Banach Limit: understanding this step in the proof In chapter 7 of the book "Functional Analysis, Spectral Theory and Application" by Einsiedler where the discussion concerns Banach Limits. The author proves the statement
$$\liminf_{n \rightarrow \infty} \leq \text{LIM}((a_n)) \leq \limsup_{n \rightarrow \infty}$$
along the following lines:
Let $I = \inf_{n \geq 1}a_{n}$ and $ S = \sup_{n \geq 1}a_{n} $, so that $|a_n-\frac{I+S}{2}| \leq \frac{S-T}{2}$ for all $ n\geq 1 $, and hence
$$|\text{LIM}((a_n))-\frac{I+S}{2}| \leq \frac{S-I}{2}$$
It is not clear me to how the last statement follows. Here the symbol $\text{LIM}$ denotes the functional $ \text{LIM}:\ell^{\infty} \rightarrow \mathbb{R} $ such that
$$\text{LIM}((a_n))=L(a_1,\frac{a_1+a_2}{2},...)$$
and $L:\ell^{\infty} \rightarrow \mathbb{R}$ is the functional extended via the Hahn Banach Theorem from $ f:c \rightarrow \mathbb{R}:(a_n)\mapsto \lim_{n \rightarrow \infty} a_n $ and so $\text{LIM}$ is the Banach Limit functional. The problem is that since we used Hahn Banach we have no information on the quantity $\text{LIM}((a_n))$ if $(a_n)$ is not convergent in some sense. Indeed we know that $| \text{LIM}((a_n)) | \leq ||(a_n)||_{\infty}$ but how does the above statement follows is still not clear to me.
Could anyone provide some clarification? Thanks!
 A: The crucial thing you need is $LIM(a_n) \geq 0$ if $a_n \geq 0$ for all $n$. To see this let $0 \leq a_n \leq M$ for all $n$. Then $LIM (M-a_n) \leq M$ because the norm of the sequence $(M-a_n)$  does not exceed $M$. Hence $LIM (M) -LIM (a_n) \leq M$. [ Here $(M)$ is the constant sequence $(M,M,...)$]. But $LIM (M) =M$ so $M-LIM(a_n) \leq M$ which gived $LIM(a_n)  \geq 0$.  This implies that if $a \leq a_n \leq b$ for all $n$ then $a \leq LIM(a_n) \leq b$. Just take $a=\frac {I+S} 2 -\frac {S-I} 2$ and $b=\frac {I+S} 2 +\frac {S-I} 2$ to complete the argument. 
A: Banach limits are shift invariant $LIM(\{a_n\})=LIM(\{a_{n+1}\})$, and at the same time bounded linear functional, with respect to $\|\{a_n\}\|=\sup |a_n|$, with norm equal to one.
So 
$$
|LIM (\{a_n\})|\le \sup_{n\in\mathbb N}|a_{n+k}|,
$$
for all $k\in\mathbb N$. Thus 
$$
|LIM (\{a_n\})|\le \limsup_{n\in\mathbb N}|a_{n}|,
$$
Say that $a_n+c\ge 0$, for all $n$. Then 
$$
LIM(\{a_n\})+c=LIM(\{a_n+c\})\le \limsup |a_n+c|=\limsup (a_n+c)=c+\limsup a_n
$$
and hence
$$
LIM(\{a_n\})\le \limsup a_n.
$$
Similarly, if we put $\{-a_n\}$ in the place of $\{a_n\}$, we obtain
$$
LIM(\{a_n\})\ge \liminf a_n.
$$
