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Let $\{x_n \}$ be a sequence of real numbers and let $y_n = \frac{(x_1 + x_2 + ... + x_n)}{n}$.

(a) Prove that $\liminf x_n \le \liminf y_n \le \limsup y_n \le \limsup x_n$

(b) Give an example of a sequence $\{x_n\}$ for which all inequalities of part (a) are strict.

I honestly have no idea where to start on this. I can observe some of the easier things such as $\lim \inf x_n \le \lim \sup x_n$ Any hints would be appreciated.

Martin Sleziak
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2 Answers2

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First note $\limsup y_m \leq \sup_{ m\geq n} x_m $

This must be true right? Because either the right hand side is $\infty$, in which case, it is trivial, or it is not. Suppose not, let $\sum\limits_{i=1}^n x_i = M$, and $\sup_{ m\geq n} x_m= x$

$y_m = \dfrac{M + x_{n+1}+... + x_m}{m} \leq \dfrac{M+(m-n)x}{m}$ for $m\geq n$

So we have $\limsup y_m \leq \lim\limits_{m\rightarrow\infty} \dfrac{M+(m-n)x}{m} = x = \sup_{ m\geq n} x_m$

We now take the limit in $n$ obtain the right handside. If you use the fact $\limsup -y_m = - \liminf y_m$, you get the other inequality. The middle one is trivial. For examples, see the hint above.

Lost1
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Hint : a] just to give you the idea :

Think of a seq. $x_n$ whose limsup is $100$ -- how can you expect that limsup of averages

is $101$ while there are infinitely many terms in the tail of $x_n$ less than $101$?

b]try $x_n=(-1)^n (2n-1)=-1, 3,-5 ,7, -9, 11,-13,15,-17,19 $ , . .

Halil Duru
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    And for (b) $\lim \sup x_n = \infty \ \lim \inf x_n = -\infty \ \lim \sup y_n = 1 \ \lim \inf y_n = -1?$ –  Apr 11 '13 at 00:37
  • @HalilDuru hold on a second there, I don't think it is not that simple. sup is for $\sup_{m\geq n}$ and $y_n$ is the average of the first $n$ terms? how is the max of the terms after n comparable to the average of first $n$ terms. – Lost1 Apr 11 '13 at 10:28
  • @Using what you said, you can prove $y_n \leq \max_{m\leq n} x_n$, well if you take limsup on both side, the right hand side is not $\limsup x_n$, just consider the sequence $1/n$ – Lost1 Apr 11 '13 at 10:30
  • @Archie you might want to post your proof, if it is correct, it will enlighten me, but I don't think this is the correct approach. – Lost1 Apr 11 '13 at 10:32
  • This wasn't what i meant lost1--Think of a seq. $x_n$ whose limsup is $100$ - how can you expect that limsup of averages is 101 while there are infinitely many terms in the tail of $x_n$ less than 101? – Halil Duru Apr 11 '13 at 10:57
  • This version is clearer and not misleading i suppose as a hint.. – Halil Duru Apr 11 '13 at 11:20
  • yes, because it is not as simple as you put it max >= average... – Lost1 Apr 11 '13 at 23:06