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Prove that there exists a continuous linear map $A $ from the space $l^\infty $ (the space of bounded sequences) to $ \mathbb{R} $ such that:
$a) A(x_1, x_2, \dots) \geq 0 \text{ if } x = (x_1, x_2, \dots) \text{ and } x_n \geq 0 \text{ for all } n.$

$b) A(x_1, x_2, \dots) = A(x_2, x_3,...)$

$c)$ If $ x_n \to \alpha $, then $Ax= \alpha$. The operator is $\sup |x_n|$

My thought:

At first I try to construct such map, that is $A(x)=\lim x_n$. Then I realized it was not linear since $\lim x_n$ might not exist. I think we could not counstruct explicitly such map but for proving its existence. But know I have no idea how to do it. Could somebody give me a hand on this problem? Thank you very much.

Alex Nguyen
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  • Hint: Let $C = {x \in l^\infty : \lim_n x_n \text{ exists}}$ and define $B \colon C \to \Bbb R$ by $B(x) = \lim_n x_n$. Prove that $C$ is a vector subspace of $l^\infty$ and that $B$ is a continuous linear map. Then there exists... – azif00 Jun 18 '25 at 02:09
  • @azif00 acording to Hanh Banach theorem, we need to prove $C$ is closed subspace. How can I do that – Alex Nguyen Jun 18 '25 at 02:18
  • We don't need that. See this, for example. – azif00 Jun 18 '25 at 02:22
  • It is not clear what "The operator is $\sup |x_{n}|$" refers to in this question. In addition, the continuous linear extension of $B$ from the hint does not always satisfy the second property from the question. See here. – Dean Miller Jun 18 '25 at 06:32

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