Existence of Banach limits using Hahn-Banach 
Let $(x_j) \in X=l^{\infty}(\mathbb{N})$ and define the left shift $T((x_j))=(x_{j+1})$. A linear functional $l:X \to \mathbb{R}$ is called a Banach limit if the following holds
(a) $l$ is positive: if $x_1,x_2, \dots \geq 0$, then $l((x_j))\geq 0$.
(b) $l \circ T=l$
(c) $l((1,1,\dots))=1$.
Establish the existence of a Banach limit using Hahn-Banach.

I considered the subspace $Y\in X$ of constant sequences $x=(a,a,\dots )$ where $a\in \mathbb{R}$. Define $\tilde{l}(x)=a$. Then obviously $\tilde{l}:Y\to \mathbb{R}$ satisfies (a)-(c). By Hahn-Banach there exists a linear functional $l:X\to \mathbb{R}$ such that $l_Y=\tilde{l}$ and $||l||_{X^{*}}= ||\tilde{l}||_{Y^{*}}$. Then certainly condition (c) is satisfied but I am not sure how to show the other ones.
 A: Let $X= l^\infty(\mathbb N)$, let $T: X \to X$, given by $T((x_j)_{j \in \mathbb N}) = (x_{j+r})_{j \in \mathbb N}$, where $r \in \mathbb N_+$
Take $p:X \to \mathbb R$ be given by $p(x) = \limsup \frac{1}{n} \sum_{k=1}^n x_k$, that is Cesaro-limit
Check that $p$ is a banach functional $($ that is $p(x+y) \le p(x) + p(y)$ and $p(tx) = tp(x)$ for $t \ge 0$ $)$, hence ( I don't know which version of H-B Theorem you are familiar with, but $p$ is both convex and banach functional. )
Take $\phi : c \to X$, given by $\phi(x) = \lim x_n$, where $c \subset l^\infty(\mathbb N)$ is subspace of convergent sequences.
Clearly $\phi \le p$, since it is known fact from analysis, that for convergent sequences we have $\lim x_n = \lim \frac{1}{n} \sum_{k=1}^n x_k$.
By Hanh Banach theorem ($\phi$ is a linear functional ), we have extension $\psi: X \to \mathbb R$, sch that $\psi(x) = \phi(x)$ for $x \in c$.
Now we want to establish those conditions, firstly note that:
$\psi(x) \le p(x) \le \limsup x_n$
and similarly (since $\psi(-x) = -\psi(x)$)
$\psi(-x) = -\psi(x) \ge -p(x) \ge \liminf(-x)$, so $\psi(x) \ge \liminf(x)$
1) Take $x \ge 0$ (that is $x_k \ge 0$, for all $k \in \mathbb N$). Then both $\liminf, \limsup$ are between $0$ and $\infty$, so $\psi(x) \ge 0$
3) It's easy, we have $\psi((1,1,...)) = \phi((1,1,...)) = \lim (1,1,...) = 1$
2) Note that $\psi(x) - \psi(T(x)) = \psi(x - T(x)) \le p(x-T(x)) = \limsup \frac{1}{n}( \sum_{k=1}^r x_k - \sum_{k=n+1}^{n+r} x_k)$
And since we're in $l^\infty(\mathbb N)$, we have $|x| \le M$, so $p(x-T(x)) \le \limsup \frac{1}{n} \cdot 2rM$ which tends to $0$. 
And similarly:
$ \psi(T(x)) - \psi(x) \le p(T(x) - x) \le \limsup \frac{1}{n} \cdot 2rM$, so that
$ \psi(T(x)) = \psi(x)$, which is what we needed.
