The linear functional $x \mapsto \lim x$ on space $c$ (of all convergent sequences) has norm $1$ and so by Hahn-Banach theorem possesses extensions $L$, of norm $1$, defined on space $\ell_{\infty}$. We call such functional $L$ an $\textbf{extended limit}$ (if I understand correctly the set of all extended limits is just the set of all linear and bounded extensions of limit functional $x \mapsto \lim x$ to $\ell_{\infty}$). Suppose $L$ is extended limit such that $L(x) = L(\sigma(x))$ for all $x \in \ell_{\infty}$, where $\sigma(x_1,x_2,x_3,\ldots) = (x_2,x_3,x_4,\ldots)$. We call such functional $L$ a $\textbf{Banach limit}$.
The author of article Im currently studying claims that $\textbf{any}$ extended limit $L$ has two following properties: (1) $L(x) \geq 0 $ whenever $x_n \geq 0$ for all $n$, (2) $\liminf x_n \leq L(x) \leq \limsup x_n$ for every $x \in \ell_{\infty}$.
I figured out how to prove (1). I've seen proofs of second property (2) only with additional assumption that $L$ is shift-invariant (L is Banach limit) and I can't see right now why this is true for any extended limit (not necessarily Banach limit). Im able to produce a particular extended limit that satisfies (2): consider functional $l(x) = \lim x_n$ on $c$ and sublinear functional $p(x) = \limsup x_n$ defined on $\ell_{\infty}$. Clearly $l(x) = p(x)$ for every $x \in c$ so by Hahn-Banach theorem there is extension $L$ of $l$ to whole space $\ell_{\infty}$ such that $L(x) \leq p(x)$ for every $x \in X$. Thus $L$ is extended limit such that $-p(-x) \leq L(x) \leq p(x)$ which is $\liminf x_n \leq L(x) \leq \limsup x_n$, for every $x \in \ell_{\infty}$. But this doesnt solve the problem.
So how can one prove that property (2) holds for any extended limit? Many thanks
