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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.

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    $\begingroup$ I think you mean that there exist an $M\in(0,\infty)$ such that $0≤a_n,b_n≤M$ for all $n$. Otherwise $a_n=b_n=0$. $\endgroup$ – Redundant Aunt Sep 8 '16 at 2:20
  • $\begingroup$ Right. My apologies. Fixed! $\endgroup$ – dovedevic Sep 8 '16 at 2:21
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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

$$\limsup_{n\to\infty} x_n=\lim_{n\to\infty} \sup_{k\ge n}x_k$$

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  • $\begingroup$ Thank you for the hint, I added my proof into my question with edits, is that what you hinted at? $\endgroup$ – dovedevic Sep 8 '16 at 2:55
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    $\begingroup$ Pretty much, yes. One thing: you wrote $b_n \le M = \sup_{k \ge n}(b_k)$. This might not be the case; $M$ is merely an upper bound for $(b_n)$, it may not be the least upper bound. The existence of such $M$ allows you to conclude that the supremum exist. $\endgroup$ – Reveillark Sep 8 '16 at 3:03
  • $\begingroup$ Ah. I see, fixed in proof. My question now is, am I allowed to just state center line 4/5? I feel that it is not trivial enough, if that makes sense..? $\endgroup$ – dovedevic Sep 8 '16 at 3:06
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    $\begingroup$ @DoveDevic Well, $A=\sup_{k\ge n} (a_kb_k)$ is an upper bound of the set $\{a_kb_k:k\ge n\}$, so in particular $A\ge a_nb_n$. $\endgroup$ – Reveillark Sep 8 '16 at 3:09
  • $\begingroup$ Is that not what I have stated? I am sorry I am not following perfectly: is saying $a_nb_n \le \sup_{k \ge n}(a_kb_k)$ not the same as $A \ge a_nb_n$? $\endgroup$ – dovedevic Sep 8 '16 at 3:15

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