Proving Properties limsup and liminf

Question

I got stuck by proving following statement: If $c \leq0 $, prove that $ \liminf_{n \to \infty }(ca_n) = c \, \limsup_{n \to \infty }(a_n) $.

Proof:

Consider $A_n = \{a_m | m \geq n\} $. The set $A_n$ has a supremum. Let $x_n = sup(A_n)$. This applies that $\forall a \in A_n: x_n \geq a $. Next, we multiply by $c \leq 0 $ and we get $ca \geq c x_n$. So $c \,x_n$ is a lower bound for the set $cA_n$. So we get: $ \liminf_{n \to \infty }(ca_n) \leq c \, \limsup_{n \to \infty }(a_n) $.

I got stuck by proving the other inequality. I tried it next way: Take $\epsilon \geq 0 $ arbitrarily. Then $cx_n + \epsilon > cx_n$. So I just need to show that $cx_n + \epsilon$ is not a lowerbound for $cA_n$. But I cannot get there. Somebody who wants to help me?

Thanks in advance!

Answer 1

Note that if $c \ge 0$ then $\sup_k c x_k = c \sup_k x_k$, and similarly for $\inf$.

Hence $\limsup_n c a_n = \lim_n \sup_{k \ge n} c a_k = \lim_n ( c \sup_{k \ge n} a_k )= c\lim_n \sup_{k \ge n} a_k = c \limsup_n a_n$, and similarly for $\inf$.

If $c \le 0$ instead, then we have $\sup_k c x_k = c \inf_k x_k$ and similarly with the $\sup, \inf$ interchanged. Repeating the above chain mutatis mutandis gives the desired result.

Answer 2

Let $\limsup (a_n)=a$ for $c<0$,consider $-\epsilon/c$ which is positive,then we can get an $N$ in $\mathbb N$ such that $\forall n>N$ we have $a_n\leq a-\epsilon/c$ i.e. $ca_n\geq ca-\epsilon$ for all $n>N$.So it follows that $\liminf (ca_n)=c \limsup(a_n)$.