3

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!

DrinkingDonuts
  • 1,257
  • 6
  • 18
  • It is not clear what you are asking, the limit function is not defined everywhere. – copper.hat May 31 '19 at 14:05
  • @copper.hat The limit function is only considered over the space of convergent sequences (where it's well-defined and have the required properties to apply Hahn-Banach). Over $\ell^\infty$ we consider a "Hahn-Banach" extension of $\lim,$ that coincides with the limit function whenever the sequence $x \in \ell^\infty$ is convergent. I'm missing something? – DrinkingDonuts May 31 '19 at 14:09
  • Could you provide the definition of Banach Limits you're using? Whenever I've looked at them, invariance under left shift was part of the definition. – Theo Bendit May 31 '19 at 14:13
  • 2
    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
  • @TheoBendit You're correct, I've made a mistake. My professor asked me to prove the existence of a Banach Limit constructing from $c.$ So my question must be read: Prove that the "Hahn-Banach" extension $\Lambda$ of $\lim$ over $c$ is a Banach Limit. – DrinkingDonuts May 31 '19 at 14:16
  • 1
    My apologies, I need to read more carefully – copper.hat May 31 '19 at 14:16
  • This might help? https://math.stackexchange.com/questions/2764172/a-corollary-of-hahn-banach-theorem-and-a-generalized-limit-function-of-ell-in/ – Theo Bendit May 31 '19 at 14:20
  • 2
    @mihaild You can't define $x$ and $y$ independently, as $x + y \in c$. – Theo Bendit May 31 '19 at 14:21
  • @TheoBendit, But the definition is consistent, since if $\Lambda(x) = 0$, then $\Lambda(y) = \Lambda(1-x) = 1-\Lambda(x) = 1$. Indeed, such $\Lambda$ can be realized by the $\omega$-limit associated to the ultrafilter of $\mathbb{N}$ containing ${2,4,6,\cdots}$. – Sangchul Lee May 31 '19 at 14:23
  • @mihaild Actually I don't know what you are referring to. I know that I don't a unique value of $\Lambda$ for a non-convergent sequence. Maybe, here is the necessity of more assumptions. Then, what I need to suppose in order to construct a Banach Limit from $c$? – DrinkingDonuts May 31 '19 at 14:26
  • @TheoBendit The thread in the link you provided is the approach follow in Conway's book. There, the Banach Limit is constructed whithout $c$ and I'm required to start with $c.$ I don't know if you're pointing out that I can use the ideas there to conclude my attempt with $c.$ – DrinkingDonuts May 31 '19 at 14:32
  • The last part of my answer here gives an approach starting from $\lim$ defined on $c$. It does use some machinery though so I now wonder if there is a more elementary approach. – Rhys Steele May 31 '19 at 14:38
  • @TheoBendit my bad, I for some reason thought about sequences convergent to $0$, not just convergent. – mihaild May 31 '19 at 14:40
  • 1
    @mihaild I don't see how you get $||\Lambda||=1$. – David C. Ullrich May 31 '19 at 14:41

1 Answers1

8

@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 $\Lambda x\ne\Lambda Sx$.

David C. Ullrich
  • 92,839