this means (a result above real sequences) that there exists a convergent subsequence b_{n_k} so that \lim b_{n_k} = b. Notice that \lim a_{n_k} = a, thus, a_{n_k}b_{n_k} is a subsequence of a_nb_n with a limit of ab. But the limit superior is the largest of all subsequential limits and so \limsup (a_nb_n)\geq ab.

    Thus, we have proved \limsup(a_nb_n) = ab.
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  3. #3
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    Let A denote the members of the sequence a_n, and similarly for the set B. Define AB:=\{a_nb_m:a_n\in A, b_m\in B\}. It is obvious (well not difficult to prove) that \limsup AB=\limsup A\cdot\limsup B. Now since a_nb_n is a particular sequence of elements in AB we have that \limsup a_nb_n\le\limsup AB=\limsup A\cdot\limsup B=\limsup a_n\cdot\limsup b_n.
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