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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 $\lim \inf x_n \le \lim \inf y_n \le \lim \sup y_n \le \lim \sup x_n$

(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.

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2 Answers 2

up vote 1 down vote accepted

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.

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I guess I'm not seeing how $\dfrac{M+(m-n)x}{m} = x$? Doesn't that mean that $M = nx$? –  Archie Apr 11 '13 at 22:50
@Archie I forgot add lim... Notice that term clearly has a lim, so lim = limsup –  Lost1 Apr 11 '13 at 23:04
Ah! Makes sense now. I realized my previous work was wrong after finding a counter-example. Thank you very much. –  Archie Apr 11 '13 at 23:05
@Archie should have said so. –  Lost1 Apr 11 '13 at 23:06
Sorry about that! :( –  Archie Apr 11 '13 at 23:21

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 $ , . .

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And for (b) $\lim \sup x_n = \infty \\ \lim \inf x_n = -\infty \\ \lim \sup y_n = 1 \\ \lim \inf y_n = -1?$ –  Archie Apr 11 '13 at 0:37
exactly ......... –  Halil Duru Apr 11 '13 at 5:27
@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

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