# Proving Properties limsup and liminf

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

Prove:

Consider $$A_n = \{a_m | m \geq n\}$$. $$A_n$$ 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 $$\epsilon \geq 0$$ random. Than $$cx_n + \epsilon > cx_n$$. So I just need to show that $$cx_n + \epsilon$$ is not a lowerbound for $$cA_n$$. But I can't 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. – drhab Aug 28 at 13:51
• I have added an answer for $c \le 0$. – copper.hat Aug 28 at 13:56