Let $l:l^{\infty}(\mathbb R)\rightarrow \mathbb R$ be a linear map( where $l^{\infty}$ is the set of bounded sequences) such that the following holds:
(i) $l(Tx)=l(x)$, where T is the left shift operator
(ii) if $x=x_n\geq0$, then $l(x)\geq 0$
(iii) $l(e)=1$ with $e=(1,1,1,1,.....)\in l^{\infty}$
And I want to proof the following 2 (of 4, so maybe not all points above are needed now) tasks:
(a) $l\in l^{\infty}(\mathbb R)'$ and $||l||=1$
(b) For $(x_n)_n\in l^{\infty}$ holds: $lim inf_{n\rightarrow \infty}x_n\leq l(x)\leq lim sup_{n\rightarrow \infty}x_n$
As for (b), let $m = \liminf_n x_n$ and $M= \limsup_n x_n$.
For all $\varepsilon >0$ you have eventually $x_n - m + \varepsilon \geq 0$. Then, shifting enough the sequence, you get that $$l(x +(-m + \varepsilon)e) \geq 0$$ i.e. $$l(x) + (-m + \varepsilon) l(e) = l(x) + -m + \varepsilon \geq 0$$
by arbitrarity of $\varepsilon$ you get $l(x) \geq m$.
The other inequality can be proved similarly.
Note that (b) implies (a) since for all $x$ you have $$-||x||_{\infty}=- \sup_n|x_n| \leq \liminf_n x_n \leq l(x) \leq \limsup_n x_n \leq \sup_n|x_n| \leq ||x||_{\infty}$$
So $|l(x)| \leq ||x||_{\infty}$. Finally, the equality holds for the constant sequence $e$, so the norm of $l$ must be $||l||=1$.