Proof for limit superior's property: $\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n$ Let $a_n,b_n>0$ for all $n\in\mathbb N$.
Prove that 
$\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n$
I know that $\limsup (a_n+b_n ) \leq \limsup a_n + \limsup b_n$. But I don't know how to use this to get what I want to show.
 A: You can use what you know about $+$, if you know that strictly increasing continuous functions preserve $\limsup$, then argue as follows:
\begin{align*}
  \limsup a_nb_n &= \limsup\bigl(\exp\log(a_nb_n)\bigr)\\
       &= \exp\bigl(\limsup (\log a_n + \log b_n)\bigr)\\
       &\le \exp \bigl(\limsup(\log a_n) + \limsup(\log b_n)\bigr)\\
       &= \limsup (\exp\log a_n) \cdot \limsup (\exp\log b_n)\\
       &= \limsup a_n \cdot \limsup b_n
\end{align*}
But you can also follow the lines of the actual proof you did for +.
A: Let's assume that the sequences $\{a_{n}\}, \{b_{n}\}$ are bounded so as to avoid any distractions related to $\infty$. Let $A = \lim\,\sup\, a_{n},\, B = \lim\,\sup\, b_{n}$. We also need to understand very clearly what the concept of limit superior means. It means that:
1) For any number $A' > A$ after a certain value of $n$ we have $a_{n} \leq A'$
2) For any number $A' < A$ there are infinitely many values of $n$ for which $a_{n} > A'$.
and we have similar statements for $B$ and $b_{n}$. Since $a_{n}, b_{n}$ are positive we must have $A \geq 0, B \geq 0$.
What we have to establish is that $$\lim\,\sup \,a_{n}b_{n} \leq AB$$ Let $C = \lim\,\sup\, a_{n}b_{n}$. If either of $A$ or $B$ is zero then it is clear that the we must have $C = 0$ (Why? Because in that case one of the above sequences tends to zero and other remains bounded so that the product $a_{n}b_{n}$ also tends to zero). Hence let's assume that $AB > 0$.
Suppose on the contrary that $C > AB$ then we can find numbers $A', B'$ such that $A' > A, B' > B$ and $AB < A'B' < C$. We have two conclusions now:
1) After a certain value of $n$ we must have $a_{n} \leq A', b_{n} \leq B'$ so that $a_{n}b_{n} \leq A'B'$.
2) Since $C = \lim\,\sup\, a_{n}b_{n}$ and $A'B' < C$ it follows that $a_{n}b_{n} > A'B'$ for infinitely many values of $n$.
Clearly these statements contradict each other and hence we must have $C \leq AB$.
A: If $\limsup a_n=\infty$ or $\limsup b_n=\infty$, then the inequality is clearly satisfied; 
so we can assume that $(a_n)$ and $(b_n)$ are bounded.
If $N\in\mathbb{N}$, let $A_N=\{a_n:n\ge N\}$, $B_N=\{b_n:n\ge N\}$, and $P_N=\{a_n b_n:n\ge N\}$;
and let $u_N=\sup A_N$, $v_N=\sup B_N$, and $w_N=\sup P_N$.
If $n\ge N$, then $0< a_n\le u_N$ and $0< b_n\le v_N \implies a_nb_n\le u_Nv_N$;
so $u_Nv_N$ is an upper bound for $P_N$ and therefore $w_N\le u_Nv_N$.
Then $$\limsup a_nb_n=\lim_{N\to\infty}w_N\le\lim_{N\to\infty}u_Nv_N=\big(\lim_{N\to\infty}u_N\big)\big(\lim_{N\to\infty}v_N\big)=(\limsup a_n)(\limsup b_n).$$
