Proof for limit superior's property: $\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n$

Question

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.

The result about sums will not help you; but the proof of this result will. — Etienne, Sep 26 '13 at 19:33

score 5 · Accepted Answer · answered Sep 26 '13 at 19:37

5

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 +.

answered Sep 26 '13 at 19:37

martini

84,101

So I was wrong: the result for sums does help! – Etienne Sep 26 '13 at 19:41
$lim supanbn=lim sup(explog(anbn))$ which property allow you to do this? I don't think I have learn this yet – Diane Vanderwaif Sep 26 '13 at 21:16
@martini: Is it correct that increasing continuous functions preserve limsup? (ie. $x < y$ implies $f(x) \leq f(y)$, but not necessarily $f(x) < f(y)$.). I think it is possible to prove that $sup f(A) = f sup(A)$ when $f$ is continuous and increasing. – spin Oct 06 '13 at 14:40
@spin Yes, it is. – martini Oct 06 '13 at 16:22
@martini: Ok, thanks. I was wondering because you have "strictly increasing" there. – spin Oct 06 '13 at 19:15
@spin You are right, I probably should've written "increasing", but I consider invertible functions here anyway. – martini Oct 07 '13 at 07:22

Paramanand Singh · Answer 2 · 2013-10-07T04:01:42.270

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 $C \leq AB$.

Hi Paramanand. Should we relax the inequality in "1) For any number $A' > A$ after a certain value of $n$ we have $a_{n} \leq A'$", to be $a_n <A'$? Of course, if the latter is true, then the former is also. But, why did you include the equal sign here? — Mark Viola, Feb 09 '17 at 06:25
@Dr.MV: Yes we can write $a_{n} < A'$. Because we can always choose $A''$ with $A < A'' < A'$ and then $a_{n} \leq A'' < A'$. I think the $\leq$ is from the source where I read about limsup and liminf. see http://math.stackexchange.com/a/1893725/72031 — Paramanand Singh, Feb 09 '17 at 06:31

score 0 · Answer 3 · answered Oct 05 '13 at 22:04

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 $$\limsup a_nb_n=\lim_{N\to\infty}w_N\le\lim_{N\to\infty}u_Nv_N=\big(\lim_{N\to\infty}u_N\big)\big(\lim_{N\to\infty}v_N\big)=(\limsup a_n)(\limsup b_n).$$

Proof for limit superior's property: $\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n$

3 Answers3