# prove limsup-liminf<2

• December 5th 2011, 12:49 PM
wopashui
prove limsup-liminf<2
Suppose that there exists $N \in N$, such that $|a_n-a_m|<1$for all $n,m>=N$ (this will be true, in particular, if $(a_n)$ is cauchy). Prove that $(limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n)$ $<=2$.

it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.
• December 5th 2011, 01:07 PM
TheEmptySet
Re: prove limsup-liminf<2
Quote:

Originally Posted by wopashui
Suppose that there exists $N \in N$, such that $|a_n-a_m|<1$for all $n,m>=N$ (this will be true, in particular, if $(a_n)$ is cauchy). Prove that $(limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n)$ $<=2$.

it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.

I will leave the epsilons to you but here is an idea. First there exists a subsequence
$a_{n}_k$
that converges to the limit supreemum and another subsequence
$a_{n}_j$
that converges to the lim inf

Now

$|L_{\text{sup}}-L_{\text{inf}}| \le |L_{\text{sup}}-a_n+a_n-L_{\text{inf}}| \le |L_{\text{sup}}-a_n|+|a_n-L_{\text{inf}}| = |L_{\text{sup}}-a_{n}_k+a_{n}_k-a_n|+|a_n-a_{n}_j+a_{n}_j-L_{\text{inf}}|$

Now just use the triangle inequality again and find the n,j, and k's to show what you want
• December 5th 2011, 01:35 PM
Plato
Re: prove limsup-liminf<2
Quote:

Originally Posted by wopashui
Suppose that there exists $N \in N$, such that $|a_n-a_m|<1$for all $n,m>=N$ (this will be true, in particular, if $(a_n)$ is cauchy). Prove that $(limsup_{n-->\infty}a_n)-(liminf_{n-->\infty}a_n)$ $<=2$.
it seems like an easy proof, but i can't find the connection between limsup and cauchy sequence, need some help.

I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

First, you must prove the the $\liminf~\&~\limsup$ both exist.
To do that notice that from the given $\left( {\forall n \geqslant N} \right)\left[ {\left| {a_n - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N \leqslant a_n \leqslant 1 + a_N } \right]$
Thus they exist and if $\lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n}$ then $-1+a_N\le\lambda~\&~\sigma\le 1+a_N$.

Now you finish
• December 5th 2011, 09:23 PM
wopashui
Re: prove limsup-liminf<2
thanks, so is TheEmptySet's approach correct?
• December 6th 2011, 04:03 PM
wopashui
Re: prove limsup-liminf<2
Quote:

Originally Posted by Plato
I think that you have the wrong idea here. The given does not prove that the sequence is Cauchy. Rather it asks you to note that the conclusion would apply to Cauchy sequences.

First, you must prove the the $\liminf~\&~\limsup$ both exist.
To do that notice that from the given $\left( {\forall n \geqslant N} \right)\left[ {\left| {a_n - a_N } \right| \leqslant 1\, \Rightarrow \, - 1 + a_N \leqslant a_n \leqslant 1 + a_N } \right]$
Thus they exist and if $\lambda=\liminf{a_n}~\&~\sigma= \limsup{a_n}$ then $-1+a_N\le\lambda~\&~\sigma\le 1+a_N$.

Now you finish

why are you using N not m here? is it because if $a_m$is not bounded, then $a_n$ will not be bounded, but $a_N$is fixed, so $a_n$ is bounded. but why do we have <= not strictly < here?
• December 6th 2011, 04:15 PM
Plato
Re: prove limsup-liminf<2
Quote:

Originally Posted by wopashui
why are you using N not m here? is it because if $a_m$is not bounded, then $a_n$ will not be bounded, but $a_N$is fixed, so $a_n$ is bounded. but why do we have <= not strictly < here?

Do you understand any of this?
It was convenient to use a fixed $a_N~,~N=m$.

Surely you understand that if $|a-b|<1$ then it is also true that $|a-b|\le 1$.