Banach limit

Remarks

Every Banach limit satisfies $\liminf_n x_n \le \varphi(x) \le \limsup_n x_n$. Check of the left inequality: if $x_n \ge c$ for all $n \ge N$, then $S^N x - c \mathbf{1}$ has nonnegative terms, so $\varphi(x) = \varphi(S^N x) \ge c$. Hence $\varphi(x) = \lim_n x_n$ for every convergent $x$, and $\abs{\varphi(x)} \le \sup_n \abs{x_n}$, so $\varphi$ is continuous of norm $1$. Banach limits exist by the Hahn–Banach theorem, applied to the sublinear functional $p(x) = \limsup_{m} \frac{1}{m} \sum_{k=1}^{m} x_k$ or a similar one.

Non-example: $\varphi(x) = \limsup_n x_n$ satisfies clauses 1, 2 and 3, but it is not linear. For $x = (1, 0, 1, 0, \dots)$ and $y = Sx$ one has $\varphi(x) + \varphi(y) = 2$ but $\varphi(x + y) = \varphi(\mathbf{1}) = 1$.

Non-example: $\varphi(x) = x_1$ is linear and positive with $\varphi(\mathbf{1}) = 1$, but it fails clause 2.