Are $(\ell^\infty(\mathbb{Z}))^*$ and $ (\ell^\infty(\mathbb{N}))^*$ isomorphs?
I think that I could establish the next function $\Phi:(\ell^\infty(\mathbb{N}))^*\to(\ell^\infty(\mathbb{Z}))^*$ such that for all $g\in (\ell^\infty(\mathbb{N}))^*$, $\Phi_g(x)=\Phi_g((x_n)_{n\in\mathbb{Z}})=g((x_n)_{n\in\mathbb{N}})$ and $0$ if $n\in\{...,-1,0\}$. Is correct the function?
Let $a:\mathbb{N}\rightarrow\mathbb{Z}$ be a bijection, we claim that $$\hat{a}:\ell^{\infty}(\mathbb{Z})\rightarrow \ell^{\infty}(\mathbb{N}),f\mapsto f\circ a$$ is an isomorphic isomorphism. First of all note that $$\hat{a}^{-1}:\ell^{\infty}(\mathbb{N})\rightarrow \ell^{\infty}(\mathbb{Z}),f\mapsto f\circ a^{-1}$$ is the inverse of $\hat{a}$, so $\hat{a}$ is bijective. Clearly it is also linear, so $\hat{a}$ is an isomorphism.
Finally let $f\in \ell^{\infty}(\mathbb{Z})$ and note that $$\|f\|_{\infty}=\sup_{z\in\mathbb{Z}}|f(z)|=\sup_{n\in\mathbb{N}}|f(a(n))|=\|\hat{a}(f)\|_{\infty}$$ so $\hat{a}$ is isomorphic.
As $\ell^{\infty}(\mathbb{Z})$ and $\ell^{\infty}(\mathbb{N})$ are isomorphic isomorph their dual spaces coincide.