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Let $\{a_n\}$ be a bounded sequence of real numbers. Let $s_n=\dfrac{a_1+a_2+\cdots+a_n}{n}~\forall~n\in\mathbb N.$ Show that $\lim \inf a_n\le\lim\inf s_n.$

The only definition I know of limit infimum is the lowest subsequencial limit. I've first prechoosed a convergent subsequence $\{s_{n_k}\}$ converging to $l$ of $\{s_n\}$ and then tried to find a convergent subsequence of $\{a_n\}$ whose limit is $\le l.$

But it didn't help.

Martin Sleziak
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sujan
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  • Even better, you could find a subsequence which is either monotone increasing or monotone decreasing. That would break your proof into two cases –  Feb 22 '14 at 07:34
  • @Bryan I have checked the logic and found that it still works (without any modification as remarks below) for unbounded sequences, and I added some explanation on the difference of the definition. – Yai0Phah Feb 22 '14 at 08:25
  • See http://math.stackexchange.com/questions/193157/if-sigma-n-fracs-1s-2-cdotss-nn-then-operatornamelim-sup-sigma and http://math.stackexchange.com/questions/317750/is-it-true-that-limsup-phi-le-limsup-a-n – Martin Sleziak Feb 22 '14 at 11:30
  • See also: http://math.stackexchange.com/questions/1455916/let-s-n-be-a-sequence-of-nonnegative-numbers-and-sigma-n-frac1ns-1 – Martin Sleziak Oct 10 '15 at 12:23

1 Answers1

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Let $\alpha=\liminf_{n\to\infty}a_n$, by the Remark2 below, there exists $N>0$, such that for each $n>N$, we have $a_n\ge\beta$ (*), thus $$s_n=\frac{a_1+\dotsb+a_n}n=\frac{a_1+\dotsb+a_{N+1}+\dotsb+a_n}n\ge\frac{a_1+\dotsb+a_N+(n-N)\beta}n$$ Take limit as $n\to\infty$, we have $\liminf_{n\to\infty}s_n\ge\beta$ for each $\beta<\alpha$, hence $\liminf s_n\ge\liminf a_n$.

Remark1 The condition that $\{a_n\}$ is bounded implies that $\alpha\in\mathbb R$. But even if $\{a_n\}$ is unbounded, $\alpha\in[-\infty,+\infty]$, the preceding argument also works.

Especially, even if $\alpha=-\infty$, there's no $\beta<\alpha$, thus $\forall\beta<\alpha\colon\liminf s_n\ge\beta$ is vacuously true, which also implies that $\liminf s_n\ge\alpha$.

Generally, if $S$ is a totally-ordered set, and $x\ge\beta$ for all $\beta<\alpha$, then $x\ge\alpha$, no matter whether $\alpha$ is the minimum.

Remark2 $\alpha=\liminf a_n$ if and only $\alpha$ is a subsequential limit and $\forall\beta<\alpha,\exists N,\forall n>N\colon a_n\ge\beta$.

If $\alpha=\liminf a_n$, we have $\alpha$ is a subsequential limit. Suppose $\exists\beta_0<\alpha,\forall N,\exists n>N\colon\alpha<\beta_0$, we can choose a subsequence $\{a_{n_k}\}$, such that $a_{n_k}<\beta_0$, which contradicts the assumption that $\alpha$ is the lowest subsequential limit (choose a proper subsequence). For the reverse deduction, note that $\forall\beta<\alpha,\exists\gamma(\beta<\gamma<\alpha),\exists N>0,\forall n>N\colon a_n\le\gamma$, thus $\beta$ isn't a subsequential limit. The argument works even if $\alpha=-\infty$, by the same reason of vacuous truth.

Yai0Phah
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