I'm given the space $c$ of convergent complex sequences. I consider the linear functional $\lambda$ defined by $$\lambda : c \to \mathbb{C} \\ \quad \quad \quad x \mapsto \lim_{n \to \infty} x_n.$$ By Hahn-Banach, I find a linear functional $\Lambda \in (\ell^\infty)^\ast$ such that $\lambda = \Lambda \circ i_{c \hookrightarrow \ell^\infty}$ and $\lVert\Lambda \rVert = \lVert\lambda \rVert = 1.$ I've proved that $\Lambda x \geq 0 \text{ if } x_n \geq 0 \ \forall n \in \mathbb{N}.$
I have to prove that the functional $\Lambda$ constructed above is a Banach Limit, so I need to prove the property of invariance under the shift operator, i.e., $$\Lambda(Sx) = \Lambda x, \text{ where } S(x_1,x_2,x_3, \dots) = (x_2,x_3,\dots).$$
I've seen a proof in Conway's book (but he didn't construct $\Lambda$ from $c$) and in other questions of MSE site it is proved constructing the functional $\Lambda$ from the space of Cesàro convergent sequences. My professor said we have to prove the theorem with $\Lambda$ constructed from $c.$
Could someone help me with the problem? I made some attempts (translating the proofs I've read) but always get stuck. Thanks to everyone!
@mihaild is right that this can't be done. I don't follow how we get $||\Lambda||=1$ from his comment. Here's a simple counterexample:
Proof: Let $E$ be the space of all $x\in\ell_\infty$ such that $\lim x_{2n}$ exists. Hahn-Banach gives us $\Lambda\in\ell_\infty^*$ with $||\Lambda||=1$ such that $\Lambda x=\lim x_{2n}$ for every $x\in E$.
So we certainly have $\Lambda x=\lim x_n$ for every $x\in c$. But if $x=(0,1,0,1,\dots)$ then $\Lambda x\ne\Lambda Sx$.