Prove that $\liminf (-a_n) = -\limsup (a_n)$.
by $\liminf$ and $\limsup$ I mean the limit supremum and limit infimum of a sequence $a_n$.
do you guys have any hints?
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3$\begingroup$ Have you tried using the definition of lim inf and lim sup? $\endgroup$– JustpassingbyNov 26, 2015 at 21:01
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$\begingroup$ your'e right, I deleted the example. $\endgroup$– F1sargyanNov 26, 2015 at 21:05
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$\begingroup$ $-\liminf (-a_n)=-\lim_{n\to\infty} \inf_{m \geq n} (-a_m) = \lim_{n\to\infty} - \inf_{m\geq n} (-a_m) = \lim_{n \to\infty} \sup_{m \geq n} -(-a_m) = \lim_{n\to\infty} \sup_{m \geq n} (a_m) = \limsup (a_n)$. The only step that really needs justification is switching from the $\inf$ to the $\sup$. Can you justify that step? $\endgroup$– kccuNov 26, 2015 at 21:07
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$\begingroup$ @Justpassingby what do you mean by using the definition? the definition I know is that lim sup is the supremum of the partial limits set of a sequence $a_n$ and for lim inf a similar definition, did u mean something else? $\endgroup$– F1sargyanNov 26, 2015 at 21:08
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$\begingroup$ @F1sargyan Working from the definitions is essentially what I did in my comment. $\endgroup$– kccuNov 26, 2015 at 21:09
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1 Answer
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I assume that the sequence $(a_n)$ is bounded. You want to show $$ \sup_m\inf_{n\geq m}(-a_n)=-\inf_m\sup_{n\geq m}(a_n)\tag{*} $$
Show first that for any bounded (real) sequence $(b_n)$: $$ \sup_n (-b_n)=-\inf_n b_n\tag{**} $$ Now use (**) to show (*).
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1$\begingroup$ en.wikipedia.org/wiki/Limit_superior_and_limit_inferior $\endgroup$– user9464Nov 27, 2015 at 12:42