$\liminf$ and $\limsup$ of real sequences and their Cesaro averages

Question

Let $(x_n)$ a real bounded sequence and define $$ \sigma_n:=\frac{x_1+\cdots+x_n}{n} $$ for all $n$. Show that $$ \liminf x_n \le \liminf \sigma_n \le \limsup \sigma_n \le \limsup x_n. $$

Answer 1

I prove only the last inequality (the first one is symmetric, the middle one is trivial). Set $\ell:=\limsup x_n$ and $s:=\sup x_n$. Hence, for all $\varepsilon>0$, there exists $n_\varepsilon$ such that $$ x_n \le \ell+\varepsilon $$ for all $n \ge n_\varepsilon$. Now, for all $n\ge n_\varepsilon$ large, it holds \begin{align} \sigma_n&=\frac{x_1+\cdots+x_{n_\varepsilon}+x_{n_\varepsilon+1}+\cdots+x_n}{n}\\ &\le \frac{s\cdot n_\varepsilon+x_{n_\varepsilon+1}+\cdots+x_n}{n}\le \ell+2\varepsilon. \end{align} Therefore, since $\varepsilon$ is arbitrary, $\limsup \sigma_n \le \ell$.

Ps. In particular, if $(x_n)$ converges, then also $(\sigma_n)$ converges.