How to prove an average inequality about limsup\liminf

Question

Suppose $an$ is an infinite sequence in $\Bbb R$ and $bn$ is the arithmetic average of $an$, or $bk = \frac{(a_1 + ... + a_k)}k$

Why is it true that: $\lim$(inf $an) \le$ $\lim$(inf $bn) \le \lim$(sup $bn) \le \lim$(sup $an$)

Answer 1

For ease of notation let $$ \ell_n = \inf_{n\leq m < \infty}a_m $$

For the first inequality use the $\epsilon-\delta$ definition of $\inf$: For any $\epsilon > 0$ there exists some $N$ (possibly depending on $n$) so that $\ell_n < a_m+\epsilon $ for all $m > N$, hence

\begin{align} \ell_n = \frac{(m-N)\ell_n}{m-N} % &< \frac{(a_{N+1}+\epsilon)+\dotsb+ (a_m+\epsilon) }{m-N}\\ % &= \frac{a_{N+1}+\dotsb+ a_m}{m}\cdot \frac{m}{m-N} + \epsilon\\ % &= \left(b_n - \frac{a_1+\dotsb+ a_N}{m}\right) \cdot \frac{m}{m-N} + \epsilon \end{align}

Given that it holds for all $m > N$,

\begin{align} \ell_n &\leq \lim_{m\to\infty}\bigg(b_n - \underbrace{(a_{1}+\dotsb+ a_N)}_{\text{constant in $m$}}\cdot\underbrace{\frac{1}{m}}_{\to\,0}\bigg)\cdot \underbrace{\frac{m}{m-N}}_{\to\,1} + \epsilon = b_n+\epsilon \end{align}

This holds for all $\epsilon > 0$, so let $\epsilon \to 0^+$, giving $\ell_n \leq b_n$. The first inequality follows:

\begin{align}\operatorname{lim\,inf}_{n\to\infty}a_n &= \operatorname{lim\,inf}_{n\to\infty}\inf_{n\leq m < \infty}a_m % = \operatorname{lim\,inf}_{n\to\infty} \ell_n\\ % &\leq \operatorname{lim\,inf}_{n\to\infty} b_n \end{align}

For the second inequality, trivially $$\operatorname{lim\,inf}_{n\to\infty} b_n \leq \operatorname{lim\,sup}_{n\to\infty} b_n$$

And finally, for the third inequality consider $(-a_n)$ and $(-b_n)$. Applying the first inequality to these (noting that the arithmetic mean of negation is the negation of the arithmetic mean) we have $$\operatorname{lim\,inf}_{n\to\infty} (-a_n) \leq \operatorname{lim\,inf}_{n\to\infty} (-b_n)$$ and given that $\operatorname{lim\,sup}_{n\to\infty}(-x_n) = -\operatorname{lim\,inf}_{n\to\infty}x_n$ we conclude

$$-\operatorname{lim\,sup}_{n\to\infty}a_n \leq -\operatorname{lim\,sup}_{n\to\infty} b_n$$