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I'd like help answering two questions.

1) Prove that there is a continuous linear functional on $\ell^\infty$ such that $f(e_n)=0 \ \forall n \in \Bbb{N}$ and $f(a)=5$ where $a=(1,1,1,1,1,1,\ldots)$.

2) Prove that there is not a continuous linear functional on $\ell^\infty$ such that $f(e_n)=0 \ \forall n \in \Bbb{N}$ and $f(a)=4$ for $a=(1,1/2,1/3,1/4,\ldots)$.

Note: $e_n$ stand for the point $(0,0,\ldots,0,1,0,\ldots,0,0)$ with $1$ in the $n$-th position and the rest $0$'s, i.e $(e_n)=(\delta_{mn})$.

Thanks in advance.

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Raul C. de Assis
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3 Answers3

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The key observation (for both questions) is that the Banach subspace of $\ell^\infty$ generated by $\{e_1,e_2,\ldots\}$ is $c_0$.

Martin Argerami
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There are positive continuous functionals on $\ell_\infty$, which map every convergent sequence to its limit. Banach limits are one class of such functionals, see fore example this question. Another example is a limit of a bounded sequence along an ultrafilter, see Wikipedia, this answer or this question.

For the second part just notice that if $f(e_n)=0$, then also each $x_n=(1,1/2,1/3,\dots,1/n,0,0,\dots,0,\dots)$ is mapped to zero. Since $f(x_n)=0$ and $a_n\to a$ in $\ell_\infty$, we get $f(a)=0$.

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Martin Sleziak
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    Martin, while that is true, nothing that sophisticated is needed for question 1. You just use that $c_0$ is a proper closed subspace and apply Hahn-Banach. – Martin Argerami Dec 22 '13 at 12:55
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For question 1), you could provide an explicit example of such a functional. One such example would be $f:\ell^\infty\to \mathbb{R}$ given by $$ f(\{x_n\}) = 5 \cdot \lim_{n\to\infty}\frac 1n \sum_{k=1}^n x_k $$ For the purposes of your question, it suffices to prove that $f$ is a bounded (i.e. continuous) linear functional satisfying the necessary criteria where it is defined. From there, one must use the HB theorem to extend $f$ to give a functional over $\ell^\infty$.

Ben Grossmann
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    The function you suggested is not defined for all bounded sequences; the limit $\lim \frac1n \sum\limits_{k=1}^n x_k$ does not always exist. Some such examples are given in answers to this question: http://math.stackexchange.com/questions/444889/bounded-sequence-with-divergent-cesaro-means However, your function can be extended to a linear continuous functional on $\ell_\infty$. (For example using Hahn-Banach, or using ultrafilters; whichever you prefer.) – Martin Sleziak Dec 23 '13 at 09:13
  • @MartinSleziak duly noted. – Ben Grossmann Dec 23 '13 at 15:24
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    I think it would work if I define for $x=(x_1,x_1,...)$, $f(x)=5 \cdot \lim_{n \to \infty} x_n$ on $c$, the space of convergent sequences in $\Bbb{R^\infty}$. So $f(e_n)=0 \ \forall n \in \Bbb{N}$ and $f(a)=5$, then find use a sublinear function such as $p(x)=5 \cdot ||x||_{l^\infty}$ that fits the conditions for Hanh-Banch and extend it to $l^\infty$. – Raul C. de Assis Dec 23 '13 at 21:58
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    @Raul that works too. I suppose it is a better choice than mine. – Ben Grossmann Dec 24 '13 at 07:24