# Proof checking $\liminf cx_n = c \limsup x_n$

I just started to work with $$\limsup$$'s and $$\liminf$$'s and I would like to know if my proof of the identity

$$\begin{equation} \liminf cx_n = c \limsup x_n \end{equation}$$

with $$x_n$$ a bounded sequence and $$c\leq0$$ is correct.

Let $$a = \limsup x_n$$ and $$\epsilon>0$$. Then

$$\begin{equation} x_n < a+\epsilon \end{equation}$$

for $$n$$ sufficiently large. Multiplying by $$c$$ we get the inequeality

$$\begin{equation} c x_n >ca + c\epsilon \end{equation}$$

or

$$\begin{equation} cx_n>ca-|c|\epsilon. \end{equation}$$

That is $$\liminf cx_n = ca$$ which implies $$\liminf cx_n = c\limsup x_n$$.

• An easier way: $\liminf x_n = - \limsup (-x_n)$, and constant multiplication is clear. – FearfulSymmetry Jun 5 '20 at 16:27
• I know, but if $c=-1$, is my proof correct? – user2820579 Jun 5 '20 at 16:30
• Last inequality doesn't show exactly that $\lim inf cx_n = ca$. You need to prove that $ca$ is a limiting point – Peanut Jun 5 '20 at 17:01

## 1 Answer

What you did only proves that $$ca\leqslant\liminf_ncx_n$$. You can prove that you actually have an equality by proving that some subsequence of the sequence $$(cx_n)_{n\in\Bbb N}$$ converges to $$ca$$ But that is easy. Take a subsequence $$(x_{n_k})_{k\in\Bbb N}$$ whose limit is $$a$$ and then $$\lim_kcx_{n_k}=ca$$.

• Correct me if I am wrong, for $\liminf x_n$ to exist, it is enough to prove that for sufficiently large $n$, $\liminf x_n - \epsilon\leq x_n$. That's why I didn't have to consider the inequality $\liminf cx_n \leq ca$. – user2820579 Jun 5 '20 at 20:41
• So for reference, the full theorem is here: $a=\liminf x_n$ if and only if whenever $\alpha < a$, $\lbrace n\in\mathbb{N}: x_n < \alpha\rbrace$ is finite and whenever $a<\beta$ we have $\lbrace n\in\mathbb{N} : x_n < \beta\rbrace$ is infinite. – user2820579 Jun 5 '20 at 20:49
• The statement from your first comment makes no sense, since it is circular. On the other hand, the “full theorem” of your second comment it correct indeed. – José Carlos Santos Jun 5 '20 at 21:11
• I can't see how is circular. First I proved that for $\epsilon>0$ and sufficiently large $n$, $x_n < \limsup x_n + \epsilon$ (this the alternative characterisation of $\limsup$). Then I can multiply by $c \leq 0$ to get $cx_n > c\limsup x_n - |c| \epsilon$. This is the "$\alpha$-statement" in my theorem, so I can conclude $\liminf cx_n = c\limsup x_n$. – user2820579 Jun 5 '20 at 21:17
• Indeed, for every $\varepsilon>0$, there are only finitely many $n$'s such that $\liminf_nx_n-\varepsilon\leqslant x_n$. But it is not true that $\liminf_nx_n$ is the only number with that property. Actually, every number smaller than $\liminf_nx_n$ has that property too. – José Carlos Santos Jun 5 '20 at 21:25