# Proving the “shift” property in order to prove that I have a Banach Limit

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!

• It is not clear what you are asking, the limit function is not defined everywhere. – copper.hat May 31 '19 at 14:05
• Without some addition requirement it's not necessary true: $x = (1, 0, 1, 0, \ldots)$ isn't in $c$ and $y = (0, 1, 0, 1, \ldots)$ can't be approximated with $x$ and $c$, so you can define, say, $\Lambda(x) = 0$, $\Lambda(y) = 1$ and still have $\|\Lambda\| = 1$. – mihaild May 31 '19 at 14:15
• My apologies, I need to read more carefully – copper.hat May 31 '19 at 14:16
• @mihaild You can't define $x$ and $y$ independently, as $x + y \in c$. – Theo Bendit May 31 '19 at 14:21
• @mihaild I don't see how you get $||\Lambda||=1$. – David C. Ullrich May 31 '19 at 14:41

@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:
There exists $$\Lambda\in\ell_\infty^*$$ such that $$||\Lambda||=1$$, $$\Lambda x=\lim x_n$$ for every $$x\in c$$, but $$\Lambda\circ S\ne\Lambda$$.
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 $$\Lambda x\ne\Lambda Sx$$.