Let us consider the space of convergent sequences which is denoted by $c$. The space of all sequences $(x_n)\in c$ with $\lim x_n=0$ is also denoted by $c_0$. Clearly $c_0$ is a proper closed subspace of $c$ under the uniform topology.

These two Banach spaces are the same if and only they are the same as two commutative C*-algebras. But $c_0$ is not unital, however $c$ is unital.

Q1. How one may prove this fact just using functional analysis point of view (elementary tools)?

There exists a functional $\phi\in c^*$ with $\phi(1)=1$ and vanishing on $c_0$ where $1$ is the constant sequence $1,1,\cdots$. The functional $\phi$ corresponds a bounded sequence $(t_n)$.

Q2. How could we determine (explicitly) all bounded sequences $(t_n)$ vanishing on $c_0$?

  • $\begingroup$ $c_0$ is isometrically isomorphic to $c$ as Banach spaces, but the isomorphism does not preserve the $C^*$-algebra structures. Also, one choice for your functional $\phi$ is given by $\phi(x_n)=\lim_{n\to\infty} x_n$. This does not "correspond to a bounded sequence $(t_n)$." $\endgroup$ – Teri May 13 '18 at 11:51
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    $\begingroup$ The proof that there is no linear isometry between these two spaces can be found here: Are these two Banach spaces isometrically isomorphic? and Linear isometry between $c_0$ and $c$. $\endgroup$ – Martin Sleziak May 13 '18 at 12:14
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    $\begingroup$ "are the same" means nothing (or means everything). $\endgroup$ – YCor May 13 '18 at 12:29
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    $\begingroup$ @Martin Sleziak Ahh, my bad, not isometrically isomorphic. $\endgroup$ – Teri May 13 '18 at 12:35

The (multiplicative) Banach–Mazur distance between $c$ and $c_0$ is exactly 3:

M. Cambern, On mappings of sequence spaces, Studia Math. 30. (1968), 73-77.

Let me take this opportunity to advertise a rather crazy conjecture (due to Pełczyński, I think):

Suppose that two Banach spaces $C(K)$ and $C(L)$ are isomorphic. Is the Banach–Mazur distance between $C(K)$ and $C(L)$ an integer?

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    $\begingroup$ I'd never heard of this conjecture and am surprised it hasn't been investigated/publicised more! $\endgroup$ – Yemon Choi May 13 '18 at 18:53
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    $\begingroup$ Interesting conjecture, indeed! i wasn't aware of it either. I am inclined to believe that this is true for countable compact $K$ and $L$ (based on some rough calculations i did in the past.) $\endgroup$ – Bunyamin Sari May 13 '18 at 22:11

Here is a simple proof that the Banach-Mazur distance between the spaces $c$ and $c_0$ is equal to $3$.

First the map $T:c \to c_0$ defined by \begin{align*} T(x)=\left(3x(\omega) , \frac{3}{2}\big(x(1)- x(\omega)\big), \frac{3}{2}\big(x(2)- x(\omega)\big),\ldots \right) \end{align*} is an isomorphism with $\|T\|\|T^{-1}\|\le 3$.

To show that we can't do better, suppose there is an isomorphism $T:c \to c_0$ with $\|T^{-1}\|\le 1$ and $\|T\|\le K<3$. Let $e_0=(1,1,1,\ldots)$, and $(e_i)_{i\ge 1}$ be the standard unit vectors. Let $\varepsilon=\frac{3-K}{2}$. Let $N$ be such that $|Te_0(t)|<\varepsilon$ for all $t>N$. By pigeonhole principle there exists (in fact, infinitely many) $i_0$ such that $|T(e_{i_0})(t)|<\varepsilon/2$ for all $t\le N$. Consider the vector $T(e_0+2e_{i_0})$. Since $\|e_0+2e_{i_0}\|=3$ we have $\|T(e_0+2e_{i_0})\|\ge 3$. The norm must be 'attained' somewhere so let's check if it is attained at some $t\le N$. $$3\le |T(e_0+2e_{i_0})(t)|\le |T(e_0-e_{i_0})(t)|+3|T(e_{i_0}(t))|<|T(e_0-e_{i_0})(t)|+3\varepsilon/2$$ So $|T(e_0-e_{i_0})(t)|>3-3\varepsilon/2$. But this is impossible since $|T(e_0-e_{i_0})(t)|\le K$ (check that $K>3-\frac{3\varepsilon}{2}$ not possible for $K<3$).

On the other hand, if the norm is attained at some $t>N$ we have $$3\le |T(e_0+2e_{i_0})(t)|<\varepsilon +2|T(e_{i_0})(t)|$$ so $$|T(e_{i_0})(t)|>\frac{3-\varepsilon}{2}$$ But since $\|e_0-2e_{i_0}\|=1$ we have $$K\ge \|T(e_0-2e_{i_0})\|\ge 2|T(e_{i_0})(t)-|T(e_0(t))|\ge 2\frac{3-\varepsilon}{2}-\varepsilon=3-2\varepsilon$$ again leads to a contradiction

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We may simply prove that $\ c\ $ and $\ c_0\ $ are not isometric.

Indeed, constant sequences $\ (0)\ $ and $\ (2)\ $ belong to $\ c,\ $ and $\ (1)\in c\ $ is the only point in $\ c\ $ which is half-way from both, this is the unique center.

However, it's a simple exercise to show that for arbitrary $\ x\ z\in c_0\ $ such that $\ ||x-z|| =2\ $ there is a continuum of different points $\ y\in c_0\ $ such that $\ ||x-y|| =||y-z|| = 1.$

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    $\begingroup$ Very nice argument. $\endgroup$ – Ali Bagheri May 14 '18 at 17:16

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