# Average limit superior [duplicate]

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 ?

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## marked as duplicate by Martin Sleziak, John, Claude Leibovici, Najib Idrissi, egregApr 17 '14 at 10:05

Use this thread to get the second inequality. Apply this one to $\{-x_n\}$ to get the other inequality.
Your other post gives that $$\sup_{n\ge k} \sigma_n\le \sup_{l\ge k} s_k+\sup_{n\ge k} \frac{1}{n}\sum_{j=1}^k s_j,$$ but then, what do you do with the last term ? –  Klaus Dec 26 '12 at 21:35
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 $\epsilon \to 0+$.