1. ## 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.

2. Originally Posted by Showcase_22
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 $\limsup(a_nb_n) = ab$.

3. 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 $\limsup a_nb_n\le\limsup AB=\limsup A\cdot\limsup B=\limsup a_n\cdot\limsup b_n$.