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Math Help - On inifinite product of vector spaces

  1. #1
    Senior Member bkarpuz's Avatar
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    On inifinite product of vector spaces

    Dear MHF members,

    I have the following problem.

    Problem. For a family of Banach spaces \{A_{n}\}_{n\in\mathbb{N}}, let A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1  },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\},
    where \|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|.

    1. Show that A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1  },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\} is the closure of the algebraic direct sum A_{\Sigma} in the Banach space A_{\Pi}.
    2. If \pi denotes the quotient map onto the quotient space A_{\Pi}/A_{\oplus}, then show that the norm \|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|.


    Thanks for being interested.
    bkarpuz
    Last edited by bkarpuz; March 24th 2011 at 07:35 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Dear MHF members,

    I have the following problem.

    Problem. For a family of Banach spaces \{A_{n}\}_{n\in\mathbb{N}}, let A_{\Pi}:=\prod_{k\in\mathbb{N}}A_{k}=\big\{a=(a_{1  },a_{2},\ldots,a_{n},\ldots):\ \|a\|<\infty\big\},
    where \|a\|:=\sup_{n\in\mathbb{N}}\|a_{n}\|.

    1. Show that A_{\oplus}:=\oplus_{k\in\mathbb{N}}A_{k}=\{a=(a_{1  },a_{2},\ldots,a_{n},\ldots):\ \lim_{n\to\infty}\|a_{n}\|=0\} is the closure of the algebraic direct sum A_{\Sigma}:=\sum_{k\in\mathbb{N}}A_{k} in the Banach space A_{\Pi}.
    2. If \pi denotes the quotient map onto the quotient space A_{\Pi}/A_{\oplus}, then show that the norm \|\pi(a)\|=\limsup_{n\to\infty}\|a_{n}\|.


    Thanks for being interested.
    bkarpuz
    Presumably you mean that you're considering A_k embedded into A_\Pi in the natural way (all coordinates zero, except the k^{\text{th}}) and by the 'algebraic direct sum' A_\Sigma you mean \displaystyle \left\{\sum_{k\in S}A_k:S\subseteq\mathbb{N}\text{ and }\#(S)<\infty\right\}. If so, let's show that \overline{A_\Sigma}=A_\oplus. To see that A_\oplus\subseteq\overline{A_\Sigma} let (a_n)_{n\in\mathbb{N}}\in A_\oplus. Then, for any \varepsilon>0 we have, since \|a_n\|_n\to 0 there exists some N\in\mathbb{N} such that n\geqslant N\implies \|a_n\|_n<\varepsilon. Define then (b_n)_{n\in\mathbb{N}} by


    b_n=\begin{cases}a_n & \mbox{if}\quad n\leqslant 0\\ 0 & \mbox{if}\quad n>N\end{cases}


    Evidently then b_n\in A_\Sigma and moreover \displaystyle \|a_n-b_n\|=\sup_{\substack{n\in\mathbb{N}\\ n>N}}\|a_n\|\leqslant \varepsilon. Since \varepsilon was arbitrary the conclusion follows (technically you have to adjust the above argument so that b_n\ne a_n but that's easy enough).

    Conversely, suppose that (a_n)_{n\in\mathbb{N}}\notin A_\oplus. Then, there exists some \varepsilon>0 such that for every N\in\mathbb{N} there exists some N'\geqslant N such that \|a_{N'}\|_{N'}\geqslant \varepsilon. Evidently this implies that for every N\in\mathbb{N} one has that \displaystyle \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_{n}\geqslant \varepsilon. So, let (b_n)_{n\in\mathbb{N}}\in A_\Sigma be arbitrary. Then, by definition there exists some N\in\mathbb{N} such that for every n\geqslant N one has that b_n=0. Consequently


    \displaystyle \begin{aligned}\|(a_n)-(b_n)\| &=\sup_{n\in\mathbb{N}}\|a_n-b_n\|_n\\ &\geqslant \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n-b_n\|_n\\ &= \sup_{\substack{n\in\mathbb{N}\\ n\geqslant N}}\|a_n\|_n\\ &\geqslant \varepsilon\end{aligned}


    Thus, (b_n)\notin B_{\varepsilon}((a_n)). Since (b_n)\in A_\Sigma was arbitrary it follows that (a_n)\notin \overline{A_\Sigma}.


    That should give you the basic idea...I might have skipped a minor detail in the case (a_n)\in A_\Sigma but it's easily fixable.
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