I am attempting to prove the following problem:
If $0\leq a_n$, $b_n \leq M$ for all $n$, for some $M \in (0,\infty) $, then $\limsup(a_nb_n)\leq \limsup(a_n)\limsup(b_n)$.
My initial intuition for a formulation is as follows:
Show $\sup(a_nb_n) \leq \sup(a_n)\sup(b_n)$
But this is where I am struggling. My intuition tells me to go about this in a similar fashion to this proof. I am unsure even why $M$ is necessary for this formulation.
But once I prove that, then I know $\sup(a_nb_n) \leq \sup(a_n)\sup(b_n)$ is true...
I know that applying a limit over an inequality does not affect an inequality, strict or not. This means I can say $\limsup(a_nb_n) \leq \lim(\sup(a_n)\sup(b_n))$ which is equivalent to $\limsup(a_nb_n)\leq \limsup(a_n)\limsup(b_n)$.
Which is what I wanted...
I am not looking for a solution to this, but some hints or suggestions would be much appreciated!
Just also a note, I found this proof here on MSE but that formulation is confusing. Maybe a better question would be a clarification on that?
Edit and addition of my Proof
Proof: $$\sup(a_nb_n) \leq \sup(a_n)\sup(b_n)$$
Since we know that $b_n$ is bounded by $M$, then we know that $b_n$'s supremum exist. I.e:
$$b_n \le \sup_{k \ge n}(b_k) \le M$$
Since we know that $a_n \ge 0$ $\forall n$, multipying by it on both sides does not affect the inequality. So now we have:
$$a_nb_n \le a_n \sup_{k \ge n}(b_k) \le \sup_{k \ge n}(a_k) \sup_{k \ge n}(b_k)$$
Notice that $\sup_{k \ge n}(a_kb_k)$ will always be the least upper bound of $a_nb_n$. Therefore:
$$a_nb_n \le \sup_{k \ge n}(a_kb_k) \le \sup_{k \ge n}(a_k) \sup_{k \ge n}(b_k)$$
Or simply:
$$\sup_{k \ge n}(a_kb_k) \le \sup_{k \ge n}(a_k) \sup_{k \ge n}(b_k)$$
We note that taking a limit on both sides of an equality does not affect it. Therefore we can say:
$$\lim_{n}(\sup_{k \ge n}(a_kb_k)) \le \lim_{n}(\sup_{k \ge n}(a_k) \sup_{k \ge n}(b_k)) = \lim_{n}(\sup_{k \ge n}(a_k)) \lim_{n}( \sup_{k \ge n}(b_k))$$
Which is the same as:
$$\limsup(a_nb_n)\leq \limsup(a_n)\limsup(b_n)$$
Q.E.D.
Hint:
To show that
$$\sup(a_nb_n)\le \sup(a_n)\sup(b_n)$$
it suffices to show that $\sup(a_n)\sup(b_n)$ is an upper bound for $(a_nb_n)$.
Also, recall that
$$\limsup_{n\to\infty} x_n=\lim_{n\to\infty} \sup_{k\ge n}x_k$$