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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!

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  • $\begingroup$ It is not clear what you are asking, the limit function is not defined everywhere. $\endgroup$ – copper.hat May 31 at 14:05
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    $\begingroup$ 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 $\|\Lambda\| = 1$. $\endgroup$ – mihaild May 31 at 14:15
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    $\begingroup$ My apologies, I need to read more carefully $\endgroup$ – copper.hat May 31 at 14:16
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    $\begingroup$ @mihaild You can't define $x$ and $y$ independently, as $x + y \in c$. $\endgroup$ – Theo Bendit May 31 at 14:21
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    $\begingroup$ @mihaild I don't see how you get $||\Lambda||=1$. $\endgroup$ – David C. Ullrich May 31 at 14:41
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@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 $\Lambda x\ne\Lambda Sx$.

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