Let $X=l^2$ and $A$ be the set of all positve elements of $X$, Then show that $Cone(A-x)$ is not a subspace.

Question

Let $X=l^2 = \{x=\{x_i\}_{i=1}^{\infty} ~ | \quad \sum_{i=1}^{\infty} |x_i|^2 < \infty \}$ and $A = \{ x=\{x_i\}_{i=1}^{\infty} \in X ~ | \quad x_i >0 ~ \forall i \}$ and take $a \in A.$

My questions (I have conjectured that) :

$1-$ Is $ C:= \text{Cone}(A-a)$ a subspace of $X$? **This part was solved by Demophilus **

$2-$ $\bar{C} $ is indeed a subspace and it is indeed the whole space $X$. In other words, $ C:= \text{Cone}(A-a)$ is dense in $X$ for all $a \in A$

P.S. $\text{Cone}(A-a) = \{\lambda (x-a ) ~ | \quad x \in A ~, \lambda \ge 0\}$

Answer 1

2nd Edit: Again you were completely right, I edited my answer. Hopefully I got it right this time. 3d Edit: I added an answer to the second question.

First question

Let $y \in C$. We try to look for a sufficient and necessary condition that $-y \in C$. Without loss of generality we assume that $y = x-a$ for some $x \in A$. Now take any $\lambda' \geq 0 $. Then define for every $n \in \mathbb{N}$ $$ x'_n = a_n\left(1+ \frac{1}{\lambda'}\right)- \frac{1}{\lambda'}x_n . $$ Note that $x_n' > 0$ if and only if $\lambda' > \frac{x_n}{a_n} - 1$. So it seems that $a-x \in C$ if and only if $\sup\{ \frac{x_n}{a_n} \mid n \in \mathbb{N} \}$ is bounded. This reminded me of the following proposition:

There does not exist a sequence $(a_n)$ of positive reals such that for all sequences $(b_n)$, $\sum_{n} \lvert b_n \rvert a_n$ converges if and only if $(b_n)$ is bounded.

So it's seems that for any $a \in A$, you could always construct a $x \in A$ such that $\sup_n (x_n a_n^{-1}) < +\infty$ making sure that $a-x \not \in C$ and as a result $C$ can't be a subspace.

To illustrate this for $a_n = \frac{1}{n}$, consider $x_n = \frac{\sqrt{\log(n)}}{n}$. Note that $\frac{x_n}{a_n}$ is cleraly unbounded. Now suppose $a-x \in C$. So there's a $\lambda \geq 0$ and $x' \in A$ such that $a-x = \lambda(x'-a)$. Now we need to have for all $n \in \mathbb{N}$ that $$ x'_n = \frac{a_n-x_n}{\lambda}+a_n >0 $$

Or in other words $\frac{1-\log(n)}{\lambda} > 1$, but this is obviously false.

Second question

Let $e_n$ be the standard basis on $l^2$. It's easy to prove that $e_n \in C$ for all $n \in \mathbb{N}$. Simply take $$ x_k = \left\{ \begin{matrix}\frac{1}{k} & k \neq n \\ 1+\frac{1}{k} & k =n\end{matrix} \right. $$ Then we have $x \in A$ and $e_n = x-a \in C$. We can also prove that $- e_n \in C$ for all $n \in \mathbb{N}$. We take $$ y_k = \left\{ \begin{matrix}\frac{1}{k} & k \neq n \\ \frac{1}{2k} & k =n\end{matrix} \right. $$ and $\lambda = 2n$. Then we have $-e_n = \lambda(y-a)$. Since $C$ is a cone it contains all positive linear combinations of $e_1,-e_1, \ldots, e_n,-e_n, \ldots$. But this is simply the set $\text{span}\{e_n \mid n \in \mathbb{N} \}$, which is clearly dense in $l^2$. So $C$ must be dense in $l^2$ too.

Answer 2

The second question is easy to answer. Let us define the finite sequences $$c_c := \{v \in \ell^2 \mid \exists N > 0 : \forall n > N: v(n) = 0\}.$$ Then, it is easy to check $c_c \subset C \subset \ell^2$. Since $c_c$ is dense in $\ell^2$, you get $\ell^2 = \overline{c_c} \subset \overline C \subset \ell^2$.