Prove that $c^*$ is isomorphic to $\ell^1$

Question

If I've shown that $c$ is isomorphic to $c_0$ and that $c_0^*$ is isometrically isomorphic to $\ell^1$ can I conclude that $c^*$ is also isometrically isomorphic to $\ell^1$.

$$c \cong c_0 \space\ \text{and} \space\ c_0^* \cong \ell^1 \Rightarrow^{?} c^* \cong \ell^1 $$

If this is the case how would I go about proving it?

Answer 1

Yes. If $c \cong c_0$, then $c^* \cong c_0^*$. In particular: take $\phi:c \to c_0$ to be an isomorphism. Its adjoint $\phi^*:c_0^* \to c^*$ must also be an isomorphism.