5

Let $l^\infty = \{x\in \mathbb{R}^\mathbb{N}\colon \sup_{n\in \mathbb{N}}|x_n|<\infty\}$ and the subspace $C \subseteq l^\infty$ given by the convergent sequences. We consider the linear operator $L$ in $C$ given by $$C \ni x \mapsto L(x) =\lim_nx_n$$ Is easy to see that $L$ is continuous with supremum norm hence by Hahn Banach extension theorem there exists $L^* \in (l^\infty)^*$ which is an extension of $L$.

There exists a explicit form for such extension?

Thanks!

Martin Sleziak
  • 53,687
user90803
  • 1,139
  • You may want to look also at Banach limits here and here, which are a special class of bounded linear functionals on $\ell^{\infty}$ whose restriction to $c$ gives the limit operator. Banach limits are not unique, but there are certain non-convergent sequences whose values are uniquely pinned down under any Banach limit (“almost convergent”). In this sense, you can obtain at least a partial description for an extension. – triple_sec Oct 29 '14 at 05:40
  • There is not. Such an extension is guaranteed to exist only by the Axiom of Choice, which is very non constructive. Such a functional provides an element of $\ell_\infty^$ which does not arise from $\ell_1$. In some models of set theory $\ell_\infty^$ IS $\ell_1$. Alas. – James Kilbane Oct 30 '14 at 09:34
  • This answer mentions some reasons why this cannot be done explicitly (more precisely in ZF): http://math.stackexchange.com/questions/55651/nonnegative-linear-functionals-over-l-infty/55664#55664 You can also find some useful other links there. – Martin Sleziak Jun 15 '16 at 08:19

1 Answers1

3

There is not. The closest you can come to explicitness, so far as I know, is to let $p$ be a free ultrafilter on $\Bbb N$ and extend $L$ to the $p$-limit: $p$-$\lim_nx_n=a$ iff for all $\epsilon>0$ $\{n\in\Bbb N:|x_n-a|<\epsilon\}\in p$.

Brian M. Scott
  • 616,228