# Proving Properties limsup and liminf

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?

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.
• I suspect that $c\geq0$ in the first line of the question is a typo. The OP continues with: "....Next we multiply by $c\leq0$..." and I think that there $c\leq0$ is not a typo. Aug 28, 2019 at 13:51
• I have added an answer for $c \le 0$. Aug 28, 2019 at 13:56
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 $$\liminf (ca_n)=c \limsup(a_n)$$.