1. ## Banach Limit

Define $T: \ell_\infty \to \ell_\infty$ by putting:
$Tx = (x(1),x(2)-x(1),x(3)-x(2),...)$
Let $M$ be range of $T$, $u=(1,1,...)$

Then we have:
(1) $d(u,M) = 1$
(2) There exists a continuous functional $\phi : \ell_\infty \to \mathbb{R}$ such that $\|\phi\| = 1$ and $\phi(m+\lambda u) = \lambda$ for all $m \in M$ (This functional is called a Banach limit)

I would like to know how to show:
(3) $x(i) \geq 0$ for all $i \implies \phi(x) \geq 0$
(4) $\phi(x) = \lim_n x(n)$ when $x(n)$ converge

2. ## Re: Banach Limit

Originally Posted by dotalam
Define $T: \ell_\infty \to \ell_\infty$ by putting:
$Tx = (x(1),x(2)-x(1),x(3)-x(2),...)$
Let $M$ be range of $T$, $u=(1,1,...)$

Then we have:
(1) $d(u,M) = 1$
(2) There exists a continuous functional $\phi : \ell_\infty \to \mathbb{R}$ such that $\|\phi\| = 1$ and $\phi(m+\lambda u) = \lambda$ for all $m \in M$ (This functional is called a Banach limit)

I would like to know how to show:
(3) $x(i) \geq 0$ for all $i \implies \phi(x) \geq 0$
(4) $\phi(x) = \lim_n x(n)$ when $x(n)$ converge
Let $k = \tfrac12\|x\|_\infty.$ If $x_i \geqslant 0$ for all $i,$ then $|x_i-k|\leqslant k$ for all $i.$ Therefore $\|x-ku\|_\infty\leqslant k.$ Since $\|\phi\| = 1,$ it follows that $|\phi(x-ku)|\leqslant k.$ So $|\phi(x) - k|\leqslant k,$ which implies that $\phi(x)\geqslant0.$

For (4), I think you need to show that every element of $\ell_\infty$ with only finitely many nonzero coordinates is in the range of $T.$ It follows that $\phi(x)$ is unchanged if finitely many coordinates of $x$ are changed. Now suppose that $x_n\to c$ as $n\to\infty.$ Given $\varepsilon>0$ we can assume (by changing finitely many coordinates of $x$) that $c-\varepsilon\leqslant x_n\leqslant c+\varepsilon$ for all n. It will then follow from (3) that $c-\varepsilon\leqslant \phi(x)\leqslant c+\varepsilon.$ Now let $\varepsilon\searrow0$ to conclude that $\phi(x) = c.$

3. ## Re: Banach Limit

Originally Posted by Opalg
Let $k = \tfrac12\|x\|_\infty.$ If $x_i \geqslant 0$ for all $i,$ then $|x_i-k|\leqslant k$ for all $i.$ Therefore $\|x-ku\|_\infty\leqslant k.$ Since $\|\phi\| = 1,$ it follows that $|\phi(x-ku)|\leqslant k.$ So $|\phi(x) - k|\leqslant k,$ which implies that $\phi(x)\geqslant0.$

For (4), I think you need to show that every element of $\ell_\infty$ with only finitely many nonzero coordinates is in the range of $T.$ It follows that $\phi(x)$ is unchanged if finitely many coordinates of $x$ are changed. Now suppose that $x_n\to c$ as $n\to\infty.$ Given $\varepsilon>0$ we can assume (by changing finitely many coordinates of $x$) that $c-\varepsilon\leqslant x_n\leqslant c+\varepsilon$ for all n. It will then follow from (3) that $c-\varepsilon\leqslant \phi(x)\leqslant c+\varepsilon.$ Now let $\varepsilon\searrow0$ to conclude that $\phi(x) = c.$
Thank you!