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  1. #1
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    Banach spaces help

    Hello!

    Could you please help a bit with this problem?

    In the Banach space  l^{1} the sequence  (x_{k}) of elements is defined by

     x_{k} = e_{k} -2e_{k+1} + e_{k+2}

    where  e_{k} is the co-ordinate sequence   ( \delta_{kj} ) . Let M be the closed linear span of the elements  x_{1}, x_{2}, ... . By considering the liner functional f on  l^{1} defined by

     f((\alpha _{j} )) = \sum_{j=1}^{\infty} \alpha _{j}

    or otherwise, prove that M does not coincide with   l^{1} .

    Now, suppose that  s \in \mathbb{R} , and
     x_{k}=se_{k} - 2e_{k+1} + e_{k+2}
    and M is as before. Prove that  M=l^{1} iff s>1 or s<-3.

    I have no idea where to start... any hints would be appreciated.
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  2. #2
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    Quote Originally Posted by Mimi89 View Post
    In the Banach space  l^{1} the sequence  (x_{k}) of elements is defined by

     x_{k} = e_{k} -2e_{k+1} + e_{k+2}

    where  e_{k} is the co-ordinate sequence   ( \delta_{kj} ) . Let M be the closed linear span of the elements  x_{1}, x_{2}, ... . By considering the liner functional f on  l^{1} defined by

     f((\alpha _{j} )) = \sum_{j=1}^{\infty} \alpha _{j}

    or otherwise, prove that M does not coincide with   l^{1} .
    Notice that f(x_k) = 0 for all k, hence M\subseteq \ker(f), which is a proper subspace of   l^{1} .

    Quote Originally Posted by Mimi89 View Post
    Now, suppose that  s \in \mathbb{R} , and
     x_{k}=se_{k} - 2e_{k+1} + e_{k+2}
    and M is as before. Prove that  M=l^{1} iff s>1 or s<-3.
    Try to do this by a similar approach to the first part of the question. Can you find a linear functional f on   l^{1} such that M\subseteq \ker(f)?

    A linear functional on  l^{1} is of the form  f((\alpha _{j} )) = \sum_{j=1}^{\infty} c_j\alpha _{j} , where (c_j) is a bounded sequence of scalars. For f of that form, the condition f(x_k)=0 tells you that sc_k - 2c_{k+1} + c_{k+2} = 0 for all k. You solve a recurrence relation of that type by using the auxiliary equation s-2\lambda + \lambda^2 = 0, which has the solutions \lambda = 1\pm\sqrt{1-s}. Then c_j = \lambda^j. In order for the sequence (c_j) to be bounded, we must take the negative square root, getting c_j = \bigl(1-\sqrt{1-s}\bigr)^j. For that solution to be viable, we need two conditions. First, we must have s\leqslant1 (for the square root to be real). Second, we need \bigl|1-\sqrt{1-s}\bigr|\leqslant1 (for the sequence (c_j) to be bounded). That second condition reduces to  s\geqslant-3.

    Therefore if -3\leqslant s\leqslant1, we can conclude (as in the first part of the question) that M\subseteq\ker (f) and hence  M is a proper subspace. With a little care, you can reverse this argument to prove the converse: if  M is a proper subspace, then (by the Hahn–Banach theorem) there is a bounded linear functional which vanishes on  M. That functional must be of the form  f((\alpha _{j} )) = \sum_{j=1}^{\infty} c_j\alpha _{j} , where (c_j) is a bounded sequence of scalars, and it follows as above that s must lie in the interval [–3,1].
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