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Math Help - Banach Limit

  1. #1
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    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
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  2. #2
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    Re: Banach Limit

    Quote Originally Posted by dotalam View Post
    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.
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  3. #3
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    Re: Banach Limit

    Quote Originally Posted by Opalg View Post
    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!
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