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?


  • $\begingroup$ 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. $\endgroup$ – triple_sec Oct 29 '14 at 5:40
  • $\begingroup$ 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. $\endgroup$ – James Kilbane Oct 30 '14 at 9:34
  • $\begingroup$ This answer mentions some reasons why this cannot be done explicitly (more precisely in ZF): math.stackexchange.com/questions/55651/… You can also find some useful other links there. $\endgroup$ – Martin Sleziak Jun 15 '16 at 8:19

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 $\epsilon>0$ $\{n\in\Bbb N:|x_n-a|<\epsilon\}\in p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.