$c_0$ is not isometrically isomorphic to $c$

Question

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 $c_0$?

Answer 1

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?

Answer 2

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

Answer 3

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 $\ ||x-y|| =||y-z|| = 1.$