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Thread: Banach spaces help

  1. #1
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    Banach spaces help

    Hello!

    Could you please help a bit with this problem?

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

    $\displaystyle x_{k} = e_{k} -2e_{k+1} + e_{k+2} $

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

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

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

    Now, suppose that $\displaystyle s \in \mathbb{R} $, and
    $\displaystyle x_{k}=se_{k} - 2e_{k+1} + e_{k+2} $
    and M is as before. Prove that $\displaystyle 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 $\displaystyle l^{1} $ the sequence $\displaystyle (x_{k})$ of elements is defined by

    $\displaystyle x_{k} = e_{k} -2e_{k+1} + e_{k+2} $

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

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

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

    Quote Originally Posted by Mimi89 View Post
    Now, suppose that $\displaystyle s \in \mathbb{R} $, and
    $\displaystyle x_{k}=se_{k} - 2e_{k+1} + e_{k+2} $
    and M is as before. Prove that $\displaystyle 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 $\displaystyle l^{1} $ such that $\displaystyle M\subseteq \ker(f)$?

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

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