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Math Help - Annihilator of subspace

  1. #1
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    Annihilator of subspace

    Dear friends,
    can u help me to solve the following problem

    Show that Y={x|x=(\lambda_j \in l^2, \lambda_2n=0, n\in N} is closed subspace of l^2 and find annihilator Y*. What is the annihilator Y* if Y=span{\epsilon_1,......\epsilon_n}\subset \subset l^2, where \epsilon_j = (\delta_jk))?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kinkong View Post
    Dear friends,
    can u help me to solve the following problem

    Show that Y={x|x=(\lambda_j \in l^2, \lambda_2n=0, n\in N} is closed subspace of l^2 and find annihilator Y*. What is the annihilator Y* if Y=span{\epsilon_1,......\epsilon_n}\subset \subset l^2, where \epsilon_j = (\delta_jk))?
    You have to try friend? The first becomes clear if one notes that f_n:\ell^2\to\mathbb{R} given by f((\lambda_j))=\lambda_n is continuous because then \displaystyle Y=\bigcap_{n\in\mathbb{N}}\ker f_{2n}. To find \text{Ann}(Y) note that \left\{e_{2n}\right\}_{n\in\mathbb{N}} is a basis for Y and so some \varphi\in\text{Hom}\left(\ell^2,\mathbb{R}\right) will be identically zero if and only if it annihilates that basis. Thus \text{Ann}\left(Y\right)=\left\{\varphi\in\text{Ho  m}\left(\ell^2,\mathbb{R}\right):\varphi(e_{2n})=0  \text{ for all }n\in\mathbb{N}\right\}.
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  3. #3
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    thanks a lot...i tried it but couldnt get the right way....
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