Is the metric space c0 of sequences of real numbers converging to zero is isometric to the metric space $\ell^1(\mathbb{N})$
To begin this proof do I need to show that:
If I'm wrong in some of the steps let me know!
On
By the Mazur-Ulam theorem every isometry from a real normed space $V$ onto a real normed space $W$ is an affine transformation, i.e. $Tx=Ax+b,$ where $A$ is a bounded linear transformation. Thus $A$ is an isometric isomorphism from $V$ onto $W.$
We may apply the above to $V=c_0$ and $W=\ell^1,$ sequences with real entries. If these space were isometrically isomorphic the dual spaces would be isometrically isomorphic. But $c_0^*=\ell^1$ is separable while $(\ell^1)^*=\ell^\infty$ is not.
The similar argument can be applied to $V=c$ and $W=\ell^1,$ where the first space consists of convergent sequences with real terms. This time the unit balls in both spaces admit extreme points.
On
[Previous answer was nonsense and has been expunged. Here is a correct argument.]
If you don't know Mazur-Ulam, you can instead argue directly that in $\ell^1$, the sequences $(0)$ and $(\delta_1^i)$ have a unique midpoint at $(\frac{1}{2}\delta_1^i)$. (Here $\delta_j^i=1$ if $i=j$, $0$ otherwise).
In $c_0$, on the other hand, no two points have unique midpoint, since if $(a_i)\in c_0$ is a nonzero sequence with $|a_k|$ realizing the supremum, then there is some $j\neq k$ where $|a_j|<|a_k|$, and then $(\frac{a_i}{2}(1-\delta_j^i)+\frac{a_k}{2} \delta_j^i)$ will be an alternative midpoint between $(a_i)$ and $(0)$, provided $\frac{a_j}{a_k}\geq 0$, otherwise $(\frac{a_i}{2}(1-\delta_j^i)-\frac{a_k}{2} \delta_j^i)$ is an alternative midpoint. By translation invariance it follows that no two points have a unique midpoint.
Since having a unique midpoint is preserved under isometries, there can be no isometry between the spaces.
A closed ball of positive radius in $c_0$ has no extreme points, but a closed ball of positive radius in $\ell^1$ has infinitely many.