[Math] Show that $(c_{0})’$ and $(c)’$ are isometrically isomorphic.

Question

We consider $c_{0}$ the set of all complex convergent sequences to $0$, and $c$ the set of all complex convergent sequences. Show that $(c_{0})'$ and $(c)'$ are isometrically isomorphic. (where $'$ means dual)

Remark: My idea consists of the following:

We know that $c_{0}\subset c$, then we consider the map
$$\begin{array}{rcl}\Gamma:(c)'&\rightarrow&(c_{0})'\\ g&\mapsto & \left.g\right|_{c_{0}} \end{array}$$
where $|_{c_{0}}$ means restrction to $c_{0}$. I managed to show that $\Gamma$ is a linear continuous bijection with $\left\|\Gamma(g)\right\|\leq \left\|g\right\|$.

The problem: I have not been able to show the other inequality, that is,$\left\|\Gamma(g)\right\|\geq \left\|g\right\|$.

Answer 1

The dual of $c$ is isometrically isomorphic to $\ell^1$. This is already written in a bunch of places, so I won't repeat the proof:

• Isometric isomorphism from $\ell^1$ to $c^*$
• Isometrical isomorphism $T : \ell _1\to c^*$
• $c^*$ isometrically isomorphic to $\ell_1$
• The dual space of $c$ is $\ell^1$

It is also true (and easier to show) that the dual of $c_0$ is isometrically isomorphic to $\ell^1$:

• Isometric identification of $c_0^*$ and $ \ell^1$
• linear operator from $\ell^1$ to $(c_0)'$
• Show a map from $(c_0)^*$ to $\ell^1$ is an isometry