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:

$$\limsup(a_nb_n)\leq \limsup(a_n)\limsup(b_n)$$


  • 1
    $\begingroup$ 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


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

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.