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I have the following theorem:

Suppose $a_n$ is a bounded sequence and $u=\limsup(a_n)$. Then:

i) There is a subsequence $a_{n_k}$ with $a_{n_k}\rightarrow u$

ii) If $a_{m_k}$ is convergent subsequence with limit x, then $x\leq u$

Do we have an analogous result for $\liminf (a_n)$? (I'm pretty sure we do but it's not in my notes which seems strange)

Thanks very much for any help

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Of course, thanks –  hmmmm Apr 7 '12 at 10:06

1 Answer 1

up vote 3 down vote accepted

Yes. One can even derive one pair of consequences from the other pair via an inherent symmetry:

$$\limsup (-a_n) =-\liminf a_n$$ $$\liminf (-a_n)=-\limsup a_n$$

For example: Suppose $\lim\inf a_n=u$. Then $\limsup(-a_n)=-u$, thus there is a subsequence of $-a_n$ converging to $-u$ (negate all signs in this subsequence, and that of $-u$, and we have our analogue of result #1). Furthermore if a subsequence of $-a_n$ converges to $-x$, we have $-x\le -u$ by (ii) for the $\limsup$ case, whence $x\ge u$ (notice the negation of the subsequence converges to $x$).

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Yeah- being a bit stupid- thanks very much help! –  hmmmm Apr 7 '12 at 10:07

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