I have learned the definition of Banach Limit:
If $F$ is a bounded functional on $l^{\infty}$ , $\forall \{a_n\},\{b_n\} \in l^\infty$,
(1) $F(\{a_n\})=F(\{a_{n+1}\})$;
(2) If $a_n \ge 0, n=1,2,\cdots$, then $F(\{a_n\})\ge 0$;
(3) $F(\alpha\{a_n\}+\beta \{b_n\})=\alpha F(\{a_n\})+\beta F(\{b_n\})$, $\alpha,\beta$ are numbers;
(4) $\liminf_\limits{n \to \infty} a_n \le F(\{a_n\}) \le \limsup_\limits{n\to \infty} a_n$.
Then $F(\{a_n\})$ is the Banach limit of $\{a_n\}$, and $F(\{a_n\}):=\mathrm{Lim}\{a_n\}$ .
I have proved that such an $F$ does exists.
My question is:
how to prove that $\mathrm{Lim} \ y$ has different values, where $y = (1,0,0,1,1,0,0,0,0,1,1,1,0,0,0,0,0,0,\cdots) \in l^\infty$.
Edit:
I tried to construct a functional on a subspace of $l^\infty$ that satisfies (1)(2)(3)(4) and the value $F(y)$ is given.
I defined $M$ the subspace of $l^\infty$ generated by $\{a_n-a_{n+1}\}$, where $\{a_n\} \in l^\infty$ . And $e_1:=(1,1,\cdots,1,\cdots)$. I have proved that $d(e_1,\overline{M})=1$.
Let $M_1:=\mathrm{span}(M \cup \{e_1\})$, let $M_2 :=\mathrm{span}(M_1 \cup \{y\})$. Define $f(z)=\lambda d(e_1,M)$ for $z = m + \lambda e_1 \in M_1$. Then $f(e_1)=1$. Then Can I just say that let $f(y)=0$, $f(z)=\lambda d(e_1,M)$ for $z = m + \lambda e_1 + \mu y \in M_2$? If this is true, then I can use the Hahn-Banach to extend $f$.
I don't know how to prove that $y \not\in M_1$ and I am not sure whether my thinking is right.
