Consider the vector space $c$ of sequences $x=\{x_n\}$ that converges in
$\mathbb{F}$. Show that there is a linear functional $L \in (\ell^\infty)′$ so that $L(x)=\lim_n{x_n}$ for each $x=\{x_n\} \in c$.
My attempt:
Let $L$ be the linear functional on c. Given that $p(x) = \limsup_n{x_n}$ and $p$ is sublinear, let $\ell$ be the linear functional on $\{0\}$ given by the zero functional. Then $p(0) = 0 = \limsup_n{0_n}$ and by Hahn-Banach $\ell$ can be extended to a linear functional $L$ on $(\ell^\infty)'$ such that $L(x) \leq p(x) = \limsup_n{x_n}$ for all $x \in \ell^\infty $.
Applying this to $-x$ we get $-L(x)=L(-x) \leq \limsup_n{(-x_n)} = -\liminf_n{x_n}$, and, therefore, $\liminf_n{x_n}\leq L(x)\leq \limsup_n{x_n}$.
Since $\limsup_n{x_n}\leq \sup_nx_n \leq \sup_n|x_n|$ and also $\liminf_n{x_n} \geq \inf_n{x_n} \geq -\sup_n|x_n|$, we have $|L(x)| \leq ||x||_\infty$ and $L$ is bounded with norm $\leq 1$ for each $x=\{x_n\} \in c$.
Is my conclusion correct?
Best Answer
Choose your linear subspace to be $c$ with $L(x)= \lim_{n}x_n$ and then extend your $L$ by Hahn--Banach since it will be dominated by $p$. It will then have the desired properties.