Real Analysis, liminf/limsup Equality and Multiplication

[joypav]

Here are a couple more problems I am working on!

Problem 1:
Prove that,
$limsupa_n+liminfb_n \leq limsup(a_n+b_n) \leq limsupa_n+limsupb_n$
Provided that the right and the left sides are not of the form $\infty - \infty$.

Proof:
Consider $(a_n)$ and $(b_n)$, sequences of real numbers.
Observe that for $n,m,l \in \Bbb{N}$,
$ \{ a_m+b_m : m > n\} \subseteq \{ a_m+b_l : m,l > n\} $
$ \implies sup\{ a_m+b_m : m > n\} \le sup\{ a_m+b_l : m,l > n\} $
Also, recall that,
$ \{ a_m+b_l : m,l > n\} = \{ a_m : m>n\} + \{ b_l : l > n\} $

Claim 1: $ sup(\{ a_m : m>n\} + \{ b_l : l > n\}) = sup\{ a_m : m>n\} + sup \{ b_l : l > n\} $

Proof of Claim:
"$\le$"
Take some $ a_m \in \{ a_m : m>n\} $ and $ b_l \in \{ b_l : l > n\} $.
Then, of course,
$ a_m + b_l \le sup\{ a_m : m>n\} + sup \{ b_l : l > n\} $
Using the definition of upper bound we can conclude that $ sup\{ a_m : m>n\} + sup \{ b_l : l > n\} $ is an upper bound for $ \{ a_m : m>n\} + \{ b_l : l > n\} $. (because $a_m$ and $b_l$ were chosen arbitrarily)
$ \implies sup(\{ a_m : m>n\} + \{ b_l : l > n\}) \leq sup\{ a_m : m>n\} + sup \{ b_l : l > n\} $

"$\ge$"
Choose $ \epsilon > 0 $.
Then $ \exists a,b, a \in \{ a_m : m>n\}, b \in \{ b_l : l > n\} $ such that
$ a > sup\{ a_m : m>n\} - \frac{\epsilon}{2} $
$ b > sup\{ b_l : l > n\} - \frac{\epsilon}{2} $
Adding the inequalities gives,
$ a + b > sup\{ a_m : m>n\} + sup\{ b_l : l > n\} - \epsilon $
This is for any $\epsilon$. So let $ \epsilon \rightarrow 0 $. Then,
$ sup(\{ a_m : m>n\} + \{ b_l : l > n\}) > a + b > sup\{ a_m : m>n\} + sup\{ b_l : l > n\} $

$ \implies $ Claim 1 is true.

Now consider the following sequences:
$ (sup_{m \geq n}(a_m+b_m)) $
$ (sup_{m \geq n}a_m + sup_{m \geq n}b_m) $

$ \forall m \in \Bbb{N} , a_m+b_m \leq supa_m + supb_m $
$ \implies lim_{n \rightarrow \infty}sup_{m \geq n}(a_m + b_m) \leq lim_{n \rightarrow \infty}(sup_{m \geq n}a_m + sup_{m \geq n}b_m) = lim_{n \rightarrow \infty}sup_{m \geq n}a_m + lim_{n \rightarrow \infty}sup_{m \geq n}b_m $
$ \implies $ 1. $ lim_{n \rightarrow \infty}sup(a_n+b_n) \leq lim_{n \rightarrow \infty}supa_n + lim_{n \rightarrow \infty}supb_n $
So we have proven the right side of the inequality.

Now we need to prove the left side. This part of my proof is, I think... wrong. Haha.
So we need to show now that,
$ limsupa_n+liminfb_n \leq limsup(a_n+b_n) $
$ \iff liminfb_n \leq limsup(a_n+b_n) - limsupa_n $
Using 1., $ \iff liminfb_n \leq limsup(a_n + b_n- a_n) = limsupb_n $
This inequality is obviously true.
$ \implies $ 2. $ limsupa_n+liminfb_n \leq limsup(a_n+b_n) $

Then, by 1. and 2. the proof is complete.Problem 2:
Prove that if, for all $ n > 0$, $ a_n > 0 $ and $ b_n > 0 $, then
$ limsup(a_nb_n) \le limsupa_n \cdot limsupb_n $
Show that the inequality may be strict.

I haven't had a chance to work on this much. Can I approach it similarly to Problem 1?

 

[jvicens]

Yes, you can approach Problem 2 similarly to Problem 1. Here is a possible proof:

Proof:
Consider $(a_n)$ and $(b_n)$, sequences of positive real numbers.
Let $x = limsupa_n$ and $y = limsupb_n$.
Then, by definition of limit superior, we have:
$ x = sup\{a_n : n > N\} $ for some $N \in \Bbb{N} $
$ y = sup\{b_n : n > M\} $ for some $M \in \Bbb{N} $

Now, for any $n > max(N,M)$, we have:
$ a_nb_n \le sup\{a_n : n > N\} \cdot sup\{b_n : n > M\} = x \cdot y $

This means that $x \cdot y$ is an upper bound for the set $\{a_nb_n : n > max(N,M)\}$.
Therefore, $limsup(a_nb_n) \le x \cdot y = limsupa_n \cdot limsupb_n$.

To show that the inequality may be strict, consider the following example:
$a_n = \frac{1}{n}$ and $b_n = n$
In this case, $limsupa_n = 0$ and $limsupb_n = \infty$.
But $a_nb_n = 1$, so $limsup(a_nb_n) = 1$, which is strictly less than $limsupa_n \cdot limsupb_n = 0 \cdot \infty = 0$.
Therefore, the inequality is strict in this case.