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


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?

Thanks in advance!


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.

  • $\begingroup$ 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. $\endgroup$ – drhab Aug 28 at 13:51
  • $\begingroup$ Yeah, my mistake! Edited it! $\endgroup$ – Kabouter9 Aug 28 at 13:54
  • $\begingroup$ @drhab: Thanks for catching that. I suspect you are correct. $\endgroup$ – copper.hat Aug 28 at 13:54
  • $\begingroup$ I have added an answer for $c \le 0$. $\endgroup$ – copper.hat Aug 28 at 13:56

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