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}


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

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

  • 2
    $\begingroup$ An easier way: $\liminf x_n = - \limsup (-x_n)$, and constant multiplication is clear. $\endgroup$ – FearfulSymmetry Jun 5 '20 at 16:27
  • $\begingroup$ I know, but if $c=-1$, is my proof correct? $\endgroup$ – user2820579 Jun 5 '20 at 16:30
  • $\begingroup$ Last inequality doesn't show exactly that $\lim inf cx_n = ca$. You need to prove that $ca$ is a limiting point $\endgroup$ – Peanut Jun 5 '20 at 17:01

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 $\lim_kcx_{n_k}=ca$.

  • $\begingroup$ 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 $\liminf cx_n \leq ca$. $\endgroup$ – user2820579 Jun 5 '20 at 20:41
  • $\begingroup$ 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. $\endgroup$ – user2820579 Jun 5 '20 at 20:49
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Jun 5 '20 at 21:11
  • $\begingroup$ 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 $\liminf cx_n = c\limsup x_n$. $\endgroup$ – user2820579 Jun 5 '20 at 21:17
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Jun 5 '20 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.