A corollary of Hahn–Banach theorem and a generalized limit function of $\ell_\infty$ A corollary of Hahn–Banach theorem states that

Let $E$ be a normed vector space, $M$ a proper closed subspace and $x \in E$. If $d(x,M) = \delta > 0$, so exists $f \in E'$ such that $\|f\|=1$, $f(x)=\delta$ and $f(m)=0$  $\forall m \in M $.

Consider $T: \ell_\infty \rightarrow \ell_\infty$ a bounded linear operator defined by $$T(x_1,x_2,x_3,\dots) = (x_2,x_3,\dots).$$
Let $M=\{ x-T(x) : x \in \ell_\infty\}$, so $M$ is a subspace of $\ell_\infty$. If $e=(1,1,1,\dots) \in \ell_\infty$, so $d(e,M)=1>0$. Then, applying the corollary above in $\overline{M} $, exists $f \in \ell_\infty'$ such that $\|f\|=1$, $f(e)=d(x, \overline{M}) = d(e, M) =1$ and $f(x)=0~\forall x \in M\subset \overline{M} $.
I was able to show that $$f(x_1,x_2,x_3,\dots) = f(x_2,x_3,\dots) ~~\forall (x_n) \in \ell_\infty.$$ Besides that, we have that $\forall x = (x_n) \in c = \{ (x_n) \in \ell_\infty : x_n \text{ is convergent} \}$ $$f(x) = \lim_{n \rightarrow \infty} x_n$$
Now, let $x=(x_n), y=(y_n) \in \ell_\infty$ such that $x_n \geq y_n$ $\forall n \in \mathbb{N}$. How can I show that $f(x) \geq f(y)$?
 A: I'm going to try to answer this from the real $\ell^\infty$ perspective. Say we take a positive sequence $(x_n)$. Then, consider the sequence
$$y_n = e_n - \frac{x_n}{\|x_n\|},$$
where $e_n = 1$ for all $n$. Note that $0 \le y_n \le 1$ for all $n$, so by definition of the norm of $f$,
$$f(y_n) \le 1 = f(e_n).$$
Rearranging,
\begin{align*}
&f\left(e_n - \frac{x_n}{\|x_n\|}\right) \le f(e_n) \\
\implies \, &f(e_n) - \frac{f(x_n)}{\|x_n\|} \le f(e_n) \\
\implies \, &\frac{f(x_n)}{\|x_n\|} \ge 0 \\
\implies \, &f(x_n) \ge 0.
\end{align*}
By linearity, this implies what you want shown.
A: The fact that $\|f\|=f(e)$ implies positivity. The proof is not specific to $\ell^\infty$, it works on any C$^*$-algebra $A$. 
So assume that $A$ is a C$^*$-algebra, and $f:A\to\mathbb C$ a bounded linear functional, with $\|f\|=f(e)$, where $e$ is the unit of the algebra. 
First, let $a\in A$ be selfadjoint ($a^*=a$) with $\|a\|=1$. For any $n\in \mathbb N$, 
\begin{align}\tag1
|f(a)+in|&=|f(a+ine)|\leq\|a+ine\|=\|(a-ine)^*(a-ine)\|^{1/2}\\ \ \\
&=\|(a+ine)(a-ine)\|^{1/2}=\|a^2+n^2e\|^{1/2}\\ \ \\
&=(1+n^2)^{1/2}.
\end{align}
The last equality is due to $a=a^*$, which implies that $a^2\geq0$. In particular
$$
|\operatorname{Im} f(a)+n|=|\operatorname{Im}(f(a)+in)|\leq|f(a)+in|\leq(1+n^2)^{1/2}.
$$
So 
$$
n-(1+n^2)^{1/2}\leq\operatorname{Im}f(a)\leq (1+n^2)^{1/2}-n.
$$
By taking $n$ arbitrarily large, we obtain $\operatorname{Im}f(a)=0$. If we revisit $(1)$ with this knowledge, we get 
$$
(f(a)^2+n^2)^{1/2}\leq(1+n^2)^{1/2},
$$
so $f(a)^2\leq1$. So $f(a)$ is real, and $-1\leq f(a)\leq 1$. 
If now $b$ is positive with $0\leq b\leq e$, the elements $2b-e$ is selfadjoint and $-e\leq 2b-e\leq e$, so $\|2b-e\|\leq 1$. By the above $$-1\leq f(2b-e)=2f(b)-1\leq 1,$$ which we may rewrite as 
$$
0\leq f(b)\leq 1. 
$$
In particular $f(b)\geq0$, so $f$ is positive. 
