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 ?

Let $A = \liminf_{n \to \infty} x_n$ and $B = \limsup_{n \to \infty} x_n$.  For any $\epsilon > 0$, there is $N$ such that for all $k > N$, $A - \epsilon \le x_k \le B + \epsilon$.  Let $S_n = \displaystyle \sum_{k=1}^n x_k$.  Then for $n > N$,
$$ S_N + (n-N) (A - \epsilon) \le S_n \le S_N + (n-N) (B + \epsilon) $$
and so
$$ \eqalign{ A - \epsilon &= \lim_{n \to \infty} \frac{S_N + (n-N) (A - \epsilon)}{n}
\le \liminf_{n \to \infty} \dfrac{S_n}{n}\cr  
\limsup_{n \to \infty} \dfrac{S_n}{n} &\le \lim_{n \to \infty} \frac{ S_N + (n-N) (B + \epsilon)}{n} = B + \epsilon} $$
Now take $\epsilon \to 0+$.
