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Thread: Limsups

  1. #1
    Super Member Showcase_22's Avatar
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    Limsups

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

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

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

    __________________________________________________ ______________

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

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

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

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

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

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