1. ## Banach spaces help

Hello!

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.

2. Originally Posted by Mimi89
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 $l^{1}$.

Originally Posted by Mimi89
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].