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I have been struggling with this computation for a while now. I thought I was almost there, but it now results I still have nothing. So here is the initial problem:

Let $c=\left\{ (x_j)_j \subset \mathbb{C}: (x_j)_j \ \text{is a convergent sequence }\right\}$ equipped with the supremum norm $\|\cdot\|_\infty$, and $Y=\left\{ (x_j)_j \in c : (x_j)_j \ \text{is a constant sequence }\right\}$. Compute $c/Y$, that is find the Banach space to whom $c/Y$ is isometrically isomorphic.

Three things one must know to tackle this problem:

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1)The equivalence relation consider here is $x \sim y$ iff $\ x-y \in Y$

2)The quotient space is $c/Y = \{ [x]: x \in c \}$

3)The norm in $c/Y$ is given by $\|[x]\|=\inf_{y \in [x]} \{\|y\|_\infty\}$

**My advances:** I already have proved that $Y \subset c$ is a closed subspace, so indeed $c/Y$ is a Banach space. Lets now consider $c_0=\left\{ (x_j)_j \in c: x_j \to 0 \text{ as } j \to \infty \right\}$, and define $S \subset c_0$ as follows

$$

S= \left\{ (z_j)_j \in c_0 : \sum_{j=1}^{\infty} z_j \ \text{ converges in } \mathbb{C} \right\}.

$$

I have fisrt conjectured that $c/Y \cong S$, and by defining $\Phi: c/Y \to S$ as

$$

\Phi([(x_j)_j]) : = (x_j-x_{j+1})_j \ \ \ \text{for } \ \ (x_j)_j \in c

$$

I successfully showed that $\Phi$ is **i)Linear**, **ii)Injective** and **iii) Onto**, however, I could not prove the **iv)Isometry** part.

**EDIT:** I now know thanks to @Jochen comment that proving **iv)** is impossible since $S$ is not it self a Banach space, and I thought it was, so $c/Y$ and $S$ are only isomorphic, but they are not the same Banach space.

So my question now changes, since $S$ is not isometrically isomorphic to $c/Y$ which Banach space must be? It is correct to still looking for a $c_0$ subspace ot it might be something totally different.I am clueless since all my bets where on $S$. I would appreciate it very much any help given here.

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It is well-known (and easy to see) that $c/Y$ is isomorphic to $c_0$. So we just need to renorm $c_0$ so that the natural map between them is an isometry. In fact, the norm $

\cdot

$ will do, where

\begin{equation*}

(x_n)

=\frac{1}{2}\sup_{m,n}|x_n-x_m|.\end{equation*}

Notice that for any $(x_n)\in c_0$ we have

\begin{equation*}\|(x_n)\|_{c/Y}=\inf_{x\in\mathbb{C}}\|(x_n)-x\|_\infty=\inf_{r\in\mathbb{C}}\sup_n|x_n-r|.\end{equation*}

For any $m,n$, let $r(m,n)$ be the midpoint between $x_m$ and $x_n$, so that

\begin{equation*}\sup_{m,n}|x_n-x_m|=2\sup_{m,n}|x_n-r(m,n)|\geq 2\inf_{r\in\mathbb{C}}\sup_n|x_n-r|.\end{equation*}

For the reverse inequality, notice that for any $r\in\mathbb{C}$ we have

\begin{equation*}\sup_{m,n}|x_n-x_m|\leq\sup_{m,n}(|x_n-r|+|x_m-r|)\leq 2\sup_n|x_n-r|\end{equation*}

and hence

\begin{equation*}\sup_{m,n}|x_n-x_m|\leq 2\inf_{r\in\mathbb{C}}\sup_n|x_n-r|.\end{equation*}

Thus, $c/Y$ is isometric to $(c_0,

\cdot

)$ via the natural map.

Let $S$ be the subspace of $c$ of convergent sequences $x = (x_1,x_2,\dotsc)$ satisfying $\sup x + \inf x = 0$. Let $y_\alpha$ denote the constant sequence $\alpha, \alpha, \dotsc$. Let $h(x)$ denote the number $\sup(x) + \inf(x) \over 2$ Then the map $c/Y \rightarrow S$ given by $[x] \mapsto x – y_{h(x)}$ is an isometry onto $S$.

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