2
$\begingroup$

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 $c_0$?

$\endgroup$
  • $\begingroup$ $c_0$ 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
  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ "are the same" means nothing (or means everything). $\endgroup$ – YCor May 13 '18 at 12:29
  • 1
    $\begingroup$ @Martin Sleziak Ahh, my bad, not isometrically isomorphic. $\endgroup$ – Teri May 13 '18 at 12:35
13
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Very nice argument. $\endgroup$ – Ali Bagheri May 14 '18 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.