Set closed. Hahn Banach theorem. Banach limits

Question

I have a question: why is this set $A$ closed? $$A=\{x-x' \mid x \in l_\infty \}$$ Where $l_\infty$ is the set of bounded sequences, $$x=( x (1),x (2), x (3),\dots),$$ and $$x'=(x (2),x (3),x (4),\dots),$$ the same sequence without the first element.

So, $$A=\{ (x (1)-x (2), x (2)-x (3), x (3)- x (4),\dots) \mid x \in l_\infty \}$$

I think in a sequence of element of $A$ convergence to another element of $A$ (for example call $z$), but I dont know because $z \in A$

If $$z=(z (1),z (2),z (3),\dots),$$ then \begin{align} z (1)&=x (1) - x (2)\\ z (2)&=x (2)- x (3)\\ z (3)&=x (3)-x (4) \end{align}

The sequence $x$ must be $$x=(x (1),x (1)-z (1),x (1)-z (1)-z (2),x (1)-z (1)-z (2)-z (3),\dots).$$

Why this sequence $x$ in $l_\infty$??

Thanks

Answer 1

Let $(z^{(k)})$ be a convergent sequence in $A$ with limit $z^{(0)}$. Then there is a sequence $(x^{(k)})$ in $l_{\infty}$ such that

$z^{(k)}=x^{(k)}-x'$ for all $k$.

Hence

$x^{(k)}=z^{(k)}+x' \to z^{(0)}+x'$ for $k \to \infty$.

Let $x^{(0)}:=z^{(0)}+x'$. Then

$$z^{(0)}=x^{(0)}-x' \in A.$$

Thus we have shown: the limit of each convergent sequence from $A$ belongs to $A$.

Therefore $A$ is closed.

Answer 2

Let $\phi = I-L$, where $L(x(1),x(2),...) = (x(2),x(3),...)$, then $A= {\cal R} \phi$.

Suppose $y=\phi x$. Then $y(1) =x(1)-x(2), y(2) = x(2)-x(3),...$. Rewriting, we see that $x(2) = x(1)-y(1)$, $x(3) = x(1)-(y(1)+y(2)),..., x(k+1) = x(1)-(y(1)+\cdots + y(k))$.

Let $y_n(k) = \begin{cases} {1 \over k}, & k \le n \\ 0, & \text{otherwise} \end{cases}$, and $y(k) = {1 \over k}$. Note that $y_n \to y$.

It is clear from the above formulae that we can find $x_n \in l_\infty$ such that $y_n = \phi x_n$, however, there is no $x \in l_\infty$ such that $y = \phi x$ (since $k \mapsto x(k)$ is not summable).

In particular, $y_n \in A, y_n \to y$ but $y \notin A$, hence $A$ is not closed.