First, the following easy corollary of Hahn-Banach for normed spaces might be useful:
Proposition: Let $X$ be a normed space, and let $Y$ be a subspace with $\phi\in Y^*$. Then $\phi$ has an extension to $\tilde{\phi}\in X^*$ such that $\|\phi\|=\|\tilde{\phi}\|$.
To prove this, apply the usual Hahn Banach theorem with $p(x)=\|\phi\|\|x\|$
Now, as you did before, define
$\psi_0:W\to\mathbb{R}$ by
$\psi_0((x_n)_n)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}x_n$
By Hahn-Banach, we can extend this to a function $Lim:l^{\infty}\to\mathbb{R}$ such that
$\|Lim\|=\|\psi_0\|$
Let's prove part c) first now. In light of the above, we need only show that
$\|\psi_0\|=1$. Indeed, we have $\frac{1}{n}\sum_{k=1}^{n}x_n\leq \frac{n\|(x_n)_n\|_{\infty}}{n}=\|(x_n)_n\|_{\infty}$
Thus $\|\psi_0\|\leq 1$. To see that $\|\psi_o\|=1$, consider the sequence $x\in W$ with only ones as its entries.
This proves c).
For part a)
Note that by linearity:
$Lim(x_1,x_2,...)=Lim(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)$
$=\psi_0(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)=Lim(x_2,x_3,x_4,...)$
This proves a)
Finally, for part b), let $x=(x_n)\in l^{\infty}$ Choose $a,b$ so that
$a<\liminf_{n\to\infty}(x_n)$ and
$b>\limsup_{n\to\infty}(x_n)$
Then there is $N$ such that for $n>N$, we have $a<x_n<b$.
Denote now 1 by the sequence with only ones as its entries and denote $y=(y_n)_n$ as the sequence such that $y_n=x_{n+N}$
Then $w:=y-a\cdot$1 and $z:=b\cdot$1$-y$ are positive and bounded sequences.
Thus,
$0\leq Lim(w)=Lim(y)-a=Lim(x)-a$
So that, $a\leq Lim(x)$
where the last equality was obtained using the shift invariance proven in a)
Similarly, we obtain $Lim(x)\leq b$.
Since, $a$ and $b$ were arbitrary, b) follows.
The one step that needs justification is that if $x=(x_n)_n$ is positive and bounded, then $Lim(x)\geq 0$
Indeed, without loss of generality, $\|x\|\leq1$(otherwise, rescale x)
Now, $1-Lim(x)=Lim($1$-x)\leq\|$1$-x\|\leq 1$
Best Answer
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.