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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,$\left\|\Gamma(g)\right\|\geq \left\|g\right\|$.

Diego Fonseca
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  • The inequality you want to prove is false - consider the functional taking a sequence to its limit. – Wojowu Nov 11 '17 at 21:36
  • You probably know the isometric isomorphism $(c_0)' \to \ell^1(\mathbb{N})$. So try to find an isometric isomorphism $\ell^1(\mathbb{N}) \to (c)'$. – Daniel Fischer Nov 11 '17 at 21:39

1 Answers1

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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:

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