1. ## Banach Limit

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

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

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

2. ## Re: Banach Limit

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

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

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

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

3. ## Re: Banach Limit

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

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