Average limit superior Let $\mathcal{l}_\mathbb{R}^\infty$ be the space of bounded sequences in $\mathbb{R}$. We define a map $p: \mathcal{l}_\mathbb{R}^\infty\to\mathbb{R}$ by
$$p(\underline x)=\limsup_{n\to\infty} \frac{1}{n}\sum_{k=1}^n x_k.$$
My notes claim that
$$\liminf_{n\to\infty} x_n\le p(\underline x)\le \limsup_{n\to\infty} x_n.$$
I haven't found a neat way to show that this holds (only a rather complicated argument). Is there an easy, intuitive way ?

Use this thread to get the second inequality. Apply this one to $\{-x_n\}$ to get the other inequality.
